Measure Theory, Volume 5, Part 1

ω1 . Show that if A ⊆ R is a ∆12 set (that is, both A and it complement are PCA sets), then A is universally measurable. ´ ski & Judah 95, 522 Notes and comments All the significant ideas of this section may be found in Bartoszyn with a good deal more. For many years it appeared that ‘measure’ and ‘category’ on the real line, or at least the structures (R, B, N ) and (R, B, M) where B is the Borel σ-algebra of R, were in a symmetric duality. It was perfectly well understood that the algebras A = B/B ∩ N and G = B/B ∩ M – what in this book I call the ‘Lebesgue measure algebra’ and the ‘category algebra of R’ – are very different, but their complexities seemed to be balanced, and such results as 522G encouraged us to suppose that anything provable in ZFC relating measure to category ought to respect the symmetry. It therefore came as a surprise to most of us when Bartoszy´ nski and Raisonnier & Stern (independently, but both drawing inspiration from ideas of Shelah 84, themselves responding to a difficulty noted in Solovay 70) showed that add N ≤ add M in all models of set theory. (It was already known that add N could be strictly less than add M.) The diagram in its present form emphasizes a new dual symmetry, corresponding to the duality of Galois-Tukey connections (512Ab). No doubt this also is only part of the true picture. It gives no hint, for instance, of a striking difference between cov M and cov N . While cov M = mcountable must have uncountable cofinality (522Uf), cov N can be ωω (Shelah 00). I have hardly mentioned shrinking numbers here. This is because while shr M and shr N can be located in Cicho´ n’s diagram (we have non M ≤ shr M ≤ cf M and non N ≤ shr N ≤ cf N , by 511Kc), they are not known to be connected organically with the rest of the diagram. I will return to them in a more general context in 523M. In 522S I give two classic ‘Martin’s axiom’ arguments. They are typical in that the structure of the proof is to establish that there is a suitable partially ordered set for which a ‘generic’ upwards-directed subset will provide an object to witness the truth of some assertion. ‘Generic’, in this context, means ‘meeting sufficiently many cofinal sets’.

§523 intro.

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If there were any more definite method of finding the object sought, we would use it; these constructions are always even more ethereal than those which depend on unscrupulous use of the axiom of choice. ‘Really’ they are names in a suitable forcing language, since (as a rule) we can lift Martin numbers above ω1 only by entering a universe created by forcing. But in this chapter, at least, I will try to avoid such considerations, and use arguments which are expressible in the ordinary language of ZFC, even though their non-trivial applications depend on assumptions beyond ZFC. Of the partially ordered sets S ∞ and P in the proof of 522S, the former comes readily to hand as soon as we cast the problem in terms of the supported relation (N N , ⊆∗ , S); we need only realize that we can express members of S as limits of upwards-directed subsets of a subfamily in which there is some room to manoeuvre, so that we have enough cofinal sets. The latter is more interesting. It belongs to one of the standard types in that the partially ordered set is made up of pairs (σ, F ) in which σ is the ‘working part’, from which the desired object T S M = R \ n∈N σ∈R,i≥n σ(i) will be constructed, and F is a ‘side condition’, designed to ensure that the partial order of P interacts correctly with the problem. In such cases, there is generally a not-quite-trivial step to be made in proving that the ordering is transitive ((b-i) of the proof of 522S). Note that we have two classes of cofinal set to declare in (b-iii) of the proof here; the QnV are there to ensure that M is meager, and the Q0H to ensure that it includes every member of H. And a final element which must appear in every proof of this kind, is the check that the partial order found is of the correct type, σ-linked in (a) and σ-centered in (b). In 522T I suggest that it is natural to try to locate any newly defined cardinal among those displayed in Cicho´ n’s diagram. Of course there is no presumption that it will be possible to do this tidily, or that we can expect any final structure to be low-dimensional; the picture in 522S is already neater than we are entitled to expect, and the complications in 522T (and 522Ye-522Yh) are a warning that our luck may be running out. However, we can surprisingly often find relationships like the ones between FN(PN), b, shr M and cov M here, which is one of my reasons for using this approach. It is very remarkable that under fairly weak assumptions on cardinal arithmetic (the hypothesis ‘cov M < ωω ’ in 522Td is much stronger than is necessary, since in ‘ordinary’ models of set theory we have cf[κ]≤ω = κ whenever cf κ > ω – see 5A1Qc and 5A1R), the axiom ‘FN(PN) = ω1 ’ splits Cicho´ n’s diagram neatly into two halves. For an explanation of why it was worth looking for such a split, see Fuchino Geschke & Soukup 01. For the sake of exactness and simplicity, I have maintained rigorously the convention that M and N are the ideals of meager and negligible sets in R with Lebesgue measure. But from the point of view of the diagram, they are ‘really’ representatives of classes of ideals defined on non-empty Polish spaces without isolated points, on the one hand, and on atomless countably separated σ-finite perfect measure spaces of non-zero measure on the other (522V). The most natural expressions of the duality between the supported relations (R, ∈, M) and (R, ∈, N ) (522G, 522Xc) depend, of course, on the fact that both structures are invariant under translation; but even this is duplicated in R r and in infinite compact metrizable groups like {0, 1}N . At some stage I ought to mention a point concerning the language of this chapter. It is natural to think of such expressions as add N as names for objects which exist in some ideal universe. Starting from such a position, the sentence ‘it is possible that add N < add M’ has to be interpreted as ‘there is a possible mathematical universe in which add N < add M’. But this can make sense only if ‘add N ’ can refer to different objects in different universes, and has a meaning independent of any particular incarnation. I think that in fact we have to start again, and say that the expression add N is not a name for an object, but an abbreviation of a definition. We can then speak of the interpretations of that definition in different worlds. In fact we have to go much farther back than the names for cardinals in this section. PN and R also have to be considered primarily as definitions. The set N itself has a relatively privileged position; but even here it is perhaps safest to regard the symbol N as a name for a formula in the language of set theory rather than anything else. Fortunately, one can do mathematics without aiming at perfect consistency or logical purity, and I will make no attempt to disinfect my own language beyond what seems to be demanded by the ideas I am trying to express at each moment; but you should be aware that there are possibilities for confusion here, and that at some point you will need to find your own way of balancing among them. My own practice, when the path does not seem clear, is to re-read Kunen 80.

523 The measure of {0, 1}I In §522 I tried to give an account of current knowledge concerning the most important cardinals associated with Lebesgue measure. The next step is to investigate the usual measure νI on {0, 1}I for an arbitrary set I. Here I discuss the cardinals associated with these measures. Obviously they depend only on #(I), and are trivial if I is finite. I start with the basic diagram relating the cardinal functions of νκ and νλ for different cardinals κ and λ (523B). I take the opportunity to mention some simple facts about the measures νI (523C-523D). Then I look at additivities (523E), covering numbers (523F-523G), uniformities (523H-523L), shrinking numbers (523M) and cofinalities (523N). I end with a description of these cardinals under the generalized continuum hypothesis (523P).

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523A Notation For any measure µ, write N (µ) for the null ideal of µ. For any set I, I will write νI for the usual measure on {0, 1}I and NI = N (νI ) for its null ideal. Recall that ({0, 1}ω , Nω ) is isomorphic to (R, N ), where N is the Lebesgue null ideal (522Va). I use the notions of Galois-Tukey connection and Tukey function as described in §§512-513. 523B The basic diagram Suppose that κ and λ are infinite cardinals, with κ ≤ λ. Then we have the following diagram for the additivity, covering number, uniformity, shrinking number and cofinality of the ideals Nκ and Nλ : cov Nλ

ω1

add Nλ

cov Nκ

add Nκ

cf Nκ

cf Nλ

shr Nκ

shr Nλ

non Nκ

non Nλ

λω

(As in 522B, the cardinals here increase from bottom left to top right.) proof For the inequalities relating two cardinals associated with the same ideal, see 511Kc; all we need to know is that Nκ and Nλ are proper ideals containing singletons. For the inequalities relating the cardinal functions of the two different ideals, use 521F; νκ is the image of νλ under the map x 7→ xκ : {0, 1}λ → {0, 1}κ , by 254Oa. Of course ω1 ≤ add Nλ . I leave the final inequality cf Nλ ≤ λω for the moment, since this will be part of Theorem 523N below. 523C In the next few paragraphs I will say what is known about the cardinals here. It will be convenient to begin with two easy lemmas. Lemma Let I be any set, and J any family of subsets of I such that every countable subset of I is included in some member of J . Then a subset A of {0, 1}I belongs to NI iff there is some J ∈ J such that {xJ : x ∈ A} ∈ NJ . proof For J ⊆ I, x ∈ {0, 1}I set πJ (x) = xI. Then νJ = νI πJ−1 (254Oa), so A ∈ NI whenever there is some J ∈ J such that πJ [A] ∈ NJ . On the other hand, if A ∈ NI , there is a countable set K ⊆ I such that πK [A] ∈ NK (254Od). −1 Now there is a J ∈ J such that K ⊆ J, so that πJ−1 [πJ [A]] ⊆ πK [πK [A]] ∈ NI and πJ [A] ∈ NJ . 523D Because the measures νI are homogeneous in a strong sense, we have the following facts which are occasionally useful. Proposition Let κ be an infinite cardinal, and T the domain of νκ . For A ⊆ {0, 1}κ write TA for the subspace σ-algebra on A. (a) If E ⊆ {0, 1}κ is measurable and not negligible, then (E, TE , Nκ ∩ PE) is isomorphic to ({0, 1}κ , T, Nκ ). S (b) If E ⊆ Nκ and #(E) < cov Nκ , then (νκ )∗ ( E) = 0. (c) If A ⊆ {0, 1}κ is non-negligible, then there is a set B ⊆ {0, 1}κ , of full outer measure, such that (A, TA , Nκ ∩ PA) is isomorphic to (B, TB , Nκ ∩ PB). (d) There is a set A ⊆ {0, 1}κ of cardinal non Nκ which has full outer measure. proof (a) In fact the subspace measure on E is isomorphic to a scalar multiple of νI (344L2 ). S (b) ?? Otherwise, let F ⊆ E be a non-negligible measurable set; then {F ∩ E : E ∈ E} witnesses that cov(F, Nκ ∩ PF ) < cov Nκ , which contradicts (a). X X (c) Let E be a measurable envelope of A. By (a), there is a bijection f : E → {0, 1}κ which is an isomorphism of the structures (E, TE , Nκ ∩ PE) and ({0, 1}κ , T, Nκ ). Set B = f [A]. Then f A is an isomorphism of the structures (A, TA , Nκ ∩ PA) and (B, TB , Nκ ∩ PB). Moreover, since A meets every member of TE \ Nκ , B meets every member of T \ Nκ , that is, B is of full outer measure. (d) Let A0 ⊆ {0, 1}κ be a non-negligible set of cardinal non Nκ . By (c), there is a set A of full outer measure which is isomorphic to A0 in the sense described there; in particular, #(A) = non Nκ . 523E Additivities Because the function κ 7→ add Nκ is non-increasing, it must stabilize, that is, there is some κ0 such that add Nκ = add Nκ0 for every κ ≥ κ0 . But in fact it stabilizes almost immediately. If κ is any uncountable cardinal, then add Nκ = add νκ = ω1 , by 521Hb. Thus among the additivities add Nκ , only add Nω = add N , the additivity of Lebesgue measure, can have any surprises for us. 2 Later

editions only.

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523F Covering numbers Still on the left-hand side of the diagram, we again have a non-increasing function κ 7→ cov Nκ . We therefore have a critical value after which it is constant. We can locate this value to some extent through the following simple fact. If θ = min{cov Nκ : κ is a cardinal}, then cov Nθ = θ. P P Let κ be such that cov Nκ = θ. For I ⊆ κ, set πI (x) = xI for x ∈ {0, 1}κ . Let E ⊆ Nκ be S a cover of {0, 1}κ of cardinality θ. For each E ∈ E, let IE ⊆ κ be a countable set such that πIE [E] ∈ NIE . Set I = E∈E IE , so that #(I) ≤ θ and πI [E] ∈ NI for every E ∈ E. Then {πI [E] : E ∈ E} is a cover of {0, 1}I by at most cov Nκ sets, and cov NI ≤ cov Nκ . Since ({0, 1}I , NI ) is isomorphic to ({0, 1}#(I) , N#(I) ), we also have cov Nθ ≤ cov N#(I) ≤ cov Nκ = θ ≤ cov Nθ , and cov Nθ = θ. Q Q What this means is that if κc is the critical value, so that cov Nκ = θ iff κ ≥ κc , then ω ≤ κc ≤ θ ≤ cov Nω = cov N ≤ c. Another way of putting the same idea is to say that if λ is a cardinal such that cov Nλ ≥ λ then cov Nκ ≥ λ for every κ (since θ ≥ λ). 523G When the additivity of Lebesgue measure is large we have a further constraint on covering numbers. Proposition (Kraszewski 01) If κ is an infinite cardinal and cov Nκ < add N , then cov Nκ ≤ cf[κ]≤ω . proof Let E be a subset of Nκ of size cov Nκ with union {0, 1}κ , and J a cofinal subset of [κ]ω of size cf[κ]≤ω . For κ J J ∈ J and x ∈ {0, 1}κ set πJ (x) = xJ, S so that πJ : {0, 1} → {0, 1} is inverse-measure-preserving. For J ∈ J set EJ = {E : E ∈ E, πJ [E] ∈ NJ }, HJ = EJ . Since #(EJ ) ≤ #(E) = cov Nκ < add N = add Nω = add NJ , S FJ = {πJ [E] : E ∈ EJ } ∈ NJ and HJ ⊆ πJ−1 [FJ ] ∈ Nκ . Since EJ = E (523C) covers {0, 1}κ , {HJ : J ∈ J } covers κ and cov Nκ ≤ #(J ) = cf[κ]≤ω . S

523H Uniformities On the other side of the diagram we have non-decreasing functions. To get lower bounds for non Nκ we have the following method. Lemma (Kraszewski 01) Suppose that I is a set and F a family of functions with domain I such that for every countable J ⊆ I there is an f ∈ F such that f J is injective. Then non NI ≤ max(#(F ), supf ∈F non Nf [I] ). proof If I is finite the result is trivial. Otherwise, for each f ∈ F take a non-negligible subset Af of {0, 1}f [I] of size non Nf [I] . For y ∈ Af set xf y = yf ∈ {0, 1}I . Set A = {xf y : f ∈ F, y ∈ Af }. ?? If A ∈ NI , there is a countable set J ⊆ I such that {xJ : x ∈ A} ∈ NJ . Let f ∈ F be such that f J is injective. Then we have a function φ : {0, 1}f [I] → {0, 1}J defined by saying that φ(z) = zf J for every z ∈ {0, 1}f [I] , and (because f J is injective) φ is inverse-measure-preserving for νf [I] and νJ , so φ[Af ] cannot be νJ -negligible. But if y ∈ Af then φ(y)(ξ) = y(f (ξ)) = xf y (ξ) for every ξ ∈ J, so φ[Af ] ⊆ {xJ : x ∈ A}, which is supposed to be negligible. X X Thus A is not negligible, and non NI ≤ #(A) ≤ max(ω, #(F ), supf ∈F #(Af )) = max(#(F ), supf ∈F non Nf [I] ) because we are supposing that I is infinite, so there is some f ∈ F such that f [I] is infinite. 523I Theorem (a) For any cardinal κ, non Nκ ≤ max(non N , cf[κ]≤ω ). (b) For any cardinal κ > 0, non N2κ ≤ max(c, cf[κ]≤ω ) is at most the cardinal power κω . (c) If κ is a cardinal of uncountable cofinality then non Nκ+ ≤ max(κ, non Nκ ). (d) If κ is a limit cardinal of uncountable cofinality, then non Nκ+ ≤ max(cf κ, supλ<κ non Nλ ). (e) If κ is any infinite cardinal then non N2κ+ ≤ max(κ+ , non N2κ ). +

Remark TEX, for once, is obscure; 2κ here is #(P(κ+ )), not (2κ )+ .

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proof (a) Let J ⊆ [κ]≤ω be a cofinal set of size cf[κ]≤ω , and for J ∈ J let fJ be the identity function on J. Applying 523H with F = {fJ : J ∈ J } we get non NI ≤ max(#(J ), sup non NJ ) J∈J

≤ max(non Nω , cf[κ]≤ω ) = max(non N , cf[κ]≤ω ). (b) Take J as in (a). This time, for J ∈ J , define fJ : Pκ → PJ by setting fJ (A) = A ∩ J for every A ⊆ κ. Applying 523H with F = {fJ : J ∈ J } we get non N2κ = non NPκ ≤ max(#(J ), sup non NPJ ) J∈J

≤ max(cf[κ]

≤ω

, non Nc )

≤ max(cf[κ]

≤ω

, non N , cf[c]≤ω )

= max(cf[κ]≤ω , c) (5A1E(c-ii)). Now non N2κ ≤ max(c, cf[κ]≤ω ) ≤ max(c, κω ) = κω if κ ≥ 2; for κ = 1, we have non N2 = 1 = 1ω . (c) Start by fixing an injective function hζ : ζ → κ for each ζ < κ+ . For ξ < κ define fξ : κ+ → κ by saying that fξ (ζ) = min(κ \ {fξ (η) : η < ζ, hζ (η) ≤ ξ}) +

+

for ζ < κ . If J ⊆ I is countable, then ξ = supη,ζ∈J,η<ζ hζ (η) is less than κ, because cf κ > ω, and fξ (η) 6= fξ (ζ) for all distinct η, ζ ∈ K. Applying 523H with F = {fξ : ξ < κ}, we get non Nκ+ ≤ max(#(F ), supf ∈F non Nf [κ+ ] ) ≤ max(κ, non Nκ ). (d) Again choose an injective function hζ : ζ → κ for each ζ < κ+ . This time, let K ⊆ κ be a cofinal set of size cf κ consisting of cardinals, and for λ ∈ K define fλ : κ+ → λ+ by the formula fλ (ζ) = min{λ+ \ {fλ (η) : η < ζ, hζ (η) ≤ λ}) for ζ < κ+ . If J ⊆ κ+ is countable, then there is a λ ∈ K such that λ ≥ supη,ζ∈J,η<ζ hζ (η), and fλ (η) 6= fλ (ζ) for all distinct η, ζ ∈ J. Applying 523H with F = {fλ : λ ∈ K}, we get non Nκ+ ≤ max(#(F ), supf ∈F non Nf [κ+ ] ) ≤ max(cf κ, supλ<κ non Nλ ). (e) For A ⊆ κ+ , ξ < κ+ set fξ (A) = A ∩ ξ. If J ⊆ Pκ+ is countable, there is a ξ < κ+ such that A ∩ ξ 6= A0 ∩ ξ for all distinct A, A0 ∈ J , that is, fξ J is injective. So 523H tells us that non N2κ+ = non NP(κ+ ) ≤ max(κ+ , sup non Nfξ [κ+ ] ) ξ<κ+

≤ max(κ+ , sup non NPξ ) = max(κ+ , non N2κ ). ξ<κ+

523J Corollary (Kraszewski 01) (a) non Nω2 = non Nω1 = non N . (b) For any n ∈ N, non Nωn+1 ≤ max(ωn , non N ). (c) non N2ω1 = non Nc . (d) If n ∈ N then non N2ωn ≤ max(ωn , non Nc ). proof (a) We have

non N = non Nω ≤ non Nω2 (523B) ≤ max(ω1 , non Nω1 ) (523Ic) ≤ max(ω1 , non N , cf[ω1 ]≤ω ) (523Ia)

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= max(ω1 , non N ) (5A1E(e-iv)) = non N . (b) Induce on n, using 523Ic for the inductive step. (c) By 523Ie, non N2ω1 ≤ max(ω1 , non Nc ); since ω1 ≤ non Nc ≤ non N2ω1 , we have the result. (d) Induce on n, using 523Ie for the inductive step. 523K Corollary (Burke n05) For any sets I, K let Υω (I, K) be the least cardinal of any family F of functions from I to K such that for every countable J ⊆ I there is an f ∈ F which is injective on J. (If #(K) < max(ω, #(I)) take Υω (I, K) = ∞.) Then (a) non NI ≤ max(Υω (I, K), non NK ) for all sets I and K; (b) if κ ≥ c is a cardinal, then non Nκ = max(Υω (κ, c), non Nc ). proof (a) This is just a slightly weaker version of 523H. (b) The point is that Υω (κ, c) ≤ non Nκ . P P Let A ⊆ {0, 1}κ×ω be a non-negligible set of cardinal non Nκ×ω . For κ×ω ω x ∈ {0, 1} define fx : κ → {0, 1} by setting fx (ξ) = hx(ξ, n)in∈N for each ξ < κ. If ξ, η < κ are distinct, then {x : fx (ξ) = fx (η)} is negligible, so if J ⊆ κ is countable then {x : fx J is injective} is conegligible and meets A. Accordingly {fx : x ∈ A} witnesses that Υω (κ, c) = Υω (κ, {0, 1}ω ) ≤ #(A) = non Nκ×ω = non Nκ . Q Q Since we already know that non Nc ≤ non Nκ ≤ max(Υω (κ, c), non Nc ), we have the result. 523L On the other side we can find lower bounds which give a notion of the rate of growth of the numbers non Nκ as κ increases. Proposition (a) If λ and κ are infinite cardinals with κ > 2λ , then non Nκ > λ. (b) If κ is a strong limit cardinal of countable cofinality then non Nκ > κ. proof (a) Let A ⊆ {0, 1}κ be any set of size at most λ. For ξ < κ set Bξ = {x : x ∈ A, x(ξ) = 1}. Because κ > 2#(A) , there is some B ⊆ A such that I = {ξ : Bξ = B} is infinite. But what this means is that if ξ ∈ I then x(ξ) = 1 for every ξ ∈ B and x(ξ) = 0 for every x ∈ A \ B, and A ⊆ {x : x is constant on I} is negligible. As A is arbitrary, non Nκ > λ. (d) By (a), non Nκ > λ for every λ < κ, so non Nκ ≥ κ; but also cf(non Nκ ) ≥ add Nκ (513C(b-ii)), so non Nκ has uncountable cofinality and must be greater than κ. 523M Shrinking numbers As with non N• , the functions κ 7→ shr Nκ and κ 7→ shr Nκ are non-decreasing, by 521Fb. Some of the ideas used in 523I can be adapted to this context, but the pattern as a whole is rather different. Proposition (a)(i) For any non-zero cardinals κ and λ, shr Nκ ≤ max(covSh (κ, λ, ω1 , 2), supθ<λ shr Nθ ). (ii) For any infinite cardinal κ, shr Nκ ≤ max(shr N , cf[κ]≤ω ). (iii) If cf κ > ω, then shr Nκ ≤ max(κ, supθ<κ shr Nθ ). (b) For any infinite cardinal κ, (i) shr Nκ ≥ κ; (ii) cf(shr Nκ ) > ω; (iii) cf(shr+ Nκ ) > κ. Remark For the definition of covSh , see 5A2Da. proof (a)(i) If covSh (κ, λ, ω1 , 2) = ∞ or κ is finite this is trivial. Otherwise, take a non-negligible A ⊆ {0, 1}κ .SLet J ⊆ [κ]<λ be a set of size covSh (κ, λ, ω1 , 2) such that for every I ∈ [κ]<ω1 there is a D ∈ [J ]<2 such that I ⊆ D, that is, there is a J ∈ J such that I ⊆ J. For each J ∈ J , AJ = {xJ : x ∈ A} is non-negligible; let BJ ⊆ AJ be a

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non-negligible set of size at most shr NJ . Let B ⊆ A be a set of size at most max(ω, #(J ), supJ∈J shr NJ ) such that BJ ⊆ {xJ : x ∈ B} for every J ∈ J . If I ⊆ κ is countable, there is a J ∈ J such that I ⊆ J, so {xI : x ∈ B} ⊇ {yI : y ∈ BJ } is non-negligible; it follows that B is non-negligible, while #(B) ≤ max(covSh (κ, λ, ω1 , 2), supθ<λ shr Nθ ). (ii) Taking λ = ω1 in (i), shr Nκ ≤ max(covSh (κ, ω1 , ω1 , 2), shr Nω ) = max(cf[κ]≤ω , shr N ). S (iii) Take λ = κ in (a); as [κ]≤ω = ξ<κ [ξ]≤ω , shr Nκ ≤ max(covSh (κ, κ, ω1 , 2), supθ<κ shr Nθ ) = max(κ, supθ<κ shr Nθ ). (b)(i) Induce on κ. If κ = ω the result is trivial. For the inductive step to κ+ , consider the set +

A = {x : x ∈ {0, 1}κ , ∃ ξ < κ+ , x(η) = 0 for every η ≥ ξ}. +

Then the only set which includes A and is determined by a countable set of coordinates is {0, 1}κ , so A has full outer measure. On the other hand, if B ⊆ A and #(B) ≤ κ, then there is some ζ < κ+ such that x(ξ) = 0 for every x ∈ B and every ξ ≥ ζ, so B is negligible. Thus A witnesses that shr Nκ+ ≥ κ+ . Because κ 7→ shr Nκ is non-decreasing (523B), the inductive step to limit cardinals κ is trivial. (ii) ?? Now suppose, if possible, that cf(shr Nκ ) = ω. Then there is a sequence hλn in∈N of cardinals less than shr Nκ with supremum shr Nκ . For each n ∈ N set In = κ × {n}, and let An ⊆ {0, 1}In be a non-negligible set such that every non-negligible subset of An has more than λn members. By 523Dc, there is a set Bn ⊆ {0, 1}In of full outer measure such that every non-negligible subset of Bn has more than λn members. Set B = {x : x ∈ {0, 1}κ×N , xIn ∈ Bn for every n ∈ N}. Q Q Then the natural isomorphism between {0, 1}κ×N and n∈N {0, 1}In identifies B with n∈N Bn , so B has full outer measure in {0,S1}κ×N (254Lb). There must therefore be a set C ⊆ B, of non-zero measure, such that #(C) ≤ shr Nκ . Express C as n∈N Cn where #(Cn ) ≤ λn for every n. Then there is an n ∈ N such that Cn is not negligible, in which case Dn = {xIn : x ∈ Cn } is non-negligible. But Dn ⊆ Bn and #(Dn ) ≤ λn , so this is impossible. X X (iii) The argument of (i) shows that if κ is a successor cardinal, then shr+ Nκ > κ. So we need consider only the case in which κ is a limit cardinal. ?? If cf(shr+ Nκ ) ≤ κ, then there is a family hλξ iξ<κ of cardinals less than shr+ Nκ with supremum shr+ Nκ . I use the same method as in (ii). For each ξ < κ set Iξ = κ × {ξ}, and let Bξ ⊆ {0, 1}Iξ be a set of full outer measure such that every non-negligible subset of Bξ has at least λξ members. Set B = {x : x ∈ {0, 1}κ×κ , xIξ ∈ Bξ for every ξ < κ. Then B has full outer measure in {0, 1}κ×κ . There must therefore be a set C ⊆ B, of non-zero measure, such that #(C) < shr+ Nκ . Let ξ < κ be such that #(C) < λξ . Then D = {xIξ : x ∈ C} is non-negligible. But D ⊆ Bξ and #(Dξ ) < λξ , so this is impossible. X X 523N Cofinalities For the cardinals cf Nκ the pattern from 523Ia and 523Mb continues, and indeed we have an exact formula. Theorem For any infinite cardinal κ, κ ≤ cf Nκ = max(cf N , cf[κ]≤ω ) ≤ κω . proof (a) cf Nκ ≤ max(cf N , cf[κ]≤ω ). P P Let J be a cofinal family in [κ]ω of cardinal cf[κ]≤ω . For each J ∈ J , κ write πJ (x) = xJ for x ∈ {0, 1} . Let EJ be a cofinal subset of NJ of cardinal cf NJ = cf Nω = cf N . Consider F = {πJ−1 [E] : J ∈ J , E ∈ EJ }. By 523C, F is cofinal with Nκ , so that cf Nκ ≤ #(F) ≤ max(cf N , cf[κ]≤ω ). Q Q (b) We know that cf[κ]≤ω ≤ cf Nκ (521Hb) and that cf N = cf Nω ≤ cf Nκ (523B). So cf Nκ = max(cf N , cf[κ]≤ω ). (c) For the inequalities, note that if κ is uncountable then cf[κ]≤ω ≥ cov(κ, [κ]≤ω ) = κ. On the other side, cf N ≤ c ≤ κω and cf[κ]≤ω ≤ #([κ]≤ω ) = κω . 523O Cofinalities of the cardinals In 523Mb I have shown that shr Nκ has uncountable cofinality for infinite κ, and rather more about shr+ Nκ . From 513Cb we have a little information concerning the cofinalities of add Nκ , cov Nκ , non Nκ and cf Nκ ; but except when κ = ω we learn only that cf Nκ and non Nκ have uncountable cofinality, and that if cov Nκ = cf Nκ then their common cofinality is at least non Nκ . This last remark can apply only to ‘small’ κ, since cf Nκ ≥ κ (if κ is infinite) and cov Nκ ≤ cov N .

The measure of {0, 1}I

523Xd

115

523P The generalized continuum hypothesis In this chapter I am trying to present arguments in forms which show their full strength and are not tied to particular axioms beyond those of ZFC. However it is perhaps worth mentioning that in one of the standard universes the pattern is particularly simple. Proposition Suppose that the generalized continuum hypothesis is true. Then, for any infinite cardinal κ, add Nκ = add νκ = cov Nκ = ω1 ; non Nκ = λ if κ = λ+ where cf λ > ω, = κ+ if cf κ = ω, = κ otherwise; shr Nκ = cf Nκ = κ+ if cf κ = ω, = κ otherwise; shr+ Nκ = (shr Nκ )+ = κ++ if cf κ = ω, = κ+ otherwise; proof Since ω1 ≤ add Nκ = add νκ ≤ cov Nκ ≤ cov N ≤ c = ω1 , the additivity and covering number are always ω1 . If cf κ = ω, then κ+ ≤ shr Nκ ≤ cf Nκ = max(ω1 , cf[κ]≤ω ) ≤ 2κ = κ+ by 523Mb and 523N. If cf κ > ω then κ ≤ shr Nκ ≤ cf Nκ ≤ κ by 523M(b-i), 523N and 5A1Pb. This deals with shr Nκ and cf Nκ . For the augmented shrinking numbers, we know that κ < shr+ Nκ ≤ (shr Nκ )+ (523M(b-iii)), with equality if shr Nκ is a successor cardinal, so in the present case we must have shr+ Nκ = (shr Nκ )+ . As for non Nκ , if κ = λ+ where cf λ > ω, then κ > 2θ for every θ < λ, so we have λ ≤ non Nκ ≤ cf[λ]ω = λ (523Ib, 523La, 5A1Pb). If κ = λ+ where cf λ = ω, then λ ≤ non Nκ ≤ λω ≤ 2λ = κ; but as non Nκ has uncountable cofinality (513C(b-ii)), non Nκ must be κ. If κ is a limit cardinal, then κ > 2θ for every θ < κ, so κ ≤ non Nκ ≤ max(ω1 , cf[κ]≤ω ); if cf κ > ω this is already enough to show that non Nκ = κ; if cf κ = ω then non Nκ cannot be κ so must be κ+ = κω . 523X Basic exercises (a) Show that (Nκ , 63, {0, 1}κ ) 4GT ([κ]≤ω , ⊆, [κ]≤ω ) n (N , 63, R) for every infinite cardinal κ. (See 512I for the definition of n.) Use this to prove 523Ia. (b) Let κ and λ be cardinals, and I a proper σ-ideal of subsets of κ. Show that non NTrI (κ;λ) ≤ max(κ, non Nλ ). (c) Let κ be an infinite cardinal, and J a family of subsets of κ such that every countable subset of κ is included in some member of J . Show that non Nκ ≤ max(cf J , supJ∈J non NJ ), shr Nκ ≤ max(cf J , supJ∈J shr NJ ), cf Nκ ≤ max(cf J , supJ∈J cf NJ ). (d) Show that (Nκ , ⊆, Nκ ) 4GT ([κ]≤ω , ⊆, [κ]≤ω ) n (N , ⊆, N ) for every infinite cardinal κ. Use this to prove 523N.

116

Cardinal functions of measure theory

523Xe

(µ)

(e) Let (X, Σ, µ) be any probability space, and for a set I write NI for the null ideal of the product measure on (µ) (µ) I X . Show that all the results of 523E-523J, 523L-523N are valid with NI in place of NI and Nω in place of N , except that (µ) —– in 523F we can no longer be sure that cov Nω ≤ c; (µ) —– in 523Ib we need to write ‘non N2κ ≤ max(c, non Nω , cf[κ]≤ω )’; —– in 523L and 523Mb we have to assume that the measure algebra of µ is not {0, 1}, so that the product measure on X N is atomless; (µ) —– in 523N we can no longer be sure that cf Nω ≤ κω . 523Y Further exercises (a) Set A = PR/N . (i) Show that c ≤ c(A) ≤ π(A) ≤ 2shr N . (ii) Show that if shr+ N ≥ c and 2λ ≤ c for every λ < c, then c(A) = 2c . (b) Let κ be an infinite cardinal. Show that there is a family J ⊆ [κ]≤ω such that #(J ) ≤ shr Nκ and every infinite subset of κ meets some member of J in an infinite set. (c) Suppose that κ ≥ ω and that [κ]≤ω has bursting number at most add N . Show that Nκ ≡T [κ]≤ω × N . (Hint: Fremlin 91a.) (d) Show that (ω1 , ≤, ω1 ) n (ω1 , ≤, ω1 ) 64GT (ω1 , ≤, ω1 ) × (ω1 , ≤, ω1 ). (e) For infinite cardinals κ, write Mκ for the ideal of meager subsets of {0, 1}κ . Show that under the same conventions as in 522B and 523B we have the diagrams cov Mλ

ω1

cov Mκ

cf Mκ

cf Mλ

shr Mκ

shr Mλ

add Mλ

add Mκ

non Mκ

non Mλ

cov Nκ

non Mκ

cf Mκ

cf Nκ

add Nκ

add Mκ

cov Mκ

non Nκ

λω

and

ω1

κω

whenever ω ≤ κ ≤ λ. Show moreover that all the results of 523E-523P have parallel forms referring to Mκ . (f ) In the language of 523Ye, show that (i) mpcω1 ≤ cov Mκ ≤ non Nκ for every infinite κ (ii) mσ−linked ≤ cov Nκ ≤ non Mκ if ω ≤ κ ≤ c. (Hint: ω1 is a precaliber of RO({0, 1}κ ), and the measure algebra of νκ is σ-linked if κ ≤ c.) (g) Show that ♣ implies that cov Nω1 = cov Mω1 = ω1 . 523Z Problem Is there a proof in ZFC that cf[κ]≤ω ≤ shr Nκ for every cardinal κ? 523 Notes and comments The basic diagram 523B is natural and easy to establish. Of course it leaves a great deal of room, especially on the right-hand side, where we have the increasing functions non N• , shr N• and cf N• , and rather weak constraints λ < non Nκ ≤ shr Nκ ≤ cf Nκ ≤ κω whenever 2λ < κ to control them. However the generalized continuum hypothesis is sufficient to determine exact values for all the cardinals considered here (523P). The combinatorics of cf[κ]≤ω and almost-disjoint families of functions are extremely complex, and depend in surprising ways on special axioms; I think it possible that the results of 523I-523J can be usefully extended. However 523N at least reduces the measure-theoretic problem of determining cf Nκ to a standard, if difficult, question in infinitary combinatorics. I do not know if there are corresponding results concerning non Nκ and shr Nκ (see 523Kb and 523Z). All the ideas in this section up to and including 523P can be applied to ideals of meager sets (523Ye) and indeed to other classes of ideals satisfying the fundamental lemma 523C; see Kraszewski 01.

524C

Radon measures

117

524 Radon measures It is a remarkable fact that for a Radon measure the principal cardinal functions are determined by its measure algebra (524J), so can in most cases be calculated in terms of the cardinals of the last section (524P-524Q). The proof of this seems to require a substantial excursion involving not only measure algebras but also the Banach lattices `1 (κ) and/or the κ-localization relation (524D, 524E). The same machinery gives us formulae for the cardinal functions of measurable algebras (524M). The results of §518 can be translated directly to give partial information on the FreeseNation numbers of measurable algebras (524O). For covering number and uniformity, we can see from 521J that strictly localizable compact measures follow Radon measures. I know of no such general results for any other class of measure, but there are some bounds for cardinal functions of countably compact and quasi-Radon measures, which I give in 524R-524T. 524A Notation If (X, Σ, µ) is a measure space, N (µ) will be the null ideal of µ. For any cardinal κ, νκ will be the usual measure on {0, 1}κ , and (Bκ , ν¯κ ) its measure algebra. As in §§522-523, I will write Nκ for N (νκ ) and N for the null ideal of Lebesgue measure on R, so that (R, N ) and ({0, 1}ω , Nω ) are isomorphic (522Va). If A is any Boolean algebra, I write A+ for A \ {0} and A− for A \ {1}. If (A, R, B) is a supported relation, R 0 is the relation {(a, I) : a ∈ R−1 [I]} (see 512F). For any cardinal κ, (κN , ⊆∗ , Sκ ) will be the κ-localization relation (522K). 524B Proposition Let (X, T, Σ, µ) be a σ-finite Radon measure space with Maharam type κ. Then N (µ) 4T B− κ. proof (a) Suppose, to begin with, that µX = 1 and κ ≥ ω. Let A be the measure algebra of the Radon product ˜ on Y = X N . Then A ∼ measure λ P By 417E(ii), A is isomorphic to the measure algebra of the usual product = Bκ . P measure λ on Y , which by 334E is isomorphic to Bκ . Q Q For E ∈ N (µ), let hFEi ii∈N be a sequence of closed subsets of X such that E ∩ FEi = ∅ and µFEi ≥ 1 − 2−i−1 for every n ∈ N. Then Q Q ˜ Q λ( FEi ) = λ( FEi ) ≥ (1 − 2−i−1) > 0; i∈N

i∈N

i∈N

set φ(E) = (Y \

Q

i∈N

FEi )• ∈ A− .

For b ∈ A− let Kb ⊆ Y be a non-empty compact self-supporting set such that Kb• ∩ b = 0. Set πi (y) = y(i) for i ∈ N Q Q and y ∈ Y . Then each πi [Kb ] ⊆ X is compact and Kb ⊆ i∈N πi−1 [πi [Kb ]], so i∈N µπi [Kb ] > 0 and supi∈N µπi [Kb ] = 1; set S ψ(b) = X \ i∈N πi [Kb ] ∈ N (µ). If E ∈ N (µ) and b ∈ A− and φ(E) ⊆ b and j ∈ N, then Kb \ πj−1 [FEj ] ⊆ Kb \

Q

i∈N

FEi

is negligible. As Kb is self-supporting, Kb \ πj−1 [FEj ] is empty and πj [Kb ] ⊆ FEj . But this means that πj [Kb ] ∩ E = ∅ for every j ∈ N, so that E ⊆ ψ(b). This shows that φ is a Tukey function, so that N (µ) 4T A− ∼ = B− . (b) If κ is finite, N (µ) has a greatest member and the constant function with value 0 is a Tukey function from N (µ) to B− κ and the result is trivial. R If κ is infinite and µX 6= 1, then, because µ is σ-finite and not trivial, there is a function f : X → ]0, ∞[ such that f dµ = 1 (215B(ix)). Let ν be the corresponding indefinite-integral measure; then ν is a Radon probability measure (416S) with the same measurable sets and the same negligible sets as µ (234L3 ), so has the same measure algebra and the same Maharam type κ. In this case, (a) tells us that N (µ) = N (ν) 4T B− κ. 524C Lemma Let P be a partially ordered set such that p ∨ q = sup{p, q} is defined for all p, q ∈ P . Suppose that ρ is a metric on P such that P is complete (as a metric space) and ∨ : P × P → P is uniformly continuous with respect to ρ. Let Q ⊆ P be an open set, and κ ≥ d(Q) a cardinal. Then (Q, ≤ 0 , [Q]<ω ) 4GT (`1 (κ), ≤, `1 (κ)). If Q is upwards-directed, then Q 4T `1 (κ). proof (a) If Q is finite, then we can set φ(q) = 0 for every q ∈ Q, ψ(x) = Q for every x ∈ `1 (κ) and (φ, ψ) will be a Galois-Tukey connection from (Q, ≤ 0 , [Q]<ω ) to (`1 (κ), ≤, `1 (κ)). So let us suppose that Q and κ are infinite. (b) Let hqξ iξ<κ run over a dense subset of Q. For each q ∈ Q let m(q) ∈ N be such that {p : p ∈ P , ρ(p, q) ≤ 2−m(q) } ⊆ Q. For each n ∈ N, let δn > 0 be such that ρ(sup I, sup J) ≤ 2−n whenever ∅ = 6 I ⊆ J ⊆ P and #(J) ≤ 2n k and maxq∈J minp∈I ρ(p, q) ≤ 2δn ; such exists because hpi ii
234D.

118

Cardinal functions of measure theory

524C

k > 0, and in particular when k = 2n . Reducing the δn if necessary, we may suppose that δn+1 ≤ δn ≤ 2−n for every n. (c) Define φ : Q → `1 (κ) as follows. Given p ∈ Q, choose a sequence hξ(p, n)in∈N in κ such that ρ(p, qξ(p,n) ) ≤ δn+1 for every n. Take φ(p) ∈ `1 (κ) such that φ(p)(m(p)) ≥ 1,

φ(p)(ξ(p, n)) ≥ 2−n for every n ∈ N

(regarding m(p) as a finite ordinal). (d) Define ψ : `1 (κ) → [Q]<ω as follows. Given x ∈ `1 (κ), set Kn (x) = {qξ : ξ < κ, x(ξ) ≥ 2−n } for n ∈ N. Then P∞ −n P P #(Kn (x)) ≤ ξ<κ {2−n : x(ξ) ≥ 2−n } ≤ 2kxk1 < ∞, n=0 2 ˜ so there is a k(x) ∈ N such that x(n) < 1 for n ∈ ω \ k(x) and also #(Kn (x)) ≤ 2n for n ≥ k(x). Set K(x) = Kk(x) (x). ˜ For s ∈ K(x) set I(x, s, k(x)) = {s},

I(x, s, n + 1) = {q : q ∈ Kn+1 (x), ρ(q, I(x, s, n)) ≤ 2δn+1 }

for n ≥ k(x), writing ρ(q, I) for inf q0 ∈I ρ(q, q 0 ). Because hKn (x)in∈N is non-decreasing, so is hI(x, s, n)in≥k(x) . Set rxsn = sup I(x, s, n) in P for n ≥ k(x); then ρ(rx,s,n+1 , rxsn ) ≤ 2−n−1 for every n ≥ k(x), by the choice of δn+1 , so ˜ rxs = limn→∞ rxsn is defined in P . Set ψ(x) = Q ∩ {rxs : s ∈ K(x)}. (e) Now (φ, ψ) is a Galois-Tukey connection from (Q, ≤ 0 , [Q]<ω ) to (`1 (κ), ≤, `1 (κ)). P P Suppose that p ∈ Q and 1 ˜ x ∈ ` (κ) are such that φ(p) ≤ x. Then qξ(p,n) ∈ Kn (x) for every n, so s = qξ(p,k(x)) ∈ K(x). Also qξ(p,n) ∈ I(x, s, n) for every n ≥ k(x), because ρ(qξ(p,n+1) , qξ(p,n) ) ≤ δn+2 + δn+1 ≤ 2δn+1 for every n. So qξ(p,n) ≤ rxsn for every n ≥ k(x). It follows that p ∨ rxs = limn→∞ qξ(p,n) ∨ rxsn = limn→∞ rxsn = rxs and p ≤ rxs . By the choice of k(x), we also have f (p)(n) < 1 for n ≥ k(x), so that m(p) < k(x). We therefore have

ρ(rxs , p) ≤ ρ(qξ(p,k(x)) , p) +

∞ X

ρ(rx,s,n+1 , rxsn )

n=k(x)

(because s = qξ(p,k(x)) is the unique member of I(x, s, k(x)), so is equal to rx,s,k(x) ) ∞ X ≤ δk(x)+1 + 2−n−1 ≤ 2−k(x)−1 + 2−k(x) ≤ 2−m(p) n=k(x)

and rxs ∈ Q. So p ≤ rxs ∈ ψ(x) and p ≤ 0 ψ(x). As p and x are arbitrary, (φ, ψ) is a Galois-Tukey connection. Q Q (f ) So (Q, ≤ 0 , [Q]<ω ) 4GT (`1 (κ), ≤, `1 (κ)), as claimed. Finally, if Q is upwards-directed, then add Q ≥ ω, so (Q, ≤, Q) ≡GT (Q, ≤ 0 , [Q]<ω ) (513Id) and (Q, ≤, Q) 4GT 1 (` (κ), ≤, `1 (κ)), that is, Q 4T `1 (κ). 524D Proposition If κ is any cardinal, 0 − <ω (B− ) 4GT (`1 (κ), ≤, `1 (κ)). κ , ⊆ , [Bκ ]

proof If κ is finite then B− κ is finite and the result is trivial. Otherwise, if we give Bκ its measure metric ρ (323Ad), then it is a complete metric space in which ∪ is uniformly continuous (323Gc, 323B) and B− κ = Bκ \ {1} is an open set. Now the topological density of B− and B is κ, by 521O; so 524C gives the result. κ κ 524E Proposition Let κ be an infinite cardinal. Then (`1 (κ), ≤ 0 , [`1 (κ)]≤ω ) 4GT (κN , ⊆∗ , Sκ ). proof (a) For each i ∈ N, let hziξ iξ<κ run over a norm-dense subset of {x : x ∈ `1 (κ)+ , kxk1 ≤ 4−i }. Now there is a function φ : `1 (κ) → κN such that P∞ for every x ∈ `1 (κ), n ∈ N there is a k ∈ N such that x ≤ k i=n zi,φ(x)(i) .

524F

Radon measures

119

P P Given x ∈ `1 (κ), choose hxn in∈N , hξn in∈N , hkn in∈N inductively, as follows. Take k0 ≥ 1 such that kx+ k1 ≤ k0 ; set x0 = k0−1 x+ and take ξ0 < κ such that kx0 − z0ξ0 k1 < 14 . Given that xn ∈ `1 (κ)+ , ξn < κ are such that −1 kxn − znξn k1 < 4−n−1 , let kn+1 ≥ 1 be such that kxn+1 k1 ≤ 4−n−1 where xn+1 = (xn − znξn )+ + kn+1 x+ , and take −n−2 ξn+1 < κ such that kxn+1 − zn+1,ξn+1 k1 < 4 ; continue. At the end of the process, set φ(x) = hξn in∈N . Now, for any n ∈ N, we have x ≤ x+ ≤ kn xn . But we also have, for any m ≥ n, xm+1 ≥ xm − zmξm , so that Pm−1 xn ≤ xm + i=n ziξi for every m ≥ n. Since kxm k1 ≤ 4−m for every m, limm→∞ xm = 0 and P∞ x ≤ kn xn ≤ kn i=n zi,φ(x)(i) . As x and n are arbitrary, φ is a suitable function. Q Q P 1 (b) Define ψ0 : Sκ → ` (κ) by setting ψ0 (S) = (i,ξ)∈S ziξ ; because P∞ −i P P∞ −i i=0 2 i=0 4 #(S[{i}]) ≤ (i,ξ)∈S kziξ k1 ≤ is finite, ψ0 (S) is well defined in `1 (κ) for every S ∈ Sκ (4A4Ie). Now define ψ : Sκ → [`1 (κ)]≤ω by setting ψ(S) = {kψ0 (S) : k ∈ N} for S ∈ Sκ . (c) If x ∈ `1 and S ∈ Sκ are such that φ(x) ⊆∗ S, then x ≤ 0 ψ(S). P P Let n ∈ N be such that (i, φ(x)(i)) ∈ S for every i ≥ n. Then there is a k ∈ N such that P∞ P x ≤ k i=n zi,φ(x)(i) ≤ k (i,ξ)∈S ziξ = kψ0 (S) ∈ ψ(S). Q Q (d) Thus (φ, ψ) is a Galois-Tukey connection and (`1 (κ), ≤ 0 , [`1 (κ)]≤ω ) 4GT (κN , ⊆∗ , Sκ ). 524F Lemma Let (X, Σ, µ) be a countably compact measure space with Maharam type κ. S (a) If µ is a Maharam-type-homogeneous probability measure, there is a family hEξ iξ<κ in N (µ) such that ξ∈A Eξ has full outer measure for every uncountable A ⊆ κ. S (b) If µ is σ-finite, there is a family hEξ iξ<κ in N (µ) such that ξ∈A Eξ is non-negligible for every uncountable A ⊆ κ. proof Let A be the measure algebra of (X, Σ, µ). (a) A is τ -generated by a stochastically independent family haξ iξ<κ of elements of measure 21 , and for every G ∈ Σ there is a smallest countable set IG ⊆ κ such that G• is in the closed subalgebra of A generated by {aξ : ξ ∈ IG } (254Rd or 325Mb). For each ξ < κ choose Fξ ∈ Σ such that Fξ• = aξ . Let K be a countably compact class such that µ is inner regular with respect to K. Let hJξ iξ<κ be a disjoint family of subsets of κ all of cardinal ω1 . For each ξ < κ choose hKξn in∈N , hαξn in∈N inductively, as follows. αξ0 is any point of Jξ . Given αξn and hKξi ii 0; let ξ ∈ A be such that IK ∩Jξ = ∅; let n ∈ N be such that µK ≥ 3−n−1 . Set Gm = K ∩ n≤i<m Kξi S for m ≥ n. Then IGm ⊆ IK ∪ i<m IKξi does not contain αξm , for any m. This means that 1 2

µGm+1 = µ(Gm ∩ Kξm ) ≥ µ(Gm \ Fαξm ) − 3−m−2 /2 = (µGm − 3−m−2 ) for every m ≥ n, and an easy induction shows that µGm ≥ 3−m−1 for every m. But this tells us that every Gm is T non-empty; because K is a countably compact class, K ∩ Eξ ⊇ m≥n Gm is non-empty, and F meets Eξ . S As F is arbitrary, ξ∈A Eξ has full outer measure. (b) For the general case, because µ is σ-finite, there is a countable partition of unity hai ii∈I in A such that all the principal ideals Aai are totally finite and Maharam-type-homogeneous (use 332A), and we can find a partition hXi ii∈I of X into measurable sets such that Xi• = ai for each i. Moreover, the subspace measure µXi on Xi is countably compact (451Db). Writing κi for the Maharam type of Aai , there is a family hEiξ iξ<κi of negligible subsets of Xi such that {ξ : ξ < κi , Eiξ ⊆ E} is countable for every negligible set E. (This is trivial if κi is countable, and otherwise we can apply (a) to a scalar multiple of µXi .) Now we know from 332S that κ = supi∈I κi = #({(i, ξ) : i ∈ I, ξ < κi }). On the other hand, for any negligible set E ⊆ X, {(i, ξ) : i ∈ I, ξ < κi , Eiξ ⊆ E} is countable. So if we re-enumerate hEiξ ii∈I,ξ<κi as hEξ iξ<κ we shall have an appropriate family.

120

Cardinal functions of measure theory

524G

524G Proposition Let (X, T, Σ, µ) be a Maharam-type-homogeneous Radon probability space with Maharam type κ ≥ ω. Then (κN , ⊆∗ , Sκ ) 4GT (N (µ), ⊆, N (µ)). proof (Compare 522M.) (a) By 524F, there is a family hEξ iξ<κ in N (µ) such that {ξ : Eξ ⊆ E} is countable for every E ∈ N (µ). Next, because the measure algebra of µ is isomorphic to the measure algebra of the usual measure on [0, 1]N×κ , there is a stochastically independent family hGiξ ii∈N,ξ<κ in Σ such that µGiξ = 2−i for every i ∈ N and ξ < κ. For f ∈ κN set S T S φ(f ) = n∈N Ef (n) ∪ n∈N m≥n Gm,f (m) ∈ N (µ). (b) Take E ∈ N (µ) and set IE = {ξ : Eξ ⊆ E}, so that IE is countable. Then there is a non-empty compact self-supporting set KE such that KE ∩ E = ∅ and KE ∩ Giξ , KE \ Giξ are compact for all i ∈ N, ξ ∈ IE . P P Define πE : X → {0, 1}N×IE by setting πE (x)(i, ξ) = 1 if x ∈ Giξ , 0 otherwise. Then πE is measurable, therefore almost continuous (418J), and there is a non-negligible measurable set H ⊆ X \ E such that πE H is continuous. Because µ is inner regular with respect to the compact self-supporting sets, there is a non-negligible compact self-supporting KE ⊆ H, and this has the required property. Q Q πE [KE ] is compact. Let hWn (E)in∈N run over the family of open-and-closed subsets W of {0, 1}N×IE meeting −1 πE [KE ]. Then πE [Wn (E)] is a non-empty relatively open subset of KE for every n; because KE is self-supporting, −1 πE [Wn (E)] is never negligible. Set −1 J(E, n, i) = {ξ : ξ ∈ IE , πE [Wn (E)] ∩ Giξ = ∅}

for n, i ∈ N. Because hGiξ ii∈N,ξ∈IE is stochastically independent, P∞ −i P −1 {µGiξ : i ∈ N, ξ ∈ IE , Giξ ∩ πE [Wn (E)] = ∅} i=0 2 #(J(E, n, i)) = is finite, by the Borel-Cantelli lemma (273K). For each n, let k(E, n) ∈ N be such that 2−i #(J(E, n, i)) ≤ 2−n−1 for i ≥ k(E, n), and set S ψ(E) = n∈N {(i, ξ) : i ≥ k(E, n), ξ ∈ J(E, n, i)} ⊆ N × κ. Then #({ξ : (i, ξ) ∈ ψ(E)} ≤

X

#(J(E, n, i))

n∈N,k(E,n)≤i

≤

X

2−n−1 2i ≤ 2i

n∈N,k(E,n)≤i

for every i ∈ N, and ψ(E) ∈ Sκ . (c) Now (φ, ψ) is a Galois-Tukey connection from (κN , ⊆∗ , Sκ ) to (N (µ), ⊆, N (µ)). P P Suppose that f ∈ κN and E ∈ N (µ) T are such S that φ(f ) ⊆ E. Because Ef (n) ⊆ φ(f ), f (n) ∈ IE for every n ∈ N. Next, KE does not meet φ(f ), so KE ∩ n∈N m≥n Gm,f (m) is empty, that is, T S πE [KE ] ∩ n∈N m≥n {w : w ∈ {0, 1}N×IE , w(m, f (m)) = 1} = ∅. By Baire’s theorem, there is some m ∈ N such that S πE [KE ] ∩ i≥m {w : w ∈ {0, 1}N×IE , w(i, f (i)) = 1} is not dense in πE [KE ], and there is an n ∈ N such that S Wn (E) ∩ i≥m {w : w ∈ {0, 1}N×IE , w(i, f (i)) = 1} = ∅. In this case, f (i) ∈ J(E, n, i) for every i ≥ m. But this means that (i, f (i)) ∈ ψ(E) for every i ≥ max(m, k(E, n)), so that f ⊆∗ ψ(E). As f and E are arbitrary, (φ, ψ) is a Galois-Tukey connection. Q Q (d) Thus φ and ψ witness that (κN , ⊆∗ , Sκ ) 4GT (N (µ), ⊆, N (µ)), as claimed. 524H Corollary Let κ be an infinite cardinal, and µ a Maharam-type-homogeneous Radon probability measure with 0 + ≤ω Maharam type κ. Then (B+ ), (`1 (κ), ≤ 0 , [`1 (κ)]≤ω ), (κN , ⊆∗ , Sκ ) and (N (µ), ⊆, N (µ)) are Galois-Tukey κ , ⊇ , [Bκ ] equivalent. proof By 524D and 524B, 0 − <ω (B− ) 4GT (`1 (κ), ≤, `1 (κ)), κ , ⊆ , [Bκ ] − (N (µ), ⊆, N (µ)) 4GT (B− κ , ⊆ , Bκ ).

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121

So 0 + ≤ω ∼ 0 − ≤ω 0 − <ω1 ) (B+ ) = (B− ) = (B− κ , ⊇ , [Bκ ] κ , ⊆ , [Bκ ] κ , ⊆ , [Bκ ]

4GT (`1 (κ), ≤ 0 , [`1 (κ)]<ω1 ) (512Gb, 512Gd) = (`1 (κ), ≤ 0 , [`1 (κ)]≤ω ) 4GT (κN , ⊆∗ , Sκ ) (524E) 4GT (N (µ), ⊆, N (µ)) (524G) ≡GT (N (µ), ⊆ 0 , [N (µ)]≤ω ) (513Id) 0 − ≤ω 4GT (B− ) κ , ⊆ , [Bκ ]

by 512Gb again. 524I Corollary Let µ be a Maharam-type-homogeneous Radon probability measure with infinite Maharam type κ. Then add N (µ) = add Nκ = addω `1 (κ), cf N (µ) = cf Nκ = cf `1 (κ). proof By 524H and 512Db, add(`1 (κ), ≤ 0 , [`1 (κ)]≤ω ) = add(N (µ), ⊆, N (µ)) = add(N (νκ ), ⊆, N (νκ )) = add(Nκ , ⊆, Nκ ). But add(`1 (κ), ≤ 0 , [`1 (κ)]≤ω ) = addω `1 (κ) (513Ia), while add(N (µ), ⊆, N (µ)) = add N (µ) and add(Nκ , ⊆, Nκ ) = add Nκ (512Ea). So addω `1 (κ) = add N (µ) = add Nκ . On the other side, 512Da tells us that cov(Nκ , ⊆, Nκ ) = cov(N (µ), ⊆, N (µ)) = cov(`1 (κ), ≤ 0 , [`1 (κ)]≤ω ). But cov(Nκ , ⊆, Nκ ) = cf Nκ ,

cov(N (µ), ⊆, N (µ)) = cf N (µ)

1

(512Ea). Next, cf ` (κ) > ω. P P If hxn in∈N is any sequence in `1 (κ), then (because κ is infinite) Fn = {x : x ≤ xn } is nowhere dense (for the norm topology) for any n ∈ N, so hFn in∈N cannot cover `1 (κ) (4A2Ma) and {xn : n ∈ N} cannot be cofinal. Q Q So 512Gf tells us that cov(`1 (κ), ≤ 0 , [`1 (κ)]≤ω ) = cov(`1 (κ), ≤, `1 (κ)) = cf `1 (κ). Putting these together, cf N (µ) = cf Nκ = cf `1 (κ) as required. 524J Theorem Let (X, T, Σ, µ) and (Y, S, T, ν) be Radon measure spaces with non-zero measure and isomorphic measure algebras. (a) N (µ) and N (ν) are Tukey equivalent, so add µ = add N (µ) = add N (ν) = add ν and cf N (µ) = cf N (ν). (b) (X, ∈, N (µ)) and (Y, ∈, N (ν)) are Galois-Tukey equivalent, so cov N (µ) = cov N (ν) and non N (µ) = non N (ν). proof (a) Let A, B be the measure algebras of µ and ν. Let hai ii∈I be a partition of unity in A+ such that all the principal ideals Aai are homogeneous and totally finite, and hbi ii∈I a matching family in B, so that Aai ∼ = Bbi for every i. Because (X, Σ, µ) and (Y, T, ν) are strictly localizable (416B), there are decompositions hXi ii∈I and hYi ii∈I of X, Y respectively such that Xi• = ai and Yi• = bi for every i (322M). Write µXi , νYi for the corresponding subspace measures; of course these are Radon measures (416Rb). Then N (µXi ) and N (νYi ) are Tukey equivalent for every i. P P If the common Maharam type of Aai and Bbi is infinite, this is a consequence of 524H. If Aai = {0, ai }, then µXi

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is purely atomic and there is a single point x of Xi such that µ{x} = µXi (414G). In this case N (µXi ) has a greatest member Xi \ {x}, and similarly N (νXi ) has a greatest member, so they have Tukey equivalent cofinal subsets and are Tukey equivalent (513Ed). Q Q Q Now E 7→ hE Q ∩ Xi ii∈I is a partially-ordered-set isomorphism between N (µ) and i∈I N (µXi ). Similarly, N (ν) is isomorphic to i∈I N (νYi ). It now follows from 513Eg that N (µ) and N (ν) are Tukey equivalent. Accordingly add N (µ) = add N (ν) and cf N (µ) = cf N (ν). By 521Ad, add µ = add N (µ) and add ν = add N (ν). (b) By 521Ja, cov N (µ) = cov N (ν) and non N (µ) = non N (ν). 524K Corollary Let (X, T, Σ, µ) and (Y, S, T, ν) be Radon measure spaces with measure algebras A, B respectively. If A can be regularly embedded in B, then N (µ) 4T N (ν). proof As usual, write µ ¯ and ν¯ for the functionals on A, B respectively defined from µ and ν, and let π : A → B be a regular embedding, that is, an order-continuous injective Boolean homomorphism. ˜ Σ, ˜ T, ˜ µ (a) Consider first the case in which µ is totally finite and π is measure-preserving for µ ¯ and ν¯. Let (X, ˜) ˜ T, ˜ ν˜) be the Stone spaces of (A, µ and (Y˜ , S, ¯) and (B, ν¯) respectively. Then π corresponds to a continuous function ˜ (312Q). By 418I, the image measure ν˜f −1 is a Radon measure on X. ˜ If a ∈ A and b f : Y˜ → X a is the corresponding ˜ open-and-closed set in X, then ν˜f −1 [b a] = ν˜(c π a) = ν¯(πa) = µ ¯a = µ ˜b a. By 415H, ν˜f −1 = µ ˜. By 521Fb, N (˜ µ) 4T N (˜ ν ). But now 524Ja tells us that N (µ) ≡T N (˜ µ) 4T N (˜ ν ) ≡T N (ν). (b) Next, consider the case in which µ and ν are totally finite but π is not necessarily measure-preserving. As it is (sequentially) order-continuous, we have a measure µ0 on X defined by saying that µ0 E = ν¯(πE • ) for E ∈ Σ, and N (µ0 ) = N (µ). Because µ0 is absolutely continuous with respect to µ, it is an indefinite-integral measure over µ (234O4 ) and is a Radon measure on X (416S). Taking µ ¯0 to be the corresponding functional on A, (A, µ ¯0 ) is the 0 0 measure algebra of µ and π is measure-preserving for µ ¯ and ν¯. So (a) tells us that N (µ) = N (µ0 ) 4T N (ν). (c) Thirdly, suppose that µ is totally finite, but ν might not be. Set Bf = {b : b ∈ B, ν¯b < ∞}. For b ∈ Bf , set cb = sup{a : a ∈ A, b ∩ πa = 0}; then b ∩ πcb = 0, because π is order-continuous. If a ∈ A \ {0}, there is a b ∈ Bf such that b ∩ πa 6= 0, so that a 6⊆ cb . Accordingly supb∈Bf 1 \ cb = 1 in A; as A is ccc, there is a sequence hbn in∈N in Bf such that supn∈N 1 \ cbn = 1, that is, ν¯(a ∩ supn∈N bn ) > 0 for every non-zeroSa ∈ A. For each n ∈ N, choose Fn ∈ T such that Fn• = bn in B, and set Y 0 = n∈N Fn . The subspace measure νY 0 is σ-finite, so there is a totally finite measure ν 0 on Y 0 , an indefinite-integral measure over νY 0 , with the same null ideal as νY 0 (use 215B(ix)). The measures νY 0 and ν 0 are both Radon measures (416Rb, 416S). Setting b = supn∈N bn in B, the principal ideal Bb can be identified with the measure algebra of νY 0 (322I) and ν 0 . Moreover, the map a 7→ b ∩ πa : A → Bb is an injective order-continuous Boolean homomorphism. By (b) and 521Da, N (µ) 4T N (ν 0 ) = N (νY 0 ) 4T N (ν). (d) If µX = ∞, let hXi ii∈I be a decomposition of X; for each i, let µi be the subspace measure on Xi and ai = Xi• ∈ A; set bi = πai . Then hbi ii∈I is a partition of unity in B, so there is a partition hYi ii∈I of Y such that Yi ∈ T for each i and ν can be identified with the direct sum of the measures νYi (322M). Now, for each i, we can identify the principal ideals Aai , Bbi with the measure algebras of the subspace measures µXi and νYi , and π Aai is an order-continuous embedding of Aai in Bbi . So (c) tells us that N (µXi ) 4T N (νYi ). Accordingly Q Q N (µ) ∼ N (µX ) 4T N (νY ) ∼ = = N (ν) i∈I

i

i∈I

i

(513Eg), and the proof is complete. 524L So far we have been looking at cardinals defined from null ideals. Of course there is an equally important series based on measurable algebras, which turns out to be similarly strongly associated with the cardinal functions of the ideals Nκ . I have already developed a good deal of the machinery in the arguments of this section. But for ‘linking numbers’ we need a new idea, which is most clearly expressed in the context of homogeneous algebras. ¯ ns 94) Let κ be an infinite cardinal. Then for any n ≥ 2 the n-linking number linkn (Bκ ) Proposition (Dow & Stepra is the least λ such that κ ≤ 2λ . 4 Formerly

234G.

524L

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proof Let λ be the least cardinal such that κ ≤ 2λ . (a) By 514Cb, Bκ is isomorphic, as partially ordered set, to a subset of P(link(Bκ )), so we must have 2link(Bκ ) ≥ #(Bκ ) ≥ κ and link(Bκ ) ≥ λ. It follows at once that linkn (Bκ ) ≥ λ for every n ≥ 2 (511Ja). (b) Now let n ≥ 2, and take an injective function φ : κ → {0, 1}λ . Fix E ∈ Σκ \ Nκ for the moment. (i) Let C be the family of measurable cylinders in {0, 1}κ , that is, sets of the form {x : x ∈ {0, 1}κ , xI = z}, where I ⊆ κ is finite and z ∈ {0, 1}I . Then we can find a CE ∈ C such that νκ (CE \ E) ≤

1 νκ CE . 4n

there is a set W , expressible as the union of finitely many measurable cylinders, such that νκ (E4W ) ≤ νκ W ≥

9 10 νκ E

so νκ (W \ E) ≤

1 νκ W . 4n

P P By 254Fe, 1 νκ E. 5n

Now

W is determined by coordinates in a finite set, so is expressible as a disjoint

union of non-empty measurable cylinders, and for at least one of these we must have νκ (C \ E) ≤

1 νκ C; 4n

take such a

one for CE . Q Q 0 Express CE as {x : xIE = zE }, where IE ⊆ κ is finite and zE ∈ {0, 1}IE , and set kE = #(IE ), IE = {ξ : ξ ∈ 00 IE , zE (ξ) = 0} and IE = {ξ : ξ ∈ IE , zE (ξ) = 1}. (ii) Now there is a set GE ⊆ {0, 1}κ , determined by coordinates in a finite subset JE of κ \ IE , such that νκ (CE ∩ (E 4 GE )) ≤

1 4n

· 2−nkE . P P This time, take a set W ⊆ {0, 1}κ , determined by coordinates in a finite subset J

of κ, such that νκ (E4W ) ≤

1 4n

· 2−nkE . Set

GE = {x : x ∈ {0, 1}κ , ∃ y ∈ W ∩ CE , xκ \ IE = yκ \ IE }, so that GE is determined by coordinates in JE = J \ IE and GE ∩ CE = W ∩ CE ; accordingly νκ (CE ∩ (E4GE )) = νκ (CE ∩ (E4W )) ≤ νκ (E4W ) ≤

1 4n

· 2−nkE . Q Q

Note that GE and CE are stochastically independent, so that νκ CE (1 − νκ GE ) = νκ (CE \ GE ) ≤ νκ (CE \ E) + νκ (CE ∩ (E \ GE )) ≤ and νκ GE ≥ 1 −

1 νκ CE 4n

+

1 (νκ CE )n 4n

≤

1 νκ CE 2n

1 . 2n

(c) Let Q be the set of all quadruples (k, U, V, W ) where k ∈ N and U , V , W are disjoint open-and-closed subsets of {0, 1}λ in its usual topology. For q = (k, U, V, W ) ∈ Q, set 0 00 Eq = {E : E ∈ Σκ \ Nκ , kE = k, φ[IE ] ⊆ U, φ[IE ] ⊆ V, φ[JE ] ⊆ W }. 0 00 0 00 For any E ∈ Σκ \ Nκ , IE , IE and JE are disjoint finite sets, so φ[IE ], φ[IE ] andTφ[JE ] also are, and there isSa q ∈ Q 0 such that E ∈ E . Now if q = (k, U, V, W ) ∈ Q and E ∈ E for i < n, then νκ ( i 0. P P Set I 0 = i
νκ G ≥ 1 −

Pn−1 i=0

1 2

(1 − νκ GEi ) ≥ ,

and G is stochastically independent of C, so that νκ (C ∩ G) ≥ 2−nk−1 . Finally, νκ (C ∩ G \ Ei ) ≤ νκ (CEi ∩ GEi \ Ei ) ≤

1 4n

· 2−nk

for each i, so νκ (C ∩ G \ T and νκ ( i 0. Q Q

T

i
Ei ) ≤ 2−nk−2 < νκ (C ∩ G)

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that if we set Aq = {E • : E ∈ Eq } for each q ∈ Q, then every Aq is an n-linked set in Bκ and S (d) This means + λ q∈Q Aq = Bκ . Because {0, 1} is a compact topological space with a subbase of size λ ≥ ω, it has λ open-and-closed sets and #(Q) = λ. So hAq iq∈Q witnesses that linkn (Bκ ) ≤ λ, and the proof is complete. 524M Theorem Let (A, µ ¯) be a semi-finite measure algebra. Let K be the set of infinite cardinals κ such that A has a homogeneous principal ideal with Maharam type κ.

(a)

#(A) = 2c(A) if A is finite, = τ (A)ω if A is ccc and infinite.

(b)

wdistr(A) = ∞ if A is purely atomic, = add N if K = {ω}, = ω1 otherwise.

(c)

π(A) = c(A) if A is purely atomic, = max(c(A), cf N , sup cf[κ]≤ω ) otherwise. κ∈K

(d)

m(A) = ∞ if A is purely atomic, = min cov Nκ otherwise. κ∈K

(e)

d(A) = c(A) if A is purely atomic, = max(c(A), sup non Nκ ) otherwise. κ∈K

(f) For 2 ≤ n < ω, linkn (A) = c(A) if A is purely atomic, = max(c(A), min{λ : τ (A) ≤ 2λ }) otherwise. proof The case A = {0} is trivial, so I shall assume henceforth that A 6= {0}. Let hai ii∈I be a partition of unity in A+ such that all the principal ideals Aai are homogeneous and totally finite. For each i ∈ I, set κi = τ (Aai ), so that b µ Aai ∼ ¯ Aai ). Let (A, ˜) be the localization of (A, µ ¯) (322Q). A can = Bκi , and let (Zi , λi ) be the Stone space of (Aai , µ b b Because be identified with an order-dense Boolean subalgebra of A, so that hai ii∈I is still a partition of unity in A. Q f f b (322P), Aa is still a principal ideal of A, b and A b can be identified with the simple product A =A i i∈I Aai (315F). (a) This is elementary if A is finite (see 511Jc). If A is infinite, then 515M tells us that #(A) = τ (A)ω . (b)

b wdistr(A) = wdistr(A) (514Ee) = min wdistr(Aai ) i∈I

(514Ef) = min wdistr(Bκi ) = min add(N (λi )) i∈I

i∈I

(514Be, because N (λi ) is the ideal of nowhere dense subsets of Z) = min add(Nκi ) i∈I

(524Ja) = ∞ if K = ∅, = add N if K = {ω}, = ω1 otherwise (523E). (c)(i) Consider first an algebra Bκ , where κ ≥ ω. Then ci B+ P If hbn in∈N is any sequence in B+ κ > ω. P κ , then (because Bκ is atomless) we can choose cn ⊆ bn such that 0 < cn ≤ 2−n−2 for each n ∈ N. Set c = supn∈N cn , b = 1 \ c; then b 6= 0 and bn 6⊆ b for every n, so {bn : n ∈ N} is not coinitial with B+ Q κ. Q

524M

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125

It follows that + + + 0 + ≤ω ci B+ ) κ = cov(Bκ , ⊇, Bκ ) = cov(Bκ , ⊇ , [Bκ ]

(512Gf) = cov(Nκ , ⊆, Nκ ) (524H) = cf Nκ . (ii) If A is purely atomic, then Aai = {0, ai } for every i, and π(A) = #(I) = c(A). Otherwise, max(c(A), sup π(Aai )) ≤ π(A) i∈I

(514Da, 514Ed) ≤ max(ω, c(A), sup π(Aai )) i∈I

(514Ef) = max(c(A), sup π(Bκ )) = max(c(A), sup cf Nκ ) κ∈K

κ∈K

(by (i)) = max(c(A), cf N , sup cf[κ]≤ω )) κ∈K

by 523N. (d) If A is purely atomic, then m(A) = ∞ (511Jf). Otherwise, b m(A) = m(A) (517Id) = min m(Aai ) = min n(Zi ) i∈I

i∈I

(517N) = min cov N (λi ) i∈I

(because N (λi ) is just the ideal of nowhere dense subsets of Zi , by 322Ra) = min cov Nκi i∈I

(524Jb) = min cov Nκ , κ∈K

as claimed. (e)(i) I note first that d(Aai ) = non Nκi for each i. P P Let A ∈ PZi \ N (λi ) be a set of cardinal non N (λi ). Then be such that the corresponding open-and-closed set b a is included in H. Then b a H = int A is not empty. Let a ∈ A+ ai can be identified with the Stone space of Aa (312T); because Aai is homogeneous, and A ∩ b a is dense in b a, d(Aai ) = d(Aa ) (because Aai is homogeneous) = d(b a) (514Bd) ≤ #(A ∩ b a) ≤ non N (λi ) = non Nκi (524Jb) ≤ d(Zi ) (because N (λi ) is the ideal of nowhere dense subsets of Zi , so surely contains no dense set) = d(Aai ) by 514Bd again. Q Q

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(ii) If A is purely atomic, d(A) = c(A). Otherwise, max(c(A), sup d(Aai ) ≤ d(A) i∈I

(514Da, 514Ed) b = d(A) (514Ee) ≤ max(ω, c(A), sup d(Aai )) i∈I

(514Ef) = max(c(A), sup non Nκi ) = max(c(A), sup non Nκ ). i∈I

κ∈K

(f ) If A is purely atomic, this is elementary, since any linked subset of A+ can contain at most one atom. Otherwise, set θ = max(c(A), min{λ : τ (A) ≤ 2λ }),

θ0 = linkn (A).

For any i ∈ I, κi ≤ τ (A) (514Ed), so κi ≤ 2θ and linkn (Aai ) = linkn (Bκi ) ≤ θ (524L; of course the case κi = 0 is trivial here). Accordingly b θ0 = linkn (A) (514Ee) ≤ max(ω, c(A), sup linkn (Aai )) i∈I

(514Ef) ≤ θ. 0

0

On the other hand, c(A) ≤ θ (514Da). For each i ∈ I, linkn (Ai ) ≤ θ0 (514Ed), so κi ≤ 2θ (524L, in the other direction). Let AiSbe a τ -generating subset of Aai of size κi . Now the order-closed subalgebra of A generated by A = {ai : i ∈ I} ∪ i∈I Ai is A, so 0

0

τ (A) ≤ #(A) = max(c(A), supi∈I κi ) ≤ max(θ0 , 2θ ) = 2θ . But this means that θ ≤ θ0 and the two are equal. Remark For the corresponding calculation of τ (A), when (A, µ ¯) is localizable, see 332S. 524N Corollary (a) If (X, Σ, µ) is a semi-finite locally compact measure space, with µX > 0, then cov N (µ) ≥ mσ-linked . (b) If A is any measurable algebra, then m(A) ≥ mσ-linked . proof (a) Because µ is semi-finite and µX > 0, there is an E ∈ Σ such that 0 < µE < ∞. The subspace measure µE on E is compact, so ν =

1 µE µE

is a compact probability measure. Set κ = max(ω, τ (ν)). Because ν is a compact

measure, there is a function f : {0, 1}κ → E which is inverse-measure-preserving for νκ and ν (343Cd5 ). Now mσ-linked ≤ m(Bc ) (because Bc is σ-linked, by 524Mf) = cov Nc (524Md) ≤ cov Nκ (523F) ≤ cov N (ν) (521Fa) ≤ cov N (µ) (521Db). (b) This is now immediate from 524Md. 5 Later

editions only.

524P

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127

524O Freese-Nation numbers I spell out those facts about Freese-Nation numbers of measure algebras which can be read off from the results in §518. Proposition (a) Let (A, µ ¯) be an infinite measure algebra. Then FN(A) ≥ FN(PN). (b) Let A be a measurable algebra. (i) FN(A) ≤ c+ . (ii) If τ (A) ≤ c then FN(A) ≤ FN(PN). (iii) If (α) cf([λ]≤ω ) ≤ λ+ for every cardinal λ ≤ τ (A), (β) λ is true for every uncountable cardinal λ ≤ τ (A) of countable cofinality, then FN(A) ≤ FN∗ (PN). (c) Suppose that the continuum hypothesis and CTP(ωω+1 , ωω ) are both true. If A is a measurable algebra, then FN(A) = c = ω1 if ω ≤ τ (A) < ωω , = c+ = ω2 otherwise . proof (a) This is a special case of 518Ca. (b)(i) Consider first the case A = Bκ for some cardinal κ. For I ⊆ κ, let CI be the closed subalgebra of Bκ consisting of those a ∈ Bκ expressible in the form E • for some measurable E ⊆ {0, 1}κ determined by coordinates in I. For a ∈ Bκ , there is a smallest subset Ia of λ such that a ∈ CI (325M); Ia is always countable. For each a ∈ Bκ , set f (a) = {b : Ib ⊆ Ia }. Then #(f (a)) ≤ c. If a ⊆ b, then there is a c ∈ Bκ such that a ⊆ c ⊆ b and Ic ⊆ Ia ∩ Ib (325M(b-ii)5 ). So f is a Freese-Nation function. This shows that FN(Bκ ) ≤ c+ . In general, A is either {0} or isomorphic to a closed subalgebra of Bκ where κ = max(ω, τ (A)), so FN(A) ≤ FN(Bκ ) ≤ c+ by 518Cc. (ii) A is σ-linked (524Mf), so 518D(iii) tells us that FN(A) ≤ FN(PN). (iii) If B ⊆ A is a countably generated order-closed subalgebra, then FN(B) ≤ FN(PN), by (i); so 518I tells us that FN(A) ≤ FN∗ (PN). (c) If τ (A) < ωω then #(A) ≤ max(4, τ (A)ω ) ≤ max(c, τ (A)) < ωω , so we can use (a) and (b-iii); otherwise use (b-i) and 518K. 524P The Maharam classification If the cardinal functions of a Radon measure space are determined by its measure algebra, there ought to be some way of calculating them directly from the classification of measure algebras in §332. In many cases this is straightforward. Theorem Let (X, T, Σ, µ) be a Radon measure space, and A its measure algebra. Let K be the set of infinite cardinals κ such that the Maharam-type-κ component of A is non-zero.

(a)

add µ = add N (µ) = ∞ if K = ∅, = add N if K = {ω}, = ω1 otherwise.

(b)

ci(Σ \ N (µ)) = c(A) if K = ∅, = max(c(A), cf N , sup cf[κ]≤ω ) otherwise. κ∈K

(c)

cov N (µ) = 1 if A = {0}, = ∞ if A has an atom, = cov Nmin K otherwise.

(d)

non N (µ) = ∞ if A = {0}, = 1 if A has an atom, = non Nmin K otherwise.

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524P

shr N (µ) = 0 if A = {0},

(e)

= 1 if A has an atom, ≥ shr N otherwise. (f) If µ is σ-finite, cf N (µ) = 1 if K = ∅, = max(cf N , cf[τ (A)]≤ω ) otherwise. proof If µX = 0 all these results are trivial, so let us suppose henceforth that µX > 0. As in part (a) of the proof of 524J, there is a decomposition hXi ii∈I of X such that the subspace measures µXi are all Maharam-type-homogeneous and non-zero. Note that max(ω, #(I)) = max(ω, c(A)) (332E). For each i ∈ I, let κi be the Maharam type of µXi . (a) By 521Ad, add µ = add N (µ). The map E 7→ hE ∩ Xi ii∈I identifies N (µ), as partially ordered set, with the product of the family hN (µXi )ii∈I . So add N (µ) = mini∈I add N (µXi ) (512Hc). Now if i ∈ I and κi = 0, Xi is an atom of (X, Σ, µ), so there is an xi ∈ Xi such that µ(Xi \ {xi }) = 0 (414G). In this case, Xi \ {xi } is the largest member of N (µXi ) and add N (µXi ) = ∞. If κi is infinite, then add N (µXi ) = add Nκi , by 524I applied to a scalar multiple of µXi . So add N (µ) = minκ∈K add Nκ , interpreting this as ∞ if K = ∅. But we know from 523E that add Nκ = ω1 if κ > ω, while of course add Nω = add N . It follows at once that

add N (µ) = min add N (µXi ) = ∞ if K = ∅, i∈I

= add N if K = {ω}, = ω1 otherwise. (b)(i) For a ∈ A+ choose φ(a) ∈ Σ such that φ(a)• = a, and for E ∈ Σ \ N (µ) set ψ(E) = E • . Then (φ, ψ) is a Galois-Tukey connection from (A+ , ⊇, A+ ) to (Σ \ N (µ), ⊇, Σ \ N (µ)). So ci(Σ \ N (µ)) = cov(Σ \ N (µ), ⊇, Σ \ N (µ)) ≥ cov(A+ , ⊇, A+ ) = ci A+ = π(A) = c(A) if K = ∅, = max(c(A), cf N , sup cf[κ]≤ω ) otherwise κ∈K

by 524Mc. (ii) For i ∈ I, let ΣXi be the subspace σ-algebra on Xi , and set λi = cf Nκi . Then ΣXi \ N (µXi ) has a coinitial set Ei of size at most λi . P P If κi = 0 then Xi is an atom for µ and includes a singleton {x} of non-zero measure, so we can take Ei = {{x}}. Otherwise, we know that π(ΣXi /N (µXi )) = π(Bκi ) = max(cf N , cf[κ]≤ω ) = λi = cf N (µXi ) by 523N, 524Mc and 524I. Let A ⊆ (ΣXi /N (µXi ))+ be a coinitial set of size λi , and for a ∈ A choose Ea ∈ ΣXi such that Ea• = a; let H ⊆ N (µXi ) be a cofinal set of size λi . Set Ei = {Ea \ H : a ∈ A, H ∈ H}. Then Ei ⊆ ΣXi \ N (µXi ) and #(Ei ) ≤ λi . If F ∈ ΣXi \ N (µXi ) there is an a ∈ A such that a ⊆ F • , so Ea \ F is negligible and is covered by some H ∈ H; now F ⊇ Ea \ H ∈ Ei . Thus Ei is coinitial with ΣXi \ N (µXi ). Q Q S S (iii) Now i∈I Ei is coinitial with Σ \ N (µ), so ci(Σ \ N (µ)) ≤ #( i∈I Ei ) is at most #(I) = c(A) if K is empty, and otherwise is less than or equal to max(ω, #(I), sup λi ) = max(c(A), cf N , sup cf[κi ]≤ω ) i∈I

i∈I

= max(c(A), cf N , sup cf[κ]≤ω ). κ∈K

(c) If E is a cover of X by negligible sets, and i ∈ I, then {E ∩ Xi : E ∈ E} is a cover of Xi by negligible sets; thus cov N (µ) ≥ supi∈I cov N (µXi ). By 524Jb, cov N (µ) ≥ supi∈I cov Nκi . If any of the κi is zero, that is, if A has an atom, this is ∞, and we can stop. Otherwise, for each i ∈ I, cov N (µXi ) = cov Nκi ≤ cov Nmin K = λ S say, by 523B. So we have a family hEiξ iξ<λ of negligible subsets of Xi covering Xi ; setting Eξ = i∈I Eiξ for each ξ, we have a family hEξ iξ<λ in N (µ) covering X, so cov N (µ) ≤ cov Nmin K . But we already know that

524P

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129

cov N (µ) ≥ supi∈I cov Nκi ≥ cov Nmin K , so cov N (µ) = cov Nmin K . (d) A set A ⊆ X is non-negligible iff A ∩ Xi is non-negligible for some i ∈ I. It follows at once that non N (µ) = mini∈I non N (µXi ). If any of the Xi is an atom, it contains a point of non-zero measure, so that non N (µ) = 1. If κi ≥ ω for every i, then we have non N (µ) = mini∈I non Nκi = non Nmin K by 524Jb and 523B again. (e) If A is purely atomic, then µ is point-supported, so shr N (µ) = 1. Otherwise, let E be a measurable set of non-zero finite measure such that the subspace measure µE is atomless; let ν be the normalized subspace measure 1 µE ; µE

then ν, like µE , is a Radon measure. By 343Cb, there is a function f : E → {0, 1}ω which is inverse-measure-

preserving for ν and νω ; because {0, 1}ω is separable and metrizable, νf −1 is a Radon measure (451O6 , or 418I-418J) and must be equal to νω (416Eb). By 521Dd and 521Fb, shr N (µ) ≥ shr N (µE ) = shr N (ν) ≥ shr N (νω ) = shr N . (f )(i) If K = ∅ then (a) tells us that N (µ) has a greatest member, so that cf N (µ) = 1. (ii) Now suppose that K is not empty. Then 524F tells us that there is a family hEξ iξ<τ (A) in N (µ) such that S {ξ : Eξ ⊆ E} is countable for every E ∈ N (µ). In this case, J 7→ ξ∈J Eξ : [τ (A)]≤ω → N (µ) is a Tukey function, so cf N (µ) ≥ cf[τ (A)]≤ω . At the same time, if i ∈ I, the identity map from N (µXi ) to N (µ) is a Tukey function; but this means that cf N (µ) ≥ cf N (µXi ) = cf Nκi (524I again) ≥ cf Nω = cf N ≤ω

(523B). Thus cf N (µ) ≥ max(cf N , cf[τ (A)]

).

(iii) In the other direction, we know from 524H (again, applied to a scalar multiple of µXi ) that (N (µXi ), ⊆ ∗ , N (µXi ) ≡GT (κN i , ⊆ , Sκi ) whenever κi is infinite. Now the maps N identity: κN i → τ (A) ,

form a Galois-Tukey connection from

∗ (κN i , ⊆ , Sκi )

S 7→ S ∩ (N × κi ) : Sτ (A) → Sκi

to (τ (A)N , ⊆∗ , Sτ (A) ). Accordingly we have

∗ (N (µXi ), ⊆, N (µXi )) ≡GT (κN i , ⊆ , Sκi )

4GT (τ (A)N , ⊆∗ , Sτ (A) ) ≡GT (Nτ (A) , ⊆, Nτ (A) ), and N (µXi ) 4T Nτ (A) . The arguments quoted assume that κi is infinite; but of course it is still true that N (µXi ) 4T Nτ (A) when κi = 0, since then any constant function from N (µXi ) to Nτ (A) is a Tukey function. It follows that ∼ Q N (µX ) 4T N I N (µ) = i i∈I τ (A) (513Eg). (iv) At this point observe that as we are assuming that K 6= ∅, τ (A) is infinite; and as µ is supposed to be σ-finite, I is countable. So we can find a disjoint family hFi ii∈I of measurable subsets of {0, 1}τ (A) such that all the subspace measures (ντ (A) )Fi are isomorphic to scalar multiples of ντ (A) . (Take Fi = {x : x(ni ) = 1, x(m) = 0 for m < ni } where i 7→ ni : I → N is injective.) In this case, the map S Q hEi ii∈I 7→ i∈I Ei : i∈I N ((ντ (A) )Fi ) → N (ντ (A) ) Q is a Tukey function, while NτI(A) is isomorphic to i∈I N ((ντ (A) )Fi ). Putting these together, Q N (µ) 4T NτI(A) ∼ = i∈I N ((ντ (A ))Fi ) 4T Nτ (A) . It follows that cf N (µ) ≤ cf Nτ (A) = max(N , cf[τ (A)]≤ω ). So we have inequalities in both directions and cf N (µ) = max(N , cf[τ (A)]≤ω ), as claimed. 6 Formerly

451N.

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*524Q

*524Q I do not know how to calculate cf N (µ) for general Radon measures µ without special assumptions. In the presence of GCH, however, we have the following result. Proposition Suppose that the generalized continuum hypothesis is true. Let (X, T, Σ, µ) be a Radon measure space and (A, µ ¯) its measure algebra. For each cardinal κ, write eκ for the Maharam-type-κ component of A, and Cκ for the principal ideal of A generated by supκ0 >κ eκ0 ; set λ = sup{κ : eκ 6= 0}. Then cf N (µ) = max(c(C0 )+ , λ+ ) unless λ > c(C0 ) and there is some γ < λ such that cf λ > c(Cγ ), in which case cf N (µ) = λ. proof (a) Write θ = λ if λ > c(C0 ) and cf λ > min c(Cγ ), γ<λ

= max(λ+ , c(C0 )+ ) otherwise. If µ is purely atomic, it is point-supported, so λ = 0 and C0 = {0} and θ = 1 = cf N (µ). So let us suppose henceforth that µ is not purely atomic, that is, C0 6= {0} and λ ≥ ω. As in the proof of 524P, there is a decomposition hXi ii∈I of X such that the subspace measures µXi are all Maharam-type-homogeneous and non-zero. Let κi be the Maharam type Q of µXi for each i, so that λ = supi∈I κi . Now N (µ) ∼ N (µ ) (see the proof of 524Ja). For i ∈ I, cf N (µXi ) = 1 = i∈I Xi ≤ω if κi = 0, and otherwise is max(cf N , cf[κi ] ) (524Ja, 523N). By 5A1Pb, cf N (µXi ) = 1 if κi = 0, = κi if cf κi > ω, = κ+ i if cf κi = ω. (b) For each cardinal κ, set Jκ = {i : i ∈ I, cf N (µXi ) > κ}, and set λ1 = supi∈I cf N (µXi ). Then 511I tells us that if λ1 > #(J1 ) and there is some γ < λ1 such that cf λ1 > #(Jγ ), then cf N (µ) = λ1 , and that otherwise cf N (µ) = max(#(J1 )+ , λ+ 1 ). As we are supposing that µ is not purely atomic, c(C0 ) ≥ ω and c(C0 ) = max(ω, #(J1 )); also λ+ ≥ λ1 ≥ λ ≥ ω. case 1 Suppose λ1 ≤ #(J1 ). Then J1 is infinite, so c(C0 ) = #(J1 ) ≥ λ, and cf N (µ) = #(J1 )+ = c(C0 )+ = θ as required. case 2 Suppose λ1 > max(λ, #(J1 )). Then there must be some i ∈ I such that cf N (µXi ) > λ ≥ ω, in which case κi = λ has countable cofinality and λ1 = λ+ . In this case, cf λ1 = λ1 > #(J1 ), so cf N (µ) = λ1 . If γ < λ, then Cγ ⊇ Cκi is non-trivial, and cf λ = ω ≤ c(Cγ ); so θ = max(λ+ , #(J1 )+ ) = λ1 = cf N (µ). case 3 Suppose λ1 = λ > #(J1 ) has countable cofinality. In this case we must have κi < λ1 for every i, so #(Jγ ) ≥ ω = cf λ1 for every γ < λ1 , and cf N (µ) = λ+ 1 . At the same time, cf λ = ω ≤ c(Cγ ) for every γ < λ, so + θ = max(λ+ , #(J1 )+ ) = max(λ+ 1 , #(J1 ) ) = cf N (µ).

case 4 Suppose λ1 = λ > #(J1 ) has uncountable cofinality. In this case we have λ > max(ω, #(J1 )) = c(C0 ), so cf N (µ) = λ1 ⇐⇒ #(Jγ ) < cf λ1 for some γ < λ1 ⇐⇒ max(ω, #(Jγ )) < cf λ1 for some γ < λ1 ⇐⇒ c(Cγ ) < cf λ for some γ < λ ⇐⇒ θ = λ ⇐⇒ θ = λ1 , and otherwise + + cf N (µ) = λ+ 1 = max(λ , c(C0 ) ) = θ.

Thus cf N (µ) = θ in all cases. 524R The results above show that most of the most important cardinal functions of measurable algebras and Radon measures are readily calculable from the cardinal functions of the ideals Nκ studied in §523. There are no such simple formulae for other classes of space such as compact or quasi-Radon measures (524Xi). However I can give a handful of partial results, as follows.

524T

Radon measures

131

Proposition Let (X, Σ, µ) be a countably compact σ-finite measure space with Maharam type κ. Then [κ]≤ω 4T N (µ). Consequently cf[κ]≤ω ≤ cf N (µ), and if κ is uncountable then add N (µ) = ω1 and cf N (µ) ≥ cf Nκ . S proof If hEξ iξ<κ is a family as in 524F, then I 7→ ξ∈I Eξ : [κ]≤ω → N (µ) is a Tukey function, if both [κ]≤ω and N (µ) are given their natural partial orderings of inclusion. By 513Ee, cf[κ]≤ω ≤ cf N (µ) and add[κ]≤ω ≥ add N (µ). But if κ is uncountable, add[κ]≤ω = ω1 so add N (µ) is also ω1 . At the same time, cf N (µ) ≥ cf N (521I), so cf N (µ) ≥ max(cf N , cf[κ]≤ω ) = cf Nκ . 524S In a different direction, there is something we can say about quasi-Radon measures. Proposition Let (X, T, Σ, µ) be a Radon measure space, with µX > 0, and (Y, S, T, ν) a quasi-Radon measure space such that the measure algebras of µ and ν are isomorphic. Then (a) N (ν) 4T N (µ), so add ν = add N (ν) ≥ add N (µ) = add µ and cf N (ν) ≤ cf N (µ); (b) (Y, ∈, N (ν)) 4GT (X, ∈, N (µ)), so cov N (ν) ≤ cov N (µ) and non N (ν) ≥ non N (µ). proof (a) Let (Z, U, Λ, λ) be the Stone space of the measure algebra B of (Y, T, ν), and R ⊆ Z × Y the relation described in 415Q, so that R−1 [F ] ∈ N (λ) for every F ∈ N (ν). Let W ⊆ Z be the union of the open sets of finite measure. Then the subspace measure λW is a Radon measure and its measure algebra is isomorphic to the measure algebras of ν and µ (411Pf 7 ). −1 : N (ν) → N (λW ) is a Tukey function. P P?? Otherwise, there is a family A ⊆ NS (ν) such that S Now F 7→ W ∩ R [F ] −1 −1 A∈ / N (ν) but {W ∩ R [A] : A ∈ A} is bounded above in N (λ ). Because W is conegligible, B = [A] W A∈A R S ∗ is negligible in Z. Let E ∈ T be a measurable envelope of A (213J/213L). Then the open-and-closed set E ⊆ Z corresponding to E • ∈ B is not negligible; as λ is inner regular with respect S to the open-and-closed sets (411Pb), there must be a non-empty open-and-closedSset V ⊆ E ∗ which is disjoint from A∈A R−1 [A]. Express V as F ∗ where F ∈ T. Then R[V ] = R[F ∗ ] is disjoint from A. But R[F ∗ ] is measurable and F \ R[F ∗ ]Sis negligible (415Qb), while F \ E must also be negligible, so E ∩ R[F ∗ ] is a non-negligible measurable subset of E \ A, which is impossible. X XQ Q This shows that N (ν) 4T N (λW ). But λW and µ are Radon measures with isomorphic non-zero measure algebras, so N (λW ) ≡T N (µ) (524J) and N (ν) 4T N (µ). Accordingly add N (ν) ≥ add N (µ) and cf N (ν) ≤ cf N (µ) (b) This is a special case of 521Ja. 524T Corollary Let (Y, S, T, ν) be a quasi-Radon measure space, and B its measure algebra. Let K be the set of infinite cardinals κ such that the Maharam-type-κ component of B is non-zero. (a) add ν = add N (ν) = ∞ if K = ∅, ≥ add N if K = {ω}. (b) ci(T \ N (ν)) = c(B) if K = ∅, = max(c(B), cf N , sup cf[κ]≤ω ) otherwise. κ∈K

(c) cov N (ν) = 1 if B = {0}, = ∞ if B has an atom, ≤ cov Nmin K otherwise. (d) non N (ν) = ∞ if B = {0}, = 1 if B has an atom, ≥ non Nmin K otherwise. (e) If ν is σ-finite, cf N (ν) = 1 if K = ∅, ≤ max(cf N , cf[τ (B)]≤ω ) otherwise. 7 Later

editions only.

132

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524T

proof Parts (a), (c), (d) and (e) are just a matter of putting 524P and 524S together. If there are atoms for µ, they may no longer include singletons of non-zero measure; but they do include minimal non-negligible closed sets, so there are non-negligible singletons and cov N (µ), non N (µ) are ∞ and 1 respectively. As for (b), the argument of part (b) of the proof of 524P still applies, with minor modifications. When we come to estimate cf N (νYi ) where Yi ∈ T and the subspace measure νYi is Maharam-type-homogeneous with Maharam type κi , we know that this cofinality is at most λi = cf Nκi , by 524Sa; and this is sufficient for the argument. 524X Basic exercises (a) Suppose that (A, µ ¯) is a probability algebra and that κ = linkn (A), where 2 ≤ n < ω. Show that there are families hA i in A \ {0} and hξ iξ<κ in ]0, 1] such that µ ¯(inf I) ≥ ξ whenever I ∈ [Aξ ]n and ξ ξ<κ S ξ<κ Aξ = A \ {0}. (b) Let (X, Σ, µ) be a localizable measure space with measure algebra A, and A a family of non-negligible (not necessarily measurable) subsets of X such that every non-negligible member of Σ includes a member of A. Show that #(A) ≥ π(A). (c) Show that if κ is uncountable, there is no function f : [0, 1]κ → {0, 1}κ which is almost continuous and inversemeasure-preserving for the usual measures on these spaces. (Hint: if K ⊆ [0, 1]κ is a zero set, any continuous function from K to {0, 1}κ is determined by coordinates in a countable set.) (d) Let I k be the split interval and µ its usual measure (343J). Show that there are f : {0, 1}ω → I k and g : I k → {0, 1}ω such that µ = νω f −1 and νω = µg −1 . (Hint: let A ⊆ [0, 1] be a non-measurable set; define f0 : [0, 1]2 → I k by setting f0 (x, y) = y + if x ∈ A, y − otherwise.) (e) Let (Z, µ) be the Stone space of (Bω , ν¯ω ). Show that there is no f : {0, 1}ω → Z such that µ = νω f −1 . (Hint: 515J.) (f ) Let X be a Hausdorff space with a compact topological probability measure µ with Maharam type κ, and suppose that w(X) < cov Nκ . (i) Show that there is an equidistributed sequence for µ. (Hint: 491Eb.) (ii) Show that if µ is strictly positive then X is separable. (g) Let (X, T, Σ, µ) be a Radon measure space, and K the set of infinite cardinals κ such that the Maharam-type-κ component of its measure algebra A is non-zero. Show that min{#(A) : A ⊆ X has full outer measure} = sup({c(A)} ∪ {non Nκ : κ ∈ K}). (h)SShow that for any σ-ideal I of sets there is a compact probability measure µ such that I = N (µ). (Hint: set X = I ∪ {x0 }.) (i) Show that for any non-zero measurable algebra B and any cardinal κ, there is a complete compact probability measure µ such that the measure algebra of µ is isomorphic to B, add N (µ) = ω1 and cf N (µ) ≥ κ. (Hint: 524Xh.) (j) Let (X, T, Σ, µ) be a Radon probability space with a strong lifting, and (Z, ν) the Stone space of its measure algebra. Show that shr N (µ) ≤ shr N (ν) and shr+ N (µ) ≤ shr+ N (ν). (Hint: 453Mb.) (k) Suppose that non N = cov Nω1 = 2ω1 = ω2 = c. Show that there is a quasi-Radon probability measure µ with Maharam type ω1 such that add N (µ) = ω2 . (l) Show that if m ≥ 2 and hAi ii∈I is a family of σ-m-linked Boolean algebras, with #(I) ≤ c, then the free product of hAi ii∈I is σ-m-linked. 524Y Further exercises (a) Let A be a measurable algebra, λ = τ (A) and κ = FN∗ (A). Show that there is a family V ⊆ [λ]≤c , cofinal with [λ]≤c , such that #({A ∩ V : V ∈ V}) < κ for every countable set A ⊆ λ. (Hint: 518J.) (b) For a Boolean algebra A and a cardinal θ, write ψθ (A) for the smallest size of any subalgebra C of A such that d(C) ≥ θ. (If θ > d(A) set ψθ (A) = ∞.) (i) Show that if Z is the Stone space of A, I is the ideal of nowhere dense sets in Z, and θ ≥ 2 then ψθ (A) ≤ cov([Z]<θ , ⊆, I). (ii) Show that if (X, Σ, µ) is a Maharam-type-homogeneous compact probability space with Maharam type κ, and θ is infinite, then ψθ (Bκ ) = cov([X]<θ , ⊆, N (µ)) = add(Σ \ N (µ), meet, [X]<θ ), where meet is the relation {(A, B) : A ∩ B 6= ∅}. (Hint: start with µ = νκ .) (iii) Show that if (X, Σ, µ) is a semi-finite locally compact measure space with measure algebra A then ψω1 (A) ≤ cf([cov N (µ)]≤ω ). (iv) Show that if (X, Σ, µ) is any probability space, with measure algebra A, and λ is the product probability measure on X N , then cov N (λ) ≤ ψω1 (A). (v) Show that ψadd M (Bω ) ≤ non M, where M is the ideal of meager subsets of R.

525D

Precalibers

133

524Z Problems (a) Let (Z, µ) be the Stone space of (Bω , ν¯ω ). Is shr N (µ) necessarily equal to shr N ? (b) Can there be a quasi-Radon probability measure µ with Maharam type greater than c such that add N (µ) > ω1 ? ´ ski 84, Fremlin 84b 524 Notes and comments The ideas of this section are derived primarily from Bartoszyn and Fremlin 91a. Of course it is not necessary to pass through both `1 (κ) and the κ-localization relation (κN , ⊆∗ , Sκ ). ´ ski 84) because it will be useful when we come to look at other I bring `1 (κ) into the argument (following Bartoszyn structures in later in the chapter, and Sκ because it echoes the ideas of §522. But note that 524G seems to need a new idea (the family hEξ iξ<κ from 524F) not required in 522M. The difficulties of the work above arise from the fact that while there are many inverse-measure-preserving functions between Radon measure spaces, immediately linking covering numbers and uniformities, there are far fewer continuous inverse-measure-preserving functions; for instance, there is no almost continuous inverse-measure-preserving function from the unit interval to the split interval, let alone to the Stone space of its measure algebra. And the straightforward Tukey functions between the ideals Nκ of §523 depend on measures being images of each other, which is something we can rely on only when our functions are almost continuous. (But see 524Xd.) I do not know of any direct construction of a Tukey function from the null ideal of the Stone space of the Lebesgue measure algebra to N , for instance. This is why I have almost had to omit shrinking numbers from this section (see 524Za). There is a significant gap in the calculations in 524P; for the cofinality of the null ideal I need to assume that the measure is σ-finite. I have no useful general recipe for cf N (µ), valid in ZFC, when µ is a non-σ-finite Radon measure. The point is that although we can identify N (µ) with the product of a family N (µXi ) of partially ordered sets to which the arguments of this section apply (524Q), this is not in itself enough to determine its cofinality in the absence of special axioms.

525 Precalibers I continue the discussion of precalibers in §516 with results applying to measure algebras. I start with connexions between measure spaces and precalibers of their measure algebras (525B-525D). The next step is to look at measureprecalibers. Elementary facts are in 525E-525H. When we come to ask which cardinals are precalibers of which measure algebras, there seem to be real difficulties; partial answers, largely based on infinitary combinatorics, are in 525J-525P. 525Q is a note on a particular pair of cardinals. Finally, 525U deals with precaliber triples (κ, κ, k) where k is finite; I approach it through a general result on correlations in uniformly bounded families of random variables (525T). 525A Notation If (X, Σ, µ) is a measure space, N (µ) will be the null ideal of µ. For any set I, νI will be the usual measure on {0, 1}I , TI its domain, NI = N (νI ) its null ideal and (BI , ν¯I ) its measure algebra. In this context, set ei = {x : x ∈ {0, 1}I , x(i) = 1}• in BI for i ∈ I. Then hei ii∈I is a stochastically independent family of elements of measure 21 in BI , and {ei : i ∈ I} τ -generates BI ; I will say that hei ii∈I is the standard generating family in BI . 525B Proposition Let (X, T, Σ, µ) be a quasi-Radon measure space, and A its measure algebra. Then the partially ordered sets (Σ \ N (µ), ⊇) and (A+ , ⊇) are Tukey equivalent. proof For E ∈ Σ \ N (µ) let FE be a closed non-negligible subset of E and set φ(E) = FE• ∈ A+ ; for a ∈ A+ , let ψ(a) be a self-supporting measurable set such that ψ(a)• = a (414F). Then if φ(E) ⊇ a, ψ(a) \ FE is negligible so E ⊇ FE ⊇ ψ(a). Thus (φ, ψ) is a Galois-Tukey connection and (Σ \ N (µ), ⊇, Σ \ N (µ)) 4GT (A+ , ⊇, A+ ). Moreover, if ψ(a) ⊇ E, then a ⊇ φ(E), so (ψ, φ) also is a Galois-Tukey connection and (A+ , ⊇, A+ ) 4GT (Σ \ N (µ), ⊇ , Σ \ N (µ)). 525C Corollary Let (X, T, Σ, µ) be a quasi-Radon measure space, and A its measure algebra. Then the downwards precaliber triples of the partially ordered set (Σ \ N (µ), ⊆) are just the precaliber triples of the Boolean algebra A. proof 516C. 525D Theorem Let (X, T, Σ, µ) be a Radon measure space and (A, µ ¯) its measure algebra. (a) A pair (κ, λ) of cardinals is a precaliber pair of A iff whenever hEξ iξ<κ is a family in Σ \ N (µ) there is an x ∈ X such that #({ξ : x ∈ Eξ }) ≥ λ. (b) A pair (κ, λ) of cardinals is a measure-precaliber pair of (A, µ ¯) iff whenever hEξ iξ<κ is a family in Σ \ N (µ) such that inf ξ<κ µEξ > 0 then there is an x ∈ X such that #({ξ : x ∈ Eξ }) ≥ λ. (c) Suppose that κ ≥ sat(A) is an infinite regular cardinal. Then the following are equiveridical: (i) κ is S a precaliber of A; (ii) µ∗ ( ξ<κ Eξ ) = 0 whenever hEξ iξ<κ is a non-decreasing family in N (µ);

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525D

S

Aξ = X, then there is some ξ < κ such

(iii) whenever hAξ iξ<κ is a non-decreasing family of sets such that that Aξ has full outer measure in X.

ξ<κ

proof (a)(i) Suppose that (κ, λ) is a precaliber pair of A and hEξ iξ<κ is a family in Σ \ N (µ). For each ξ < κ, let Kξ ⊆ Eξ be a non-negligible compact set. Then there is a Γ ∈ [κ]λ such that {Kξ• : ξ ∈ Γ} is centered in A. But in this case {X} ∪ {Kξ : ξ ∈ Γ} has the finite intersection property, and must have non-empty intersection. If x is any point of this intersection, {ξ : x ∈ Eξ } includes Γ and has size at least λ. (ii) Suppose that whenever hEξ iξ<κ is a family in Σ \ N (µ) there is an x ∈ X such that #({ξ : x ∈ Eξ }) ≥ λ. Because µ is complete and strictly localizable (416B), it has a lifting ψ : A → Σ (341K). Let haξ iξ<κ be a family in A \ {0}; then there is an x ∈ X such that Γ = {ξ : x ∈ ψaξ } has cardinal at least λ. But now {ψaξ : ξ ∈ Γ} is centered in Σ so {aξ : ξ ∈ Γ} is centered in A. As haξ iξ<κ is arbitrary, (κ, λ) is a precaliber pair of A. (b) We can use exactly the same argument, provided that in part (i) we make sure that µKξ ≥ inf ξ<κ µ ¯Kξ• > 0.

1 2 µEξ ,

so that

(c)(i)⇒(iii) Suppose that κ is a precaliber of A and hAξ iξ<κ is a non-decreasing family of sets with union X. ?? If no Aξ has full outer measure, then we can choose, for each ξ < κ, a non-negligible compact set Kξ ⊆ X \ Aξ . Because κ is a precaliber of A, there is a set Γ ∈ [κ]κ such that {Kξ• : ξ ∈ Γ} is centered. Now {Kξ : ξ ∈ Γ} has the finite T S intersection property and there is some x ∈ ξ∈Γ Kξ , in which case x ∈ / ξ∈Γ Aξ . But since Γ must be cofinal with κ, S X As hAξ iξ<κ is arbitrary, (iii) is true. ξ∈Γ Aξ = X. X S (iii)⇒(ii) Suppose that (iii) is true, and that hEξ iξ<κ is a non-decreasing family in N (µ). ?? If ξ<κ Eξ has S non-zero inner measure, let E ⊆ ξ<κ Eξ be a non-negligible measurable set. Set Aξ = Eξ ∪ (X \ E) for each ξ; then hAξ iξ<κ is a non-decreasing family with union X, so there is some ξ such that Aξ has full outer measure. But E \ Eξ is a non-negligible measurable set disjoint from Aξ . X X As hEξ iξ<κ is arbitrary, (ii) is true. (ii)⇒(i) Let Z be the Stone space of A and ν its usual measure (411P). Because µ has a lifting, there is an inversemeasure-preserving function f : X → Z (341P). Let hFξ iξ<κ be a non-decreasing family of nowhere dense subsets of Z. Then they are all negligible (411Pa) so S S hf −1 [Fξ ]iξ<κ is a non-decreasing family in N (µ) and µ∗ ( ξ<κ f −1 [Fξ ]) = 0. But this means that if G = int( ξ<κ Fξ ), νG = µf −1 [G] = 0 and G is empty. By 516Rb, κ is a precaliber of A. 525E Proposition Let (A, µ ¯) be a measure algebra. (a) Any precaliber triple of A is a measure-precaliber triple of (A, µ ¯). (b) If (κ, λ, <θ) is a measure-precaliber triple of (A, µ ¯) and κ has uncountable cofinality, then (κ, λ, <θ) is a precaliber triple of A. (c) If κ is a measure-precaliber of (A, µ ¯), so is cf κ. proof (a) is immediate from the definitions. (b) If haξ iξ<κ is any family in A+ , then there is a δ > 0 such that Γ = {ξ : µ ¯aξ ≥ δ} has cardinal κ; and now there is a Γ0 ∈ [Γ]λ such that {aξ : ξ ∈ I} has a non-zero lower bound for every I ∈ [Γ0 ]<θ . (c) The point is that κ is a measure-precaliber of (A, µ ¯) iff it is a precaliber of the supported relation (Aδ , ⊇, A+ ) for every δ > 0, where Aδ = {a : a ∈ A, µ ¯a ≥ δ}; so this is just a special case of 516Bd. 525F Proposition (a) Let (A, µ ¯) be a probability algebra and κ an infinite cardinal. Then κ is a precaliber of A iff either A is finite or κ is a measure-precaliber of (A, µ ¯) and cf κ > ω. (b) An infinite cardinal κ is a precaliber of every measurable algebra iff it is a measure-precaliber of every probability algebra and has uncountable cofinality. proof (a) If κ is a precaliber of A, of course κ is a measure-precaliber of (A, µ ¯). Also cf κ is a precaliber of A (516Bd), so cf κ ≥ sat(A) (516Ja); and if A is infinite, cf κ > ω. If A is finite, then any infinite cardinal is a precaliber of A (516Lc). If κ is a measure-precaliber of (A, µ ¯) and cf κ > ω, then κ is a precaliber of A by 525Eb. (b) Recall that an algebra A is ‘measurable’ iff either A = {0} or there is a functional µ ¯ such that (A, µ ¯) is a probability algebra (391B). So the result follows directly from (a). 525G Proposition Let (A, µ ¯) be a probability algebra. (a) ω is a measure-precaliber of (A, µ ¯). (b) If ω ≤ κ < m(A), then κ is a measure-precaliber of (A, µ ¯).

525I

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135

proof (a) Let han in∈N be a sequence in A such that inf n∈N µ ¯an = δ > 0. Set a = inf n∈N supm≥n am ; then µ ¯a = inf n∈N µ ¯(supm≥n am ) ≥ δ > 0, so a 6= 0. If 0 6= b ⊆ a and n ∈ N, there is an m ≥ n such that b ∩ am 6= 0. We can therefore choose inductively a strictly increasing sequence hni ii∈N such that a ∩ inf j≤i anj 6= 0 for every i, so that hani ii∈N is centered. As han in∈N is arbitrary, ω is a measure-precaliber of (A, µ ¯). (b) If κ = ω, this is (a). Otherwise, let haξ iξ<κ be a family in A with inf ξ<κ µ ¯aξ = δ > 0. Set c = inf J⊆κ,#(J)<κ supξ∈κ\J aξ ; then µ ¯c = inf J⊆κ,#(J)<κ µ ¯(supξ∈κ\J aξ ) ≥ δ. Choose hIξ iξ<κ inductively so that, for each ξ < κ, Iξ is a countable subset of κ \ For ξ < κ, set

S

η<ξ Iη

and c ⊆ supη∈Iξ aη .

Qξ = {b : 0 6= b ⊆ c, ∃ η ∈ Iξ , b ⊆ aη }. + Then Qξ is coinitial with A+ c . Because κ < m(A) ≤ m(Ac ), there is a centered R ⊆ Ac meeting every Qξ . Now

Γ = {η : η < κ, ∃ b ∈ R, b ⊆ aη } meets every Iξ so has cardinal κ, and {aη : η ∈ Γ} is centered. As haξ iξ<κ is arbitrary, κ is a measure-precaliber of (A, µ ¯). 525H As is surely to be expected, questions about precalibers of measurable algebras can generally be reduced to questions about precalibers of the algebras Bκ . Some of these can be quickly answered in terms of the cardinals examined earlier in this chapter. Proposition (a) Let (A, µ ¯) be a totally finite measure algebra. Let K be the set of infinite cardinals κ0 such that 0 the Maharam-type-κ component of A is non-zero (cf. 524M). If κ, λ and θ are cardinals, of which κ is infinite, then (κ, λ, <θ) is a measure-precaliber triple of (A, µ ¯) iff it is a measure-precaliber triple of (Bκ0 , ν¯κ0 ) for every κ0 ∈ K. (b) Suppose that ω ≤ κ < cov Nκ0 . Then κ is a measure-precaliber of Bκ0 . (c) For any cardinal κ0 , ω1 is a precaliber of Bκ0 iff cov Nκ0 > ω1 . (d) If κ, κ0 are cardinals such that non Nκ0 < cf κ, then κ is a precaliber of Bκ0 . proof (a)(i) Suppose that (κ, λ, <θ) is a measure-precaliber triple of (A, µ ¯), κ0 ∈ K and hbξ iξ<κ is a family in Bκ0 0 with inf ξ<κ ν¯κ bξ = δ > 0. Let Aa be a principal ideal of A such that there is an isomorphism π : Bκ0 → Aa with µ ¯(πb) = µ ¯a · ν¯κ0 b for every b ∈ Bκ0 . Then inf ξ<κ µ ¯(πbξ ) = δ µ ¯a > 0, so there is a Γ ∈ [κ]λ such that inf ξ∈I πbξ and <θ inf ξ∈I bξ are non-zero for every I ∈ [Γ] . As hbξ iξ<κ is arbitrary, (κ, λ, <θ) is a measure-precaliber triple of (Bκ0 , ν¯κ0 ). (ii) Suppose that (κ, λ, <θ) is a measure-precaliber pair of (Bκ0 , ν¯κ0 ) for every κ0 ∈ K and haξ iξ<κ is a family in A with inf ξ<κ µ ¯aξ = δ > 0. Let D ⊆ A \ {0} be a partition ofP unity in A such that all the principal ideals Ad , for d ∈ D, are homogeneous. Let C ⊆ D be a finite set such that d∈D\C µ ¯d ≤ 21 δ. Then for every ξ < κ there is a c ∈ C such that µ ¯(aξ ∩ c) ≥ 21 δ µ ¯c, so (because κ is infinite) there are c ∈ C and Γ0 ∈ [κ]κ such that µ ¯(a ∩ c) ≥ 12 δ µ ¯c for every ξ ∈ Γ0 . If c is an atom then inf ξ∈I aξ is non-zero for every I ⊆ Γ0 . Otherwise, the Maharam type κ0 of Ac belongs to K. Let π : Bκ0 → Ac be an isomorphism with µ ¯(πb) = µ ¯c · ν¯κ0 b for every b ∈ Bκ0 . Set bξ = π −1 (aξ ∩ c); 1 λ then ν¯κ0 bξ ≥ 2 δ for every ξ ∈ Γ0 . There is therefore a Γ ∈ [Γ0 ] such that inf ξ∈I bξ and inf ξ∈I aξ are non-zero for every I ∈ [Γ]<θ . As haξ iξ<κ is arbitrary, (κ, λ, <θ) is a measure-precaliber triple of (A, µ ¯). (b) We have cov Nκ0 = m(Bκ0 ) (524Md); so we can use 525Gb. (c) If cov Nκ0 > ω1 then (b) tells us that ω1 is a precaliber of Bκ0 . If cov Nκ0 = ω1 , let hEξ iξ<ω1 be a cover of S 0 {0, 1}κ by negligible sets; then h η<ξ Eη iξ<ω1 is a non-decreasing family in Nκ0 with union of non-zero inner measure, so 525Dc tells us that ω1 is not a precaliber of Bκ0 . (d) If κ0 is finite this is elementary. Otherwise, d(Bκ0 ) = non Nκ0 (524Me). By 516Lc, κ is a precaliber of Bκ0 . 525I The structure of BI Several of the arguments below will depend on the following ideas. Let I be any set and hei ii∈I the standard generating family in BI . If a ∈ BI , there is a smallest countable set J ⊆ I such that a belongs to the closed subalgebra CJ of BI generated by {ei : i ∈ J} (254Rd, 325Mb). (Of course CJ is canonically isomorphic to BJ ; see 325Ma.) Now suppose that haξ iξ∈Γ is a family in BJ , that for each ξ ∈ Γ we are given a set Iξ ⊆ I such that aξ ∈ CIξ , and that J ⊆ I is such that Iξ ∩ Iη ⊆ J for all distinct ξ, η ∈ Γ. Then haξ iξ∈Γ is relatively stochastically independent over CJ (458Ld). It follows that if ∆ ⊆ Γ is finite and inf ξ∈∆ upr(aξ , CJ ) 6= 0, then inf ξ∈∆ aξ 6= 0 (458Lf); in particular, if hupr(aξ , CJ )iξ∈Γ is centered, so is haξ iξ∈Γ .

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Cardinal functions of measure theory

525J

525J Theorem (a)(i) If κ > 0 and (κ, λ, <θ) is a measure-precaliber triple of (Bκ , ν¯κ ), then it is a measure-precaliber triple of every probability algebra. (ii) If κ > 0 and (κ, λ, <θ) is a precaliber triple of Bκ , then it is a precaliber triple of every measurable algebra. (b) Suppose that cf κ ≥ ω2 . If (κ, λ) is a precaliber pair of Bκ0 for every κ0 < κ, then it is a precaliber pair of every measurable algebra. (c) Suppose that (κ, λ, <θ) is a measure-precaliber triple of (Bω , ν¯ω ) and that κ0 is such that cf[κ0 ]≤ω < cf κ. Then (κ, λ, <θ) is a measure-precaliber triple of (Bκ0 , ν¯κ0 ). proof (a)(i) Let (A, µ ¯) be any probability algebra and haξ iξ<κ a family in A+ such that inf ξ<κ µ ¯aξ > 0. Let B be the closed subalgebra of A generated by {aξ : ξ < κ}. Then (B, µ ¯ B) is a probability algebra with Maharam type at most κ, so is isomorphic to a closed subalgebra of (Bκ , ν¯κ ) (332N). Since (κ, λ, <θ) is a measure-precaliber triple of (Bκ , ν¯κ ) it is a measure-precaliber triple of (B, µ ¯ B) (cf. 516Sb), and there is a Γ ∈ [κ]λ such that {aξ : ξ ∈ I} is bounded + + below in B and therefore in A for every I ∈ [Γ]<θ . As haξ iξ<κ is arbitrary, (κ, λ, <θ) is a measure-precaliber triple of (A, µ ¯). (ii) The same argument applies, deleting the phrase ‘inf ξ<κ µ ¯aξ > 0’, since if A is a measurable algebra other than {0} there is a functional µ ¯ such that (A, µ ¯) is a probability algebra. (b) By (a)(ii), it is enough to prove that (κ, λ) is a precaliber pair of Bκ . Let haξ iξ<κ be a family in B+ κ . For each I ⊆ κ, let CI be the closed subalgebra of Bκ generated by {ei : i ∈ I}, as in 525I. Then for each ξ < κ we have a countable set Iξ ⊆ κ such that aξ ∈ CIξ . Because cf κ ≥ ω2 , there are a Γ ∈ [κ]κ and a J ∈ [κ]<κ such that Iξ ∩ Iη ⊆ J for all distinct ξ, η ∈ Γ (5A1Ia). Because #(J) < κ, (κ, λ) is a precaliber pair of BJ ∼ = CJ , so there is a Γ0 ∈ [Γ]λ such that hupr(aξ , CJ )iξ∈Γ0 is centered. It follows that haξ iξ∈Γ0 is centered (525I). As haξ iξ<κ is arbitrary, we have the result. (c) Let haξ iξ<κ be a family in Bκ0 such that ν¯κ0 aξ ≥ δ > 0 for every ξ < κ. Fix a cofinal family J in [κ0 ]≤ω of cardinal less than cf κ. For each ξ < κ let Jξ ∈ J be such that aξ ∈ CJξ , where this time CJξ is interpreted as a subalgebra of Bκ0 . Then there must be some J ∈ J such that A = {ξ : Jξ = J} has cardinal κ. Now (CJ , ν¯κ0  CJ ) is isomorphic to a subalgebra of (Bω , ν¯ω ), so has (κ, λ, <θ) as a measure-precaliber triple, and there is a Γ ∈ [A]λ such that {aξ : ξ ∈ I} has a non-zero lower bound for every I ∈ [Γ]<θ . As haξ iξ<κ is arbitrary, (κ, λ, <θ) is a measure-precaliber triple of (Bκ0 , ν¯κ0 ). 525K Corollary Suppose that κ is an infinite cardinal and κ < cov Nκ . Then κ is a measure-precaliber of every probability algebra. proof By 525Hb, κ is a measure-precaliber of (Bκ , ν¯κ ); by 525Ja, it is a measure-precaliber of every probability algebra. 525L Proposition Let κ > non Nω be a regular cardinal such that cf[λ]≤ω < κ for every λ < κ (e.g., κ = c+ , (c ) , etc.). Then κ is a precaliber of every measurable algebra. + +

proof The point is that κ is a precaliber of Bλ for every λ < κ. P P If λ is finite, this is trivial. Otherwise, d(Bλ ) = non Nλ ≤ max(non Nω , cf[λ]≤ω ) < κ = cf κ by 524Me and 523Ia; it follows that κ is a precaliber of Bλ (516Lc). Q Q By 525Jb, κ is a precaliber of all measurable algebras. 525M If κ > c is not a strong limit cardinal we can do a little better than 525L. Proposition (Dˇ zamonja & Plebanek 04) Suppose that λ and κ are infinite cardinals such that λω < cf κ ≤ κ ≤ 2λ , ω where λ is the cardinal power. Then κ is a precaliber of every measurable algebra. proof By 525Fb and 525J, it is enough to show that κ is a precaliber of Bκ . Let haξ iξ<κ be a family in Bκ \ {0}. Let θ : Bκ → Tκ be a lifting, and for each ξ < κ let Kξ be a non-empty closed subset of θaξ which is determined by coordinates in a countable set Iξ . We may suppose that each Iξ is infinite; let hξ : N → Iξ be a bijection, and set gξ (x) = xhξ for x ∈ {0, 1}κ , so that gξ : {0, 1}κ → {0, 1}N is continuous and Kξ = gξ−1 [gξ [Kξ ]]. As c < cf κ, there is an L ⊆ {0, 1}N such that Γ0 = {ξ : ξ < κ, gξ [Kξ ] = L} has cardinal κ. Because κ ≤ 2λ , there is an f : κ × λω → N such that whenever hξn in∈N is a sequence of distinct elements of κ there is an η < λω such that For each η < λω , set Aη = {ξ : ξ < κ, f (hξ (n), η) = n for S f (ξn , η) = n for every n (5A1Eg). ω every n ∈ N}; then η<λω Aη = κ, while cf κ > λ , so there is an η ∗ < λω such that Γ = Γ0 ∩ Aη∗ has κ members. Fix z ∈ L. For ξ, η ∈ Γ and i, j ∈ N, hξ (i) = hη (j) =⇒ i = f (hξ (i), η ∗ ) = f (hη (j), η ∗ ) = j.

525P

Precalibers

137

So we can find an x ∈ {0, 1}κ such that x(hξ (i)) = z(i) whenever ξ ∈ Γ and i ∈ N; that is, gξ (x) = z for every ξ ∈ Γ. T But this means that x ∈ gξ−1 [L] = Kξ for every ξ ∈ Γ. It follows that whenever I ∈ [Γ]<ω then ξ∈I θaξ 6= ∅ and inf ξ∈I aξ 6= 0; that is, that {aξ : ξ ∈ Γ} is centered. As haξ iξ<κ is arbitrary, κ is a precaliber of Bκ . 525N Proposition Let (A, µ ¯) be a probability algebra and κ an infinite cardinal such that cf κ is a measureprecaliber of (A, µ ¯) and λω < κ for every λ < κ. Then κ is a measure-precaliber of (A, µ ¯). proof (a) Suppose first that A = BI for some set I; let hei ii∈I be the standard generating family in BI . If κ is regular, the result is trivial. Otherwise, let haξ iξ<κ be a family in A+ such that inf ξ<κ ν¯I aξ = δ > 0. There is a strictly increasing family hκα iα cf κ and if α < cf κ and λ < κα then λω < κα . P P All we need to know is that if θ < κ there is a regular uncountable cardinal θ0 such that 0 ω 0 θ ≤ θ < κ and λ < θ whenever λ < θ0 ; and θ0 = (θω )+ has this property. Q Q For each ξ < κ, let Iξ ⊆ I be a countable set such that aξ belongs to the closed subalgebra of A generated by {ei : i ∈ Iξ }. By the ∆-system Lemma (5A1I(a-ii)), thereis for each S α < cf κ a set Γα ⊆ κα+1 \ κα such that #(Γα ) = κα+1 and hIξ iξ∈Γα is a ∆-system with root Jα say. Set J = α
whenever α < cf κ and ξ ∈ cf κ Now recall that we are supposing that such that S cf κ is00 a measure-precaliber of A. So there is a ∆ ∈ [cf κ] 00 {cα : α ∈ ∆} is centered in A. Now Γ = α∈∆ Γα has cardinal κ, and hbξ iξ∈Γ00 is centered. It follows that haξ iξ∈Γ00 is centered (525I). As haξ iξ<κ is arbitrary, κ is a precaliber of A. Q (b) For the general case, observe that by Maharam’s theorem (332B) A is isomorphic to the simple product k∈K Adk of a countable family of homogeneous principal ideals, where hdk ik∈K is a partition Let haξ iξ<κ be a P of unity in A. family in A such that inf ξ<κ µ ¯aξ = δ > 0. Let L ⊆ K be a finite set such that k∈K\L µ ¯dk = δ 0 < δ. Then there is some k ∈ L such that Γk = {ξ : ξ < κ, µ ¯(aξ ∩ dk ) ≥

δ−δ 0 } #(L)

has cardinal κ. Since cf κ is a measure-precaliber of A, it is also a measure-precaliber of Adk (cf. 516Sc). Since (Adk , µ ¯ Adk ) is isomorphic, up to a scalar multiple of the measure, to (BI , ν¯I ) for some I, (a) tells us that κ is a measure-precaliber of Adk . There is therefore a set Γ ∈ [Γk ]κ such that haξ ∩ dk iξ∈Γ and haξ iξ∈Γ are centered. As haξ iξ<κ is arbitrary, κ is a measure-precaliber of A. 525O Proposition (Argyros & Tsarpalias 82) Let κ be either ω or a strong limit cardinal of countable cofinality, and suppose that 2κ = κ+ . Then κ+ is not a precaliber of Bκ . proof By 523Lb, non Nκ > κ. So if we enumerate {0, 1}κ as hxξ iξ<κ+ and set Eξ = {xη : η < ξ} for ξ < κ+ , hEξ iξ<κ+ is an increasing family in Nκ with union {0, 1}κ . By 525Dc, κ+ is not a precaliber of Bκ . 525P As in 523P, GCH decides the most important questions. Proposition Suppose that the generalized continuum hypothesis is true. (a) An infinite cardinal κ is a measure-precaliber of every probability algebra iff cf κ is not the successor of a cardinal of countable cofinality. (b) An infinite cardinal κ is a precaliber of every measurable algebra iff cf κ is neither ω nor the successor of a cardinal of countable cofinality. proof (a)(i) If κ is a measure-precaliber of every probability algebra, so is cf κ (525Ec). By 525O, cf κ cannot be the successor of a cardinal of countable cofinality. (ii) Now suppose that cf κ is not the successor of a cardinal of countable cofinality. Then κ > λω for every λ < κ and cf κ > λω for every λ < cf κ (5A1Pc). By 525L, cf κ is a measure-precaliber of every probability algebra; by 525N, so is κ. (b) Put (a) and 525Fb together.

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Cardinal functions of measure theory

*525Q

*525Q As in 522T, the Freese-Nation number of PN is relevant to the questions here. Proposition (mcountable , FN∗ (PN)) is not a precaliber pair of Bω . proof Fix a nowhere dense compact set K ⊆ {0, 1}ω of non-zero measure. By 518D, the Freese-Nation number of the topology of {0, 1}ω is FN(PN); the regular Freese-Nation numbers are therefore also equal. By 518E and 522Ra, there is a set A ⊆ {0, 1}ω , of cardinal mcountable , such that #(A ∩ F ) < FN∗ (PN) for every nowhere dense set F ⊆ {0, 1}ω . For each x ∈ A, set ax = (K + x)• in Bω , where + here is the usual group operation corresponding to the identification {0, 1}ω ≡ Zω 2 . Then every ax is non-zero. If B ⊆ A and {ax : x ∈ B} is centered, then {K + x : x ∈ B} has the finite intersection property, so there is a y in its intersection; now B ⊆ A ∩ (K + y), and K + y is nowhere dense, so #(B) < FN∗ (PN). Thus hax ix∈A has no centered subfamily of size FN∗ (PN) and witnesses that (mcountable , FN∗ (PN)) is not a precaliber pair of Bω . 525R I turn now to some results which may be interpreted as information on precaliber triples in which the third cardinal is finite. Lemma Let (A, µ ¯) be a semi-finite measure algebra, hun in∈N a k k2 -bounded sequence in L2 = L2 (A, µ ¯)+ , and F a non-principal ultrafilter on N. Suppose that p ∈ [0, ∞[ is such that supn∈N kupn k2 is finite, and set v = limn→F un , w = limn→F upn , the limits being taken for the weak topology in L2 . Then v p ≤ w. proof Of course the positive cone of L2 is closed for the weak topology so v ≥ 0 and we can speak of v p . ?? If v p 6≤ w, there are α, β ≥ 0 such that αp > β and a = [[v > α]] \ [[w > β]] 6= 0. Let b ⊆ a be such that 0 < µ ¯b < ∞ and consider u =

1 χb. µ ¯b

Then, setting q = p/(p − 1) (of course p > 1),

Z Z p (α¯ µb) ≤ ( v) = lim ( un × χb × χb)p ≤ lim kun × χbkp kχbkq )p p

n→F

b

n→F

(by H¨ older’s inequality, 244Eb) p/q

Z

upn

= lim (¯ µb) n→F

p−1

Z

w ≤ β(¯ µb)p < αp (¯ µb)p ,

= (¯ µb)

b

b

which is absurd. X X 525S Lemma Let (A, µ ¯) be a probability algebra and hun in∈N a k k∞ -bounded sequence in L∞ (A, µ ¯)+ such that R k δ = inf n∈N un > 0. Let k0 , . . . , km be strictly positive integers with sum k. Suppose that γ < δ . R Qm kj (a) There are integers n0 < n1 < . . . < nm such that unj ≥ γ. R Qm j=0kj (b) In fact, there is an infinite set I ⊆ N such that j=0 unj ≥ γ whenever n0 , . . . , nm belong to I and n0 < n1 < . . . < nm . k

proof (a) Let F be a non-principal ultrafilter on N. For each j ≤ m, let vj be the limit limn→F unj for the weak topology on L2 (A, µ ¯); let v be the limit limn→F un . By 525R, Z Y m

vj ≥

Z Y m

j=0

(where q =

k , k−1

v

kj

Z =

v k = (kχ1kq kvkk )k

j=0

or ∞ if k = 1) Z ≥


k v × χ1

(by H¨ older’s inequality again, if k > 1) Z =

v


k

Z = lim

n→F

un


k

≥ δ k > γ.

R v k ≥ ( v)k .) We can therefore choose n0 , . . . , nm inductively so that R Qs kj Qm j=0 unj × j=s+1 vj > γ Qm for each s ≤ m (interpreting the final product j=m+1 vj as χ1), since when we come to choose ns we shall be able to use any member of

(Or use 244Xd to show more directly that

R

525T

Precalibers

{n : n > nj for j < s,

R

ukns ×

Qs−1 j=0

139 k

unjj ×

Qm

j=s+1

vj > γ},

which belongs to F so is not empty. At the end of the induction we shall have a sequence n0 < . . . < nm such that R Qm kj j=0 unj ≥ γ, as required. R Qm kj (b) Let J ⊆ [N]m+1 be the family of all sets of the form {n0 , . . . , nm } where n0 < . . . < nm and j=0 unj ≥ γ. By (a), applied to subsequences of hun in∈N , every infinite subset of N includes some member of J . By Ramsey’s theorem (4A1G), there is an infinite I ⊆ N such that [I]m+1 ⊆ J , which is what we need. 525T Theorem (Fremlin 88) Let (A, µ ¯) beR a probability algebra and κ an infinite cardinal. Let huξ iξ<κ be a k k∞ -bounded family in L∞ (A)+ . Set δ = inf ξ<κ uξ . Then for any k ∈ N and γ < δ k+1 there is a Γ ∈ [κ]κ such that R Qk i=0 uξi ≥ γ for all ξ0 , . . . , ξk ∈ Γ. proof (a) It will be helpful to note straight away that it will be enough to consider the case (A, µ ¯) = (BI , ν¯I ) for some set I. P P There is always a (BI , ν¯I ) in which (A, µ ¯) can be embedded. In this case, L∞ (A) can be identified, as f -algebra, with a subspace of L∞ (BI ), and the embedding respects integrals. So we can regard huξ iξ<κ as a family in L∞ (BI ) and perform all calculations there. Q Q At the same time, the case δ = 0 is trivial, so let us suppose henceforth that δ > 0. (b) Next, having fixed on a suitable set I, let hei ii∈I be the standard generating family in BI , and for J ⊆ I let CJ be the closed subalgebra of BI generated by {ei : i ∈ J}; following 325N, I will say that a member of CJ is ‘determined by coordinates in J’. For J ⊆ I let PJ : L1 (BI , ν¯I ) → L1 (CJ , ν¯I CJ ) be the conditional expectation operator. Note that if J, K ⊆ I then PJ PK = PJ∩K (254Ra). ∞ PnIt will be useful to start by looking at a particular subset W of L (BI ), being the set of linear combinations i=0 αi χci where every αi is rational and every ci is determined by coordinates in a finite set. Now PJ [W ] ⊆ W for every J ⊆ I. P P If c ∈ CK where K ⊆ I is finite, then P ν¯ (c ∩ d) χd ∈ W . PJ (χc) = PJ PK (χc) = PJ∩K (χc) = d is an atom of CJ∩K I ν¯I d

As PJ is linear, this is enough. Q Q Observe also that if K ⊆ I is finite, then PK [W ] is countable, being the set of rational linear combinations of {χc : c ∈ CK }. R (c) Suppose, therefore, that we have a set I, a k k∞ -bounded family huξ iξ<κ in L∞ (BI )+ with inf ξ<κ uξ = δ > 0, a k ∈ N and a γ < δ k+1 . To begin with, let us suppose further that (α) uξ ≤ χ1 for every ξ < κ, (β) uξ ∈ W , as described in (b), for each ξ < κ; for each ξ < κ, let Iξ ∈ [I]<ω be such that PIξ uξ = uξ . (i) Suppose that κ = ω. Because there are only finitely many sequences k0 , . . . , km of strictly positive integers R Qm kj with sum equal to k + 1, we can use 525Sb a finite number of times to find an infinite Γ ⊆ ω such that j=0 unj ≥ γ R Qk Pm whenever j=0 kj = k + 1 and n0 < . . . < nm belong to Γ. But in this case we surely have i=0 uni ≥ γ for all n0 , . . . , nk ∈ Γ. (ii) Next, suppose that κ > ω is regular. By the ∆-system Lemma (4A1Db) there is a ∆ ∈ [κ]κ such that hIξ iξ∈∆ is a ∆-system with root J say. Since PJ [W ] is countable, there is a v such that Γ = {ξ : ξ ∈ ∆, PJ uξ = v} has cardinal κ. Of course

R

v=

R

uξ ≥ δ

for every ξ ∈ Γ. By 458Ld, hCIξ iξ∈∆ is relatively independent over CJ . Now suppose that ξ0 , . . . , ξk belong to Γ. Then Z Y k

uξi ≥

Z Y k

i=0

PJ uξi

i=0

(458Lh8 ) Z =

v

k+1

Z ≥

(as in the proof of 525Sa) ≥ δ k+1 ≥ γ, 8 Later

editions only.

v


k+1

140

Cardinal functions of measure theory

525T

and this is what we need to know. (iii) Finally, suppose that κ > cf κ ≥ ω. Set λ = cf κ and let hκζ iζ<λ be a strictly increasing family of regular cardinals greater than λ and with supremum κ. S For each ζ < λ let ∆ζ ⊆ κζ+1 \ κζ be a set of size κζ+1 such that hIξ iξ∈∆ζ is a ∆-system with root Jζ say. Set J = ζ<λ Jζ ; note that #(J) ≤ λ < κζ+1 for every ζ, so that S ∆0ζ = {ξ : ξ ∈ ∆ζ , (Iξ \ Jζ ) ∩ (J ∪ η<κζ Iη ) = ∅} still has cardinal κζ+1 . If ζ < λ and ξ ∈ ∆0ζ , Iξ ∩ J is included in the finite set Jζ ; so {PJ uξ : ξ ∈ ∆0ζ } is countable, and there is a vζ such that ∆00ζ = {ξ : ξ ∈ ∆0ζ , PJ uξ = vζ } has cardinal κζ+1 . Note that

R

vζ =

R

uξ ≥ δ

whenever ζ < λ and ξ ∈ ∆00ζ . R Qk Because λ is regular,Swe can apply (i) or (ii) above to find an A ∈ [λ]λ such that i=0 vζi ≥ γ whenever ζ0 , . . . , ζk ∈ A. Set Γ = ζ∈A ∆00ζ ; because A must be cofinal with λ, #(Γ) = κ. If ξ, η ∈ Γ are distinct, then Iξ ∩ Iη ⊆ J. So hCIξ iξ∈Γ is relatively independent over C. Take any ξ0 , . . . , ξk ∈ Γ; for each i ≤ k, let ζi ∈ A be such that ξj ∈ ∆00ζi . By 458Lh again, R Qn R Qn R Qn i=0 PJ uξi = i=0 vζi ≥ γ, i=0 uξi ≥ so we are done (provided (α)-(β) are true). (c) Now let us unwind these conditions from the bottom. (i) If (α) is true, but (β) is not, take  ∈ ]0, δ[ such that (δ − )k+1 > γ + (k + 1). For each ξ < κ, let u0ξ ∈ W R S be such that u0ξ ≤ χ1 and |uξ − u0ξ | ≤ . (Such a u0ξ exists because {CK : K ∈ [I]<ω } is topologically dense in BI R 0 and u ∧ χ1 ∈ W for every u ∈ W .) Then uξ ≥ δ −  for each ξ, so we can apply (a) to hu0ξ iξ<κ to see that there is a R Qk 0 0 Γ ∈ [κ]κ such that i=0 uξi ≥ γ + (k + 1) for all ξ0 , . . . , ξk ∈ Γ. Now (because uξ and uξ all lie between 0 and χ1) we have Qk Pk Qk uξ − u0 ≤ |uξ − u0 | i=0

i

i=0

ξi

i=0

i

ξi

(see 285O), so that

R Qk

i=0

uξi ≥

R Qk

i=0

u0ξi −

Pk

i=0

R

|uξi − u0ξi | ≥ γ

whenever ξ0 , . . . , ξn ∈ Γ, and the theorem is still true. 1

(ii) Finally, for the general case, set M = 1 + supξ<κ kuξ k∞ , and u0ξ = uξ for ξ < κ. Then every u0ξ belongs M R 0 R Qk δ γ κ 0 to [0, χ1] and uξ ≥ . By (i) there is a Γ ∈ [κ] such that i=0 uξi ≥ M k+1 for all ξ0 , . . . , ξk ∈ Γ; in which case M R Qk i=0 uξi ≥ γ for all ξ0 , . . . , ξk ∈ Γ. This completes the proof. 525U Corollary (Argyros & Kalamidas 82) (a) If κ is an infinite cardinal and k ∈ N, (κ, κ, k) is a measureprecaliber triple of every probability algebra. (b) If κ is a cardinal of uncountable cofinality and k ∈ N, (κ, κ, k) is a precaliber triple of every measurable algebra. In particular, every measurable algebra satisfies Knaster’s condition. (c) If κ is a cardinal of uncountable cofinality, (A, µ ¯) is a probability algebra, k ≥ 1 and haξ iξ<κ is a family in A \ {0}, then there are a δ > 0 and a Γ ∈ [κ]κ such that µ ¯(inf ξ∈I aξ ) ≥ δ for every I ∈ [Γ]k . (d) For any measurable algebra A, m(A) ≥ mK ; and if m(A) > ω1 , then m(A) ≥ mpcω1 . proof Really this is just the special case of 525T in which every uξ is a characteristic function.   (a) If (A, µ ¯) is a probability algebra and haξ iξ<κ is a family in A such that inf ξ<κ µ ¯aξ = δ > 0, take any γ ∈ 0, δ k . R R Q k Setting uξ = χaξ for each ξ, uξ ≥ δ for each ξ, so there is a Γ ∈ [κ]κ such that i=1 uξi ≥ γ for every ξ1 , . . . , ξk ∈ Γ; k in which case inf ξ∈J aξ 6= 0 for every J ∈ [Γ] . As haξ iξ<κ is arbitrary, (κ, κ, k) is a measure-precaliber triple of (A, µ ¯). (b) This now follows at once from 525Eb, since any non-zero measurable algebra can be given a probability measure. Taking κ = ω1 and k = 2, we have Knaster’s condition. (c) For the quantitative version, we have only to note that there must be some α > 0 such that #({ξ : µ ¯aξ ≥ α} has cardinal κ, and take δ < αk . (d) By (b), A satisfies Knaster’s condition; it follows at once that m(A) ≥ mK . If m(A) > ω1 , then ω1 is a precaliber of A (517Ig) so m(A) ≥ mpcω1 .

526A

Asymptotic density zero

141

525X Basic exercises (a) Let (X, Σ, µ) be any measure space and A its measure algebra. (i) Show that (A+ , ⊇) 4T (Σ \ N (µ), ⊇). (ii) Show that a pair (κ, λ) is a downwards precaliber pair of Σ \ N (µ) iff it is a precaliber pair of A. > (b) Let A be a measurable algebra. Show that ω1 is a precaliber of A iff either A is purely atomic or τ (A) ≤ ω and cov Nω > ω1 or cov Nω1 > ω1 . (Hint: 525H, 523F.) > (c) Suppose that add Nω = cov Nω = κ. Show that κ is not a precaliber of Bω . (d) Let (X, Σ, µ) be a complete strictly localizable measure space and A its measure algebra. Show that the supported relation (Σ \ N (µ), 3, X) has the same precaliber pairs as the Boolean algebra A. (e) Suppose that (κ, λ) is a precaliber pair of every measurable algebra, that I is a set, and that X ⊆ R I is a compact set such that #({i : x(i) 6= 0}) < λ for every x ∈ X. Show that #({i : x(i) 6= 0}) < κ for every x belonging to the closed convex hull of X in R I . (Hint: 461I.) (f ) Let κ be a cardinal such that (α) λω < κ for every λ < κ (β) λω < cf κ for every λ < cf κ. Show that κ is a measure-precaliber of every probability algebra. 525Z Problem Can we, in ZFC, find an infinite cardinal κ which is not a measure-precaliber of all probability algebras? From 525O we see that a negative answer will require a model of set theory in which 2κ > κ+ for all strong limit cardinals κ of countable cofinality; for such models see Foreman & Woodin 91, Cummings 92. 525 Notes and comments There seem to be three methods of proving that a cardinal is a precaliber of a measure algebra. First, we have the counting arguments of 516L; since we know something about the centering numbers of measure algebras (524Me), this gives us a start (see the proof of 525L). Next, we can try to use Martin numbers, as in 517Ig and 525G; since we can relate the Martin number of a measure algebra to the cardinals of §523 (524Md), we get the formulation 525K. In third place, we have arguments based on the special structure of measure algebras, using 525I to apply ∆-system theorems from infinitary combinatorics. Subject to the generalized continuum hypothesis, these ideas are enough to answer the most natural questions (525P). Without this simplification, they leave conspicuous gaps. The most important seems to be 525Z. Even if we know all the cardinals add Nκ , cov Nκ , non Nκ and cf Nκ of §523, we may still not be able to determine which cardinals are precalibers; 525Xb is an exceptional special case. I have presented this section with a bias towards measure-precalibers rather than precalibers. When there is a difference, the former search deeper. ‘Cofinality ω1 ’ has a rather special position in this theory (525Jb), deriving from the combinatorial arguments of 5A1H.

526 Asymptotic density zero In §491, I devoted several paragraphs to the ideal Z of subsets of N with asymptotic density zero, as part of an investigation into equidistributed sequences in topological measure spaces. Here I return to Z to examine its place in the Tukey ordering of partially ordered sets. We find that it lies strictly between N N and `1 (526B, 526K) but in some sense is closer to `1 (526F-526G). On the way, I mention the ideal of nowhere dense subsets of N N (526H, 526K) and ideals of sets with negligible closures (526I-526J). 526A Proposition For I ⊆ N, set νI = supn≥1 n1 #(I ∩ n). Then ν is a strictly positive submeasure on PN. We have a metric ρ on PN defined by setting ρ(I, J) = ν(I4J) for all I, J ⊆ N, under which the Boolean operations ∪, ∩, 4 and \ and upper asymptotic density d∗ : PN → [0, 1] are uniformly continuous, PN is complete and Z is a separable closed subset. With the subspace topology, (Z, ⊆) is a metrizably compactly based directed set (definition: 513J). proof (a) It is elementary to check that ν is a strictly positive submeasure. By 392H, ρ is a metric under which the Boolean operations are uniformly continuous. Since |d∗ (I) − d∗ (J)| ≤ d∗ (I4J) ≤ ν(I4J) ∗ −j for all I, J ⊆ N, T d isSuniformly continuous. Let hIj ij∈N be a sequence in PN such that ρ(Ij , Ij+1 ) ≤ 2 for every j ∈ N. Set I = m∈N j≥m Ij . For any m ∈ N and n ≥ 1,

1 #(n ∩ (Im 4I)) n

≤

1 P∞ #(n ∩ (Ij 4Ij+1 )) n j=m

≤

P∞

j=m

ρ(Ij , Ij+1 ) ≤ 2−m+1 ,

so ρ(Im , I) ≤ 2−m+1 . Thus hIj ij∈N is convergent to I; as hIj ij∈N is arbitrary, PN is complete (cf. 2A4E).

142

Cardinal functions of measure theory

526A

(b) Z is a closed subset of PN. P P If I belongs to the closure Z of Z, and  > 0, let J ∈ Z be such that ρ(I, J) ≤ 12 , 1 and let m ≥ 1 be such that n #(J ∩ n) ≤ 12  for every n ≥ m; then n1 #(I ∩ n) ≤  for every n ≥ m. As  is arbitrary, I ∈ Z; as I is arbitrary, Z is closed. Q Q 1 Z is separable because [N]<ω is a countable dense set. (If I ∈ Z and n ∈ N, ρ(a, a ∩ n) ≤ supm>n m #(m ∩ I).) (c) Z is closed under ∪, so is a directed set under ⊆, and ∪ : Z × Z → Z is continuous. If a ∈ Z, then on {b : b ⊆ a} the topology Tρ induced by ρ agrees with the usual compact Hausdorff topology S of PN ∼ P If n ∈ N and = {0, 1}N . P 1 ρ(b, c) < n+1 , then b ∩ n = c ∩ n; so Tρ is finer than S on PN. If  > 0, there is an m ∈ N such that #(a ∩ n) ≤ n whenever n ≥ m; now ρ(b, c) ≤  whenever b, c ⊆ a and b ∩ m = c ∩ m. So S is finer than Tρ on {b : b ⊆ a}. Q Q Since {b : b ⊆ a} is S-compact, it is Tρ -compact. Now suppose that han in∈N is a sequence S in Z with Tρ -limit a. Then it has a subsequence hank ik∈N such that ρ(a, ank ) ≤S2−k for every k. Set b = k∈N ank . Then, given  > 0, let r, m ∈ N be such that 2−r ≤  and #(n ∩ (a ∪ k≤r ank )) ≤ n for every n ≥ m; then [

#(n ∩ b) ≤ #(n ∩ (a ∪

k≤r

≤ n +

∞ X

∞ X

ank )) +

#(ank \ a)

k=r+1

2−k n ≤ 2n

k=r+1

for every n ≥ m. So b ∈ Z and {ank : k ∈ N} is bounded above in Z. 526B Proposition (Fremlin 91a) N N 4T Z 4T `1 . proof (a) For α ∈ N N , set φ(α) = {2n i : n ∈ N, i ≤ α(n)}. Then φ(α) ∈ Z, because if n ∈ N then #(m ∩ φ(α)) ≤ (n + 1)α(n) + d2−n me for every m. Also φ : N N → Z is a Tukey function, because if φ(α) ⊆ a ∈ Z then α(n) ≤ min{i : 2n i ∈ / a} for every n ∈ N. 1 n

(b) For a, b ∈ Z set ρ(a, b) = supn≥1 #((a4b) ∩ n). Then Z is complete and separable and the lattice operation ∪ is uniformly continuous (526A). By 524C, (Z, ⊆ 0 , [Z]<ω ) 4GT (`1 (ω), ≤, `1 (ω)). Since Z is upwards-directed, (Z, ⊆ , Z) ≡GT (Z, ⊆ 0 , [Z]<ω ) (513Id) and (Z, ⊆, Z) 4GT (`1 , ≤, `1 ), that is, Z 4T `1 . 526C The next three lemmas are steps on the way to Theorem 526F. I give them in much more generality than is required by that theorem because a couple of them will be useful later, and I think they are interesting in themselves. But if you are reading this primarily for the sake of 526F, you might save time by looking ahead to the proof there and working backwards, extracting arguments adequate for the special case of 526E which is actually required. Q Lemma Let h(An , µ ¯n )in∈N be a sequence of purely atomic probability algebras, and A = n∈N An the simple product algebra. Then there is an order-continuous Boolean homomorphism π : A → PN such that lim supn→∞ µ ¯n a(n) is the upper asymptotic density d∗ (πa) for every a ∈ A; consequently, limn→∞ µ ¯n a(n) is the asymptotic density d(πa) of πa if either is defined. proof (a) For each ¯n c Pn ∈ N, let Cn be the set of atoms of An , and choose rational numbers αn (c) such that αn (c) ≤ µ for each c ∈ Cn , c∈Cn αn (c) > 1 − 2−n , and {c : c ∈ Cn , αn (c) > 0} is finite. Express αn (c) as rn (c)/sn for each c ∈ Cn , where rn (c) S ∈ N and sn ≥ 1; let hIn (c)ic∈Cn be a disjoint family of subsets of N with #(In (c)) = rn (c) for each c, and set Jn = c∈Cn In (c); let πn : An → PJn be the Boolean homomorphism such that πn (c) = In (c) for each c ∈ Cn . Then P P (1 − 2−n )sn < sn c∈Cn αn (c) = c∈Cn rn (c) = #(Jn ) ≤ sn . Note that #(Jn ) > 0. Also X

#(Jn )(¯ µn d − 2−n ) ≤ #(Jn )

αn (c)

c∈Cn ,c⊆d

≤ sn

X c∈Cn ,c⊆d

αn (c) =

X c∈Cn ,c⊆d

rn (c) = #(πn d)

526D

Asymptotic density zero

143

and (1 − 2−n )#(πn d) = (1 − 2−n )sn

P

αn (c) ≤ #(Jn ) · µ ¯n d

c∈Cn ,c ⊆ d

for every d ∈ An . So, for a ∈ A, lim supn→∞ µ ¯n a(n) = lim supn→∞

#(πn a(n)) . #(Jn )

(b) Let hmn in∈N be a sequence in N such that P mn #(Jn ) ≥ 2n max(#(Jn+1 ), i
#(πn a(n)) , #(Jn )

1 l

γ 0 = lim supl→∞ #(l ∩ πa). P 0 0 Then γ ≤ γ 0 . P P Setting ln0 = i
1 0 #(ln+1 0 ln+1

∩ πa) ≥ lim sup n→∞

mn #(πn a(n)) = γ. Q Q (1 + 2−n )mn #(Jn )

∗

Also γ 0 ≤ γ. P P Let  > 0. Let n∗ ≥ 1 be such that 2−n ≤  and #(πn a(n)) ≤ (γ + )#(Jn ) for every n ≥ n∗ . 0 Suppose that l ≥ ln0 ∗ +1 . Then l is of the form ln+1 + j#(Jn+1 ) + i where n ≥ n∗ , j < mn+1 and i < #(Jn+1 ). Now 0 0 ln+1 = ln + mn #(Jn ), so 0 #(ln+1 ∩ πa) ≤ ln0 + mn #(πn a(n)) ≤ ln0 + mn #(Jn )(γ + )

≤ mn #(Jn )(γ +  + 2−n ) ≤ mn #(Jn )(γ + 2) by the choice of mn . Accordingly #(l ∩ πa) ≤ mn #(Jn )(γ + 2) + (j + 1)#(πn+1 a(n + 1)) ≤ mn #(Jn )(γ + 2) + j(γ + )#(Jn+1 ) + #(Jn+1 ) ≤ (γ + 2)l + #(Jn+1 ) ≤ (γ + 2)l + 2−n mn #(Jn ) (by the choice of mn ) ≤ (γ + 3)l. As this is true for any l ≥ ln0 ∗ +1 , γ 0 ≤ γ + 3; as  is arbitrary, γ 0 ≤ γ. Q Q (d) Thus 1 n

lim supn→∞ µ ¯n a(n) = lim supn→∞ #(n ∩ πa) for every a ∈ A. But as π is a Boolean homomorphism, it follows at once that 1 n

lim inf n→∞ µ ¯n a(n) = lim inf n→∞ #(n ∩ πa) for every a, so that the limits are equal if either is defined. 526D Lemma Let (A, µ ¯) be a semi-finite measure algebra, and κ ≥ max(ω, c(A), τ (A)) a cardinal. Let (Bκ , ν¯κ ) be the measure algebra of the usual measure on {0, 1}κ , and γ > 0. Then there is a function f : A → Bκ such that (i) f (sup A) = sup f [A] for every non-empty A ⊆ A such that sup A is defined in A; (ii) ν¯κ f (a) = 1 − e−γ µ¯a for every a ∈ A, interpreting e−∞ as 0;

144

Cardinal functions of measure theory

526D

(iii) if hai ii∈I is a disjoint family in A, and Ci is the closed subalgebra of Bκ generated by {f (a) : a ⊆ ai } for each i, then hCi ii∈I is stochastically independent. ¯ in place of (Bκ , ν¯κ ). Set Af = {a : proof (a) By 495J, we have exactly this result for some probability algebra (B, λ) f f a ∈ A, µ ¯a < ∞}, and give A its measure metric ρ (323Ad). Then f  A is uniformly continuous for ρ and the measure metric σ of B. P P If  > 0, there is a δ > 0 such that |e−γs − e−γt | ≤ 21  whenever s, t ∈ [0, ∞[ and |s − t| ≤ δ. Now if 0 a, a ∈ A and µ ¯(a 4 a0 ) ≤ δ, set b = a ∩ a0 ; then f (b) ⊆ f (a) and µ ¯a − µ ¯b ≤ δ, so ¯ (a) \ f (b)) = λf ¯ (a) − λf ¯ (b) = e−γ µ¯b − e−γ µ¯a ≤ 1 . σ(f (a), f (b)) = λ(f 2

Similarly, σ(f (a0 ), f (b)) ≤ 12  so σ(f (a), f (b)) ≤ . As  is arbitrary, this gives the result. Q Q (b) By 521Ob, there is a set B ⊆ Af , of cardinal at most κ, which is dense for ρ. Accordingly f [B] is dense in f [Af ] for σ. Taking D to be the closed subalgebra of B generated by f [B], τ (D) ≤ κ and f [Af ] ⊆ D. But if a ∈ A \ Af then f (a) = 1, so f [A] ⊆ D. Now there is a measure-preserving Boolean homomorphism g : D → Bκ (332N), and gf : A → Bκ has the properties we need. 526E Lemma Let h(An , µ ¯n )in∈N be a sequence of finite probability algebras and hγn in∈N a sequence in ]0, ∞[. Write P for the set Q {p : p ∈ n∈N An , limn→∞ γn µ ¯n p(n) = 0}, Q with the ordering inherited from the product partial order on n∈N An . Then P 4T Z. proof (a) By 526D, we can find for each n a probability algebra (Bn , ν¯n ) and a function fn : An → Bn such that, for all a, a0 ∈ An , fn (a ∪ a0 ) = fn (a) ∪ fn (a0 ), ν¯n fn (a) = 1 − exp(−γn µ ¯an ). Q Q We may suppose that Bn is generated by fn [An ], so is itself finite. Set A = n∈N An , B = n∈N Bn , f (p) = hfn (p(n))in∈N for p ∈ A; then f (sup A) = sup f [A] for any non-empty subset A of A. Set Q Q = {q : q ∈ n∈N Bn , limn→∞ ν¯n q(n) = 0}. Then f P is a Tukey function from P to Q. P P P = f −1 [Q], because limn→∞ γn ξn = 0 iff limn→∞ 1 − e−γn ξn = 0. So f P is a function from P to Q. If q ∈ Q, A = {p : p ∈ A, f (p) ⊆ q} has a supremum p0 ∈ A; now f (p0 ) = sup f [A] ⊆ q, so f (p0 ) ∈ Q and p0 ∈ P is an upper bound for A in P . Q Q (b) By 526C, we have an order-continuous Boolean homomorphism π : B → PN such that π(q) ∈ Z iff q ∈ Q. Now πQ is a Tukey function from Q to Z. P P If d ∈ Z, set B = {q : q ∈ B, π(q) ⊆ d}. Because π is an order-continuous Boolean homomorphism, B contains its supremum, and B is bounded above in Q. Q Q (c) Thus πf P : P → Z is a Tukey function and P 4GT Z. 526F Theorem (`1 , ≤, `1 ) 4GT (N N , ≤, N N ) n (Z, ⊆, Z). proof (a) Let Q ⊆ N N be the set of strictly increasing sequences α such that α(0) > 0. For α ∈ Q, set ∞ X

Pα = {x : x ∈ `1 , kxk∞ ≤ α(0), lim 2n n→∞

x(i)+ = 0}

i=α(n) α(n+1)−1

X

= {x : x ∈ `∞ , kxk∞ ≤ α(0), lim 2n

x(i)+ = 0}

n→∞

i=α(n)

because 2n

∞ X i=α(n)

x(i)+ =

∞ X

α(m+1)−1

X

2n−m 2m

m=n

x(i)+

i=α(m) α(m+1)−1

≤ 2 sup 2m m≥n

for every n and x.

X i=α(m)

x(i)+

526G

Asymptotic density zero

145

The point is that Pα 4T Z. P P For each n ∈ N set kn = 22n (α(n + 1) − α(n)), Vn = (α(n + 1) \ α(n)) × kn α(0) ⊆ N × N,

An = PVn ,

and let µ ¯n be the uniform probability measure on An , so that µ ¯n d = #(d)/#(Vn ) for d ⊆ Vn . For x ∈ `∞ , n ∈ N set fn (x) = {(i, j) : α(n) ≤ i < α(n + 1), j < kn min(α(0), x(i))} ⊆ Vn . Then |#(fn (x)) − kn

P

α(n)≤i<α(n+1)

min(α(0), x(i)+ )| ≤ α(n + 1) − α(n),

so if kxk∞ ≤ α(0) then α(n+1)−1 X n 2 α(0)(α(n + 1) − α(n))¯ µfn (x) − 2n x(i)+ ≤ 2n (α(n + 1) − α(n))/kn i=α(n)

= 2−n . Accordingly Pα = {x : x ∈ `∞ , kxk∞ ≤ α(0), limn→∞ 2n α(0)(α(n + 1) − α(n))¯ µn fn (x) = 0}. Let A =

Q

n∈N

An be the simple product of the Boolean algebras An , and let I be the ideal {a : a ∈ A, limn→∞ 2n α(0)(α(n + 1) − α(n))¯ µn a(n) = 0}.

For x ∈ `∞ , set f (x) = hfn (x)in∈N . Observe that f : `∞ → A is supremum-preserving in the sense that f (sup A) = sup f [A] for any non-empty bounded subset A of `∞ . The last formula for Pα shows that f (x) ∈ I for every x ∈ Pα . But if a ∈ I, A = {x : x ∈ `∞ , f (x) ⊆ a} is upwards-directed and has a supremum x0 ∈ `∞ , with kx0 k∞ ≤ α(0). Now f (x0 ) = supx∈A f (x) ⊆ a, so x0 ∈ Pα and is an upper bound for A in Pα . Thus f Pα is a Tukey function from Pα to I. By 526E, there is a Tukey function g : I → Z. Now gf Pα is a Tukey function from Pα to Z. Q Q Thus (Pα , ≤, Pα ) 4GT (Z, ⊆, Z); it follows at once that (Pα , ≤, `1 ) 4GT (Z, ⊆, Z). (b) Now, for α ∈ N N , take Pα0 = Pβ where β(n) = 1 + n + maxi≤n α(i) for n ∈ N. Then Pα ⊆ Pα0 whenever α ≤ α0 S N in N , α∈N N Pα0 = `1 and (Pα0 , ≤, `1 ) 4GT (Z, ⊆, Z) for every α; so (`1 , ≤, `1 ) 4GT (N N , ≤, N N ) n (Z, ⊆, Z), by 512K. 526G Corollary Let N be the ideal of Lebesgue negligible subsets of R. (a) addω Z = add N and cf Z = cf N . (b) If A ⊆ Z and #(A) < add N , there is a J ∈ Z such that I \ J is finite for every I ∈ A. proof (a)(i) Putting 526B and 513Ie together, we see that addω N N ≥ addω Z ≥ addω `1 , that is, b ≥ addω Z ≥ add N (522A, 524I). Next, we can deduce from 526F that addω `1 ≥ min(addω N N , addω Z). P P Let (φ, ψ) be a Galois-Tukey connection from `1 to N

(N N , ≤, N N ) n (Z, ⊆, Z) = (N N × Z N , T, N N × Z), where T = {((p, f ), (q, a)) : p ≤ q in N N , f (q) ⊆ a ∈ Z}. We can interpret φ as a pair (φ1 , φ2 ) where φ1 is a function from `1 to N N and φ2 is a function from `1 × N N to Z, and saying that (φ, ψ) is a Galois-Tukey connection means just that if φ1 (x) ≤ q and φ2 (x, q) ⊆ a then x ≤ ψ(q, a). 1

Now suppose that A ⊆ ` and #(A) < min(addω N N , addω Z). Then there is a sequence hqn in∈N in N N such that for every x ∈ A there is an n ∈ N such that φ1 (x) ≤ qn . Next, for each n ∈ N there is a sequence hanm im∈N in Z such that for every x ∈ A there is an m ∈ N such that φ2 (x, qn ) ⊆ anm . In this case, B = {ψ(anm ) : m, n ∈ N} is a countable subset of Z, and for every x ∈ A there are m, n ∈ N such that φ1 (x) ≤ qn and φ2 (x, qn ) ⊆ anm , so that x ≤ ψ(anm ) ∈ B. As A is arbitrary, addω `1 ≥ min(addω N N , addω Z). Q Q Thus we have add N ≥ min(b, addω Z) = addω Z, and addω Z = add N . (ii) On the other hand, 524I, 526F, 512Da and 512Jb tell us that

146

Cardinal functions of measure theory

526G

cf N = cf `1 = cov(`1 , ⊆, `1 ) ≤ cov((N N , ≤, N N ) n (Z, ⊆, Z)) = max(cov(N N , ≤, N N ), cov(Z, ⊆, Z)) = max(d, cf Z). But from 526B we see that d ≤ cf Z ≤ cf `1 , so cf Z = cf N . (b) By (a), there is a countable set D ⊆ Z such that every member of A is included in a member of D. By 491Ae, there is a J ∈ Z such that I \ J is finite for every I ∈ D; this J serves. 526H For a theorem in §527 it will be useful to know a little about the Tukey classification of a familiar ideal. Lemma Let I be the ideal of nowhere dense subsets of N N and M the ideal of meager subsets of N N . (a) I is isomorphic, as partially ordered set, to I N . (b) (I, ⊆ 0 , [I]≤ω ) ≡GT (M, ⊆, M). (c) Let X be a set and V a countable family of subsets of X. Set D = {D : D ⊆ X, for every V ∈ V there is a V 0 ∈ V such that V 0 ⊆ V \ D}. Then D 4T I. (d) If X is any non-empty Polish space without isolated points, and IX is the ideal of nowhere dense subsets of X, then I ≡T IX . (e) If X is a compact metrizable space and Cnwd is the family of closed nowhere dense subsets of X with the Fell (or Vietoris) topology (4A2T), then (Cnwd , ⊆) is a metrizably compactly based directed set. Remark Recall that if R is any relation then R 0 is the relation {(x, B) : (x, y) ∈ B for some y ∈ B} (see 512F). S proof Enumerate S ∗ = n∈N N n as hσn in∈N . For σ ∈ S ∗ write Iσ = {α : σ ⊆ α ∈ N N }. (a) Define φ : I → I N by setting φ(F )(n) = {α : a α ∈ F }, where a α = (n, α(0), α(1), . . . ) for n ∈ N and α ∈ N N . Then φ is an isomorphism between I and I N . S (b)(i) Choose φ : M → I N such that M ⊆ n∈N φ(M )(n) for every m ∈ M. Then φ is a Tukey function so M 4T I N ∼ = I, that is, (M, ⊆, M) 4GT (I, ⊆, I). By 513Id and 512Gb, (M, ⊆, M) ≡GT (M, ⊆ 0 , [M]≤ω ) 4GT (I, ⊆ 0 , [I]≤ω ). (ii) For n ∈ N and τ ∈ N n , define gτ : N N → N N by saying that gτ (α) = τ a α, that is, gτ (α)(i) = τ (i) if i < n, = α(i − n) otherwise. Note that gτ is a homeomorphism between N N and Iτ , so that gτ [F ] and gτ−1 [F ] are nowhere dense whenever F is. Now for any F ∈ I we can find a φ(F ) ∈ I such that F ⊆ φ(F ) and for every σ ∈ S ∗ either Iσ ∩ φ(F ) = ∅ or there is a τ ∈ S ∗ , extending σ, such that gτ [F ] ⊆ φ(F ). P P Choose S hτn in∈N , hυn in∈N inductively, as follows. Given that Iυi ∩ (F ∪ gτj [F ]) = ∅ for all i, j < n, set E = Iσn ∩ (F ∪ j
526I

Asymptotic density zero

147

(c) If V = ∅ then D = PX has a greatest element and the result is trivial (any function from D to F will be a Tukey function). Otherwise, choose a function h : S ∗ → V ∪ {X} such that h(∅) = X and hh(σ a )ii∈N runs over {V : h(σ) ⊇ V ∈ V} for every σ ∈ S ∗ . Note that h(τ ) ⊆ h(σ) whenever τ ⊇ σ, and that {h(τ ) : σ ⊆ τ ∈ N n } = {V : V ∈ V, ∗ V ⊆ h(σ)} whenever m ∈ N, σ ∈ N m and n > m. For each D ∈ D we can choose a sequence S hτDn in∈N in S such that N τDn ⊇ σn and D ∩ h(τDn ) is empty and #(τDn ) ≥ n for every n ∈ N. Set φ(D) = N \ n∈N IτDn , so that φ(D) ∈ I. S Take any F ∈ I, and set D0 = {D : D ∈ D, φ(D) ⊆ F }. Then D0 ∈ D. P P Let V ∈ V. Let υ ∈ N 1 be such that h(υ) = V . Take τ ⊇ υ such that F ∩ Iτ = ∅. ?? S If D0 ∩ h(τ ) 6= ∅, then there is a D ∈ D such that φ(D) ⊆ F and D ∩ h(τ ) 6= ∅. Iτ ∩ φ(D) is empty, that is, Iτ ⊆ n∈N IτDn ; because #(τDn ) ≥ n for every n, this can happen only because there is some n ∈ N such that τDn ⊆ τ . But this means that D ∩ h(τ ) ⊆ D ∩ h(τDn ) = ∅, which is impossible. X X Thus D0 ∩ h(τ ) is empty, and h(τ ) is a member of V included in V \ D0 . As V is arbitrary, D0 ∈ V. Q Q As F is arbitrary, φ is a Tukey function and D 4T I, as claimed. (d)(i) Taking V to be a countable base for the topology of X not containing ∅, we have IX = {F : F ⊆ X, for every V ∈ V there is a V 0 ∈ V such that V 0 ⊆ V \ F }, so (c) tells us that IX 4T I. (ii) X has a dense subset Y which is homeomorphic to N N (5A4Je). Let IY be the family of nowhere dense subsets of Y . For F ∈ IY let φ(F ) be its closure in X. Then φ is a Tukey function from IY to IX , so I ∼ = IY 4T IX . (e) By 4A2Tg9 , the Fell topology on the family C of all closed subsets of X is compact and metrizable. E ∪ F ∈ Cnwd for all E, F ∈ Cnwd , and ∪ : Cnwd × Cnwd → Cnwd is continuous (4A2T(b-ii)). If F ∈ Cnwd , the set {E : E ∈ Cnwd , E ⊆ F } = {E : E ∈ C, E ∪ F = F } is closed in C, therefore compact. Now suppose that hEk ik∈N is a sequence in Cnwd converging to E ∈ Cnwd . If X = ∅ then of course {Ek : k ∈ N} is bounded above in Cnwd . Otherwise, let hUn in∈N run over a base for the topology of X not containing ∅, and choose hkn in∈N S, hVn in∈N inductively, as follows. Given ki ∈ N for i < n, let Vn ⊆ Un be a non-empty open set such that V n ∩ (E ∪ i 1 − 2−k }. Set Ek = {E : E ⊆ X, for every U ∈ Vk there is a V ∈ Vk such that V ⊆ U \ E}. T Then E = k∈N Ek . P P Because X ∈ Vk , µE ≤ 2−k for every E ∈ Ek , so k∈N Ek ⊆ E. On the other hand, if E ∈ E and k ∈ N and V ∈ Vk , then µ(V \ E) > 1 −S2−k and U 0 = {U : U ∈ U, U ⊆ V \ E} has union V \ E. As U 0 is countable, greater than 1 − 2−k , so that U ∈ Vk and U ⊆ V \ E. As V there is a finite U10 ⊆ U 0 such that U = U10 has measure T is arbitrary, E ∈ Vk ; as E and k are arbitrary, E ⊆ k∈N Ek . Q Q Q Q This means that the map E 7→ (E, E, E, . . . ) is a Tukey function from E to k∈N Ek , so that E 4T k∈N Ek . At the same time, Ek 4T I for every k, by 526Hc. So E 4T I N ≡ I (513Eg, 526Ha). T

(c) Let A be the countable subalgebra of PX generated by U. Then there is a Boolean homomorphism π : A → PN such that d(πE) is defined and equal to µE for every E ∈ U. P P This is easy to prove directly (see 491Xu), but we can also argue as follows. Let hAn in∈N beQa non-decreasing sequence of finite subalgebras of A with union A. By 526C, we have a Boolean homomorphism π 0 : n∈N An → PN such that d(π 0 hEn in∈N ) = limn→∞ µEn whenever En ∈ An for every n and the limit on the right is defined. For each n ∈ N let πn : A → An be a Boolean homomorphism extending the identity homomorphism on An (314K, or otherwise); set πE = π 0 hπn Ein∈N for E ∈ A; this works. Q Q Let hVn in∈N be a sequence running over the closed sets belonging to A. Let hkn in∈N be a strictly increasing sequence in N\{0} such that

1 #(kn ∩πE) kn

Define φ : E → PN by setting 9 Later

editions only.

≥ µE −2−n whenever E belongs to the subalgebra Bn of A generated by {Vi : i ≤ n}.

148

Cardinal functions of measure theory

φ(E) =

T

526I

{ki ∪ πVi : i ∈ N, E ⊆ Vi }.

Then φ is a Tukey function from E to Z. P P (i) If E ∈ E and  > 0 there is a U ∈ U such that U ⊆ X \ E and µU ≥ 1 − . Let i ∈ N be such that X \ U = Vi ; then φ(E) ⊆ ki ∪ πVi , so d∗ (φ(E)) ≤ d∗ (Vi ) = µVi ≤ . As  is arbitrary, φ(E) ∈ Z. Thus φ is a function from E to Z. S S / φ(E) for every E ∈ A, (ii) Take any A ⊆ E, and set F = A, a = E∈E φ(E). If n ∈ N and i ∈ S kn \ a, then i ∈ so for S every E ∈ A there is a j < n such that E ⊆ V and i ∈ / πV . Set F = {V : j < n, i ∈ / πVj }, so that i ∈ / πFni , j j ni j T while A ⊆ Fni and F ⊆ Fni . Set Fn = i∈kn \a Fni , so that F ⊆ Fn and no member of kn \ a belongs fo πFn , that is, kn ∩ πFn ⊆ a. Note that Fn ∈ Bn . So we have 1 #(kn kn

∩ a) ≥

1 #(kn kn

∩ πFn ) ≥ µFn − 2−n ≥ µF − 2−n .

This means that d∗ (a) ≥ µF . So if {φ(E) : E ∈ A} is bounded above in Z, A must be bounded above in I; that is, φ is a Tukey function. Q Q Thus E 4T Z also. 526J Proposition Let ELeb be the ideal of subsets of R whose closures are Lebesgue negligible. Then N N 4T ELeb but ELeb 64T N N . proof (a) Enumerate Q ∩ [0, 1] as hqi ii∈N . Define φ : N N → ELeb by setting φ(f )(n) = {n + qi : n ∈ N, i ≤ S f (n)}. Then it is easy to see that φ is a Tukey function, because if F ⊆ N N and {f (n) : f ∈ F } is unbounded, then f ∈F φ(f ) is dense in [n, n + 1] so does not belong to ELeb . (b) Let ψ : ELeb → N N be any function. Let µ be Lebesgue measure on R, and choose hf (n)in∈N inductively in N such that µ∗ {t : t ∈ [0, 1], ψ({t})(i) ≤ f (i) for every i ≤ n} > 21 for every n. Set T An = {t : t ∈ [0, 1], ψ({t})(i) ≤ f (i) for every i ≤ n}, F = n∈N An so that µF ≥ 21 . Let hUn in∈N enumerate the set of rational open intervals of R which meet F , and for each n ∈ N choose tn ∈ An ∩ Un . Then ψ({tn })(i) ≤ f (i) whenever n ≥ i, so {ψ({tn }) : n ∈ N} is bounded above in N N ; but {tn : n ∈ N} includes F , so {{tn } : n ∈ N} is not bounded above in ELeb . Thus ψ cannot be a Tukey function. 526K Proposition Let I be the ideal of nowhere dense subsets of N N . (a) I 4 6 T NN. (b) Z 64T I, Z 64T N N and `1 64T I. ˇkovic ´ 99) `1 64T Z. (c) (Louveau & Velic proof (a) Put 526I and 526J together. (b) Let φ : Z → I be any function. Let hUn in∈N enumerate a base for the topology of N N which contains ∅ and is closed under finite unions. For each n ∈ N, set an = {i : i ∈ N, φ(a) ∩ Un 6= ∅ whenever i ∈ a ∈ Z}. Set a = {min(an \ n2 ) : n ∈ N, an 6⊆ n2 } (interpreting n2 in the formula above as a member of N rather than as a subset of N 2 ). Then a ∈ Z and a ∩ an 6= ∅ S whenever an is infinite. Set K = {n : n ∈ N, φ(a) ∩ Un } = ∅}, so that n∈K Un = N N \ φ(a) is dense, while S an is finite for every n ∈ K (since otherwise there is an i S ∈ a ∩ an , and φ(a) ∩ Un will not be empty). For n ∈ N, m∈K∩n Um belongs to U; let r(n) ∈ N be such that Ur(n) = m∈K∩n Um . Then r(n) ∈ K for every n, so ar(n) is always finite. Take a strictly increasing sequence hkn in∈N in K such that ar(n) ⊆ kn for every n. For i < k0 , set bi = {i}; for kn ≤ i < kn+1 , choose bi ∈ Z such that i ∈ bi and φ(bi ) ∩ Ur(n) is empty (such exists because i ∈ / ar(n) ). S Examine E =S i∈N φ(bi ) ⊆ N N . If m ∈ K then Um ⊆ USr(n) for every n > m so Um ∩ φ(bi ) = ∅ for every i ≥ km+1 and Um ∩ E = i
526L

Asymptotic density zero

149

(c)(i) Let φ : `1 → Z be any function. Give Z the submeasure ν and the metric ρ of 526A. By Blumberg’s theorem (5A4H), there is a dense set D ⊆ `1 such that φD is continuous. Fix x∗ ∈ D. S (ii) For any  > 0 and n ∈ N, there is a finite set I ⊆ D such that k supx∈I x+ k1 ≥ n and ν( x∈I φ(x)\φ(x∗ )) ≤ 3. P P Let δ > 0 be such that ρ(φ(x), φ(x∗ )) ≤  whenever x ∈ D and kx − x∗ k1 ≤ 2δ. For each i ∈ N, let xi ∈ D be such that kxi − x∗ k1 ≤ 2δ and xi (i) ≥ δ + x∗ (i). Consider the sequence hφ(xi )ii∈N in PN. This has a cluster point c in PN for the usual compact Hausdorff topology of PN; now ρ(φ(xi ), φ(x∗ )) ≤  for every i, so #((c \ φ(x∗ )) ∩ n) ≤ supi∈N #((φ(xi )4φ(x∗ )) ∩ n) ≤ n for every n. Choose sequences hl(i)ii∈N and hm(i)ii∈N in N as follows. Given hl(j)ij m(j) for j < i and S #( j i, so S a ∩ n ⊆ c ∪ j
150

Cardinal functions of measure theory

526L

add(E, ⊆, N (µ)) ≥ add(M, 63, N N ) = cov M (512Db). But, writing M(R) for the ideal of meager subsets of R, cov M = cov M(R) = mcountable , by 522Vb and 522Ra. Remark If X = R and µ is Lebesgue measure, then add(E, ⊆, N (µ)) = mcountable and cov(E, ⊆, N (µ)) = non M; see ´ ski & Shelah 92 or Bartoszyn ´ ski & Judah 95, 2.6.14. Bartoszyn 526X Basic exercises (a) Let ν : PN → [0, 1] be the submeasure described in 526A. Show that d∗ (I) = limn→∞ ν(I \ n) for every I ⊆ N. (b) For I, J ⊆ N say that I ⊆∗ J if I \ J is finite. Show that (Z, ⊆∗ , Z) ≡GT (Z, ⊆ 0 , [Z]≤ω ). (c) Let I be the ideal of nowhere dense subsets of N N and M the ideal of meager subsets of N N . Show that addω I = add M, non I = non M, cov I = cov M and cf I = cf M. (d) Let X be a topological space with a countable π-base, and IX the ideal of nowhere dense subsets of X. Show that IX 4T I, where I is the ideal of nowhere dense subsets of N N . (e) In 526He, show that Cnwd is a dense Gδ subset of the family C of all closed subsets of X with its Fell topology, so is a Polish space in the subspace topology. (f ) Let CLeb be the family of closed Lebesgue negligible subsets of [0, 1]. Show that CLeb with its Fell topology is a Polish space and a metrizably compactly based directed set. (g) Let ELeb be the ideal of subsets of R with negligible closures. (i) Show that it is Tukey equivalent to the partially N ordered set CLeb of 526Xf. (ii) Show that it is isomorphic to ELeb . (iii) Show that if we write Eσ for the σ-ideal of 0 ≤ω subsets of R generated by ELeb , then (ELeb , ⊆ , [ELeb ] ) ≡GT (Eσ , ⊆, Eσ ). (iv) Show that addω ELeb = add Eσ and cf ELeb = cf Eσ . 526Y Further exercises (a) Let X be a locally compact σ-compact metrizable space. Let Cnwd be the family of closed nowhere dense sets in X with its Fell topology. Show that Cnwd is a metrizably compactly based directed set. (b) Let Z be the asymptotic density algebra PN/Z and define d¯∗ : Z → [0, 1] by setting d¯∗ (I • ) = d∗ (I) for every I ⊆ N, as in 491I. Show that if A ⊆ Z is non-empty, downwards-directed and has infimum 0, and #(A) < p, then inf a∈A d¯∗ (a) = 0. (Compare 491Id.) (c) Show that wdistr(Z) = ω1 . (d) Show that m(Z) ≥ mσ-linked . (e) Show that FN(PN) ≤ FN(Z) ≤ max(FN∗ (PN), (cf N )+ ). (f ) Show that τ (Z) ≥ p. 526Z Problems Let I be the ideal of nowhere dense subsets of N N , and ELeb the ideal of subsets of R with Lebesgue negligible closures. We know that N N 4T ELeb 4T Z 4T `1 and that ELeb 4T I 4T `1 ; and in the other direction that ELeb 64T N N , that Z 4 6 T I and that `1 64T Z. But this leaves open the possibilities that I 4T Z or even that I ≡T ELeb . Is either of these true? 526 Notes and comments The ‘positive’ results of this section are straightforward enough, except perhaps for 526F. As elsewhere in this chapter, I am attempting to describe a framework which will accommodate the many arguments which have been found effective in discussing the cardinal functions of these partially ordered sets. I note that in this section I use the symbol M to represent the ideal of meager subsets of N N , rather than the ideal of meager subsets of R, as elsewhere in the chapter. If you miss this point, however, none of the formulae here are dangerous, because the two ideals are Tukey equivalent, and indeed isomorphic (522Vb). When we come to ‘negative’ results, we have problems of a new kind. The special character of Tukey functions is that they need not be of any particular type. They are not asked to be order-preserving, and even if we have partially ordered sets with natural Polish topologies (as in 526A, 526Xe and 526Xf, for instance), Tukey functions between them are not required to be Borel functions. This means that in order to show that there is no Tukey function between a given pair of partially ordered sets, we have had to consider arbitrary functions, or seek to calculate suitable invariants

527Bc

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151

which we know to be related to the Tukey ordering, like precaliber triples (516C), and show that they are incompatible with the existence of a Tukey function. I have indeed been able to find such invariants which make the distinctions listed in 526J and 526K (Fremlin n02), but the apparatus takes longer to describe than the proofs above from which it was constructed. Note that all the principal partially ordered sets of this section (Z, N N , I, ELeb , `1 ) are either themselves metrizably compactly based directed sets (526A, 513Xj, 513Xl) or are Tukey equivalent to metrizably compactly based ˇevic ´ 04 directed sets (526Hd-526He, 526Xf-526Xg). We can hope, therefore, that the methods of Solecki & Todorc (513J-513N) will lead to further results. In 526Yb-526Yf I list miscellaneous facts about the asymptotic density algebra. A remarkable description of its Dedekind completion is in 556S below.

527 Skew products of ideals The methods of this chapter can be applied to a large proportion of the partially ordered sets which arise in analysis. In this section I look at skew products of ideals, constructed by a method suggested by Fubini’s theorem and the Kuratowski-Ulam theorem (527E). 527A Notation If (X, Σ, µ) is a measure space, N (µ) will be the null ideal of µ; N will be the null ideal of Lebesgue measure on R. If X is a topological space, B(X) will be the Borel σ-algebra of X and M(X) the σ-ideal of meager subsets of X; M will be the ideal M(R) of meager subsets of R. 527B Skew products of ideals Suppose that I C PX and J C PY are ideals of subsets of sets X, Y respectively. (a) I will write I n J for their skew product {W : W ⊆ X × Y , {x : W [{x}] ∈ / J } ∈ I}. (This use of the symbol n is unconnected with the usage in §512 except by the vaguest of analogies.) It is easy to check that I n J C P(X × Y ). Similarly, I o J will be {W : W ⊆ X × Y , {y : W −1 [{y}] ∈ / I} ∈ J }. (b) Suppose that X and Y are not empty and that I and J are proper ideals. Then add(I n J ) = min(add I, add J ),

cf(I n J ) ≥ max(cf I, cf J ),

non(I n J ) = max(non I, non J ),

cov(I n J ) = min(cov I, cov J ). S P P (i) If hWξ iξ<κ is a family in I n J with κ < min(add I, add J ), set W = ξ<κ Wξ . For each ξ < κ, Hξ = {x : S S Wξ [{x}] ∈ / J } belongs to I; as κ < add I, H = ξ<κ Hξ ∈ I. For any x ∈ / H, W [{x}] = ξ<κ Wξ [{x}] ∈ J because κ < add J ; so W ∈ I n J . As hWξ iξ<κ is arbitrary, add(I n J ) ≥ min(add I, add J ). In the other direction, as X ∈ / I, J = {F : F ⊆ Y , X × F ∈ I n J }, so F 7→ X × F is a Tukey function from J to I n J and add J ≥ add(I n J ), cf J ≤ cf(I n J ). Similarly, E 7→ E × Y is a Tukey function from I to I n J and add I ≥ add(I n J ), cf I ≤ cf(I n J ). (ii) Let A ⊆ X and B ⊆ Y be such that A ∈ / I, B ∈ / J , #(A) = non I and #(B) = non J . Then A × B ∈ / I nJ, so non(I n J ) ≤ #(A × B). But note that as I and J are ideals, A and B are either singletons or infinite; so #(A × B) = max(#(A), #(B)) and non(I n J ) ≤ max(non I, non J ). In the other direction, take any W ∈ P(X × Y ) \ (I n J ). Set E = {x : W [{x}] ∈ / J }. Then #(E) ≥ non I and #(W [{x}]) ≥ non J for every x ∈ E, so #(W ) ≥ max(non I, non J ); as W is arbitrary, non(I n J ) ≥ max(non I, non J ). (iii) If A ⊆ I covers X, then {A × Y : A ∈ A} ⊆ I n J covers X × Y ; so cov(I n J ) ≤ cov I. Similarly, cov(I n J ) ≤ cov J . Now suppose that W ⊆ I n J and that #(W) S < min(cov I, cov J ). For each W ∈ W set EW = {x : W [{x}] ∈ / J }; then EW ∈ I for every W , so there is an x ∈ X \ E , because #(W) < cov I. Now W [{x}] ∈ J for every W, W W ∈W S S so there is a y ∈ Y \ W ∈W W [{x}], because #(W) < cov J . In this case (x, y) ∈ (X × Y ) \ W. As W is arbitrary, cov(I n J ) ≥ min(cov I, cov J ) and we have equality. Q Q (c) The idea of the operation n here is that we iterate notions of ‘negligible set’ in a way indicated by Fubini’s theorem: a measurable subset of R 2 is negligible iff almost every vertical section is negligible, that is, iff it belongs to N n N . However it is immediately apparent that N n N contains many non-measurable sets, and indeed many sets of full outer measure (527Xa). We are therefore led to the following idea. If Λ is a family S of subsets of X × Y , write I nΛ J for the ideal generated by (I n J ) ∩ Λ. Note that if κ ≤ min(add I, add J ) and W ∈ Λ for every W ∈ [Λ]<κ , then add(I nΛ J ) ≥ κ; in particular, I nΛ J will be a σ-ideal whenever I and J are σ-ideals and Λ is a σ-algebra of b subsets of X × Y . Typical applications will be with Λ a Borel σ-algebra or an algebra of the form Σ⊗T. Thus 252F can be restated in the form

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if (X, Σ, µ) and (Y, T, ν) are measure spaces with c.l.d. product (X × Y, Λ, λ) then N (µ) nΛ N (ν) ⊆ N (λ). If µ and ν are σ-finite then we get N (λ) = N (µ) nΣ⊗T b N (ν) 2

(252C). If we take B = B(R ) to be the Borel σ-algebra of R 2 , then all four ideals N nB(R 2 ) N , MnB(R 2 ) M, MnB(R 2 ) N and N nB(R 2 ) M become interesting. In the next few paragraphs I will sketch some of the ideas needed to deal with ideals of these kinds. 527C We are already familiar with N nB(R 2 ) N ; I begin by repeating a result from §417 in this language. Theorem Let (X, T, Σ, µ) and (Y, S, T, ν) be σ-finite effectively locally finite τ -additive topological measure spaces. ˜ be the τ -additive product measure on X × Y (417C, 417G). Then N (µ) nB(X×Y ) N (ν) = N (λ). ˜ Let λ proof Completing µ and ν does not change N (µ) or N (ν), and leaves µ and ν effectively locally finite and τ -additive; ˜ (use the uniqueness assertion in 417D). We may therefore assume that they are complete. so it also does not change λ ˜ Now 417H tells us that a Borel subset of X × Y is λ-negligible iff it belongs to N (µ) n N (ν). On the other hand, ˜ and every λ-negligible ˜ ˜ because µ and ν are σ-finite, so is λ, set is included in a λ-negligible Borel set. P P Suppose that ˜ Let hWn in∈N be a cover of X × Y by sets of finite measure. Because λ ˜ is inner regular with respect to the W ∈ N (λ). ˜ n = λW ˜ n for each n. Now Borel sets (417D), we can find Vn ∈ B(X × Y ) such that Vn ⊆ Wn \ W and λV S ˜ ∩ B(X × Y ). Q W ⊆ (X × Y ) \ n∈N Vn ∈ N (λ) Q ˜ So N (µ) nB(X×Y ) N (ν) = N (λ). 527D The case M nB(R 2 ) M is also well known. Theorem Let X and Y be topological spaces, with product X × Y . Write M∗ = M(X) nB(X×Y ) M(Y ) and b M∗1 = M(X) nB(X×Y b ) M(Y ), writing B(X × Y ) for the Baire-property algebra of X × Y . ∗ ∗ (a) If M(X × Y ) ⊆ M1 , then M = M∗1 = M(X × Y ). (b) Let C be the category algebra of Y (514I). If π(C) < add M(X) then M∗ = M(X × Y ). b × Y ) ∩ (M(X) n M(Y )) proof (a)(i) ?? If M∗1 6= M(X × Y ), there is a set W ∈ M∗ \ M(X × Y ); take W1 ∈ B(X such that W1 ⊇ W . By 4A3Ra, there is an open set V ⊆ X × Y such that W1 4V is meager and V ∩ V 0 is empty whenever V 0 ⊆ X × Y is open and V 0 ∩ W1 is meager. As W1 ∈ / M(X × Y ), V cannot be empty; let G ⊆ X, H ⊆ Y be non-empty open sets such that G × H ⊆ V . In this case, G × H cannot be meager, so neither G nor H can be meager. (If F ⊆ X is nowhere dense, then F × Y is nowhere dense in X × Y ; so M × Y ∈ M(X × Y ) whenever M ∈ M(X); as G × Y ∈ / M(X × Y ), G ∈ / M(X).) But now we see that {x : (G × H)[{x}] ∈ / M(Y )} = G ∈ / M(X), ∗

so that G × H ∈ / M ; but (G × H) \ W1 is meager, so belongs to M∗ , and W1 is also supposed to belong to M∗ . X X b × Y ). In the other direction, if (ii) So M∗1 = M(X × Y ). Of course M∗ ⊆ M∗1 just because B(X × Y ) ⊆ B(X W ∈ M(X ×Y ) there is a meager Fσ set W 0 ⊇ W . Now W 0 is a Borel set in M(X ×Y ) = M∗1 , so W 0 ∈ M(X)nM(Y ) and witnesses that W ∈ M∗ . Thus M(X × Y ) ⊆ M∗ and the three classes are equal. (b) By (a), I have only to show that W ∈ M∗ whenever W ⊆ X × Y is meager. Let D ⊆ C \ {0} be an order-dense subset of cardinal π(C). Let H be the smallest comeager regular open subset of Y , so that an open subset of Y is meager iff it is disjoint from H (4A3Ra). For each d ∈ D let Vd ⊆ Y be an open set such that Vd• = d in C; since H • = 1, we may suppose that Vd ⊆ H. Observe that if F ⊆ Y is a non-meager closed set, then there is a d ∈ D such that 0 6= d ⊆ F • in C, in which case Vd \ F is meager; as Vd ⊆ H, Vd ⊆ F . If W ⊆ X × Y is a nowhere dense closed set, it belongs to M∗ . P P Set E = {x : W [{x}] is not meager}. For each d ∈ D, the set Ed = {x : Vd ⊆ W [{x}]} = {x : (x, y) ∈ W for every y ∈ Vd } is a closed set in X and Ed × Vd ⊆ W ; so int Ed × Vd S is an open subset of W . As W is nowhere dense, and Vd 6= ∅, int Ed must be empty, and Ed ∈ M(X). Next, E = d∈D Ed and #(D) = π(C) < add M(X), so E ∈ M(X) and W ∈ M∗ . Q Q Since M∗ is a σ-ideal, it follows that every meager subset of X × Y belongs to M∗ , as required. 527E Corollary If X and Y are separable metrizable spaces, then M(X × Y ) = M(X) nB(X×Y ) M(Y ). proof π(C) ≤ π(Y ) ≤ w(Y ) ≤ ω < add M(X) (514Ja, 5A4Ba, 4A2P(a-i)). Remark The case X = Y = R is the Kuratowski-Ulam theorem.

527F

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153

527F If we mix measure and category, as in M nB(R 2 ) N and N nB(R 2 ) M, we encounter some new phenomena. To deal with the first we need the following, which is important for other reasons. ´ & Pawlikowski 86) Let X be a set, Σ a σ-algebra of subsets of X, and I a σ-ideal of subsets Theorem (see Cichon of X generated by Σ ∩ I; suppose that the quotient algebra Σ/Σ ∩ I is non-zero, atomless and has countable π-weight. Let Y be a set, T a σ-algebra of subsets of Y , and hHn in∈N a sequence of finite covers of Y by members of T. Set S Hn∗ = { m≥n Hm : Hm ∈ Hm ∪ {∅} for every m ≥ n} for each n ∈ N. Then there is a sequence hWn in∈N of subsets of N N × X × Y such that (i) for every n ∈ N, Wn is expressible as the union of a sequence of sets of the form I × E × F where I ⊆ N N is open-and-closed, E ∈ Σ and F ∈ T; N ∗ (ii) whenever n ∈ N, T α ∈ N and x ∈ X then {y : (α, x, y) ∈ Wn } ∈ Hn ; (iii) setting W = n∈N Wn , the set {(α, x) : α ∈ N N , x ∈ X, (α, x, f (x)) ∈ / W } belongs to [N N ]≤ω n I for every (Σ, T)-measurable function f : X → Y . proof If X ∈ I or Y is empty, we can take every Wn to be ∅; suppose otherwise. S (a) Set S ∗ = n∈N N n . There is a family hUσ iσ∈S ∗ such that every Uσ belongs to Σ \ I, S for every σ ∈ S ∗ , hUσa ii∈N is a disjoint sequence of subsets of Uσ and Uσ \ i∈N Uσa ∈ I, for every E ∈ Σ \ I there is a σ ∈ S ∗ such that Uσ \ E ∈ I. P P Let D be a countable order-dense set in A = Σ/Σ ∩ I. Then the subalgebra B of A generated by D is countable and atomless and non-trivial. Let E be the algebra of subsets of P(N N ) generated by the sets Iσ = {α : σ ⊆ α ∈ N N } for σ ∈ S ∗ . This is also an atomless countable Boolean algebra, and must therefore be isomorphic to B (316M10 ). Let π : E → B be an isomorphism, and set bσ = πIσ for each σ ∈ S ∗ . Set U∅ = X and for n ∈ N, σ ∈ N n choose a disjoint sequence hUσa ii∈N of subsets of Uσ such that Uσ•a = bσa for every i. This construction ensures that −1 Uσ ∈ Σ \ I for every σ. If E ∈ Σ \ I, there must be a non-zero d ∈ D such that d ⊆ E • ; now S π d ∈ E \ {∅}, so there is ∗ −1 • ∗ a σ ∈ S such that Iσ ⊆ π d, bσ ⊆ E and Uσ \ E ∈ I. Finally, if σ ∈ S , set E = Uσ \ i∈N Uσa ; then for every τ ∈ S ∗ either τ ⊆ σ and Uτ \ E ⊇ Uσa <0> ∈ / I, or Uτ ∩ Uσ = ∅ and Uτ \ E ⊇ Uτ ∈ / I. This means that E must belong to I, so that hUσ iσ∈S ∗ has all the required properties. Q Q S (b) Enumerate S ∗ as hτk ik∈N . Let hHn in∈N be a sequence running over n∈N Hn . For n ∈ N, set Kn = {(σ, k) : σ ∈ N n+2 , k < #(τσ(n) ), σ(n + 1) = #(τk ), Hτσ(n) (k) ∈ Hn }, Vn =

S

(σ,k)∈Kn {(α, x, y)

: τk ⊆ α ∈ N N , x ∈ Uσ , y ∈ Hτσ(n) (k) }.

If α ∈ N N and x ∈ X the section {y : (α, x, y) ∈ Vn } is either empty or Hτσ(n) (k) where σ ∈ N n+2 , x ∈ Uσ and S τk = ασ(n + 1); in either case it belongs to Hn ∪ {∅}. So if we set Wn = m≥n Vm , Wn satisfies (i) and (ii) for every n. T (c) Set W = n∈N Wn . ?? Suppose, if possible, that f : X → Y is a (Σ, T)-measurable function such that {(α, x) : (α, x, f (x)) ∈ / W} ∈ / [N N ]≤ω n I. Note that {V : V ⊆ N N × X × Y , {x : (α, x, f (x)) ∈ V } ∈ Σ for every α ∈ N N } is a σ-algebra of subsets of N N × X × Y containing I × E × F whenever I is open-and-closed, E ∈ Σ and F ∈ T, so contains every Vn and every Wn . Set A0 = {α : α ∈ N N , {x : (α, x, f (x)) ∈ / W} ∈ / I}, so that A0 is uncountable. For each α ∈ A0 , S / Wn } = {x : (α, x, f (x)) ∈ / W} n∈N {x : (α, x, f (x)) ∈ does not belong to I. So there is an n ∈ N such that A1 = {α : α ∈ A0 , {x : (α, x, f (x)) ∈ / Wn } ∈ / I} is uncountable. For each α ∈ A1 , set Gα = {x : (α, x, f (x)) ∈ / Wn }; then Gα ∈ Σ \ I, so there is a σ ∈ S ∗ such that ∗ Uσ \ Gα ∈ I. Let σ ∈ S be such that A2 = {α : α ∈ A1 , Uσ \ Gα ∈ I} is uncountable. Set m = max(n, #(σ)), so that Uσ ∩ {x : (α, x, f (x)) ∈ Vm } ∈ I for every α ∈ A2 . Set M = #(Hm ). 10 Formerly

393F.

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Take k ∈ N such that #({αk : α ∈ A2 }) ≥ M . Let hαi ii<M be a family in A2 such that αi k 6= αj k for distinct i, j < M ; let hri ii<M , hli ii<M be such that αi k = τri for each i and Hm = {Hli : i < M }. Let s ∈ N be such that τs (ri ) is defined and equal to li for i < M . Let σ 0 ∈ N m+2 be such that σ 0 ⊇ σ, σ 0 (m) = s and σ 0 (m + 1) = k. Then Uσ0 ∈ /I and Uσ0 \ Gα ∈ I for every α ∈ A2 . Suppose that i < M and x ∈ Uσ0 . Then {y : (αi , x, y) ∈ Vm } = Hτs (j) where τj = αi k, so j = ri and τs (j) = li and (σ 0 , j) ∈ Km ; thus {y : (αi , x, y) ∈ Vm } = Hli . But this means that, for any x ∈ Uσ0 , S S i<M {y : (αi , x, y) ∈ Vm } = i<M Hli = Y S contains f (x); that is, Uσ0 ⊆ i<M {x : (αi , x, f (x)) ∈ Vm }. On the other hand, Uσ0 ∩ {x : (αi , x, f (x)) ∈ Vm } ⊆ Uσ ∩ {x : (αi , x, f (x)) ∈ Vm } ∈ I for each i < M , while Uσ0 itself does not belong to I. So this is impossible. X X Thus hWn in∈N satisfies (iii). 527G Theorem Let X be a set, Σ a σ-algebra of subsets of X, and I a σ-ideal of subsets of X which is generated by Σ ∩ I; suppose that the quotient algebra Σ/Σ ∩ I is non-zero, atomless and has countable π-weight. Let (Y, T, ν) ≤ω 4T K, so be an atomless perfect semi-finite measure space such that νY > 0. Set K = I nΣ⊗T b N (ν). Then [c] add K = ω1 and cf K ≥ c. proof (a) To begin with (down to the end of (d)) suppose that ν is totally finite. Because ν is atomless, we can −n for each n ∈ N S find a finite cover Hn of Y by measurable sets with measures at most 2 . Let T0 be the σ-algebra generated by n∈N Hn , so that T0 is a σ-subalgebra of T. Construct hHn∗ in∈N , hWn in∈N and W from hHn in∈N as in 527F. Then if f : X → Y is (Σ, T0 )-measurable, {(α, x) : (α, x, f (x)) ∈ / W } ∈ [N N ]≤ω n I. Note that νH ≤ 2−n+1 ∗ −n+1 for every H ∈ Hn , so ν{y : (α, x, y) ∈ Wn } ≤ 2 for every α ∈ N N , x ∈ X and n ∈ N. For each α ∈ N N set b b ⊗T, b and that Kα = {(x, y) : (α, x, y) ∈ W }. Observe that Kα ∈ Σ⊗T because W ∈ B(N N )⊗Σ νKα [{x}] ≤ inf n∈N ν{y : (α, x, y) ∈ Wn } = 0 for every x ∈ X, so Kα ∈ K for every α ∈ N N . ˆ = {E4M : E ∈ Σ, M ∈ I}. Then Σ ˆ is a σ-algebra of subsets of X (cf. 212Ca) and I is a σ-ideal in Σ; ˆ (b) Set Σ ˆ ˆ also the identity embedding of Σ in Σ induces an isomorphism between Σ/Σ ∩ I and Σ/I (cf. 322Da). Consequently ˆ ˆ is closed under Souslin’s operation (431G11 ). Σ/I has countable π-weight, therefore is ccc, and Σ S b a set disjoint from α∈A Kα . (I aim to show that (X ×Y )\V ∈ (c) Let A ⊆ N N be an uncountable set, and V ∈ Σ⊗T / I n N (ν).) There must be sequences hCn in∈N in S Σ, hFn in∈N in T such that V belongs to the σ-algebra generated by {Cn × Fn : n ∈ N}; we may of course arrange that n∈N Hn ⊆ {Fn : n ∈ N}. Let T1 be the σ-subalgebra of T generated b by {F Pn∞: n ∈ N}, so that T0 ⊆ T1 and V ∈ Σ⊗T1 . Let g : Y → R be the Marczewski functional defined by setting g = n=0 3−n χFn . Because ν is perfect, there is a Borel set H ⊆ g[Y ] such that g −1 [H] is conegligible. Let h : H → Y be any function such that g(h(t)) = t for every t ∈ H; note that h is (B(H), T1 )-measurable, where B(H) is the Borel σ-algebra of H, just because g[Fn ] ∩ g[Y \ Fn ] is empty for every n. Set V0 = {(x, t) : x ∈ X, t ∈ H, (x, h(t)) ∈ V }; b then V0 ∈ Σ⊗B(H). It follows that V0 belongs to the class of sets obtainable by Souslin’s operation from sets of the ˜ = {x : V0 [{x}] 6= ∅}. Because H is form E × F where E ∈ Σ and F ⊆ H is relatively closed in H. (Use 421F.) Set E ˆ ˜ ˆ ˆ ˜→H analytic and Σ is closed under Souslin’s operation, E ∈ Σ and there is a (Σ, B(H))-measurable function f1 : E ˜ ˜ ˆ such that (x, f1 (x)) ∈ V0 for every x ∈ E (423M). Now f2 = hf1 : E → Y is (Σ, T1 )-measurable and (x, f2 (x)) ∈ V for ˜ every x ∈ E. ˆ so there is an En0 ∈ Σ such that En 4En0 ∈ I. Similarly, there is an For every n ∈ N, En = f1−1 [Fn ] belongs to Σ, 0 ˜ 0 ∈ Σ such that E4 ˜ E ˜ 0 ∈ I. Because I is generated by Σ∩I, there is an M0 ∈ Σ∩I including (E4 ˜ E ˜ 0 )∪S E n∈N (En 4En ). −1 0 0 ˜ ˜ ˜ Now E \ M0 = E \ M0 belongs to Σ. Set f3 = f2  E \ M0 ; then f3 [Fn ] = En \ M0 ∈ Σ for every n, so f3 is (Σ, T1 )˜ \ M0 , y0 for other x ∈ X; then f is (Σ, T1 )-measurable, measurable. Take any y0 ∈ Y , and set f (x) = f3 (x) if x ∈ E therefore (Σ, T0 )-measurable. The set {(α, x) : (α, x, f (x)) ∈ / W } belongs to [N N ]≤ω n I, so there must be an α ∈ A such that M1 = {x : (α, x, f (x)) ∈ / W } belongs to I. ?? Suppose, if possible, that (X × Y ) \ V ∈ I n N (ν). Then there must be an x ∈ X \ (M0 ∪ M1 ) such that V [{x}] is conegligible. In this case, V [{x}] ∩ g −1 [H] is conegligible, so is not empty, and there is a y ∈ V [{x}] ∩ g −1 [H]. Consider y 0 = h(g(y)); then g(y 0 ) = g(y), so {n : y 0 ∈ Fn } = {n : y ∈ Fn }, and {F : y ∈ F ⇐⇒ y 0 ∈ F } is a σ-algebra of subsets of Y containing every Fn and therefore containing V [{x}]. So 11 Later

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˜ as x ∈ y 0 ∈ V [{x}] and (x, g(y)) ∈ V0 . This means that x ∈ E; / M0 , f (x) = f3 (x) = f2 (x) and (x, f (x)) ∈ V . On the other hand, x ∈ / M1 , so (α, x, f (x)) ∈ W and (x, f (x)) ∈ Kα ; contradicting the choice of V as a set disjoint from Kα . X X S This shows that (X × Y ) \ V ∈ / I n N (ν). As V is arbitrary, α∈A Kα ∈ / K. S (d) This is true for every uncountable A ⊆ N N . But this means that A 7→ α∈A Kα is a Tukey function from [N N ]≤ω to K, and [c]≤ω ≡ [N N ]≤ω 4T K. (e) Thus the theorem is true if νY is finite. For the general case, let Y0 ∈ T be such that 0 < νY0 < ∞. Then the subspace measure νY0 is still atomless and perfect (214J, 451Dc), so [c]≤ω 4T K0 , where K0 = I nΣ⊗(T∩PY N (νY0 ). b 0) But K0 = K ∩ P(X × Y0 ), so the identity map from K0 to K is a Tukey function, and [c]≤ω 4T K0 4T K in this case also. It follows at once that add K ≤ add[c]≤ω = ω1 , so that add K = ω1 , and that cf K ≥ cf[c]≤ω = c. 527H Corollary M nB(R 2 ) N ≡T [c]≤ω . proof By 527G, [c]≤ω 4T M nB(R 2 ) N . In the other direction, all we need to observe is that #(B(R 2 )) = c. Let hWξ iξ
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Thus V1 [{x}] ∪ V2 [{x}] is dense for every x, and if we set V3 = (X × Y ) \ V2 we shall have V3 \ V1 ∈ L, while both b V1 and V3 belong to B(X)⊗B(Y ), and V1 ⊆ W ⊆ V3 . So W ∈ V ∗ . This is true for every open set W ⊆ X × Y , so the ∗ σ-algebra V must contain every Borel set, as required. Q Q b It follows that every member of L is included in a member of L ∩ (B(X)⊗B(Y )). P P If V ∈ L there is a Borel set b V 0 ⊇ V which belongs to L, and now there is a set V 00 ∈ B(X)⊗B(Y ) such that V 00 ⊇ V 0 and V 00 \ V 0 ∈ L, in which case V 00 ⊇ V also must belong to L. Q Q Thus L = N (µ) nB(X)⊗B(Y M(Y ). b ) (b) To begin with let us suppose that X is compact and metrizable, µ is totally finite and Y is a Baire space. (i) Taking Σ = B(X), define W, D0 and L0 as in 527I. Now let D be the family of closed subsets belonging to D0 , and L1 the σ-ideal of subsets of X × Y generated by {E × Y : E ∈ N (µ)} ∪ D. S (ii) D0 ⊆ L1 . P P If D ∈ D0 , express (X × Y ) \ D as H∈H EH × H where EH ∈ B(X) for every H ∈ H. Because X is compact and metrizable and µ is totally finite, µ is outer regular with respect to theP open sets. So for each n ∈ N we −n can find a familyShGnH iH∈H of open sets in X such that EH ⊆ GnH for every H and H∈H µ(G T nHS\ EH ) ≤ 2 . Set Dn = (X × Y ) \ H∈H (GnH × H). Then Dn is closed and Dn ⊆ D ∈ D0 so Dn ∈ D. Set E = n∈N H∈H (GnH \ EH ); then E ∈ N (µ) and S D ⊆ (E × Y ) ∪ n∈N Dn ∈ L1 . Q Q (iii) Of course every member of D belongs to L, so L1 ⊆ L. But in fact L = L1 . P P If V ∈ L, there is a b V 0 ∈ (N (µ) n M(Y )) ∩ (B(X)⊗B(Y )) such that V ⊆ V 0 , by (a). By 527I, we can express V 0 as W 4L where W ∈ W and L ∈ L0 . By (ii), L0 ⊆ L1 , so W ∈ L. There is therefore a negligible set E ⊆ X such that W [{x}] is meager for every x ∈ X \ E. But W [{x}] is always open, and Y is a Baire space, so W ⊆ E × Y ∈ L1 . Accordingly V 0 and V belong to L1 . As V is arbitrary, L ⊆ L1 . Q Q (iv) Let G be a countable base for the topology of X containing X. Let U0 be the family of those sets U ⊆ X × Y such that U is expressible as a finite union of sets of the form G × H where G ∈ G and H ∈ H, and U the set of those U ∈ U0 such that π1 [U ] = X, where π1 is the projection from X × Y onto X. Consider D0 = {D : D ⊆ X × Y , for every U0 ∈ U there is a U ∈ U such that U ⊆ U0 \ D}. D ⊆ D0 . P P Suppose that D ∈ D and U0 ∈ U, and consider U1 = {U : U ∈ U0 , U ⊆ U0 \ D}. For every x ∈ X the section U0 [{x}] is open and not empty and the section D[{x}] is nowhere dense, so there is a y such that (x, y) ∈ U0 \ D; now there are G ∈ G, containing x, and an open H containing y such that G × H ⊆ U0 \ D. Let H 0 ∈ H be such that ∅ 6= H 0 ⊆ H. Then U = G × H 0 ∈ U1 and x ∈ π1 [U ]. As x is arbitrary, {π1 [U ] : U ∈ U1 } is an open cover of X; as X is compact and U1 is upwards-directed, there is a U ∈ U1 such that π1 [U ] = X; in which case U ∈ U and U ⊆ U0 \ D. As U is arbitrary, D ∈ D0 ; as D is arbitrary, D ⊆ D0 . Q Q D is cofinal with D0 . P P Let D ∈ D0 . Take x ∈ X and a non-empty open set H ⊆ Y . Then there is a non-empty H 0 ∈ H included in H. In this case X × H 0 belongs to U, so there is a U ∈ U such that U ⊆ (X × H 0 ) \ D; in which case U is disjoint from D and U [{x}] is a non-empty open subset of H \ D[{x}]. As H is arbitrary, D[{x}] is nowhere dense; as x is arbitrary, D ∈ D; as D is arbitrary, D is cofinal with D0 . Q Q (v) Because U is countable, 527Gc tells us that D0 4T F, where F is the ideal of nowhere dense subsets of S N N ; while 0 N of course D ≡T D (513Ed). Let φ : L → N (µ)×D be such that if φ(V ) = (E, hDn in∈N ) then V ⊆ (E ×Y )∪ n∈N Dn ; such a function exists by (iii), and is evidently a Tukey function. Note that the measure algebra of µ, being a totally finite measure algebra with countable Maharam type, can be regularly embedded in the measure algebra of Lebesgue measure on either [0, 1] or on R. Consequently N (µ) 4T N (524K) and L 4T N (µ) × DN 4T N × F N 4T N × F (513Eg, 526Ha). Accordingly

(L, ⊆, L) ≡GT (L, ⊆ 0 , [L]≤ω ) (513Id) 4GT (N × F, ≤ 0 , [N × F]≤ω ) (512Gb) ≡GT (N , ⊆ 0 , [N ]≤ω ) × (F, ⊆ 0 , [F]≤ω ) (512Hd)

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≡GT (N , ⊆ N ) × (M, ⊆, M) (513Id, 526Hb, 512Hb) 4GT (N , ⊆, N ) × (N , ⊆, N ) (522O) ≡GT (N , ⊆, N ) (513Eh), and L 4T N . (c) This proves the theorem when X is compact and metrizable, µ is totally finite and Y is a Baire space. Now suppose that Y is still a Baire space, while (X, µ) is a totally finite quasi-Radon measure space with countable Maharam type. (i) There is a compact metrizable Radon measure space (Z, λ) such that λ and µ have isomorphic measure algebras. P P Because the measure algebra (A, µ ¯) of µ is totally finite, it is isomorphic to the simple product of a countable family h(Ai , µ ¯i )ii∈I of homogeneous totally finite measure algebras (332B). Because µ has countable Maharam type, every Ai is either {0}, {0, 1} or isomorphic to the measure algebra of Lebesgue measure on an interval; in any case it is isomorphic to the measure algebra of a compact Radon measure space (Zi , λi ). Take (Z 0 , λ0 ) to be the direct sum of the measure spaces h(Zi , λi )ii∈I ; then the measure algebra of (Z 0 , λ0 ) is isomorphic to A. If we give Z 0 its disjoint union topology, it is a locally compact σ-compact metrizable space, and its one-point compactification Z is second-countable, therefore metrizable; taking λ to be the trivial extension of λ0 , (Z, λ) is a compact metrizable Radon measure space ¯ ∼ with measure algebra (B, λ) ¯). Q Q = (A, µ (ii) Let f : X → Z be an inverse-measure-preserving function inducing an isomorphism π : B → A of the measure algebras (416Wb). By 418J, f is almost continuous, so there is a Borel measurable function which is equal almost everywhere to f ; this function will still represent π, so we may suppose that f itself is Borel measurable. Now if b b V ∈ B(X)⊗B(Y ), there is a V 0 ∈ B(Z)⊗B(Y ) such that {x : V [{x}] 6= V 0 [{f (x)}]} ∈ N (λ). P P Let V˜ be the family 0 0 b of subsets V of X × Y for which there is a V ∈ B(Z)⊗B(Y ) such that {x : V [{x}] 6= V [{f (x)}]} ∈ N (λ). Then V˜ is a σ-algebra. If E ∈ B(X) and H ∈ B(Y ), then there must be an F ∈ B(Z) such that F • = πE • in B, so that ˜ Accordingly V˜ must include B(X)⊗B(Y b E4f −1 [F ] ∈ N (µ); now F × H witnesses that E × H belongs to V. ). Q Q (iii) We know that N (µ) 4T N (λ) (524Sa), so there is a Tukey function θ : N (µ) → N (λ). Set L0 = 0 b N (λ) nB(Z)⊗B(Y b ) M(Y ). Define a function φ : L → L as follows. First, for V ∈ L, choose φ0 (V ) ∈ L ∩ (B(X)⊗B(Y )) b including V ((a) above). Next, by (ii) here, we can choose φ1 (V ) ∈ B(Z)⊗B(Y ) such that NV = {x : φ0 (V )[{x}] 6= φ1 (V )[{f (x)}]} belongs to N (µ). Set F = {z : z ∈ Z, φ1 (V )[{z}] is not meager}; then F is a Borel set, by 4A3Sa, and f −1 [F ] ⊆ NV ∪{x : φ0 [{x}] ∈ / M(Y )} ∈ N (µ); so F ∈ N (λ) and φ1 (V ) ∈ L0 . Finally, set φ(V ) = (θ(NV )×Y )∪φ1 (V ) ∈ 0 L. φ is a Tukey function from L to L0 . P P Take W ∈ L0 and consider E = {V : V ∈ L, φ(V ) ⊆ W }. If Y = ∅ then of course E is bounded above in L. Otherwise, N ∗ = {z : W [{z}] = Y } must be negligible, and θ(NV ) ⊆ N ∗ for every ˜ = S{NV : V ∈ E} is negligible. Take W1 ∈ L0 ∩ (B(Z)⊗B(Y b V ∈ E; because θ is a Tukey function, N )) including W , ˜ = {(x, y) : (f (x), y) ∈ W1 }; then W ˜ ∈ B(X)⊗B(Y b and set W ) because f is Borel measurable. As ˜ [{x}] ∈ {x : W / M(Y )} = f −1 {z : W1 [{z}] ∈ / M(Y )} ˜ ∈ L. So V0 = (N ∗ × Y ) ∪ W ˜ belongs to L. Now take any V ∈ E. If x ∈ X \ N ∗ , then x ∈ is negligible, W / NV , so ˜ [{x}] = V0 [{x}]. V [{x}] ⊆ φ0 (V )[{x}] = φ1 (V )[{f (x)}] ⊆ W [{f (x)}] ⊆ W1 [{f (x)}] = W This shows that V ⊆ V0 ; as V is arbitrary, V0 is an upper bound for E in L; as W is arbitrary, φ is a Tukey function. Q Q (iv) By (b), we know that L0 4T N , so (iii) tells us that L 4T N , and the theorem is true in this case also. (d) We are nearly home. If Y is a Baire space and (X, µ) is a σ-finite quasi-Radon measure space withRcountable Maharam type, which is not totally finite, then there is a measurable function f : X → ]0, ∞[ such that f dµ = 1 (215B(ix)). Let ν be the indefinite-integral measure defined by f . Then ν has the same negligible sets as µ (234Lc12 ), and is a quasi-Radon measure (415Ob), so L = N (ν) nB(X×Y ) M(Y ) 4T N , by (c). (e) Finally, suppose that Y is not a Baire space. In this case, let H ∗ be the minimal comeager regular open subset of Y , and set L∗ = N (µ) nB(X×H ∗ ) M(H ∗ ). Then L 4T L∗ . P P For every V ∈ L, let V 0 be such that 12 Formerly

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V ⊆ V 0 ∈ B(X × Y ) ∩ (N (µ) n M(Y )), and set φ(V ) = V 0 ∩ (X × H ∗ ). Then φ(V ) is a Borel subset of X × H ∗ , and φ(V )[{x}] = V 0 [{x}] ∩ H ∗ is meager in H ∗ whenever V 0 [{x}] is meager in Y , so φ(V ) ∈ L∗ . To see that φ : L :→ L∗ is a Tukey function, take any W ∈ L∗ . There is a Borel set W 0 ∈ L∗ including W , and now V 0 = W 0 ∪ (X × (Y \ H ∗ )) is a Borel subset of X × Y ; since V 0 [{x}] is meager in Y whenever W 0 [{x}] is meager in H ∗ , V 0 ∈ L. Of course V 0 is an upper bound of {V : V ∈ L, φ(V ) ⊆ W }; as W is arbitrary, φ is a Tukey function and L 4T L∗ . Q Q By (d), L 4T N in this case also, and the proof is complete. 527K Corollary N nB(R 2 ) M ≡T N . proof By 527J, N nB(R 2 ) M 4T N . On the other hand, E 7→ E × R is a Tukey function from N to N nB(R 2 ) M, so N 4T N nB(R 2 ) M. 527L There are some interesting questions concerning the saturation of skew products. Here I give two results which will be useful later. Theorem Let X be a set, Σ a σ-ideal of subsets of X, and I C PX a σ-ideal; suppose that Σ/Σ ∩ I is ccc. Let b b ∩ (I n N (ν)) is ccc. (Y, T, ν) be a σ-finite measure space. Then (Σ⊗T)/(Σ ⊗T) proof (a) The case νY = 0 is trivial, as then I n N (ν) = P(X × Y ). Otherwise, there is a probability measure on Y with the same domain and null ideal as ν (215B), so we may suppose that νY = 1. (b) The family W of sets W ⊆ X × Y such that W [{x}] ∈ T for every x ∈ X and x 7→ νW [{x}] is Σ-measurable is a Dynkin class (definition: 136A), and contains E × F whenever E ∈ Σ and F ∈ T; by the Monotone Class Theorem b (136B) it includes Σ⊗T. b (c) Now suppose that hWξ iξ<ω1 is a disjoint family in Σ⊗T. For n ∈ N and ξ < κ set Enξ = {x : νWξ [{x}] ≥ 2−n }; then #({ξ : x ∈ Enξ }) ≤ 2−n for every x ∈ X. It follows that An = {ξ : ξ < ω1 , Enξ ∈ / I} is countable. P P?? Otherwise, • write A for the ccc algebra Σ/Σ ∩ I, and aξ = Enξ for ξ < ω1 . Then A is Dedekind complete; set bξ = supξ≤η<ω1 aη for ξ < ω1 and b = inf ξ<ω1 bξ . Because A is ccc, there is a ζ < ω1 such that b = bξ for every ξ ≥ ζ; because A is uncountable, b 6= 0. Choose hci ii∈N and hξi ii∈N inductively such that c0 = b and, given that 0 6= ci ⊆ b, ξi is to be such that ci+1 = aξi ∩ ci 6= 0 and ξi > ξj for every j < i. T Now inf i≤2n aξi ⊇ c2n +1 is non-zero, so there is an x ∈ i≤2n Enξi ; but this is impossible. X XQ Q (d) This is true for every n ∈ N, so there is a ξ < ω1 such that ξ ∈ / An for every n, that is, Enξ ∈ I for every n. But in this case S {x : Wξ [{x}] ∈ / N (ν)} = n∈N Enξ b b and belongs to I and Wξ ∈ I n N (ν). As hWξ iξ<ω1 is arbitrary, (Σ⊗T) ∩ (I n N (ν)) is ω1 -saturated in Σ⊗T b b (Σ⊗T)/(Σ⊗T) ∩ (I n N (ν)) is ccc (316C). 527M The next result provides me with an opportunity to introduce a concept which will be needed in §546. Definition A Boolean algebra A is harmless (cf. Just 92) if it is ccc and whenever B is a countable subalgebra of A, there is a countable subalgebra C of A such that B ⊆ C and C is regularly embedded in A. 527N Lemma (a) If A is a Boolean algebra and D is a harmless order-dense subalgebra of A, then A is harmless. (b) If A is a Dedekind complete Boolean algebra, then it is harmless iff every order-closed subalgebra of A with countable Maharam type has countable π-weight. (c) For any set I, the regular open algebra RO({0, 1}I ) of {0, 1}I is harmless. (d) If A has countable π-weight it is harmless. (e) If A is a harmless Boolean algebra, B is a Boolean algebra and π : A → B is a surjective order-continuous Boolean homomorphism, then B is harmless. In particular, any principal ideal of a harmless Boolean algebra is harmless. proof (a) By 513Ee, A is ccc. Let B be a countable subalgebra of A. For eachSb ∈ B let Db ⊆ D be a countable set with supremum b (313K, 316E). Let D0 be the subalgebra of D generated by b∈B Db . Then D0 is countable, so there is a countable subalgebra D1 of D, including D0 , which is regularly embedded in D. Let C be the subalgebra of A generated by B ∪ D1 . Then C is countable. Now every member of C is the supremum of the members of D1 it includes. P P Set C = {c : c ∈ C, c = sup{d : d ∈ D1 , d ⊆ c} = inf{d : d ∈ D1 , c ⊆ d}}. Then C is closed under union (use 313Bd) and complementation (313A), and includes B ∪ D1 , so C = C. Q Q

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S It follows that C is regularly embedded in A, because if C ⊆ C has supremum 1 in C then c∈C {d : d ∈ D1 , d ⊆ c} must have supremum 1 in C and therefore in D1 (because D1 ⊆ C) and A (because D1 is regularly embedded in A). But this means that sup C must be 1 in A. As C is arbitrary, C is regularly embedded. As B is arbitrary, A is harmless. (b)(i) Suppose that A is harmless and that B ⊆ A is an order-closed subalgebra of countable Maharam type. Let B ⊆ B be a countable set which τ -generates B, and B0 the algebra generated by B; let C be a countable subalgebra of A, including B0 , which is regularly embedded in A. Let D be the set {d : d ∈ A, d = sup{c : c ∈ C, c ⊆ d} = inf{c : c ∈ C, d ⊆ c}}. Then D is an order-closed subalgebra of A. P P As in (a) just above, it is a subalgebra. If D ⊆ D is a non-empty set with supremum a in A, set C = {c : c ∈ C, c ⊆ a}, C 0 = {c : c ∈ C, a ⊆ c}. Then a is an upper bound for C and a lower bound for C 0 . ?? If either a is not the least upper bound of C, or a is not the greatest lower bound of C 0 , then A = {c0 \ c : c0 ∈ C 0 , c ∈ C} is a subset of C with a non-zero lower bound in A, so A has a non-zero lower bound c∗ in C. Now if d ∈ D, c ∈ C and c ⊆ d, then c ∈ C so c ∩ c∗ = 0; as d = sup{c : c ∈ C, c ⊆ d}, d ∩ c∗ = 0. This is true for every d ∈ D, so a ∩ c∗ = 0 and 1 \ c∗ ∈ C 0 ; but c∗ was chosen to be included in every member of C 0 . X X Thus a ∈ D; as D is arbitrary, D is order-closed in A. Q Q Now B ⊆ C ⊆ D. As B is regularly embedded in A (314Ga), B ∩ D is an order-closed subalgebra of B including B, so is the whole of B, and B ⊆ D. It follows that π(B) ≤ π(D) (514Eb). But C is countable order-dense in D, so π(D) and π(B) are countable. As B is arbitrary, A satisfies the declared condition. (ii) Now suppose that A satisfies the condition. Note first that A is ccc. P P?? Suppose, if possible, otherwise; let haξ iξ<ω1 be a disjoint family in A \ {0}. Replacing a0 by a0 ∪ (1 \ supξ<ω1 aξ ) if necessary, we may suppose that supξ<ω1 aξ = 1. The map I 7→ supξ∈I aξ : Pω1 → A is an injective order-continuous Boolean homomorphism, so its image B is an order-closed subalgebra of A isomorphic to Pω1 . Now τ (B) = τ (Pω1 ) = ω (514Ef, or otherwise), but π(B) = ω1 ; which is supposed to be impossible. X XQ Q If B is a countable subalgebra of A is countable, let B1 be the order-closed subalgebra of A which it generates. Then τ (B1 ) is countable, so π(B1 ) is countable, and there is a countable subalgebra C of B1 which is order-dense in B1 ; of course we may suppose that B ⊆ C. Now the identity maps from C to B1 and from B1 to A are both order-continuous, so their composition also is, and C is regularly embedded in A. As B is arbitrary, A is harmless. (c) All regular open algebras are Dedekind complete. If B ⊆ RO({0, 1}I ) is an order-closed subalgebra with countable Maharam type, let hGn in∈N be a sequence in B which τ -generates B. Every regular open subset of {0, 1}I is determined by coordinates in some countable set (4A2E(b-i)), so there is a countable J ⊆ I such that every Gn is determined by coordinates in J. Let πJ : {0, 1}I → {0, 1}J be the restriction map; then we have an injective ordercontinuous Boolean homomorphism H 7→ πJ−1 [H] : RO({0, 1}J ) → RO({0, 1}I ) (4A2Bf(iii)13 ). Let C be the image of this homomorphism, so that C is an order-closed subalgebra of RO({0, 1}I ). If Hn = πJ [Gn ] then Hn is regular and open for each n (4A2B(f-iii) again), so Gn = πJ−1 [Hn ] ∈ C; accordingly B ⊆ C. Now B is an order-closed subalgebra of C so π(B) ≤ π(C) = π({0, 1}J ) ≤ ω. As B is arbitrary, RO({0, 1}I ) satisfies the condition of (b) and is harmless. (d) Let D be a countable order-dense set in A. If B is a countable subalgebra of A, let C be the subalgebra of A generated by D ∪ B; then C is countable, includes B and is order-dense, therefore regularly embedded in A. As B is arbitrary, A is harmless. (e) Let D ⊆ B be a countable subalgebra. Because π[A] = B, there is a countable subalgebra C of A such that π[C] = D. Let C1 ⊇ C be a countable regularly embedded subalgebra of A. Then D1 = π[C1 ] is regularly embedded in B. P P Let D ⊆ D1 be a non-empty set such that 1 is not the least upper bound of D in B. Set C = C1 ∩ π −1 [D ∪ {0}]; then 1 is not the least upper bound of π[C] in B, so (because π is order-continuous) 1 is not the least upper bound of C in A. Because C1 is regularly embedded in A, there is a non-zero c0 ∈ C1 such that c0 ∩ c = 0 for every c ∈ C. In particular, c0 ∈ / C and πc0 6= 0. But we also have πc ∩ πc0 = 0 for every c ∈ C, that is, d ∩ πc0 = 0 for every d ∈ D, and 1 is not the least upper bound of D in D1 . As D is arbitrary, D1 is regularly embedded. Q Q Of course D1 is countable. As D is arbitrary, B is harmless. If c ∈ A then a 7→ a ∩ c is an order-continuous homomorphism onto the principal ideal Ac generated by c, so Ac is harmless. 527O Theorem Let (X, Σ, µ) be a σ-finite measure space and Y a topological space such that the category algebra C b b of Y is harmless. Write L for (Σ⊗B(Y )) ∩ (N (µ) n M(Y )) and A for the measure algebra of µ. Then G = (Σ⊗B(Y ))/L 13 Later

editions only.

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527O

is ccc, and is isomorphic to the Dedekind completion of the free product A ⊗ C. If neither A nor C is trivial, the isomorphism corresponds to embeddings E • 7→ (E × Y )• : A → G and F • 7→ (X × F )• : B → G. proof Write S for the topology of Y . S (a) Let W be the family of all sets of the form n∈N En × Hn , where En ∈ Σ and Hn ⊆SY is open for every n. Then for any W ∈ W there is a W 0 ∈ W such that W 0 4((X × Y ) \ W ) ∈ L. P P Express W as n∈N En × Hn where En ∈ Σ and Hn ∈ S for each n. Let D be the order-closed subalgebra of C generated by {Hn• : n ∈ N}. Because C is harmless and Dedekind complete, π(D) ≤ ω (527Nb); let hGn in∈N be a sequence in S such that {G•n : n ∈ N} is a π-base for D; we may suppose that any non-empty open subset of any Gn is non-meager. Let S1 be the second-countable topology on Y generated by {Hn : n ∈ N} ∪ {Gn : n ∈ N}, and B1 (Y ) the corresponding Borel σ-algebra. Then V • ∈ D for every V ∈ S1 , because V is the union of a countable family of sets all with images in D. If V ∈ S1 is dense for S1 , and n ∈ N is such that Gn is non-empty, V ∩ Gn 6= ∅ so V • ∩ G•n 6= 0, by the choice of the Gn . But this means that V • = 1, that is, V is comeager for the original topology of Y . b 1 (Y ). By 527I, there are W 0 and hDn in∈N such that Now W and (X × Y ) \ W belong to Σ⊗B S ((X × Y ) \ W )4W 0 S ⊆ n∈N Dn , W 0 is expressible as n∈N Fn × Vn where Fn ∈ Σ and Vn ∈ S1 for every n, b 1 (Y ), every Dn belongs to Σ⊗B for every x ∈ X and n ∈ N, Dn [{x}] is closed and nowhere dense for S1 . Evidently W 0 ∈ W; but we have just seen that sets which are closed and nowhere dense for S1 are meager for S. So every Dn belongs to L and ((X × Y ) \ W )4W 0 ∈ L. Q Q (b) It follows (as in the proof of 527I) that V = {W 4D : W ∈ W, D ∈ L} is a σ-algebra of sets, and as E × H ∈ W b for every E ∈ Σ and H ∈ S, V = Σ⊗B(Y ). (c) G is ccc. P P?? Otherwise, there is a disjoint family heξ iξ<ω1 in G \ {0}. For each ξ < ω1 , there is a Vξ ∈ S b (Σ⊗B(Y )) \ L such that Vξ• = eξ , and a Wξ ∈ W such that Vξ 4Wξ ∈ L. Express Wξ as n∈N Eξn × Hξn ; as Wξ ∈ / L, there must be an nξ such that Eξ = Eξ,nξ ∈ / N (µ) and Hξ = Hξ,nξ is non-meager. Since the measure algebra of µ satisfies Knaster’s condition (525Ub), there is an uncountable A ⊆ ω1 such that Eξ ∩ Eη ∈ / N (µ) for all ξ, η ∈ A; because C is ccc, there are distinct ξ, η ∈ A such that Hξ ∩ Hη is non-meager. But also (Eξ ∩ Eη ) × (Hξ ∩ Hη ) ⊆ Wξ ∩ Wη ∈ L because (Wξ ∩ Wη ) ] = eξ ∩ eη = 0. So this is impossible. X XQ Q Thus G is ccc. As it is Dedekind σ-complete (314C), it is Dedekind complete (316Fa). •

(d) If either µX = 0 or Y is meager, then both A ⊗ C and G are trivial; so suppose otherwise. The map E 7→ (E × Y )• : Σ → G is a Boolean homomorphism with kernel Σ ∩ N (µ), so induces a Boolean homomorphism π1 : A → G. Similarly, we have a Boolean homomorphism π2 : C → G defined by setting π2 (F • ) = (X × F )• for F ∈ B(Y ). These now give us a Boolean homomorphism φ : A ⊗ C → G defined by saying that ψ(E • ⊗ F • ) = π1 (E • ) ∩ π2 (F • ) = (E × F )• for E ∈ Σ and F ∈ B(Y ) (315Ib). If E ∈ Σ \ N (µ) and F ∈ B(Y ) \ M(X), then E × F ∈ / L; so φ is injective (315Jb). If c ∈ G is non-zero, it is expressible as W • for some W ∈ W \ L; there must now be E ∈ Σ, F ∈ B(Y ) such that E × F ⊆ W and E × F ∈ / L, so that φ(E • ⊗ F • ) is non-zero and included in w. Thus φ[A ⊗ C] is isomorphic to A ⊗ C and is an order-dense subalgebra of the Dedekind complete Boolean algebra G; it follows that G can be identified with the Dedekind completion of A ⊗ C. 527X Basic exercises > (a) Show that there is a set belonging to N nN which has full outer measure for Lebesgue measure in the plane. (Hint: enumerate the compact non-negligible subsets of the plane as hKξ iξ (b) Show that there is a unique construction of iterated skew products I0 n I1 n . . . n In such that (i) whenever X0 , . . . , Xn are sets and Ij is an ideal of subsets of Xj for every j, then I0 n . . . n In is an ideal of subsets of X0 × . . . × Xn ; (ii) whenever X0 , . . . , Xn are sets, Ij is an ideal of subsets of Xj for every j, and k < n, then the natural identification of X0 × . . . × Xn with (X0 × . . . × Xk ) × (Xk+1 × . . . × Xn ) identifies I0 n . . . n In with (I0 n . . . n Ik ) n (Ik+1 n . . . n In ) as defined in 527B. (c) Complete the analysis in 527Bb by describing what happens if one of X, Y is empty or one of the ideals is not proper.

§528 intro.

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161

> (d) Let Z be the Stone space of the measure algebra of Lebesgue measure on [0, 1], and f : Z → [0, 1] the canonical inverse-measure-preserving continuous function (416V). Let F ⊆ [0, 1] be a nowhere dense set which is not negligible, and set W = {(x, z) : x ∈ [0, 1], z ∈ Z, x + f (z) ∈ F }. Show that W is a nowhere dense closed set in [0, 1] × Z but does not belong to M([0, 1]) n M(Z). (Hint: meager subsets of Z are negligible (321K).) > (e) Suppose that X = {0, 1}I and Y = {0, 1}J . Show that M(X) nB(X×Y ) M(Y ) = M(X × Y ). (f ) Write X for the class of topological spaces which have category algebras which are atomless and with countable π-weight. (i) Show that R, with the right-facing Sorgenfrey topology, belongs to X. (ii) Show that the split interval (419L) belongs to X. (iii) Show that if the regular open algebra of a topological space X is atomless and has countable π-weight, then X ∈ X. (iv) Show that any open subspace of a space in X belongs to X. (v) Show that any dense subspace of a space in X belongs to X. (vi) Show that any comeager subspace of a space in X belongs to X. (vii) Show that the product of countably many spaces in X belongs to X. (g) Let P be a partially ordered set. Show that if κ ≥ cf P and λ ≤ add P then P 4T [κ]<λ . (h) Let X be a topological space with a σ-finite measure µ such that µ has countable Maharam type and every measurable set can be expressed as the symmetric difference of a Borel set and a negligible set. Let Y be a topological space with a countable π-base. Show that N (µ) nB(X×Y ) M(Y ) 4T N (µ) × M. 527Y Further exercises (a) Show that I n J 6= I o J for any of the four cases in which {I, J } ⊆ {M, N }. (b) Let X be a set, Σ a σ-algebra of subsets of X and I a σ-ideal of Σ such that Σ/I is ccc. Let (Y, T, ν) be a b b ∩ (I n N (ν)) is ccc. (Hint: show that if V ∈ Σ⊗T b then x 7→ νV [{x}] is probability space. Show that (Σ⊗T)/(Σ ⊗T) b Σ-measurable, and hence that there is no uncountable disjoint family in (Σ⊗T) \ (I n N (ν)).) (c) Let hAi ii∈I be any family of Boolean algebras all of countable π-weight, and A their free product (315H). Show that the Dedekind completion of A is harmless. (d) Let µ be a σ-finite Borel probability measure on a topological space X, and Y a topological such that its category algebra is harmless. Show that B(X × Y )/B(X × Y ) ∩ (N (µ) n M(Y )) can be identified with the Dedekind completion of A ⊗ C, where A is the measure algebra of µ and C is the category algebra of Y . 527 Notes and comments Skew products of ideals have been used many times for special purposes, and we are approaching the point at which it would be worth developing a general theory of such products. I am not really attempting to do this here, though the language of 527B is supposed to point to the right questions. My primary aim in this section is to show that M nB(R 2 ) N and N nB(R 2 ) M are very different (527H, 527K). Of course the difference appears only when the continuum hypothesis is false (513Xf, 527Xg). The version of the Kuratowski-Ulam theorem given in 527D is a natural one from the point of view of this chapter, but you should be aware that there are many more cases in which M∗ = M(X × Y ); see 527Xe and Fremlin Natkaniec & Reclaw 00. The statement of 527J includes the phrase ‘quasi-Radon measure’. Actually we do not really need either τ -additivity or inner regularity with respect to closed sets. What we need is a measure µ such that N (µ) 4T N and the Borel sets generate the measure algebra (527Xh). The argument for 527J betrays its origin in the case X = Y = [0, 1], which is of course also the natural home of 527C-527F. It is interesting that all four of the quotient algebras B(R 2 )/B(R 2 ) ∩ (M n M),

B(R 2 )/B(R 2 ) ∩ (M n N ),

B(R 2 )/B(R 2 ) ∩ (N n M),

B(R 2 )/B(R 2 ) ∩ (N n N )

are ccc (see 527E, 527Yb, 527O, 527Bc and also 527L). This should not be taken for granted; for a variety of examples of quotient algebras associated with σ-ideals see Fremlin 03. 528 Amoeba algebras In the course of investigating the principal consequences of Martin’s axiom, Martin & Solovay 70 introduced the partially ordered set of open subsets of R with measure strictly less than γ, for γ > 0 (528O). Elementary extensions of this idea lead us to a very interesting class of partially ordered sets, which I study here in terms of their regular open algebras, the ‘amoeba algebras’ (528A). Of course the most important ones are those associated with Lebesgue measure, and these are closely related to the ‘localization posets’ (528I), themselves intimately connected with the localization relations of 522K. As elsewhere in this chapter, I will write (Bκ , ν¯κ ) for the measure algebra of the usual measure on {0, 1}κ . In any measure algebra (A, µ ¯) I will write Af = {a : a ∈ A, µ ¯a < ∞}.

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528A Amoeba algebras Let (A, µ ¯) be a measure algebra. (a) If 0 < γ ≤ µ ¯1, the amoeba algebra AM(A, µ ¯, γ) is the regular open algebra RO↑ (P ) where P = {a : a ∈ A, µ ¯a < γ}, ordered by ⊆ . (b) The variable-measure amoeba algebra AM∗ (A, µ ¯) (Truss 88) is the regular open algebra RO↑ (P 0 ) where P 0 = {(a, α) : a ∈ A, α ∈ ]¯ µa, µ ¯1]}, ordered by saying that (a, α) ≤ (b, β) if a ⊆ b and β ≤ α. 528B Proposition Suppose that (X, Σ, µ) is a measure space, (A, µ ¯) its measure algebra and 0 < γ ≤ µX. If E ⊆ Σ is any family such that µ is outer regular with respect to E, then AM(A, µ ¯, γ) is isomorphic to RO↑ ({E : E ∈ E, µE < γ}). proof Set P = {a : a ∈ A, µ ¯a < γ}, Q = {E : E ∈ E, µE < γ}. Because µ is outer regular with respect to E, the map G 7→ G• : Q → A maps Q onto a cofinal subset P 0 of P . Moreover, two elements E0 and E1 of Q are compatible upwards in Q iff µ(E0 ∪ E1 ) < γ iff E0• and E1• are compatible upwards in P . By 514R, RO↑ (P ) and RO↑ (Q) are isomorphic. 528C Proposition Let (A, µ ¯) be a non-totally-finite atomless measure algebra such that all its principal ideals of finite measure are homogeneous. Then all the amoeba algebras AM(A, µ ¯, γ), for γ > 0, are isomorphic. proof Suppose that β, γ > 0. As in Lemma 332I, we have a partition D of unity in A such that µ ¯a = β for every a ∈ D. Similarly, we have a partition D0 of unity such that µ ¯a = γ for every a ∈ D0 . By 332E, #(D) = #(D0 ) = c(A). Let h : D → D0 be a bijection. Because the principal ideal generated by d ∪ h(d) is homogeneous, the principal ideals Ad , Ah(d) always have the same Maharam type, and are therefore isomorphic as measure algebras, up to a scalar factor of the measure; let πd : Ad → Ah(d) be a Boolean isomorphism such that µ ¯(πd a) = f

γ µ ¯a β

for every a ⊆ d. Now we

f

have a function π : A → A defined by saying that πa = supd∈D πd (a ∩ d) whenever µ ¯a < ∞, and π is a Boolean γ β

ring automorphism such that µ ¯πa = µ ¯a for every a ∈ Af . But now π includes an isomorphism between the partially ordered sets {a : µ ¯a < β} and {a : µ ¯a < γ}, so their regular open algebras AM(A, µ ¯, β) and AM(A, µ ¯, γ) are isomorphic. 528D Proposition (Truss 88) Let (A, µ ¯) be an atomless homogeneous probability algebra. Then the amoeba algebras AM(A, µ ¯, γ) and AM(A, µ ¯, γ 0 ) are isomorphic for all γ, γ 0 ∈ ]0, 1[. proof (a) Set P = {a : a ∈ A, µ ¯a < γ}, and let κ be the Maharam type of A. Then the upwards cellularity of P is at most κ. P P?? Otherwise, there is an up-antichain A ⊆ P with cardinality κ+ . Let  > 0 be such that A0 = {a : a ∈ A, µ ¯a ≤ γ − } has cardinal κ+ . Because the topological density of A is κ (521Oa), there must be distinct a, a0 ∈ A0 such that µ ¯(a 4 a0 ) < ; but in this case µ ¯(a ∪ a0 ) < γ, so that a ∪ a0 is an upper bound for {a, a0 } in P . X XQ Q √ (b) If 1 − 1 − γ ≤ α < γ and D is a countable subset of ]α, γ[ such that sup D = γ, then there is a maximal up-antichain hatξ i(t,ξ)∈D×κ in P such that µ ¯atξ = t for every t ∈ D, ξ < κ. P P Start with a stochastically independent √ family hctξ i(t,ξ)∈D×κ of elements of A with µ ¯ctξ = t for all t ∈ D, ξ < κ. Because α ≥ 1 − 1 − γ, A = hctξ i(t,ξ)∈D×κ is an up-antichain in P . Next, because sup D = γ, Q = {a : a ∈ P , µ ¯a ∈ D} is cofinal with P . So there is a maximal up-antichain A0 ⊇ A such that A0 ⊆ Q (513Aa). Now (because c↑ (P ) ≤ κ) {a : a ∈ A0 , µ ¯a = t} has cardinal κ for every t ∈ D, so we can enumerate A0 as hatκ i(t,ξ)∈D×κ in P where µ ¯atξ = t for every t ∈ D and ξ < κ. Q Q (c) There are α, α0 ∈ ]0, 1[ such that √ 1 − 1 − γ ≤ α < γ,

1−

√

1 − γ 0 ≤ α0 < γ 0 ,

γ−α 1−α

=

γ 0 −α0 . 1−α0

P P We need consider only the case γ ≤ γ 0 . Set β=√ Then

γ−α 1−γ

α0 ≥ 1 −

√

=β=

γ 0 −α0 . 1−γ 0

1 − γ0. Q Q

1 1−γ

− 1,

α = γ − β(1 − γ),

α0 = γ 0 − β(1 − γ 0 ).

Of course α ≤ γ and α0 ≤ γ 0 . On the other side, α = 1 −

√

1 − γ, while β ≤ √

1 1−γ 0

− 1 so

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163

(d) If a ∈ P , {b : a ⊆ b ∈ P } is isomorphic, as partially ordered set, to {b : b ∈ A, µ ¯b < A1 \ a generated by 1 \ a is isomorphic, up to a scalar multiple of the measure, to A, and as partially ordered set, to {b : b ⊆ 1 \ a, µ ¯b < γ − µ ¯a}. Q Q

γ−¯ µa }. P P The principal ideal 1−¯ µa {b : a ⊆ b ∈ P } is isomorphic,

(e) For each n ∈ N, set αn = γ − 2−n (γ − α), αn0 = γ 0 − 2−n (γ 0 − α0 ); then γ 0 −α0n 1−γ 0

= 2−n

γ 0 −α0 1−γ 0

=

γ−αn 1−γ

for every n ∈ N. Set P 0 = {a : a ∈ A, µ ¯a < γ 0 }. By (b), we have a maximal up-antichain hanξ i(n,ξ)∈N×ξ in P such that µ ¯anξ = αn for all n ∈ N and ξ < κ; similarly, there is a maximal up-antichain ha0nξ i(n,ξ)∈N×ξ in P 0 such that µ ¯a0nξ = αn0 for all n ∈ N and ξ < κ. Now, for each n ∈ N and ξ < κ, [anξ , ∞[, taken in P , is isomorphic, as partially ordered set, to [a0nξ , ∞[, taken in P 0 , by (d). So RO↑ (P ) ∼ =

Y

RO↑ ([anξ , ∞[)

n∈N,ξ<κ

(514Nf) ∼ =

Y

  RO↑ ( a0nξ , ∞ ) ∼ = RO↑ (P 0 ).

n∈N,ξ<κ

528E Proposition Let (A, µ ¯) be an atomless measure algebra, not {0}, such that all its principal ideals of finite measure are homogeneous, and 0 < γ < µ ¯1. Then the amoeba algebra AM(A, µ ¯, γ) can be regularly embedded in AM∗ (A, µ ¯). proof (a) We need to know that there is a family hc(α)iα∈[0,¯µ1[ in A such that µ ¯c(α) = α and c(α) ⊆ c(β) whenever ¯ be the measure algebra of Lebesgue measure on [0, µ ¯ is isomorphic to a 0 ≤ α ≤ β. P P Let (C, λ) ¯1[. By 332P, (C, λ) closed subalgebra of a principal ideal of (A, µ ¯); let π : C → A be such an isomorphism, and set c(α) = π[0, α]• for each α ∈ [0, µ ¯1[. Q Q (b) Set P = {(a, α) : a ∈ A, α ∈ ]α, µ ¯1]}. Let hγn in∈N be a strictly increasing sequence with supremum µ ¯1 and γ0 = 0. For each n ∈ N, set S Pn = {(a, α) : γn ≤ µ ¯a < α ≤ γn+1 } and Qn = {a : a ∈ A, µ ¯a < γn+1 }, so that Pn is an up-open set in P . Note that n∈N Pn is dense in P for the up-topology, since if (a, α) ∈ P then (a, min(α, γn+1 )) ∈ Pn where γn ≤ µ ¯a < γn+1 . Also RO↑ (Qn ) = AM(A, µ ¯, γn+1 ) ∼ ¯, γ). = AM(A, µ P P If µ ¯1 = ∞, this is 528C. If (A, µ ¯) is totally finite, then A is homogeneous, so we can apply 528D to an appropriate multiple of the measure µ ¯. Q Q For a ∈ Af , the function α 7→ µ ¯(c(α) \ a) is continuous and non-decreasing. So we can define δ(a, α), for a ∈ Af and 0≤α<µ ¯1 − µ ¯a, by saying that δ(a, α) = min{β : µ ¯(c(β) \ a) = α} = min{β : µ ¯(a ∪ c(β)) = µ ¯a + α}. Observe that δ(a, α) ≤ δ(a0 , α0 ) whenever a ⊆ a0 and α ≤ α0 . For (a, α) ∈ Pn , set fn (a, α) = a ∪ c(δ(a, γn+1 − α)), so that µ ¯fn (a, α) = µ ¯a + γn+1 − α < γn+1 and fn (a, α) ∈ Qn . (c)(i) fn : Pn → Qn is order-preserving. P P If (a, α) ≤ (a0 , α0 ) in Pn , then δ(a, γn+1 − α) ≤ δ(a0 , γn+1 − α0 ), so 0 0 Q fn (a, α) ⊆ fn (a , α ). Q (ii) If p ∈ Pn , b ∈ Qn and fn (p) ⊆ b, there is a p0 ∈ Pn such that p ≤ p0 and b ⊆ fn (p0 ). P P Express p as (a, α). Consider a0 = a ∪ (b \ fn (p)). Then µ ¯a0 = µ ¯a + µ ¯b − µ ¯fn (p) = µ ¯a + µ ¯b − µ ¯a − γn+1 + α < α, so (a0 , α) ∈ P . Of course (a, α) ≤ (a0 , α), so p0 = (a0 , α) ∈ Pn . Also fn (p0 ) ⊇ fn (p) and fn (p0 ) ⊇ a0 ⊇ b \ fn (p), so b ⊆ fn (p0 ). Q Q (iii) fn [Pn ] is cofinal with Qn . P P If b ∈ Qn , take b0 ∈ A such that b ⊆ b0 and γn ≤ µ ¯b0 < γn+1 . Then (b0 , γn+1 ) ∈ Pn and b ⊆ b0 = fn (b0 , γn+1 ) ∈ fn [Pn ]. Q Q

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Cardinal functions of measure theory

528E

(d) By 514P, RO↑ (Qn ) can be regularly embedded in RO↑ (Pn ). Now AM(A, µ ¯, γ) is isomorphic to RO↑ (Qn ), so ↑ there is an injective order-continuous Boolean homomorphism πn : AM(A, µ ¯, γ) → RO Q (Pn ). ↑ Putting these together, we have an injective order-continuous Boolean homomorphism π : AM(A, µ ¯, γ) → n∈N RO (Pn ) defined by setting πe = hπn (e)in∈N for e ∈ AM(A, µ ¯, γ). On the other hand, since hPn in∈N is a disjoint sequence of up-open subsets of P with dense union, Q ∗ ↑ ↑ ∼ ¯) n∈N RO (Pn ) = RO (P ) = AM (A, µ by 514He. So we have a regular embedding of AM(A, µ ¯, γ) into AM∗ (A, µ ¯), as claimed. 528F Proposition Let (A, µ ¯) be a measure algebra and e ∈ A an element of non-zero measure; let Ae be the principal ideal of A generated by e. If Ae is atomless and 0 < γ ≤ µ ¯e, then AM(Ae , µ ¯ Ae , γ) can be regularly embedded in AM(A, µ ¯, γ). proof Set P = {a : a ∈ A, µ ¯a < γ} and Q = P ∩ Ae . As in 528E, we have an order-preserving function α 7→ cα : [0, µ ¯e] → Ae such that µ ¯cα = α for each α. As in (b) of the proof of 528E, we can set δ(a, α) = min{β : µ ¯(cβ \ a) = α} when a ∈ A and 0 ≤ α ≤ µ ¯(e \ a). Now define f : P → Q by setting f (a) = (a ∩ e) ∪ cδ(a,¯µ(a \ e)) for a ∈ P . Then µ ¯f (a) = µ ¯a for each a and f , like δ, is order-preserving. If a ∈ P , b ∈ Q and f (a) ⊆ b, there is an a0 ∈ P such that a ⊆ a0 and b ⊆ f (a0 ). P P Set a0 = a ∪ (b \ f (a)). Then 0 0 0 0 µ ¯a = µ ¯b < γ so a ∈ P . Also b ⊆ f (a) ∪ (a ∩ e) ⊆ f (a ). Q Q Finally, f [P ] = f [Q] = Q. So by 514P, AM(Ae , µ ¯ Ae , γ) = RO↑ (Q) can be regularly embedded in AM(A, µ ¯, γ) = RO↑ (P ). 528G Proposition Let (A, µ ¯) be a measure algebra, and C a σ-subalgebra of A such that sup(C ∩ Af ) = 1 in A. ∗ Then AM (C, µ ¯ C) can be regularly embedded in AM∗ (A, µ ¯). R f proof (a) For each a ∈ A we have a ‘conditional expectation’ ua ∈ L1 (C) defined by saying that c ua = µ ¯(a ∩ c) for every c ∈ Cf . (Apply 365P to the identity map from Cf to Af .) Note that as the supremum of Cf in A is 1,

R

R

ua = supc∈Cf

c

ua = supc∈Cf µ ¯(a ∩ c) = µ ¯a.

Also, of course, 0 ≤ µ ¯(a ∩ c) ≤ µ ¯c for every c ∈ Cf , so 0 ≤ ua ≤ χ1 in L∞ (C). Next, let u∗a be the decreasing rearrangement of ua , that is, the element of L∞ (AL ) (where AL is the measure algebra of Lebesgue measure on [0, ∞[) • such that [[u∗ > α]] = [0, µ ¯[[u > α]][ for every α ≥ 0 (373D). (b) Set P = {(a, α) : a ∈ A, α ∈ ]¯ µa, µ ¯1]},

Q = {(c, α) : c ∈ C, α ∈ ]¯ µc, µ ¯1]}.

Define a function f on P by saying that f (a, α) = (c, β) if c = [[ua = 1]] = max{d : d ∈ C, d ⊆ a}, β = max{β 0 : β 0 ≥ 0, β 0 +

R∞ β0

u∗a ≤ α}.

Note that β > µ ¯c because µ ¯c +

R∞ µ ¯c

u∗a =

R

u∗a =

R

ua < α,

using 373Fa for the equality in the middle, while β ≤ α ≤ µ ¯1; so (c, β) belongs to Q. (c)(i) If p ≤ p0 in P , then f (p) ≤ f (p0 ) in Q. P P Express p, p0 , f (p) and f (p0 ) as (a, α), (a0 , α0 ), (c, β), (c0 , β 0 ) 0 0 respectively. Then c ⊆ a ⊆ a so c ⊆ c . Next, χa ≤ χa0 so ua ≤ ua0 and u∗a ≤ u∗a0 (373Db); accordingly α0 ≤ α = β +

R∞ β

u∗a ≤ β +

R∞ β

u∗a0

and β 0 ≤ β. Q Q If p ∈ P , q ∈ Q and f (p) ≤ q, then there is a p0 ≥ p such that f (p0 ) ≥ q. P P Express p, f (p) and q as (a, α), (c, β) and (d, γ) respectively. Set a0 = a ∪ d. Then Z

0

µ ¯a = µ ¯a + µ ¯d − µ ¯(a ∩ d) =

Z ua + µ ¯d −

ua ≥ d

(apply 373E with v = χd) Z

∞

=µ ¯d +

u∗a < β +

Z

µ ¯d

(because [[u∗a = 1]] = [0, µ ¯c]• and µ ¯c ≤ µ ¯d < γ ≤ β)

∞

β

u∗a

Z

u∗a

Z +µ ¯d − 0

µ ¯d

u∗a

528H

Amoeba algebras

165

= α. R So (a0 , α) ∈ P . Next, computing the integrals b ua ∨ χd for b belonging to Cf and either included in d or disjoint from it, we see that ua0 = ua ∨ χd so that [[ua0 = 1]] = [[ua = 1]] ∪ d = d. Accordingly µ ¯a0 =

R

ua0 =

R

¯d + u∗a0 = µ

R∞ µ ¯d

u∗a0 < γ +

R∞ γ

u∗a0

R∞ (as noted above for ua , we have [[u∗a0 = 1]] = [0, µ ¯d]• ), and if we set α0 = min(α, γ + γ u∗a0 ) then p0 = (a0 , α0 ) ≥ p and f (p0 ) ≥ q, as required. Q Q Since P and Q have a common least element (0, µ ¯1) which is invariant under f , f satisfies the second condition of 514P and AM∗ (C, µ ¯ C) = RO↑ (Q) is regularly embedded in AM∗ (A, µ ¯) = RO↑ (P ). 528H Proposition Let (A, µ ¯) be a semi-finite measure algebra, not {0}, and let κ ≥ max(ω, τ (A), c(A)) be a cardinal. Then AM∗ (A, µ ¯) can be regularly embedded in AM(Bκ , ν¯κ , 12 ). proof (a) To begin with (down to the end of (h) below), assume that A is atomless. Let (AN , µ ¯∞ ) be the simple P∞ a product of a sequence of copies of (A, µ ¯) (322L), so that µ ¯∞ = n=0 µan if a = han in∈N ∈ AN . Note that as A is certainly infinite, τ (AN ) = τ (A) and c(AN ) = c(A). By 526D, there is a function θ : AN → Bκ such that θ(sup A) = sup θ[A] for every non-empty A ⊆ AN with a supremum in A; a) = 1 − exp(−¯ ν¯κ θ(a µ∞a ) for every a ∈ AN ; (i) a ii∈I is a disjoint family in AN and Ci is the closed subalgebra of Bκ generated by {θ(a a) : whenever ha (i) a ⊆ a } for each i, then hCi ii∈I is stochastically independent. a : a ∈ AN , θ(a a) ⊆ b}. (b) For b ∈ Bκ , set g(b) = sup{a (i) It is immediate from its definition that g : Bκ → AN is order-preserving. (ii) Because θ is supremum-preserving, θ(g(b)) ⊆ b for every b ∈ Bκ . (iii) If b ∈ Bκ \ {1} then 1 − ν¯κ b ≤ 1 − ν¯κ θ(g(b)) = exp(−¯ µ∞ g(b)), so µ ¯∞ g(b) ≤ − ln(1 − ν¯κ b) is finite. a)) for every a ∈ AN ; and if a ∈ AN has finite measure then g(θ(a a)) = a , because if a 0 6⊆ a then (iv) a ⊆ g(θ(a 0 a ∪ a ) > ν¯κ θ(a a). ν¯κ θ(a (v) If b ∈ Bκ and  > 0, there is a δ > 0 such that µ ¯∞ (g(b0 ) \ g(b)) ≤  whenever ν¯κ (b0 \ b) ≤ δ. P P?? Otherwise, we can find a sequence hbn in∈N in Bκ such that ν¯κ (bn \ b) ≤ 2−n and µ ¯∞ (g(bn ) \ g(b)) ≥  for every n ∈ N. Set a (n) = supm≥n g(bm ); then a(n) ) = supm≥n θ(g(bm )) ⊆ supm≥n bm , θ(a a(n) ) \ b) ≤ 2−n+1 for every n, while ha a(n) in∈N is non-increasing. Set a = inf n∈N a (n) ; then µ a \ g(b)) ≥  so ν¯κ (θ(a ¯∞ (a a) \ b) = 0. But in this case θ(a a) ⊆ b and a ⊆ g(b). X and ν¯κ (θ(a XQ Q a) = supn∈N an whenever a = han in∈N ∈ AN . (c) Define ψ : AN → A by setting ψ(a (i) ψ is supremum-preserving and ψ(0) = 0. (ii) If a , a 0 ∈ AN then a)4ψ(a a0 )) ≤ µ a 4 a 0 ). µ ¯(ψ(a ¯∞ (a (iii) Now if b ∈ Bκ , a ∈ A, a ⊇ ψ(g(b)) and µ ¯a < α ∈ R, there is a b0 ⊇ b such that a ⊆ ψ(g(b0 )), µ ¯ψ(g(b0 )) ≤ α and ν¯κ (b0 \ b) ≤ 1 − exp(−¯ µ(a \ ψ(g(b))). P P By (b-v), there is a δ > 0 such that µ ¯∞ (g(b0 ) \ g(b)) ≤ α − µ ¯a whenever ν¯κ (b0 \ b) ≤ δ. Set a(n) = hani ii∈N for each n ∈ N, where ani = a \ ψ(g(b)) if i = n, 0 otherwise. For each n ∈ N, let Cn be the closed subalgebra of Bκ generated by a) : a ∈ AN , a ∩ a (m) = 0 for every m ≥ n}, {θ(a and let Tn : L1 (Bκ , ν¯κ ) → L1 (Bκ , ν¯κ ) be the corresponding conditional-expectation operator (365R). Then hCn in∈N a(n) ∩ c) = ν¯κ θ(a a(n) ) · ν¯κ c for every n ∈ N and c ∈ Cn , by the final clause of (a). By is non-decreasing; also ν¯κ θ(a L´evy’s martingale theorem (275I, 367Jb), hTn (χb)in∈N is k k1 -convergent. We can therefore find an n ∈ N such that a(n) ). Then g(b0 ) ⊇ g(b) ∪ a (n) , so ψ(g(b0 )) ⊇ ann ∪ ψ(g(b)) = a. Also kTn (χb) − Tn+1 (χb)k1 ≤ δ exp(−¯ µa). Set b0 = b ∪ θ(a a(n) ) = 1 − exp(−¯ ν¯κ (b0 \ b) ≤ ν¯κ θ(a µ∞a (n) ) = 1 − exp(−¯ µ(a \ ψ(g(b))).

166

Cardinal functions of measure theory

528H

?? If µ ¯ψ(g(b0 )) > α, set e = g(b0 ) \ (g(b) ∪ supm∈N a (m) ). Since ψ(g(b) ∪ supm∈N a (m) ) = ψ(g(b)) ∪ a = a, ψ(ee) ⊇ ψ(g(b0 )) \ a and µ ¯ ∞e > α − µ ¯a; as e ⊆ g(θ(ee)) and e ∩ g(b) = 0, ν¯κ (θ(ee) \ b) > δ. On the other hand,

a(n) ))¯ a(n) )) (1 − ν¯κ θ(a νκ (b ∩ θ(ee)) = (1 − ν¯κ θ(a

Z Tn (χb) e) θ(e

(because e ∩ a (m) = 0 for every m, so θ(ee) ∈ Cn ) Z = Z =

a(n) )) · χ(1 \ θ(a

Z Tn (χb) × χθ(ee)

a(n) )) × Tn (χb) × χθ(ee) χ(1 \ θ(a

a(n) )) are stochastically independent) (because Tn (χb) × χθ(ee) ∈ L0 (Cn ) and χ(1 \ θ(a Z = Tn (χb), e) \ θ(a a(n) ) θ(e

a(n) ))¯ a(n) ) = ν¯κ (b ∩ θ(ee) \ θ(a a(n) ) (1 − ν¯κ θ(a νκ (θ(ee)) = ν¯κ (θ(ee) \ θ(a a(n) )) (because θ(ee) ⊆ θ(g(b0 )) ⊆ b0 = b ∪ θ(a Z =

Tn+1 (χb) e) \ θ(a a(n) ) θ(e

a(n) ) both belong to Cn+1 . So because θ(ee) and θ(a δ exp(−¯ µa) = δ exp(−¯ µ∞a (n) ) < ν¯κ (θ(ee) \ b) exp(−¯ µ∞a (n) ) a(n) ))¯ = (1 − ν¯κ θ(a νκ (θ(ee) \ b)) Z = Tn+1 (χb) − Tn (χb) e) \ θ(a a(n) ) θ(e

≤ kTn (χb) − Tn+1 (χb)k1 ≤ δ exp(−¯ µa), which is impossible. X X So µ ¯ψ(g(b0 )) ≤ α, as required. Q Q (d) Fix c ∈ Bκ with measure 12 ; then the principal ideal of Bκ generated by c is isomorphic to Bκ with the measure halved. We therefore have a Boolean isomorphism π : Bκ → (Bκ )c such that ν¯κ πb = 12 ν¯κ b for every b ∈ Bκ . Set h(b) = ψ(g(π −1 (b ∩ c))) for b ∈ Bκ . Then h : Bκ → A is order-preserving and h(b) = h(b ∩ c) for every b ∈ Bκ . Translating the results of (b) and (c), we see that if b ∈ Bκ and  > 0, there is a δ > 0 such that µ ¯(h(b0 ) \ h(b)) ≤  whenever ν¯κ (b0 \ b) ≤ δ, if b ∈ Bκ , a ∈ A, a ⊇ h(b) and µ ¯a < α ∈ R, there is a b0 ⊇ b such that a ⊆ h(b0 ), µ ¯h(b0 ) ≤ α and 1 0 ν¯κ (b \ b) ≤ 2 (1 − exp(−¯ µ(a \ h(b))). Note also that µ ¯h(b) = µ ¯ψ(g(π −1 (b ∩ c))) ≤ µ ¯∞ g(π −1 (b ∩ c)) ≤ − ln(1 − ν¯κ π −1 (b ∩ c)) ≤ − ln(1 − 2¯ νκ (b ∩ c)) if we take ln(0) to be −∞. (e) Set γ0 = 12 (1 − exp(−¯ µ1)), interpreting exp(−∞) as 0, so that 0 < γ0 ≤ 12 . Let P be the partially ordered set {(a, α) : a ∈ A, α ∈ ]¯ µa, µ ¯1]} and Q the partially ordered set {b : b ∈ Bκ , ν¯κ b < γ0 }, so that AM∗ (A, µ ¯) = RO↑ (P ) and ↑ 0 0 AM(Bκ , ν¯κ , γ0 ) = RO (Q). For b ∈ Q, set αb = sup{¯ µh(b ) : b ⊆ b ∈ Q}. Then αb > µ ¯(h(b)). P P We have µ ¯h(b) ≤ − ln(1 − 2¯ νκ (b ∩ c)) < − ln(1 − 2γ0 ) = µ ¯1, so h(b) 6= 1. Because A is atomless, there is an a ∈ A, disjoint from h(b), such that 0 < µ ¯a < − ln(2¯ νκ b). Set a = han in∈N where a0 = a, an = 0 for n ≥ 1. Then a) = 1 − exp(−¯ ν¯κ θ(a µa) < 1 − 2¯ νκ b, 0

0

a) ∈ Q1 , while h(b ) ⊇ h(b) ∪ a ⊃ h(b). Q so b = b ∪ πθ(a Q (f ) We can therefore define f : Q → P by setting f (b) = (h(b), αb ) for b ∈ Q.

528J

Amoeba algebras

167

(i) f is order-preserving because h is. (ii) If P1 ⊆ P is up-open and cofinal with P , f −1 [P1 ] is cofinal with Q. P P Take any b ∈ Q. Set α = min(αb , µ ¯h(b) − ln(1 − 2γ0 + 2¯ νκ b)) > µ ¯h(b), so that f (b) ≤ (h(b), α) in P . Then there is an (a, β) ∈ P1 such that (h(b), α) ≤ (a, β), that is, h(b) ⊆ a and β ≤ α. In this case, there is a b1 ∈ Bκ such that b1 ⊇ b, h(b1 ) ⊇ a, µ ¯h(b1 ) < β and 1 2

ν¯κ (b1 \ b) ≤ (1 − exp(−¯ µ(a \ h(b)))) < γ0 − ν¯b because µ ¯(a \ h(b)) < − ln(1 − 2γ0 + 2¯ νκ b). So b1 ∈ Q. Let δ > 0 be such that µ ¯(h(b0 ) \ h(b1 )) < β − µ ¯h(b1 ) whenever 0 ν¯κ (b \ b1 ) ≤ δ. Let b2 ∈ Q be such that b1 ⊆ b2 , b1 ∩ c = b2 ∩ c and µ ¯b2 ≥ γ0 − δ. Then h(b2 ) = h(b1 ). If b0 ∈ Q and b0 ⊇ b2 , then ν¯κ (b0 ∩ c \ b1 ) ≤ δ, so µ ¯(h(b0 ) \ h(b1 )) = µ ¯(h(b0 ∩ c) \ h(b1 )) ≤ β − µ ¯h(b1 ) and µ ¯h(b0 ) ≤ β; thus αb2 ≤ β. Accordingly f (b2 ) = (h(b2 ), αb2 ) ≥ (a, β), while b2 ⊇ b. As P1 is up-open, f (b2 ) ∈ P1 ; as b is arbitrary, f −1 [P1 ] is cofinal with Q. Q Q (g) f [Q] is cofinal with P . P P Take (a, α) ∈ P . Set a = han in∈N ∈ AN where a0 = a and an = 0 for n ≥ 1, and set a). Note that b = πθ(a 1 2

µa)) < γ0 , ν¯κ b = (1 − exp(−¯ so b ∈ Q. Because µ ¯∞a < ∞, g(π −1 b) = a and h(b) = a. As in the last step in the proof of (f) just above, we can now find a b1 ⊇ b in Q such that h(b1 ) = h(b) and αb1 ≤ α. So f (b1 ) ≥ (a, α). As (a, α) is arbitrary, f [Q] is cofinal. Q Q (h) By 514O, AM∗ (A, µ ¯) = RO↑ (P ) can be regularly embedded in RO↑ (Q1 ) ∼ = RO↑ (Q) = AM(Bκ , ν¯κ , γ0 ) ∼ = AM(Bκ , νκ , 1 ) 2

by 528D. (i) All this has been done on the assumption that A is atomless, as required in (e). For the general case, consider ¯ of (A, µ the localizable measure algebra free product (C, λ) ¯) and (Bω , ν¯ω ) (325E). By 521Pa, we have max(ω, c(C), τ (C)) ≤ max(ω, c(A), c(Bω ), τ (A), τ (Bω )) ≤ κ. Also C is atomless because Bω is isomorphic to a closed subalgebra of C (325Dd) and is atomless. By (a)-(h), ¯ can be regularly embedded in AM(Bω , ν¯ω , 1 ). Now consider the canonical embedding ε1 : A → C. This AM∗ (C, λ) 2 is order-continuous and measure-preserving (325Da), so identifies the Dedekind σ-complete Boolean algebra A with a σ-subalgebra of C; also Af has supremum 1 both in A and C. By 528G, AM∗ (A, µ ¯) can be regularly embedded in ¯ and therefore in AM(Bω , ν¯ω , 1 ). AM∗ (C, λ) 2 528I Definition For any set I, the I-localization poset is the set SI∞ = {p : p ⊆ N × I, #(p[{n}]) ≤ 2n for every n, supn∈N #(p[{n}]) is finite}, ordered by ⊆. For p ∈ SI∞ set kpk = maxn∈N #(p[{n}]). I will write S ∞ for the N-localization poset SN∞ . 528J Proposition Let κ be an infinite cardinal, Sκ∞ the κ-localization poset, and (A, µ ¯) a semi-finite measure algebra, not {0}, with κ ≥ max(ω, c(A), τ (A)). Then the variable-measure amoeba algebra AM∗ (A, µ ¯) can be regularly embedded in RO↑ (Sκ∞ ). proof (a) To begin with (down to the end of (d) below), suppose that A is atomless. Let P be the partially ordered set {(a, α) : a ∈ A, α ∈ ]¯ µa, µ1]}, so that AM∗ (A, µ ¯) = RO↑ (P ). Give Af its measure metric, so that its topological 1 density is at most κ (521Ob). Set γ0 = 2 µ ¯1 and for n ≥ 1 set γn = 2−2n−1 µ ¯1 if µ ¯1 < ∞, 4−n otherwise. For each n, f let Dn be a dense subset of {a : a ∈ A , µ ¯a ≤ γn }, containing 0, of cardinal at most κ, and let hdnξ iξ<κ be a family running over Dn with cofinal repetitions. (b) If p ∈ Sκ∞ set ap = sup(n,ξ)∈p dnξ ,

αp = µ ¯ap +

P∞

n n=0 (2

− #(p[{n}]))γn .

Then µ ¯ap < αp ≤

P∞

n=0

2n γn = µ ¯1,

so we can define f : Sκ∞ → P by setting f (p) = (ap , αp ). f is order-preserving, because if p ⊆ p0 in Sκ∞ then

168

Cardinal functions of measure theory

α

p0

∞ X

=µ ¯a + p0

528J

(2n − #(p0 [{n}]))γn

n=0

X

≤µ ¯ap +

µ ¯dnξ +

(n,ξ)∈p0 \p

≤µ ¯ap +

∞ X

∞ X

(2n − #(p0 [{n}]))γn

n=0

#(p0 [{n}] \ p[{n}])γn +

n=0

∞ X

(2n − #(p0 [{n}]))γn = αp .

n=0

(c) Suppose that p ∈ Sκ∞ and f (p) ≤ (a, α) ∈ P . Take α0 ∈ ]¯ µa, α[. For n ∈ N, set kn = 2n − #(p[{n}]). Then ap ⊆ a and P∞ µ ¯a < α ≤ αp = µ ¯ap + n=0 kn γn . So there is an r ∈ N such that P∞

µ ¯a < µ ¯ap +

P∞

n

0

n=0

γn min(r, kn );

take r so large that, in addition, n=r+1 2 γn ≤ α − α . 0 For each n, set kn0 = min(r, kn ) and Cn = {sup D : D ∈ [Dn ]≤kn }. Then (because A is atomless) Cn is dense in f 0 ¯c ≤ kn γn }. We can therefore choose hcn in∈N inductively in such a way that {c : c ∈ A , µ µ ¯(a ∪ supm
for every n ∈ N. P P For the inductive step to n ≥ 0, set b = a \ (ap ∪ supm µ ¯(a ∪ supm
m=n+1

0 γm > µ ¯(b \ b0 ) + µ ¯(b0 \ cn ) ≥ µ ¯(a \ (ap ∪ supm≤n cm )) km

and the induction proceeds. Q Q For each n, we can find a set Dn0 ⊆ Dn , of size kn0 , such that cn = sup Dn0 . Because hdnξ iξ<κ runs over Dn with cofinal repetitions, we can find a set In ⊆ κ \ p[{n}] such that #(In ) = kn0 and cn = supξ∈In dnξ . Set q = p ∪ {(n, ξ) : n ∈ N, ξ ∈ In }. Then #(q[{n}]) ≤ #(p[{n}]) + kn0 ≤ min(2n , kpk + r) for every n, so q ∈ Sκ∞ and p ⊆ q. Now aq = ap ∪ supn∈N,ξ∈In dnξ = ap ∪ supn∈N cn ⊇ a because µ ¯(a \ aq ) ≤ inf n∈N µ ¯(a \ (ap ∪ supm
P∞

m=n

2m γm = 0.

Also µ ¯aq = supn∈N µ ¯(ap ∪ supm
528M

Amoeba algebras

169

528K Theorem (Truss 88) Let (A, µ ¯) be an atomless σ-finite measure algebra in which every non-zero principal ideal has Maharam type κ, and 0 < γ < µ ¯1. Then each of the algebras AM(A, µ ¯, γ),

AM∗ (A, µ ¯),

AM(Bκ , ν¯κ , 12 )

can be regularly embedded in the other two, and all three can be regularly embedded in RO↑ (Sκ∞ ). proof By 528H, AM∗ (A, µ ¯) can be regularly embedded in AM(Bκ , ν¯κ , 12 ). Take any e ∈ A such that γ < µ ¯e < ∞. Then the principal ideal (Ae , µ ¯ Ae ) is isomorphic, up to a scalar multiple of the measure, to (Bκ , ν¯κ ), so 1 2

AM(Bκ , ν¯κ , ) ∼ = AM(Bκ , ν¯κ ,

γ ) µ ¯e

(528D) ∼ ¯ Ae , γ) = AM(Ae , µ can be regularly embedded in AM(A, µ ¯, γ) (528F). By 528E, AM(A, µ ¯, γ) can be regularly embedded in AM∗ (A, µ ¯). ∗ ↑ ∞ Finally, by 528J, AM (A, µ ¯) can be regularly embedded in RO (Sκ ). Because regular embeddability is transitive (313N), these facts are enough to prove the theorem. 528L It is possible without great effort to calculate many of the cardinal functions of these algebras. Lemma m(AM(Bω , ν¯ω , 21 )) ≤ add N . proof Set P = {a : a ∈ Bω , ν¯ω a < 12 }. Then wdistr(Bω ) ≥ m↑ (P ). P P Take a family hBξ iξ<κ of maximal antichains in Bω , where κ < m↑ (P ). Let C ⊆ Bω be a maximal disjoint set such that {b : b ∈ Bξ , b ∩ c 6= 0} is finite for every ξ < κ and c ∈ C. ?? Suppose, if possible, that c0 = 1 \ sup C is not 0. Take a0 ∈ P such that ν¯ω (a0 ∪ c0 ) > 12 . (If ν¯ω c0 > 21 , take a0 = 0; otherwise, take a0 ⊆ 1 \ c0 such that 12 − ν¯ω c0 < ν¯ω a0 < 12 .) For each ξ < κ, set Qξ = {a : a ∈ P , {b : b ∈ Bξ , b 6⊆ a} is finite}; then Qξ is cofinal with P . There is therefore an upwards-directed R ⊆ P such that a0 ∈ R and R meets every Qξ . Set e = sup R; then ν¯ω e ≤ 21 so c1 = c0 \ e = (a0 ∪ c0 ) \ e is non-zero. If ξ < κ, there is an a ∈ R ∩ Qξ , so that {b : b ∈ Bξ , b ∩ c1 6= 0} ⊆ {b : b ∈ Bξ , b 6⊆ a} is finite. But this means that we ought to have added c1 to C. X X Thus C is a maximal antichain. As hBξ iξ<κ is arbitrary, wdistr(A) ≥ m↑ (P ). Q Q ↑ Now 524Mb tells us that wdistr(Bω ) = add N , so m (P ) ≤ add N . Finally, by 517Db, m(AM(A, µ ¯, γ)) = m↑ (P ) ≤ add N , as claimed. 528M Lemma m↑ (S ∞ ) ≥ add N . proof (a) Recall the definition of the supported relations (N N , ⊆∗ , S (α) ) from 522L, where S (α) = {S : S ⊆ N × N, #(S[{n}]) ≤ α(n) for every n ∈ N} for α ∈ N N . Putting 522L, 522M and 512Db together, we have add(N N , ⊆∗ , S (α) ) = add N whenever limn→∞ α(n) = ∞. (b) The core of the argument is the following fact. Suppose that Q ⊆ S ∞ is cofinal and up-open, n ∈ N and σ ∈ [N × N]<ω . Let G ⊆ N × N be a set with finite vertical sections. Then there is a k ∈ N such that whenever σ ⊆ p ∈ S ∞ , p ⊆ σ ∪ G and kpk ≤ n, there is a q ∈ Q such that p ⊆ q and kqk ≤ k. P P?? Suppose, if possible, otherwise. Then for each j ∈ N we can find pj ∈ S ∞ such that σ ⊆ pj ⊆ σ ∪ G, kpj k ≤ n and kqk > j whenever p ⊆ q ∈ Q. Let p be a cluster point of hpj ij∈N in P(σ ∪G). Then #(p[{i}]) ≤ supj∈N #(pj [{i}]) ≤ min(2i , n) for every i, so p ∈ S ∞ . Because Q is cofinal with S ∞ , there is a q ∈ Q such that p ⊆ q. Set k = n + kqk. Then (σ ∪ G) ∩ (k × N) is finite, so there is an i ≥ k such that pi ∩ (k × N) = p ∩ (k × N) ⊆ q. Set q 0 = pi ∪ q. Then #(q 0 [{j}]) = #(q[{j}]) ≤ min(kqk, 2j ) if j < k, ≤ kpi k + kqk ≤ k ≤ 2j otherwise. So q 0 ∈ S ∞ and kq 0 k ≤ k ≤ i; because Q is up-open in S ∞ , q 0 ∈ Q, while pi ⊆ q 0 . But we chose pi so that this could not happen. X XQ Q (c) We need to know that S ∞ is upwards-ccc. P P For any n ∈ N, finite σ ⊆ N × N the set {p : p ∈ S ∞ , kpk ≤ 2n−1 , p ∩ (n × N) = σ} is upwards-linked. Q Q

170

Cardinal functions of measure theory

528M

(d) Now let hQξ iξ<κ be any family of cofinal subsets of S ∞ , where κS< add N , and p0 ∈ S ∞ . For each ξ < κ let Aξ ⊆ Qξ be a maximal up-antichain; by (d), Aξ is countable. Set Q0ξ = {[q, ∞[ : q ∈ Aξ }, so that Q0ξ is an up-open S cofinal subset of S ∞ . Set A = {p0 } ∪ ξ<κ Aξ . For q ∈ A, let Fq ⊆ N N be a finite set such that (identifying each S S member of Fq with its graph) q ⊆ Fq ; set F = q∈A Fq , so that #(F ) ≤ max(ω, κ) < add N ≤ b (522B). Let g0 ∈ N N be a strictly increasing function such that {i : f (i) > g0 (i)} is finite for every f ∈ F , and also p0 [{i}] ⊆ g0 (i) for every i. Set G = {(i, j) : i ∈ N, j < g0 (i)}, so that G ⊆ N × N has finite vertical sections. Observe that if q ∈ A then q \ G is finite. For each ξ < κ, n ∈ N and finite σ ⊆ N × N, let k(ξ, σ, n) ∈ N be such that whenever p ∈ S ∞ and σ ⊆ p ⊆ σ ∪ G then there is a q ∈ Q0ξ such that p ⊆ q and kqk ≤ k(ξ, σ, n); such a k exists by (c) above. Set kξ (n) = sup{k(ξ, σ, n) : σ ⊆ n × g0 (n)}. Again because κ < b, there is a g1 ∈ N N such that {n : kξ (n) > g1 (n)} is finite for every ξ < κ. Let α ∈ N N be a non-decreasing function such that limn→∞ α(n) = ∞ and α(2g1 (n)) ≤ n,

α(n) + #(p0 [{n}]) ≤ 2n ,

2α(n) ≤ n

for every n. Because add(N N , ⊆∗ , S (α) ) = add N , there is an S0 ∈ S (α) such that f ⊆∗ S0 for every f ∈ F , so that q \ S0 is finite for every q ∈ A. Replacing S0 by S0 ∩ G if necessary, we may suppose that S0 ⊆ G. (e) Let S be the family of subsets S of N×N such that #(S[{n}]) ≤ 2n for every n, as in 522K. Note that p0 ∪S0 ∈ S, because α(n) + #(p0 (n)) ≤ 2n for every n. Let C be the family of finite subsets σ of N × N such that σ ∪ S0 ∈ S. For each ξ < κ, set Dξ = {σ : σ ∈ C, ∃ q ∈ Aξ , q ⊆ σ ∪ S0 }. Then Dξ is cofinal with C. P P Let σ ∈ C. Let n0 be so large that g1 (2n0 ) ≥ kξ (2n0 ) and σ ⊆ n0 × g0 (n0 ). Set n0 m = 2g1 (2 ), p = σ ∪ (S0 ∩ (m × N)) ∈ S ∞ . Then σ ⊆ p ⊆ σ ∪ G and kpk ≤ max(2n0 , α(m)) = 2n0 , so there is a q ∈ Q0ξ such that p ⊆ q and kqk ≤ k(ξ, σ, 2n0 ) ≤ kξ (2n0 ) ≤ g1 (2n0 ) =

m . 2

Let q 0 ∈ Aξ be such that q 0 ⊆ q. Let m0 ≥ max(m, n0 ) be such that q 0 ⊆ (m0 × N) ∪ S0 , and set τ = q ∩ (m0 × N), so that σ ⊆ τ . For n < m, we have S0 [{n}] ⊆ p[{n}] ⊆ q[{n}] = τ [{n}], so (τ ∪ S0 )[{n}] = q[{n}] has at most 2n members. For m ≤ n < m0 , we have #((τ ∪ S0 )[{n}]) ≤ #(q[{n}]) + #(S0 [{n}]) ≤ kqk + α(n) ≤

m 2

+

n 2

≤ 2n ,

while for n ≥ m0 we have #((τ ∪ S0 )[{n}]) = #(S0 [{n}]) ≤ α(n) ≤ 2n . So τ ∪ S0 ∈ S and τ ∈ C. Since q 0 ⊆ (q 0 ∩ (m0 × N)) ∪ S0 ⊆ (q ∩ (m0 × N)) ∪ S0 = τ ∪ S0 , τ ∈ Dξ . As σ is arbitrary, Dξ is cofinal with C. Q Q (f ) Because p0 ∪ S0 ∈ S, σ0 = p0 \ S0 belongs to C. Because κ < add S N ≤ mcountable ≤ m↑ (C), there is an upwardsdirected set E ⊆ C meeting every Dξ and containing σ0 . Set S1 = S0 ∪ E. Then, because E is upwards-directed, #(S1 [{n}]) = supσ∈E #((σ ∪ S0 )[{n}]) ≤ 2n for every n, and S1 ∈ S. Set R = {p : p ∈ S ∞ , p ⊆ S1 }; then R ⊆ S ∞ is upwards-directed (in fact, closed under ∪), and p0 ∈ R because σ0 ∈ E. Now R meets Qξ for every ξ < κ. P P There is a σ ∈ Dξ ∩ E. But this means that there is a q ∈ Aξ such that q ⊆ σ ∪ S0 ⊆ S1 and q ∈ R ∩ Qξ . Q Q As p0 and hQξ iξ<κ are arbitrary, m↑ (S ∞ ) ≥ add N . ´ 95) Let (A, µ 528N Theorem (Brendle 00, 2.3.10; Judah & Repicky ¯) be an atomless σ-finite measure algebra with countable Maharam type, and 0 < γ < µ ¯1. Then the algebras AM(A, µ ¯, γ) and AM∗ (A, µ ¯) and the N-localization poset S ∞ (active upwards) all have Martin numbers equal to add N . proof By 517Ia and 528K, with 517Db again,

528P

Amoeba algebras

171

1 2

m(AM(A, µ ¯, γ)) = m(AM∗ (A, µ ¯)) = m(AM(Bω , ν¯ω , )) ≥ m(RO↑ (S ∞ )) = m↑ (S ∞ ). As m(AM(Bω , ν¯ω , 12 )) ≤ add N ≤ m↑ (S ∞ ) (528L, 528M), all these are equal to add N . 528O Corollary Let γ > 0. Let G be the partially ordered set {G : G ⊆ R is open, µL G < γ}, where µL is Lebesgue measure. Then m↑ (G) = add N . proof Put 528B and 528N together. 528P Proposition Let (A, µ ¯) be an atomless semi-finite measure algebra, and 0 < γ < µ ¯1. (a) For any integer m ≥ 2, c(AM(A, µ ¯, γ)) = linkm (AM(A, µ ¯, γ)) = max(c(A), τ (A)). (b) d(AM(A, µ ¯, γ)) = π(AM(A, µ ¯, γ)) = max(cf[c(A]≤ω , π(A)). proof Set P = {a : a ∈ A, µ ¯a < γ}, so that AM(A, µ ¯, γ) = RO↑ (P ). It will be helpful to note straight away that P is separative upwards. P P Let a, b ∈ P be such that a 6⊆ b. If µ ¯(a ∪ b) ≥ γ then a and b are already incompatible upwards. Otherwise, take d ⊆ 1 \ (a ∪ b) such that µ ¯(a ∪ b ∪ d) = γ, and set c = b ∪ d; then µ ¯c = γ − µ ¯(a \ b) < γ, so c ∈ [b, ∞[ ⊆ P , while a and c are incompatible upwards in P . Q Q (a) Set κ0 = max(c(A), τ (A)), κ1 = linkm (RO↑ (P )) = link↑m (P ) and κ2 = c(RO↑ (P )) = c↑ (P ) (514N). (i) The topological density of Af for its measure metric is κ0 (521Ob), so P has a metrically dense subset D of size at most κ0 . For d ∈ D, set Ud = {a : a ∈ P , µ ¯(a \ d) <

1 (γ m

−µ ¯d)}.

Then Ud is upwards-m-linked in P . Also, if a ∈ P , there is a d ∈ D such that µ ¯(a 4 d) < So P is κ0 -m-linked upwards and κ1 ≤ κ0 .

1 ¯a), m+1 (γ − µ

and now a ∈ Ud .

(ii) By 511Hb or 511Ja, κ2 ≤ κ1 . (iii) We need to check that κ2 is infinite. P P Take a ∈ A such that µ ¯a = γ. For any n ≥ 1, we can find disjoint 1 a0 , . . . , an ⊆ a all of measure n+1 γ; now ha \ ai ii≤n is an up-antichain in P . So κ2 = c↑ (P ) ≥ n + 1; and this is true for every n. Q Q Now if (A, µ ¯) is totally finite, then c(A) = ω ≤ κ2 . Otherwise, there is a partition D of unity in A such that µ ¯d = 21 γ for every d ∈ D; now #(D) is an up-antichain in P and κ2 ≥ #(D) = c(A). So we see that in all cases κ2 ≥ c(A). P?? Otherwise, set α = µ ¯e,rκ = τ (Ae ). (iv) If e ∈ Af and the principal ideal Ae is homogeneous, then τ (Ae ) ≤ κ2 . P Because µ ¯1 > γ, there is a d ⊆ 1 \ e such that γ < µ ¯(e ∪ d) < γ + µ ¯e, that is, 0 < γ − µ ¯d < α. Set β =

1−

γ−¯ µd . α

Because Ae is isomorphic, up to a scalar multiple of the measure, to the measure algebra of the usual measure on [0, 1]κ , there is a family hcξ iξ<κ in Ae such that µ ¯cξ = βα,

µ ¯(cξ ∩ cη ) = β 2 α

whenever ξ, η < κ are distinct. Set bξ = d ∪ (e \ cξ ) for ξ < κ. Then µ ¯(bξ ∪ bη ) = µ ¯d + α − β 2 α = γ, µ ¯bξ = µ ¯d + α − βα < γ for all distinct ξ, η < κ. So hbξ iξ<κ is an up-antichain in P and witnesses that κ2 ≥ κ. X XQ Q (v) Let E be a partition of unity in A such that 0 < µ ¯e < ∞ and AeSis homogeneous for every e ∈ E. For e ∈ E, let Ae ⊆ Ae be a set of size at most κ2 which τ -generates Ae . Then A = e∈E Ae τ -generates A, so that κ0 = max(c(A), τ (A)) ≤ max(c(A), #(A)) ≤ max(c(A), κ2 ) = κ2 , and the three cardinals must be equal.

172

Cardinal functions of measure theory

528P

(b) Set κ3 = max(π(A), cf[c(A]≤ω ), κ4 = π(AM(A, µ ¯, γ)) and κ5 = d(AM(A, µ ¯, γ)). (i) By 323F, Af is complete in its measure metric. As in 323B, ∪ : Af × Af → Af is uniformly continuous for the measure metric, and P is an open subset of Af , while the topological density of Af is κ0 . By 524C, (P, ⊆ 0 , [P ]<ω ) 4GT (`1 (κ0 ), ≤, `1 (κ0 )), where ⊆ 0 is defined as in 512F. It follows that κ4 = π(RO↑ (P )) = cf P (514Nb) = cov(P, ⊆ , P ) ≤ max(ω, cov(P, ⊆ 0 , [P ]<ω )) (512Gf) ≤ max(ω, cf `1 (κ0 )) (512Da) = cf `1 (κ0 ) = cf Nκ0 (where Nκ0 is the null ideal of the usual measure on {0, 1}κ0 , as in 524I) = max(cf N , cf[κ0 ]≤ω ) (523N) = max(cf N , cf[τ (A)]≤ω , cf[c(A)]≤ω ) = max(cf N , cf[τ (A)]≤ω , c(A), cf[c(A)]≤ω ) = max(π(A), cf[c(A)]≤ω ) (524Mc) = κ3 . (ii) By 514Nd, d↑ (P ) = κ5 . Let hBξ iξ<κ5 be a family of upwards-centered sets covering P . For each ξ, bξ = sup Bξ is defined in A (counting sup ∅ as 0 if necessary), and µ ¯bξ = supI∈[Bξ ]<ω µ ¯(sup I) ≤ γ. Set D = {bξ \ bη : ξ, η < κ5 }. Then D is order-dense in A. P P If a ∈ A \ {0}, take a0 ⊆ a such that 0 < µ ¯a0 < γ. Then 0 0 0 a ∈ P , so there is some ξ < κ5 such that a ∈ Bξ and a ⊆ bξ . Next, let c ⊆ 1 \ bξ be such that γ−µ ¯bξ < µ ¯c < γ − µ ¯bξ + µ ¯a0 . Then c ∪ bξ \ a0 ∈ P , so there is an η < κ5 such that c ∪ (bξ \ a0 ) ⊆ bη . Now d = bξ \ bη ⊆ a0 ; as µ ¯(bξ ∪ c) > γ ≥ µ ¯bη , bξ ⊆ 6 bη and d 6= 0. Of course d ∈ D and d ⊆ a; as a is arbitrary, D is order-dense. Q Q Accordingly π(A) ≤ #(D) ≤ κ5 . At the same time, cf[c(A)]≤ω ≤ κ5 . P P There is a disjoint set E ⊆ A \ {0} of size c(A) (332F). For each ξ < κ5 , let Iξ be the countable set {e : e ∈ E, e ∩ bξ 6= 0}. If J ⊆ E is countable, let he ie∈J be a strictly positive family of real numbers with sum less than γ. For each e ∈ J let ae ⊆ e be such that 0 < µ ¯ a e ≤ e , and set a = supe∈J ae . Then a ∈ P so there is a ξ < κ5 such that a ⊆ bξ and J ⊆ Iξ . As J is arbitrary, {Iξ : ξ < κ5 } is cofinal with [E]≤ω , and cf[c(A)]≤ω = cf[E]≤ω ≤ κ5 . Q Q Putting these together, we see that κ3 ≤ κ5 . (iii) By 514Da, κ5 ≤ κ4 , so the three cardinals are equal. 528Q Proposition Let S ∞ be the N-localization poset. (a) π(RO↑ (S ∞ )) = cf S ∞ = c. (b) For every m ≥ 2, c(RO↑ (S ∞ )) = c↑ (S ∞ ) = linkm (RO↑ (S ∞ )) = link↑m (S ∞ ) = ω. (c) d(RO↑ (S ∞ )) = d↑ (S ∞ ) = cf N . proof (a) If p, q ∈ S ∞ and p 6⊆ q, take (n, i) ∈ p \ q; then there is a q 0 ∈ S ∞ such that q 0 ⊇ q, #(q 0 [{n}]) = 2n and (n, i) ∈ / q 0 , in which case p and q 0 are incompatible upwards in S ∞ . So S ∞ is separative upwards and 514Nb tells us that π(RO↑ (S ∞ )) = cf S ∞ . Next, there is an almost-disjoint family hhξ iξ
528R

Amoeba algebras

173

AIr = {p : p ∈ S ∞ , p ∩ (r × N) = I, kpk ≤ ∞ for any family hpi ii<m in AIr , that is, AIr is i<m pi ∈ S ↑ ∞ linkm (S ) ≤ ω. Of course c↑ (S ∞ ) is infinite, and since c↑ (S ∞ )

Then

S

so tells us that

2r }. m

upwards-m-linked in S ∞ . Also

S

(I,r)∈Q

AIr = S ∞ ,

≤ link↑m (S ∞ ) (511Hb), both must be ω. Now 514N

c(RO↑ (S ∞ )) = linkm (RO↑ (S ∞ )) = ω. (c) Consider the localization relation (N N , ⊆∗ , S) of 522K. We know from 522M and 512Da that cov(N N , ⊆∗ , S) = cov(N , ⊆, N ) = cf N . (i) Let A ⊆ S be a set of cardinality cf N such that for every f ∈ N N there is an S ∈ A such that f ⊆∗ S. Let A be S {S : S ∈ S, S \ A0 is finite for some finite A0 ⊆ A}; ∗

then every member of S ∞ is included in some member of A∗ . But if S ∈ A∗ then {p : p ∈ S ∞ , p ⊆ S} is upwardsdirected. So d ↑ (S ∞ ) ≤ #(A∗ ) ≤ cf N . S (ii) Now let Q be a family of upwards-centered subsets of S ∞ covering S ∞ . For each Q ∈ Q, SQ = Q belongs to S. Also every f ∈ N N belongs to S ∞ so is covered by some SQ . So SQ witnesses that cf N = cov(N N , ⊆∗ , S) ≤ #(Q); as Q is arbitrary, cf N ≤ d ↑ (S ∞ ). (iii) 514Nd tells us that d(RO↑ (S ∞ )) = d ↑ (S ∞ ), so we have equality throughout. 528R Theorem Let κ be any cardinal. (a) τ (RO↑ (Sκ∞ )) ≤ ω, where Sκ∞ the κ-localization poset. (b) τ (AM(Bκ , ν¯κ , 12 )) ≤ ω. proof (a)(i) If κ is finite then cf Sκ∞ is finite and the result is trivial. So let us suppose from now on that κ is infinite. P If p, q ∈ Sκ∞ and p 6⊆ q, take (n, ξ) ∈ p \ q. Let J ⊆ κ \ p[{n}] be a set of size (ii) Sκ∞ is separative upwards. P 0 2 − #(q[{n}], and set q = q ∪ ({n} × J); then q ⊆ q 0 ∈ S ∞ and p, q 0 are incompatible upwards in Sκ∞ . Q Q Accordingly [p, ∞[ ∈ RO↑ (Sκ∞ ) for every p ∈ Sκ∞ (514M). n

(iii) For n ∈ N, m < 2n and ξ < κ, set Gmnξ = sup{[p, ∞[ : p ∈ Sκ∞ , #(p[{n}]) = 2n , (n, ξ) ∈ p and #(p[{n}] ∩ ξ) = m}, the supremum being taken in RO↑ (Sκ∞ ). (iv) If n ∈ N, m < 2n and ξ < η < κ then Gmnξ ∩ Gmnη = ∅. P P If p, q ∈ Sκ∞ , #(p[{n}]) = #(q[{n}]) = 2n , (n, ξ) ∈ p, (n, η) ∈ q and #(p[{n}] ∩ ξ) = #(q[{n}]) ∩ η) = m then p[{n}] 6= q[{n}], #(p[{n}] ∪ q[{n}]) > 2n and [p, ∞[ ∩ [q, ∞[ is empty. Q Q S (v) If ξ < κ then {Gmnξ : n ∈ N, m < 2n } is dense in Sκ∞ . P P If p ∈ Sκ∞ , take n ∈ N such that #(p[{n}]) < 2n ; ∞ then there is a q ∈ Sκ such that p ⊆ q, ξ ∈ q[{n}] and #(q[{n}]) = 2n . Set m = #(q[{n}] ∩ ξ); then [q, ∞[ ⊆ [p, ∞[ ∩ Gmnξ . Q Q Thus sup{Gmnξ : n ∈ N, m < 2n } = 1 in RO↑ (S ∞ ) whenever ξ < κ. (vi) Let A be the order-closed subalgebra of RO↑ (Sκ∞ ) generated by {Gmnξ : n ∈ N, m < 2n , ξ < κ}. For n ∈ N and ξ < κ set Hnξ = [{(n, ξ)}, ∞[; then Hnξ = supm<2n Gmnξ belongs to A. P P Certainly Gmnξ ⊆ Hnξ whenever m < 2n . If {(n, ξ)} ⊆ p ∈ S Sκ∞ , let q ∈ Sκ∞ be such that p ⊆ q and #(q[{n}]) = 2n ; set m = #(q[{n}] ∩ ξ); then [q, ∞[ ⊆ Hnξ ∩ Gmnξ . Thus m<2n Gmnξ is dense in Hnξ and Hnξ = supm<2n Gmnξ ∈ A. Q Q (vii) If p ∈ Sκ∞ then [p, ∞[ = inf (n,ξ)∈p Hnξ belongs to A, by 514Ne. So A includes an order-dense subset of RO (Sκ∞ ) and must be the whole of RO↑ (Sκ∞ ); that is, RO↑ (Sκ∞ ) is τ -generated by {Gmnξ : n ∈ N, m < 2n , ξ < κ}. With (iv) and (v), we see that the conditions of 514F are satisified with J = κ and I = {(m, n) : n ∈ N, m < 2n }, so that ↑

τ (RO↑ (Sκ∞ )) ≤ max(ω, #(I)) = ω.

174

Cardinal functions of measure theory

528R

(b)(i) Let P be {p : p ∈ Bκ , ν¯κ p < 12 }, so that AM(Bκ , ν¯κ , 12 ) = RO↑ (P ). Then P is separative upwards. P P Suppose that p, q ∈ P and that p 6⊆ q; set c = 1 \ (p ∪ q); then µ ¯q + µ ¯c ≥ 12 . Take d ⊆ c such that 12 − µ ¯(p \ q) < µ ¯q + µ ¯d < 12 , and set a = q ∪ d; then q ⊆ a ∈ P and p, a are incompatible upwards in P . Q Q Let heξ iξ<κ be the standard generating family in Bκ (525A), and D the subalgebra of Bκ generated by {eξ : ξ < κ}. For J ⊆ κ, write CJ for the order-closed subalgebra of Bκ generated by {eξ : ξ ∈ J}; recall that for every a ∈ Bκ there is a countable set I ⊆ κ such that a ∈ CJ iff I ⊆ J (254Rd, 325Mb); I will call I the ‘support’ supp a of a. (ii) If κ is finite, so are P and AM(Bκ , ν¯κ , 21 ), and the result is trivial. (iii) If κ = ω we can argue as follows. Let Q be {q : q ∈ P , q = sup{d : d ∈ D, d ⊆ q}}. Then Q is cofinal with P . P P D is dense in Bκ for the measure metric (323A). If p ∈ P , choose hdn in∈N in D such that ν¯ω (dn 4 p) ≤ 2−n−2 ( 12 − ν¯ω p) for each n, and set q = supn∈N dn . Then ν¯ω (q \ p) ≤

P∞

n=0

ν¯ω (dn \ p) <

1 2

− ν¯ω p,

so q ∈ P ; ν¯ω (p \ q) ≤ inf n∈N ν¯ω (p \ dn ) = 0, so p ⊆ q; and q = supn∈N dn ⊆ sup{d : d ∈ D, d ⊆ q} ⊆ q, so q ∈ Q. Q Q Let A be the order-closed subalgebra of AM(Bκ , ν¯κ , 12 ) generated by {[d, ∞[ : d ∈ D ∩ P }. If q ∈ Q, then T [q, ∞[ = {[d, ∞[ : d ∈ D, d ⊆ q} ∈ A; as Q is cofinal with P , A includes an order-dense subset of AM(Bκ , ν¯κ , 12 ) and must be the whole of AM(Bκ , ν¯κ , 21 ). So the countable set {[d, ∞[ : d ∈ D ∩ P } τ -generates AM(Bκ , ν¯κ , 12 ) and τ (AM(Bκ , ν¯κ , 12 )) ≤ ω. (iv) We may therefore suppose henceforth that κ is uncountable. For n ∈ N let Cn be the family of cylinders of the form c = inf ξ∈I eξ \ supξ∈J eξ where I, J ⊆ κ are disjoint and #(I ∪ J) = n; for definiteness, say that C0 = {1}. Note that (α) if c ∈ Cm , c0 ∈ Cn and c0 ⊆ c, then m ≤ n (β) if a, b ∈ Bκ , supp a ∩ supp b = ∅, c ∈ Cn and c ⊆ a ∪ b then either c ⊆ a or c ⊆ b. P P If I = supp a, then we can express c as c0 ∩ c00 where supp c0 ⊆ I and supp c00 ⊆ κ \ I. Now 0 = c \ (a ∪ b) = (c0 \ a) ∩ (c00 \ b). But c0 \ a ∈ CI and c00 \ b ∈ Cκ\I are independent, so one of them must be zero. Q Q S S (v) If n ∈ N and hci ii∈N is a sequence in m≤n Cm , there is an infinite J ⊆ N such that supi∈J ci ∈ m≤n Cm . P P Set Ki = supp ci ∈ [κ]≤n for each i. Then there is an infinite I ⊆ N such that hKi ii∈I is a ∆-system with root K say (5A1Ic). For i ∈ I express ci as c0i ∩ c00i where supp c0i = K and supp c00i = Ki \ K, and let J ⊆ I be an infinite set such that c0i = c0j for all i, j ∈ J; take c0 for this common value. In this case, S supi∈J ci = c0 ∩ supi∈J c00i = c0 ∈ m≤n Cm because hc00j ii∈J is a stochastically independent family of sets all of measure at least 2−n , so that 1 \ supi∈J c00j = Q 00 inf j∈J 1 \ cj has measure i∈J 1 − ν¯κ c00j = 0. Q Q S (vi) If a ∈ Bκ and n ∈ N then there is a δ > 0 such that ν¯κ (c \ a) ≥ δ whenever c ∈ m≤n Cn and c 6⊆ a. P P Induce on n. For n = 0 the result is trivial, since we S need only consider the case c = 1. For the inductive step to nS ≥ 1, take δ 0 > 0 such that ν¯κ (c \ a) ≥ δ 0 whenever c ∈ m
P

i∈J

1 2

ν¯κ (ci \ a) ≤ δ 0 ,

which is absurd. X X Thus the induction proceeds. Q Q S (vii) If p ∈ P and n ∈ N there is a q ∈ P such that p ⊆ q and whenever c ∈ m≤n Cm and c 6⊆ p then q ∪ c ∈ / P. S P P Let δ > 0 be such that ν¯κ (c \ p) ≥ δ whenever c ∈ m≤n Cm and c 6⊆ p. Let r ∈ N be such that 2−r < 2−2n−1 δ, and let k be maximal subject to (1 − ν¯κ p)(1 − 2−r )k > 12 ; let c0 , . . . , ck−1 ∈ Cr be such that their supports are disjoint from each other and from supp p. (This is where it is helpful to suppose that κ is uncountable.) Set q = p ∪ supi
528R

Amoeba algebras

ν¯κ q <

1 2

175

≤ 2−r + ν¯κ q.

S Suppose that c ∈ m≤n Cm is such that c 6⊆ p. Express c as inf i≤k c0i where supp c0i ⊆ supp ci for i < k and S supp c0k ⊆ κ \ i
≥δ·

Y1 i∈J

≥δ·2

ν¯κ c0i 2

−#(J)

·

i∈k\J

Y

(1 − 2

−r

)

i∈k\J

ν¯κ c · (1 − 2−r )k

≥ 2−2n δ(1 − 2−r )k ≥ 2−r+1

1 2

1 − ν¯κ p

≥ 2−r ≥

1 2

− ν¯κ q,

so c ∪ q ∈ / P. Q Q (viii) For p ∈ P and n ∈ N, set An (p) = {c : c ∈

S

m≤n

Cm , c ⊆ p}.

Note that An (p) ⊆ An (q) whenever p ⊆ q in P . Let Qn be {p : An (q) = An (p) whenever p ⊆ q ∈ P }. Then for every p ∈ P there is a q ∈ Qn such P Given p ∈ P , let q ⊇ p be as in (vii). If q 0 ∈ P , q 0 ⊇ q S that p ⊆ q and An (q) = An (p). P 0 and c ∈ An (q ), then c ∈ m≤n Cm and c ∪ q ∈ P , so c ⊆ p; thus An (q 0 ) = An (p) = An (q). As q 0 is arbitrary, q ∈ Qn . Q Q In particular, Qn is cofinal with P . S (ix) Enumerate n∈N Cn as heξ iξ<κ in such a way that whenever n ∈ N and ζ < κ is a limit ordinal, heζ+i ii∈N has a subsequence consisting of elements of Cn with disjoint supports. It follows that if ξ < κ, n ∈ N and K ⊆ κ is finite, there is an i ∈ N such that eξ+i ∈ Cn and K ∩ supp eξ+i = ∅. For p ∈ P and n ∈ N, let En (p) be {ξ : ξ < κ, eξ is a maximal member of An (p)}; then En (p) is finite, by (v). For m, n, i ∈ N and ξ < κ, set Gmniξ = sup{[q, ∞[ : q ∈ Qn , ξ + i ∈ En (q), #(En (q) ∩ (ξ + i)) = m} 1 ∈ AM(Bκ , ν¯κ , ). 2 (x) If m, n, i ∈ N and ξ < η < κ then Gmniξ ∩ Gmniη = ∅. P P If q, q 0 ∈ Qn , ξ + i ∈ En (q), η + i ∈ En (q 0 ) and #(En (q) ∩ (ξ + i)) = m = #(En (q 0 ) ∩ (η + i)), then ξ + i < η + i, En (q) 6= En (q 0 ) and An (q) 6= An (q 0 ). But by the definition of Qn , this means that [q, ∞[ ∩ [q 0 , ∞[ must be empty. Q Q S 1 (xi) If ξ < κ then m,n,i∈N Gmniξ is dense in P . P P Take p ∈ P . Let n ∈ N be such that ν¯κ p + 2−n < . Let 2 S δ > 0 be such that ν¯κ (c \ p) ≥ δ whenever c ∈ k 0 be such that 2−n (δ − ) > . Let a ∈ D be such that ν¯κ (p 4 a) ≤ ; then supp a is finite. By the choice of heξ iξ<κ , there is an i ∈ N such that eξ+i ∈ Cn and supp eξ+i is disjoint from supp a. Set p0 = p ∪ eξ+i ; by the choice of n, p0 ∈ P . Of course eξ+1 ∈ An (p0 ). Let q ∈ Qn be such that p0 ⊆ q and An (q) = An (p0 ).S ?? If ξ + i ∈ / En (q), there is a c ∈ k
S

(because c0 ∈

S

k
Ck ) ≥ 2−n (δ − ) − 

k
Ck and c0 6⊆ p) > 0. X X

S So ξ + i ∈ En (q); set m = #(En (q) ∩ (ξ + i)). Then [q, ∞[ ⊆ Gmniξ , so [p, ∞[ meets Gmniξ . As p is arbitrary, Q m,n,i∈N Gmniξ is dense. Q (xii) If m, n ∈ N and ξ < κ, then Gmn0ξ ⊆ [eξ , ∞[. P P If m, n ∈ N, q ∈ Qn , ξ ∈ En (q) and #(En (q) ∩ ξ) = m, then eξ ⊆ q and [q, ∞[ ⊆ [eξ , ∞[. Q Q (xiii) Let A be the order-closed subalgebra of AM(Bκ , ν¯κ , 21 ) generated by {Gmniξ : m, n, i ∈ N, ξ < κ}. Then [c, ∞[ ∈ A whenever c ∈ P is a cylinder. P P Let n ∈ N be such that c ∈ Cn . Set

176

Cardinal functions of measure theory

H=

S

528R

{Gmn0ξ : m ∈ N, ξ < κ, Gmn0ξ ⊆ [c, ∞[};

then int H ∈ A. Of course H and int H are included in [c, ∞[. If p ∈ [c, ∞[, then c ∈ An (p); let q ∈ Qn be such that p ⊆ q and An (q) = An (p). Let ξ < κ be such that eξ is a maximal member of An (q) including c; then ξ ∈ En (q); set m = #(En (q) ∩ ξ). Now [q, ∞[ ⊆ Gmn0ξ ⊆ [eξ , ∞[ ⊆ [c, ∞[, Q so q ∈ [p, ∞[ ∩ H. Thus H is dense in [eξ , ∞[ and [eξ , ∞[ = int H belongs to A. Q S (xiv) We are nearly home. If p ∈ P , then p = sup{c : c ∈ n∈N Cn , c ⊆ p}, so [p, ∞[ = inf{[c, ∞[ : c ⊆ p is a cylinder} belongs to A. Accordingly AM(Bκ , ν¯κ , 12 ) = A is τ -generated by {Gmniξ : m, n, i ∈ N, ξ < κ}. By 514F again, AM(Bκ , ν¯κ , 21 ) has countable Maharam type. 528X Basic exercises (a) Let (A, µ ¯) be a semi-finite measure algebra such that all its principal ideals of finite measure are homogeneous. Show that AM(A, µ ¯, γ) is homogeneous for every γ < µ ¯1. (Hint: first check that A ∼ = A1 \ a whenever a ∈ A and 0 < µ ¯a < µ ¯1.) (b) (i) Let (A, µ ¯) be a totally finite measure algebra. Show that AM(A, µ ¯, µ ¯1) is isomorphic to A. (ii) Let (A, µ ¯) be an atomless measure algebra and e ∈ A a non-zero element of finite measure. Show that the principal ideal Ae can be regularly embedded in AM(A, µ ¯, µ ¯e). (c) Show that if κ ≤ λ and 0 < γ ≤ 1 then AM(Bκ , ν¯κ , γ) can be regularly embedded in AM(Bλ , ν¯λ , γ). (d) Let (A, µ ¯) be an atomless semi-finite measure algebra and 0 < γ < µ ¯1. Show that m(AM(A, µ ¯, γ)) = wdistr(A). (e) Let (A, µ ¯) be an atomless semi-finite measure algebra. (i) Show that c(AM∗ (A, µ ¯)) = linkm (AM∗ (A, µ ¯)) = max(c(A), τ (A)) for any integer m ≥ 1. (ii) Show that d(AM∗ (A, µ ¯)) = π(AM∗ (A, µ ¯)) = max(cf[c(A]≤ω , π(A)). (f ) Let (X, T, Σ, µ) be a quasi-Radon measure space and (A, µ ¯) its amoeba algebra. Show that the additivity of µ is not a precaliber of AM(A, µ ¯, γ) for any γ < µX. (g) Suppose that (X, Σ, µ) is a measure space and (A, µ ¯) its measure algebra. Let E ⊆ Σ be a family such that µ is outer regular with respect to E, and P the set {(E, α) : E ∈ E, µE < α ≤ µX}, ordered by saying that (E, α) ≤ (F, β) if E ⊆ F and β ≤ α. Show that RO↑ (P ) is isomorphic to AM∗ (A, µ ¯). (h) Show that τ (AM∗ (Bκ , ν¯κ )) ≤ ω for every cardinal κ. 528Y Further exercises (a) Let (A, µ ¯) be an atomless probability algebra and γ ∈ ]0, 1[. Show that AM(A, µ ¯, γ) is not weakly (σ, ∞)-distributive. (b) Let (A, µ ¯) be a semi-finite measure algebra such that all its principal ideals are homogeneous. Show that AM∗ (A, µ ¯) is homogeneous. (c) For α ∈ N N and any set I let (SI∞ )(α) be the partially ordered set {p : p ⊆ N × I, #(p[{n}]) ≤ α(n) for every n, supn∈N #(p[{n}]) < ∞}. ↑

Show that RO

(SI∞ )(α)

can be regularly embedded in RO↑ (SI∞ )(β) whenever I is infinite and α, β are both unbounded.

(d) Let κ be an infinite cardinal. Show that (i) π(RO↑ (Sκ∞ )) = cf Sκ∞ is the cardinal power κω ; (ii) for every m ≥ 2, c(RO↑ (Sκ∞ )) = c↑ (Sκ∞ ) = linkm (RO↑ (Sκ∞ )) = link↑m (Sκ∞ ) = κ; (iii) d(RO↑ (Sκ∞ )) = d ↑ (Sκ∞ ) = max(cf N , cf[κ]≤ω ). (e) Let (A, µ ¯) be a σ-finite measure algebra and 0 < γ < µ ¯1. Show that AM(A, µ ¯, γ) and AM∗ (A, µ ¯) both have countable Maharam type.

529B

Further partially ordered sets of measure theory

177

528Z Problems (a) Let (AL , µ ¯L ) be the measure algebra of Lebesgue measure on R. Is the amoeba algebra AM(AL , µ ¯L , 1) isomorphic to the amoeba algebra AM(Bω , ν¯ω , 12 )? (b) Let (A, µ ¯) be a probability algebra, B a closed subalgebra of A, and 0 < γ < 1. Is it necessarily true that AM(B, µ ¯ B, γ) can be regularly embedded in AM(A, µ ¯, γ)? (See 528Xc.) (c) Can Bω1 be regularly embedded in AM(Bω , ν¯ω , 12 )? (See 528Xb.) 528 Notes and comments The ideas of 528A-528K are based on Truss 88. The original amoeba algebras of Martin & Solovay 70, used in their proof that add N ≥ m (528L), are closest to 528B. For some more about the amoeba ´ ski & Judah 95, §3.4. In this section I have been willing to algebras derived from Lebesgue measure, see Bartoszyn assume that the measure algebras involved are atomless; amoeba algebras are surely still interesting for other measure algebras, but the new questions seem to be combinatoric rather than measure-theoretic. It seems still to be unknown whether the algebras AM(AL , µ ¯L , 1) and AM(Bω , ν¯ω , 12 ) are actually isomorphic, rather than just mutually embeddable (528K, 528Za). If we think of the partially ordered sets of 528A and 528I as forcing notions, we can study them in terms of the forcing universes they lead to. I will not attempt to use such methods here; but I mention them because there are general results which lead relatively quickly to Theorem 528R, in particular.

529 Further partially ordered sets of measure theory I end the chapter with notes on some more structures which can be approached by the methods used earlier. The Banach lattices of Chapter 36 are of course partially ordered sets, and many of them can easily be assigned places in the Tukey classification (529C, 529D, 529Xa). More surprising is the fact that the Nov´ak numbers of {0, 1}I , for large I, are supported by the additivity of Lebesgue measure (529F); this is associated with an interesting property of the localization poset from the last section (529E). There is a similarly unexpected connexion between the covering number of Lebesgue measure and ‘reaping numbers’ r(ω1 , λ) for large λ (529G). 529A Notation As in previous sections, I will write N (µ) for the null ideal of µ in a measure space (X, Σ, µ), and N for the null ideal of Lebesgue measure on R. 529B Proposition Let (A, µ ¯) be a semi-finite measure algebra. (a) For p ∈ [1, ∞[, give Lp = Lp (A, µ ¯) (definition: 366A) its norm topology. Then its topological density is d(Lp ) = 1 if A = {0}, = ω if 0 < #(A) < ω, = max(c(A), τ (A)) if A is infinite. (b) Give L0 = L0 (A) its topology of convergence in measure (367L). Then d(L0 ) = 1 if A = {0}, = ω if 0 < #(A) < ω, = τ (A) if A is infinite. proof (a)(i) The case in which A is finite is elementary, since in this case Lp (A, µ ¯) ∼ = R n , where n is the number of atoms of A. So henceforth let us suppose that A is infinite. (ii) If Af is the set of elements of A of finite measure, we have a natural injection a 7→ χa : Af → Lp , and kχa − χbkp = µ(a 4 b)1/p , so χ is a homeomorphism for the measure metric on Af and the norm metric on Lp . It follows that the density d(A) of A = {χa : a ∈ Af } for the norm topology is equal to the density of Af for the strong measure-algebra topology, which is max(c(A), τ (A)), by 521Ob. So max(c(A), τ (A)) = d(A) ≤ d(Lp ) by 5A4B(h-ii). In the other direction, if A0 is a dense subset of A of size d(A) and D is the set of rational linear combinations of members of A0 , D ⊇ S(Af ) is dense in Lp , so d(Lp ) ≤ #(D) ≤ max(ω, #(A0 )) = max(c(A), τ (A)).

178

Cardinal functions of measure theory

529B

(b) Again, the case of finite A is trivial, so we need consider only infinite A. In this case, τ (A) is the topological density of A with its measure-algebra topology T (521Oa). (i) Let A ⊆ A be a topologically dense set with cardinal τ (A). Set Pn D = { i=0 qi χai : q0 , . . . , qn ∈ Q, a0 , . . . , an ∈ A}, so that D ⊆ L0 has cardinal τ (A). Because a 7→ χa : A → L0 is continuous (367R), the closure D of D includes {χa : a ∈ A}. Because D is a linear subspace of L0 , it includes S(A). Because S(A) is dense in L0 (367Nc), D = L0 and d(L0 ) ≤ #(D) = τ (A). (ii) Let B ⊆ L0 be a dense set with cardinal d(L0 ). Set A = {[[u > 21 ]] : u ∈ B}, 0 so that A ⊆ P If c ∈ A, a ∈ Af and  > 0, there is a u ∈ B R A and #(A) ≤ d(L ). Then A is topologically dense in A. P such that |u − χc| ∧ χa ≤ 21 . But in this case, setting b = [[u > 12 ]], |χ(b 4 c)| ≤ 2|u − χc|, so

R

µ ¯(a ∩ (b4c)) ≤ 2 |u − χc| ∧ χa ≤ . As c, a and  are arbitrary, A is topologically dense in A. Q Q Accordingly τ (A) = dT (A) ≤ #(A) ≤ d(L0 ) and d(L0 ) = τ (A), as claimed. 529C Theorem (Fremlin 91a) Let U be an L-space. Then U ≡T `1 (κ), where κ = dim U if U is finite-dimensional, and otherwise is the topological density of U . proof (a) The finite-dimensional case is trivial, since in this case U and `1 (κ) are isomorphic as Banach lattices. So henceforth let us suppose that U is infinite-dimensional. Now ∨ : U × U → U is uniformly continuous. P P We have only to observe that u ∨ v = 21 (u + v + |u − v|) in any Riesz space, that addition and subtraction are uniformly continuous in any linear topological space, and that u 7→ |u| is uniformly continuous just because ||u| − |v|| ≤ |u − v| (see 354B). Q Q So 524C, with Q = P = U , tells us that U 4T `1 (κ). The rest of the proof will therefore be devoted to showing that `1 (κ) 4T U . (b) By Kakutani’s theorem (369E), there is a localizable measure algebra (A, µ ¯) such that U is isomorphic, as Banach lattice, to L1 (A, µ ¯). Let hai ii∈I be a partition of unity in A such that 0 < µ ¯ai < ∞ and the principal ideal Aai is homogeneous for each i. Set κi = τ (Aai ), so that κi is either 0 or infinite for every i, and κ = max(#(I), supi∈I κi ) by 529Ba. the calculations to follow ifR we arrange that all the ai have measure 1. To do this, set R ν¯a = P It will simplify −1 −1 0 (¯ µ a ) µ ¯ (a ∩ a ) for a ∈ A; that is, ν ¯ a = w d¯ µ , where w = sup (¯ µ a ) χa in L (A). In this case, v d¯ ν= i i i i i∈I i∈I a R R R 1 1 1 × u d¯ ν for every u ∈ L (A, µ ¯) (365T). But this means that v × w d¯ ν for every v ∈ L (A, ν¯), while u d¯ µ = w

u 7→ u ×

1 w

is a Banach lattice isomorphism between L1 (A, µ ¯) and L1 (A, ν¯), and U is isomorphic, as L-space, to

L1 = L1 (A, ν¯); while ν¯ai = 1 for every i. (c) There are p a set J, of cardinal κ, and a family huj ij∈J in L1 such that #(J) = κ, kuj k ≤ 2 for every j ∈ J and 1 k supj∈K uj k ≥ 2 #(K) for every finite K ⊆ J. P P Set J = {(i, 0) : i ∈ I, κi = 0} ∪ {(i, ξ) : i ∈ I, ξ < κi }. Then #(J) = κ. If i ∈ I and κi = 0, set u(i,0) = χai . If i ∈ I and κi > 0, then (Aai , ν¯ Aai ) is a homogeneous ¯ i ) of ]0, 1]κi with its usual probability algebra with Maharam type κi ≥ ω, so is isomorphic to the measure algebra (Ci , λ measure λi , the product of Lebesgue measure on each copy of ]0, 1] (334E). For ξ < κi , set 1 hiξ (t) = √

t(ξ)

κ

for t ∈ ]0, 1] i ,

¯ i ) (365B). Of course and let u(i,ξ) ∈ L1 correspond to h•iξ ∈ L1 (λi ) ≡ L1 (Ci , λ Z ku(i,ξ) k =

Z hiξ (t)λi (dt) =

(because the coordinate map t 7→ t(ξ) is inverse-measure-preserving) = 2.

1

1 √ dα α 0

529D

Further partially ordered sets of measure theory

If L ⊆ κi is finite and not empty, then k supξ∈L u(i,ξ) k = κ for t ∈ ]0, 1] i . Now, for any α ≥ 1,

R

179

g dλi where g = supξ∈L hiξ , that is, g(t) = supξ∈L t(ξ)−1/2

λi {t : g(t) ≤ α} = λi {t : α2 t(ξ) ≥ 1 for every ξ ∈ L} 1 2

1 2

= (1 − α−2 )#(L) ≤ max( , 1 − #(L)α−2 ) (see part (a) of the proof of 273K) 1 2

= 1 − α−2 #(L) if α ≥

p

#(L).

So Z k sup u(i,ξ) k =

Z

∞

λi {t : g(t) > α}dα

g dλi =

ξ∈L

0

(252O) Z ∞ ≥ √ λi {t : g(t) > α}dα #(L) Z ∞ 1 −2 1p #(L). ≥ √ α #(L)dα = #(L)

2

2

What this means p is that if K ⊆ J is finite and all the first coordinates of members of K are the same, then k supj∈K uj k ≥ 12 #(K). In general, if K ⊆ J is finite, then for each i ∈ I set Li = {ξ : (i, ξ) ∈ K}. Set vi = 0 if Li p P is empty, supξ∈Li u(i,ξ) otherwise, so that kvi k ≥ 21 #(Li ); now supj∈K uj = i∈I vi , so p P 1 pP 1p 1P #(Li ) ≥ #(K), k supj∈K uj k = i∈I kvi k ≥ i∈I #(Li ) = i∈I 2

2

2

as required. Q Q (d) We can now apply the idea of the proof of 524C, as follows. The density of `1 (κ) is of course κ, by 529Ba applied to counting measure on κ, or otherwise. Index a dense subset of `1 (κ) as hyj ij∈J . For each x ∈ `1 , choose a sequence hk(x, n)in∈N in J such that Pn kx − i=0 yk(x,i) k ≤ 8−n for every n. Note that kyk(x,n) k ≤ kx −

Pn

i=0

yk(x,i) k + kx −

for each n. Choose f (x) ∈ L1 such that kf (x)k ≥ kxk and f (x) ≥ n ∈ N} is bounded.

Pn−1

−n i=0 yk(x,i) k ≤ 9 · 8 P∞ −n uk(x,n) ; this is n=0 2

possible because {uk(x,n) :

(e) f is a Tukey function. P P Take v ∈ L1 and set A = {x : f (x) ≤ v},

Kn = {k(x, n) : x ∈ A}

for n ∈ N. Fix n for the moment. If j ∈ Kn , then there is an x ∈ A such that j = k(x, n) and uj = uk(x,n) ≤ 2n f (x) ≤ 2n v, p while kyj k = kyk(x,n) k ≤ 9 · 8−n . If K ⊆ Kn is finite, k2n vk ≥ 21 #(K), by (c); so #(Kn ) ≤ 22n+1 kvk2 . This means P that if we set zn = j∈Kn |yj | we shall have kzn k ≤ 9 · 8−n #(Kn ) ≤ 18 · 2−n kvk2 , while yk(x,n) ≤ zn for every x ∈ A. P∞ Pn Now z = n=0 zn is defined in `1 (κ), and if x ∈ A then i=0 yk(x,i) ≤ z for every n ∈ N, so that x ≤ z. Thus A is bounded above in `1 (κ). As v is arbitrary, f is a Tukey function. Q Q (g) Accordingly `1 (κ) 4T L1 ≡ U , and the proof is complete. 529D Theorem (Fremlin 91a) Let A be a homogeneous measurable algebra with Maharam type κ ≥ ω. Then L0 (A) ≡T `1 (κ). proof (a) Let µ ¯ be such that (A, µ ¯) is a probability algebra. If we give L0 = L0 (A) its topology Rof convergence in measure, its density is κ, by 529Bb. Moreover, this topology is defined by the metric (u, v) 7→ |u − v| ∧ χ1, under which the lattice operation ∨ is uniformly continuous. P P Just as in part (a) of the proof of 529C, we have

180

Cardinal functions of measure theory

529D

u ∨ v = 12 (u + v + |u − v|) for all u and v, addition and subtraction are uniformly continuous, and u 7→ |u| is uniformly continuous. Q Q So, just as in 529C, we can use 524C to see that L0 4T `1 (κ). (b) For the reverse connection, I repeat ideas from the proof of 529C. (A, µ ¯) is isomorphic to the measure algebra 1 κ κ ¯ (C, λ) of ]0, 1] with its usual measure λ. For ξ < κ, t ∈ ]0, 1] set hξ (t) = √ , and set uξ = h•ξ in L0 (λ) ≡ L0 (C) t(ξ) p (364Jc). This time, observe that if x ∈ `1 (κ)+ and α ≥ kxk then ¯ λ[[sup ξ<κ

Y p p p ¯ inf [[ x(ξ)uξ ≤ α]]) = ¯ x(ξ)uξ ≤ α]] x(ξ)uξ ≤ α]] = λ( λ[[ ξ<κ

=

Y

ξ<κ

r λ{t :

x(ξ) t(ξ)

≤ α} =

(1 − α−2 x(ξ))

ξ<κ

ξ<κ

≥1−α

Y

−2

X

x(ξ) → 0

ξ<κ

p as α → ∞. This means that supξ<κ x(ξ)uξ is defined in L0 (λ) (364M). So we can define f : `1 (κ) → L0 (λ) by saying p that f (x) = supξ<κ max(0, x(ξ))uξ for every x ∈ `1 (κ). (c) f is a Tukey function. P P Take v ∈ L0 (λ)+ , and set A = {x : f (x) ≤ v}. Note that f (x ∨ x0 ) = f (x) ∨ f (x0 ) 0 1 ¯ ≤ α]] = β > 1 . If x ∈ A and x ≥ 0 then for all x, x ∈ ` (κ), so A is upwards-directed. Take α > 0 such that λ[[v 2 p f (x) ≥ x(ξ)χ1 so x(ξ) ≤ α for every ξ. Now the calculation in (b) tells us that ¯ β ≤ λ[[sup ξ<κ

Y p x(ξ)uξ ≤ α]] = (1 − α−2 x(ξ)) ξ<κ

≤

X 1 1 max( , 1 − α−2 x(ξ)) 2 2

1 2

1 2

= max( , 1 − α−2 kxk),

ξ<κ

so kxk ≤ 2α2 (1 − β). As A is upwards-directed and norm-bounded and contains 0, it is bounded above in `1 (κ) (354N). As v is arbitrary, f is a Tukey function. Q Q (d) Accordingly `1 (κ) 4T L0 (λ) ∼ = L0 (A) and `1 (κ) and L0 (A) are Tukey equivalent. 529E Proposition Let S ∞ be the N-localization poset (528I). Then RO({0, 1}c ) can be regularly embedded in RO↑ (S ∞ ). proof (a) Let hhξ iξ
(n, h(n)) ∈ / p, (i, h(i)) ∈ p for every i > n} ∪ {(h, p) : h ∈ N N , p ∈ S ∞ , (i, h(i)) ∈ p for every i ∈ N}, W1 =

[

{(h, p) : h ∈ N N , p ∈ S ∞ , #(p[{n}]) = 2n ,

n∈N is odd

(n, h(n)) ∈ / p, (i, h(i)) ∈ p for every i > n}. Observe that (i) W0 ∩ W1 = ∅ (ii) if (h, p) ∈ Wj , where j = 0 or j = 1, and p ⊆ q ∈ S ∞ then (h, q) ∈ Wj (iii) if p ∈ S ∞ then #({ξ : (hξ , p) ∈ W0 ∪ W1 }) ≤ kpk is finite. (b) Set Q = Fn<ω (c, {0, 1}), the set of functions from finite subsets of c to {0, 1}, ordered by extension of functions, so that (Q, ⊆) is isomorphic to (U, ⊇) where U is the usual base of the topology of {0, 1}c , and RO↑ (Q) ∼ = RO↓ (U) can c ∞ be identified with the regular open algebra of {0, 1} (514Sd). Define f : S → Q by setting f (p)(ξ) = j if (hξ , p) ∈ Wj . Then f is order-preserving. (c) For p ∈ S ∞ , set

529F

Further partially ordered sets of measure theory

181

A0 (p) = {ξ : ξ < c, {n : n is even, (n, hξ (n)) ∈ / p} is finite}, A1 (p) = {ξ : ξ < c, {n : n is odd, (n, hξ (n)) ∈ / p} is finite}, A(p) = A0 (p) ∪ A1 (p), so that A(p) is finite and dom f (p) ⊆ A(p). Now P1 = {p : p ∈ S ∞ , A(p) = dom f (p)} is cofinal with S ∞ . P P Take p ∈ S ∞ . Let m be such that 2m ≥ kpk + #(A(p)), (n, hξ (n)) ∈ p whenever ξ ∈ A0 (p) and n > m is even, (n, hξ (n)) ∈ p whenever ξ ∈ A1 (p) and n > m is odd. 0

Let p ∈ S

∞

be such that for n ≤ m, p0 [{n}] ⊇ p[{n}] and #(p0 [{n}]) = 2n , for n > m, p0 [{n}] = p[{n}] ∪ {hξ (n) : ξ ∈ A(p)}.

Then p ≤ p0 and A(p0 ) = A(p). Also A(p) = dom f (p0 ), because if ξ ∈ A(p) then either (n, hξ (n)) ∈ p0 for every n and (hξ , p0 ) ∈ W0 , or there is a largest n such that (n, hξ (n)) ∈ / p0 , in which case n ≤ m and #(p0 [{n}]) = 2n , so (hξ , p0 ) belongs to W0 if n is even and W1 otherwise. Q Q (d) If p ∈ P1 and q ∈ Q extends f (p), there is a p1 ∈ S ∞ such that p1 ⊇ p and f (p0 ) = q. P P Let m be such that 2m ≥ kpk + #(dom q) and hξ (n) 6= hη (n) whenever ξ, η ∈ dom q are distinct and n ≥ m. For each ξ ∈ dom q \ dom f (p) = dom q \ A(p), {n : n is even, (n, hξ (n)) ∈ / p} and {n : n is odd, (n, hξ (n)) ∈ / p} are both infinite. So we can find m0 ≥ m such that all these sets meet m0 \ m. Set p0 = p ∪ {(n, hξ (n)) : n ∈ m0 \ m is odd, q(ξ) = 0} ∪ {(n, hξ (n)) : n ∈ m0 \ m is even, q(ξ) = 1} ∪ {(n, hξ (n)) : n ∈ N \ m0 , ξ ∈ dom q}, so that p ⊆ p0 ∈ S ∞ . Let p1 ∈ S ∞ be such that p1 ⊇ p0 , p1 \ p0 is finite, #(p1 [{n}]) = 2n for every n < m0 and (n, hξ (n)) ∈ / p1 \ p0 whenever n ∈ N and ξ ∈ dom q. Now f (p1 ) = q, while p ⊆ p1 . Q Q (e) Putting (c) and (d) together, we see that f −1 [Q0 ] must be cofinal with S ∞ for every cofinal Q0 ⊆ Q; moreover, since ∅ ∈ P1 and f (∅) is the empty function, f [S ∞ ] = Q. So f satisfies the conditions of 514O, and RO↑ (Q) ∼ = RO({0, 1}c ) can be regularly embedded in RO↑ (S ∞ ). 529F Corollary (Brendle 00, 2.3.10; Brendle 06, Theorem 1) n({0, 1}I ) ≥ add N for every set I. proof If I is finite, this is trivial. Otherwise, write λ = n({0, 1}I ). Then λ ≥ n({0, 1}c ). P P If J ⊆ I is a countably infinite set, then {{x : xJ = z} : z ∈ {0, 1}J } is a cover of {0, 1}I by continuum many nowhere dense sets, so λ ≤ c. Let hEξ iξ<λ be a cover of {0, 1}κ by nowhere dense sets. Then each Eξ is included in a nowhere dense closed set Fξ S determined by coordinates in a countable set Kξ ⊆ I (4A2E(b-i)). Set K = ξ<λ Kξ , so that #(K) ≤ c. Then all the projections Fξ0 = {xK : x ∈ Fξ } are nowhere dense in {0, 1}K (use 4A2Bf), and they cover {0, 1}K . Next, we have an injection φ : K → c, and the sets Fξ00 = {x : xφ ∈ Fξ0 } form a cover of {0, 1}c by nowhere dense sets; so n({0, 1}c ) ≤ λ. Q Q Because every non-empty open set {0, 1}c includes an open set homeomorphic to {0, 1}c ,

n({0, 1}c ) = min{n(H) : H ⊆ {0, 1}c is open and not empty} = m(RO({0, 1}c )) (517J) ≥ m(RO↑ (S ∞ )) (where S ∞ is the N-localization poset, by 529E and 517Ia) = add N by 528N.

182

Cardinal functions of measure theory

529G

529G Reaping numbers (or ‘unsplitting numbers’, ‘refining numbers’)(following Brendle 00) For cardinals θ ≤ λ let r(θ, λ) be the smallest cardinal of any set A ⊆ [λ]θ such that for every B ⊆ λ there is an A ∈ A such that either A ⊆ B or A ∩ B = ∅. 529H Proposition (Brendle 00, 2.7; Brendle 06, Theorem 5) r(ω1 , λ) ≥ cov N for all uncountable λ. proof Let hAξ iξ<κ be a family in [λ]ω1 , where κ < cov(N ). I seek a B ⊆ λ such that Aξ ∩ B and Aξ \ B are non-empty for every ξ < κ. (a) If κ ≤ ω1 , then choose hαξ iξ<κ and hβξ iξ<κ inductively so that αξ ∈ Aξ \ {βη : η < ξ},

βξ ∈ Aξ \ {αη : η ≤ ξ}

for every ξ < κ, and set B = {βξ : ξ < κ}; this serves. So henceforth let us suppose that κ > ω1 . (b) For each ξ < κ let A0ξ ⊆ Aξ be a set of order type ω1 . For each n, let Xn be a set of size n! with the discrete Q 1 topology and the uniform probability measure which gives measure to every singleton. Give X = n∈N Xn its n!

product measure µ and its product topology. Because X is a compact metrizable space and µ is a Radon measure (416U), cov N (µ) = cov N (522Va). We can therefore choose a family hxξ iξ<κ in X in such a way that each xζ is random over its predecessors in the sense that whenever ξ < κ and {xη : η ∈ A0ξ ∩ ζ} is negligible, it does not contain xζ . For x, y ∈ X, set ∆(x, y) = sup{i : x(i) = y(i)}. For x ∈ X, T S E(x) = {y : ∆(y, x) = ω} = n∈N i≥n {y : y(i) = x(i)} is negligible, because

P∞

1

n=0 n!

is finite; set B(x) = {η : ∆(xη , x) is finite and even}.

P There is a ζ < κ (c) For every ξ < κ, {x : x ∈ X, A0ξ ⊆ B(x)} and {x : x ∈ X, A0ξ ∩ B(x) = ∅} are negligible. P such that ζ ∈ A0ξ , A0ξ ∩ ζ is countable and D = {xη : η ∈ A0ξ ∩ ζ} is dense in {xη : η ∈ A0ξ }. Since xζ ∈ D, D has measure greater than 0. By 275I, it has a point w which is a density point in the sense that limn→∞

µ{y:y  n=w n, y∈D} µ{y:y  n=w n}

= 1.

Consequently, setting Jn = {y(n) : yn = wn, y ∈ D} = {y(n) : yn = wn, y ∈ D}, limn→∞

#(Jn ) n!

= 1.

Next note that, for any y ∈ X and n ∈ N, µ{x : ∃ i > n, x(i) = y(i)} ≤

P∞

1

i=n+1 i!

≤

n+2 . (n+1)(n+1)!

So µ{x : ∃ y ∈ D, x(n) = y(n), x(i) 6= y(i) for every i > n}
n+2 #(Jn ) 1− →1 ≥ n!

(n+1)(n+1)!

as n → ∞, and µ{x : ∃ y ∈ D, ∆(x, y) is even} = µ{x : ∃ y ∈ D, ∆(x, y) is odd} = 1. But if y ∈ D and ∆(x, y) is even, we have an η ∈ Aξ such that ∆(x, xη ) is even, and η ∈ Aξ ∩ B(x); similarly, if there is a y ∈ D such that ∆(x, y) is odd, there is an η ∈ Aξ \ B(x). So {x : Aξ ⊆ B(x)} and {x : Aξ ∩ B(x) = ∅} are both negligible. Q Q (d) Since cov N (µ) = cov N > κ, there is an x ∈ X such that both Aξ ∩ B(x) and Aξ \ B(x) are non-empty for every ξ < κ. So in this case also we have a suitable set B. 529X Basic exercises (a) Let (X, Σ, µ) be a measure space, and p ∈ [1, ∞[. Show that Lp (µ) ≡T `1 (κ), where κ = dim Lp (µ) if this is finite, d(Lp (µ)) otherwise. (b) Let U be an L-space. (i) Show that add U = ∞ if U = {0}, ω otherwise. (ii) Show that addω U = ∞ if U is finite-dimensional, add N if U is separable and infinite-dimensional, ω1 otherwise. (iii) Show that cf U = 1 if U = {0}, ω if 0 < dim U < ω, max(cf N , cf[d(U )]≤ω ) otherwise. (iv) Show that link↑<κ (U ) = 1 if κ ≤ ω, cf U otherwise.

529 Notes

Further partially ordered sets of measure theory

(c) Let I be the ideal of subsets I of N such that

P

1 n∈I n+1

183

is finite. (See 419A.) Show that `1 ≡T I, so that

addω I = add N and cf I = cf N . (d) Show that if λ is an infinite cardinal then r(ω, λ) ≥ max(add N , cov Nλ ), where Nλ is the null ideal of the usual measure on {0, 1}λ . (Hint: 529F.) (e) Let U be a separable L-space. Suppose that huξ iξ<κ is a family in U , where κ < add N . Show that there is a family hξ iξ<κ of strictly positive real numbers such that {ξ uξ : ξ < κ} is order-bounded in U . > (f ) Let U be a separable Banach lattice, and D ⊆ U a dense set. Let A ⊆ U be a set of size less than add N . Show that there is a w ∈ U such that for every u ∈ A and every  > 0 there is an v ∈ D such that |u − v| ≤ w. 529Y Further exercises (a) Show that, for any infinite set I, the regular open algebra RO({0, 1}I ) of {0, 1}I is homogeneous, so that m(RO({0, 1}I )) = n({0, 1})I . (b) Let X be a Polish space and Kσ the family of Kσ subsets of X. Show that (Kσ , ⊆, Kσ ) 4GT (N N , ≤∗ , N N ) (definition: 522C). (c) Show that r(ω, ω) ≥ b. (d) Let X be a topological space with a countable network, and c : PX → [0, ∞] an outer regular submodular Choquet capacityS(definitions: 432I14 ). Show that if A is an upwards-directed family of subsets of X such that #(A) < mσ-linked , then c( A) = supA∈A c(A). (e) Let r ≥ 3 be an integer, and c : PR r → [0, ∞] Choquet-NewtonScapacity (§47914 ). Show that if A is an upwards-directed family of subsets of R r such that #(A) < add N , then c( A) = supA∈A c(A). (Hint: 479Xe.) 529 Notes and comments Many of the ideas of the last two chapters were first embodied in forcing arguments. In 529E this becomes particularly transparent. If we have an upwards-directed set R ⊆ S ∞ which is ‘generic’ in the sense that it meets all the cofinal subsets of S ∞ definable in a language LSwith terms for all the functions hξ , as well as such obvious ones as {p : #(p[{n}]) = 2n } for each n, and we set S = R, then S will belong to the set S of 522K, and S we shall have hξ ⊆∗ S for every ξ; so that we have a corresponding function f˜(S) = p∈R f (p) ∈ {0, 1}c defined by setting f˜(S)(ξ) ≡ sup{i : (i, hξ (i)) ∈ / S} mod 2. Next, if G ⊆ {0, 1}c is a dense open set with a definition in L, then f˜(S) ∈ G; for, setting Uq = {φ : q ⊆ φ ∈ {0, 1}c } when q ∈ Fn<ω (c, {0, 1}), {q : Uq ⊆ G} is cofinal with Q, so {p : Uf (p) ⊆ G} is cofinal with S ∞ (part (d) of the proof of 529E) and meets R. Thus f˜(S) is ‘generic’ in the sense that it belongs to every dense open set with a name in L; and it is a commonplace in the theory of forcing that a function which transforms generic objects in one forcing extension into generic objects in another extension corresponds to a regular embedding of the corresponding regular open algebras.

14 Later

editions only.

184

Topologies and measures III

Chapter 53 Topologies and measures III In this chapter I return to the concerns of earlier volumes, looking for results which can be expressed in the language so far developed in this volume. In Chapter 43 I examined relationships between measure-theoretic and topological properties. The concepts we now have available (in particular, the notion of ‘precaliber’) make it possible to extend this work in a new direction, seeking to understand the possible Maharam types of measures on a given topological space. §531 deals with general Radon measures; new patterns arise if we restrict ourselves to completion regular Radon measures (§532). In §533 I give a brief account of some further results depending on assumptions concerning the cardinals examined in Chapter 52, including notes on uniformly regular measures and a description of the cardinals κ for which R κ is measure-compact (533J). In §534 I set out the elementary theory of ‘strong measure zero’ ideals in uniform spaces, concentrating on aspects which can be studied in terms of concepts already induced. Here there are some very natural questions which have not I think been answered (534Z). In the same section I run through the properties of Hausdorff measures when examined in the light of the concepts in Chapter 52. In §535 I look at liftings and strong liftings, extending the results of §§341 and 453; in particular, asking which non-complete probability spaces have liftings. In §536 I run over what is known about Alexandra Bellow’s problem concerning pointwise compact sets of continuous functions, mentioned in §463. With a little help from special axioms, there are some striking possibilities concerning repeated integrals, which I examine in §537. In §539, I complete my account of the result of B.Balcar, T.Jech and T.Paz´ak that it is consistent to suppose that every Dedekind complete ccc weakly (σ, ∞)-distributive Boolean algebra is a Maharam algebra, and work through applications of the methods of Chapter 52 to Maharam submeasures and algebras. Moving into new territory, I devote a section (§538) to a study of special types of filter on N associated with measure-theoretic phenomena, and to medial limits.

531 Maharam types of Radon measures In the introduction to §434 I asked What kinds of measures can arise on what kinds of topological space? In §§434-435, and again in §438, I considered a variety of topological properties and their relations with measuretheoretic properties of Borel and Baire measures. I passed over, however, some natural questions concerning possible Maharam types, to which I now return. For a given Hausdorff space X, the possible measure algebras of totally finite Radon measures on X can be described in terms of the set MahR (X) of Maharam types of Maharam-type-homogeneous Radon probability measures on X (531F). For X 6= ∅, MahR (X) is of the form {0} ∪ [ω, κ∗ [ for some infinite cardinal κ∗ (531Ef). In 531E and 531G I give basic results from which MahR (X) can often be determined; for obvious reasons we are primarily concerned with compact spaces X. In more abstract contexts, there are striking relationships between precalibers of measure algebras, the sets MahR (X) and continuous surjections onto powers of [0, 1], which I examine in 531L-531N and 531T. Intertwined with these, we have results relating the character of X to MahR (X) (531O-531P). The arguments here depend on an analysis of the structure of homogeneous measure algebras (531J, 531K, 531R). 531A Proposition Let (X, T, Σ, µ) be a quasi-Radon measure space with measure algebra (A, µ ¯). (a) The Maharam type τ (A) of A is at most the weight w(X) of X. (b) The cellularity c(A) of A is at most the hereditary Lindel¨of number hL(X) of X. If µ is locally finite, c(A) is at most the Lindel¨ of number L(X) of X. (c) #({a : a ∈ A, µ ¯a < ∞}) ≤ max(1, w(X)ω ), where w(X)ω is the cardinal power. (d) If X is Hausdorff and µ is a Radon measure, then the Maharam type τ (A) of A is at most the network weight nw(X) of X. proof (a) Let U be a base for T with #(U) = w(X). Set B = {U • : U ∈ U} and let B be the order-closed subalgebra of A generated by B; set T = {E : E ∈ Σ, E • ∈ B}. Then T is a σ-subalgebra of Σ containing every negligible set. If G ⊆ X is open, then G ∈ T. P P By 414Aa, G• = sup{U • : U ∈ U, U ⊆ G} belongs to B. Q Q So every Borel set belongs to T. If E ∈ Σ and µE < ∞, then, because µ is inner regular with respect to the Borel sets, there is a Borel subset F of E with the same measure, so F , E \ F and E belong to T. Thus {a : a ∈ A, µ ¯a < ∞} ⊆ B; because µ is semi-finite, B = A and τ (A) ≤ #(U) = w(X). (b)(i) If L(X) = S n is finite, and F0 ,S . . . , Fn ⊆ X are disjoint closed sets, then at least P For S one of them is empty. P i ≤ n, set Gi = X \ j≤n,j6=i Fj ; then i≤n Gi = X, so there is some k ≤ n such that i6=k Gi = X, and now Fk = ∅. Q Q As µ is inner regular with respect to the closed sets, c(A) ≤ n = L(X) ≤ hL(X).

531C

Maharam types of Radon measures

185

(ii) Suppose that ω ≤ L(X) ≤ hL(X). Let G be of open subsets of X of finite measure. Then there is S the family S a set H ⊆ G, of cardinal at most hL(X), such that H = G (5A4Bf). Now supH∈H H • = 1, because µ is effectively locally finite. If D ⊆ A \ {0} is disjoint, then for each d ∈ D take Hd ∈ H such that d ∩ Hd• 6= 0. If H ∈ H, then {d : Hd = H} must be countable, since µH < ∞. So #(D) ≤ max(ω, #(H)); as D is arbitrary, c(A) ≤ max(ω, hL(X)) = hL(X). S (iii) Finally, if ω ≤ L(X) and µ is locally finite, then in (ii) above we have X = G, so we can take H to have size at most L(X), and continue as before, ending with c(A) ≤ max(ω, #(H)) = L(X). (c) Again let U be a base for the topology of X with cardinal w(X). Let T be the σ-subalgebra of Σ generated by U. If E ∈ Σ and µE < ∞, then for each n ∈ N we can find an open set Gn such that µ(Gn 4E) ≤ 2−n ;Snow there T is an open set Hn , a finite union of members of U, such that Hn ⊆ Gn and µ(Gn \ Hn ) ≤ 2−n . Setting F = m∈N n∈N Hn , we see that F ∈ T and E4F is negligible. Thus {F • : F ∈ T} ⊇ {a : µ ¯a < ∞} and #({a : µ ¯a < ∞} ≤ #(T) ≤ max(1, #(U)ω ) = max(1, w(X)ω ). (d) If a ∈ A \ {0} and the principal ideal Aa is Maharam-type-homogeneous, then τ (Aa ) ≤ nw(X). P P There is a compact set K ⊆ X such that 0 6= K • ⊆ Aa ; let µK be the subspace measure on K. Then

τ (Aa ) = τ (µK ) ≤ w(K) (by (a)) = nw(K) (5A4C(a-i)) ≤ nw(X) (5A4Bb). Q Q By (b), c(A) ≤ #(T) ≤ 2nw(X) (5A4Ba); so 332S tells us that τ (A) ≤ nw(X). 531B For strictly positive measures we have some easy inequalities in the other direction. Proposition Let (X, Σ, µ) be a measure space, with measure algebra A, and T a topology on X such that Σ includes a base for T and µ is strictly positive. (a) If X is regular, then w(X) ≤ #(A). (b) If X is Hausdorff, then #(X) ≤ 2#(A) . P proof Set V = Σ ∩ T, so that V is a base for T. If V , W ∈ V and V • = W • in A, then int V = int W . P µ∗ (V \ W ) ≤ µ(V \ W ) = 0, so (because µ is strictly positive) V ⊆ W and V ⊆ W and int V ⊆ int W . Similarly, int W ⊆ int V . Q Q So if we set W = {int V : V ∈ V}, #(W) ≤ #(A). (a) If T is regular, W is a base for T, so w(X) ≤ #(W) ≤ #(A). (b) If T is Hausdorff, then for any distinct x, y ∈ X, there is a W ∈ W containing x but not y. P P Let G, H be disjoint open sets containing x, y respectively. Take V ∈ V such that x ∈ V ⊆ G, and set W = int V . Q Q So #(X) ≤ 2#(W) ≤ 2#(A) . 531C Lemma Let hXi ii∈I be a family of topological spaces with product X, and µ a totally finite quasi-Radon measure on X with Maharam type κ.PFor each i ∈ I, let µi be the marginal measure on Xi , and κi its Maharam type. Then κ is at most the cardinal sum i∈I κi . • proof For each i ∈ I, let hEiξ iξ<κi be a family in dom µi such that {Eiξ : ξ < κi } τ -generates the measure algebra P −1 of µi . Consider W = {πi [Eiξ ] : i ∈ I, ξ < κi }, so that W ⊆ dom µ and #(W) ≤ i∈I κi . Let B be the closed subalgebra of the measure algebra A of µ generated by {W • : W ∈ W}. For each i ∈ I, the canonical map πi : X → Xi induces a measure-preserving homomorphism φi from the measure • algebra Ai of µi to A (324M). Now φ−1 i [B] is a closed subalgebra of Ai containing Eiξ for every ξ < κi , so is the whole −1 • • of Ai , that is, φi [Ai ] ⊆ B. In particular, if G ⊆ Xi is open, πi [G] = φi (G ) belongs to B. Now the family V of open sets V ⊆ X such that V • ∈ B is closed under finite intersections and contains πi−1 [G] whenever i ∈ I and G ⊆ Xi is open, so V is a base for the topology of X. But also V is closed under arbitrary unions, because B is closed and µ is τ -additive (414Aa). So V • ∈ B for every open set V ⊆ X, and therefore for every Borel set V ⊆ X; as µ is inner regular with respect to the Borel sets, B = A. P Thus {W • : W ∈ W} witnesses that the Maharam type τ (A) of µ is at most i∈I κi , as claimed.

186

Topologies and measures III

531D

531D Definition If X is a Hausdorff space, I write MahR (X) for the set of Maharam types of Maharam-typehomogeneous Radon probability measures on X. Note that 0 ∈ MahR (X) iff X is non-empty, and that any member of MahR (X) is either 0 or an infinite cardinal. 531E Proposition Let X be any Hausdorff space. (a) κ ≤ w(X) for every κ ∈ MahR (X). (b) MahR (Y ) ⊆ S MahR (X) for every Y ⊆ X. (c) MahR (X) = {MahR (K) : K ⊆ X is compact}. (d) If X is compact and Y is a continuous image of X, MahR (Y ) ⊆ MahR (X). (e) ω ∈ MahR (X) iff X has a compact subset which is not scattered. (f) (Haydon 77) If ω ≤ κ0 ≤ κ ∈ MahR (X) then κ0 ∈ MahR (X). (g) If Y is another Hausdorff space, and neither X nor Y is empty, then MahR (X × YQ) = MahR (X) S ∪ MahR (Y ); generally, for any non-empty finite family hXi ii∈I of non-empty Hausdorff spaces, MahR ( i∈I Xi ) = i∈I MahR (Xi ). proof (a) This is immediate from 531Aa. (b) If κ ∈ MahR (Y ), there a Maharam-type-homogeneous Radon probability measure µ on Y with Maharam type κ. Now µ has a (unique) extension to a Radon probability measure µ0 on X, setting µ0 (Y \ X) = 0. µ0 and µ have isomorphic measure algebras, so µ0 is Maharam-type-homogeneous and has Maharam type κ, and κ ∈ MahR (X). (c) By (b), MahR (K) ⊆ MahR (X) for every compact set K ⊆ X. In the other direction, if κ ∈ MahR (X), there is a Maharam-type-homogeneous Radon probability measure µ on X with Maharam type κ. Let K ⊆ X be a compact set with µK > 0. Then the normalized subspace measure µ0 = (µK)−1 µK is a Radon probability measure on K, and its measure algebra is isomorphic to a principal ideal of the measure algebra of µ, so is Maharam-type-homogeneous with Maharam type κ. Accordingly κ ∈ MahR (K). (d) Take κ ∈ MahR (Y ); then there is a Maharam-type-homogeneous Radon probability measure µ on Y with Maharam type κ. Let f : X → Y be a continuous surjection. By 418L, there is a Radon measure µ0 on X such that f is inverse-measure-preserving for µ0 and µ and induces an isomorphism of their measure algebras. So µ0 witnesses that κ ∈ MahR (Y ). (e)(i) If X has a compact subset K which is not scattered, then there is a continuous surjection from K onto [0, 1] (4A2G(j-iv)). Of course Lebesgue measure witnesses that ω ∈ MahR ([0, 1]), so (d) and (b) tells us that ω ∈ MahR (K) ⊆ MahR (X). (ii) If every compact subset of X is scattered and µ is a Maharam-type-homogeneous Radon probability measure on X, let K be a compact set of non-zero measure and Z ⊆ K a closed self-supporting set. Then Z has an isolated point z say; in this case, µ{z} > 0 so {z} is an atom for µ and (because µ is Maharam-type-homogeneous) the Maharam type of µ is 0. As µ is arbitrary, ω ∈ / MahR (X). (f )(i) Suppose first that X is compact. Let µ be a Maharam-type-homogeneous Radon probability measure on X with Maharam type κ. Let hEξ iξ<κ be a stochastically independent family in Σ with µEξ = 21 for every ξ. For R 0 each ξ < κ0 and n ∈ N, let fξn ∈ C(X) be such that |fξn − χEξ | ≤ 2−n (416I). Define f : X → R κ ×N by setting f (x)(ξ, n) = fξn (x) for x ∈ X, ξ < κ0 and n ∈ N. Then f is continuous, so by 418I the image measure ν = µf −1 on the compact set f [X] is a Radon measure. For each ξ < κ0 , the set Fξ = {w : w ∈ f [X], hw(ξ, n)in∈N is Cauchy} is a Borel set, and f −1 [Fξ ]4Eξ is µ-negligible; so hFξ iξ<κ0 is a stochastically independent family of subsets of f [X] with measure 12 . If B is the measure algebra of ν, and C the closed subalgebra of B generated by {Fξ• : ξ < κ0 }, then C is Maharam-type-homogeneous, with Maharam type κ0 ; at the same time, 0

τ (B) ≤ w(f [X]) ≤ w(R κ ×N ) = κ0 . By 332N, B can be embedded in C; by 332Q, B and C are isomorphic, that is, B is Maharam-type-homogeneous with Maharam type κ0 , and ν witnesses that κ0 ∈ MahR (f [X]). By (d), κ0 ∈ MahR (X). (ii) In general, (c) tells us that there is a compact set K ⊆ X such that κ ∈ MahR (K), so κ0 ∈ MahR (K) ⊆ MahR (X). (g) Because neither Y nor X is empty, both X and Y are homeomorphic to subspaces of X × Y , so (b) tells us that MahR (X × Y ) ⊇ MahR (X) ∪ MahR (Y ). In the other direction, given a Maharam-type-homogeneous Radon probability measure µ on X × Y , let µ1 , µ2 be the marginal measures on X and Y respectively, so that each µk is a Radon probability measure (418I). Let hEi ii∈I , hFj ij∈J be countable partitions of X, Y into Borel sets such that all the subspace measures (µ1 )Ei and (µ2 )Fj are Maharam-type-homogeneous. Then there must be i ∈ J, j ∈ J such that

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187

µ(Ei × Fj ) > 0. Let µ0 be the subspace measure µEi ×Fj ; then the Maharam type of µ0 is κ, because µ is Maharam-typehomogeneous. Let µ01 , µ02 be the marginal measures of µ0 on Ei and Fj respectively. Then µ01 is an indefinite-integral measure over (µ1 )Ei (415Oa), so its measure algebra is isomorphic to a principal ideal of the measure algebra of (µ1 )Ej (322K), and has the same Maharam type κ1 say. As in (b) above, κ1 ∈ MahR (X). Similarly, the Maharam type κ2 of µ02 belongs to MahR (Y ). Now 531C tells us that κ ≤ κ1 + κ2 . Since κ is either zero or infinite, it must be less than or equal to at least one of them, and belongs to MahR (X) ∪ MahR (Y ) by (f) above. The result for general finite products now follows easily by induction on #(I). 531F Proposition Let X be a Hausdorff space. Then a totally finite measure algebra (A, µ ¯) is isomorphic to the measure algebra of a Radon measure on X iff (α) whenever Aa is a non-trivial homogeneous principal ideal of A then τ (Aa ) ∈ MahR (X) (β) c(A) ≤ #(X). proof (a) If µ is a totally finite Radon measure on X with measure algebra A and the principal ideal Aa generated by a ∈ A \ {0} is homogeneous, then there are an E ∈ dom µ such that E • = a and an F ⊆ E such that 0 < µF < ∞. Let ν be the Radon probability measure (µF )−1 µ F , that is, νH = µ(H ∩ F )/µF whenever H ⊆ X is such that µ measures H ∩ F . Then the measure algebra of ν is isomorphic to a principal ideal of Aa so is homogeneous with the same Maharam type, and ν witnesses that τ (Aa ) ∈ MahR (X). Thus A satisfies (α). As for (β), if X is infinite this is trivial (because (A, µ ¯) is totally finite, so A is ccc), and otherwise A is finite, with c(A) = #({a : a ∈ A is an atom}) = #({x : x ∈ X, µ{x} > 0}) ≤ #(X). (b) Now suppose that (A, µ ¯) is a totally finite measure algebra satisfying the conditions. Express it as the simple product of a countable family h(An , µ ¯0n )ii∈I of non-zero homogeneous measure algebras (332B); we may suppose that I ⊆ N. For n ∈ I, set κn = τ (An ) and γn = µ ¯0n 1An . (β) tells us that #(I) ≤ #(X); let hxn in∈I be a family of distinct elements of X. Set J = {n : n ∈ I, κn ≥ ω}. For each n ∈ J, (α) tells us that there is a Maharam-type-homogeneous Radon probability measure µn on X with Maharam type κn . Now there is a disjoint family hEn in∈I of Borel subsets of X such that µn En > 0 for every n ∈ J. P P Choose hEn in∈N , hFn in∈N inductively, as follows. F0 = X \ {xn : n ∈ I}. Given that Fn is a Borel set and µj Fn > 0 for every j ∈ J \ n, then if n ∈ / J set En = ∅ and Fn+1 = Fn . Otherwise, for each j ∈ J such that j > n, we can partition Fn into finitely many Borel sets of µn -measure less than 2−j µn Fn , because S µn is atomless; take one of these, Gnj say, such that µj Gnj > 0; now set Fn+1 = j∈J,j>n Gnj and En = Fn \ Fn+1 . Continue. Q Q Now set P P µE = n∈I\J,xn ∈E γn + n∈J (µn En )−1 γn µn (E ∩ En ) whenever E ⊆ X is such that µn measures E ∩ En for every n ∈ J. Of course µ is a measure. Because every µn is a topological measure, so is µ; because every µn is inner regular with respect to the compact sets, so is µ; because every µn is complete, so is µ; thus µ is a Radon measure. Because every subspace measure (µn )En is Maharam-typehomogeneous with Maharam type κn , the measure algebra of µ is isomorphic to (A, µ ¯). 531G Proposition Let hXi ii∈I be a family of non-empty Hausdorff spaces with product X. Then an infinite cardinal κ belongs to MahR (X) iff either κ ≤ #({i : i ∈ I, #(Xi ) ≥ 2}) or κ is expressible as supi∈I κi where κi ∈ MahR (Xi ) for every i ∈ I. proof (a)(i) Suppose that κ = supi∈I κi where κi ∈ MahR (Xi ) for each i ∈ I. For each i, let µi be a Maharamtype-homogeneous Radon probability measure on Xi with Maharam type κi and compact support (see the proof of 531Ec). Let λ be the ordinary product of the measures µi . By 325I, the measure algebra of λ can be identified with the probability algebra free product of the measure algebras of the Pµi . It is therefore isomorphic to the measure algebra of 0 the usual measure on {0, 1}κ , where κ0 is the cardinal sum i∈I κi ; in particular, it is homogeneous with Maharam type κ0 (since we are supposing that κ ≥ ω). By 417E(ii), the measure algebra of the τ -additive product µ of hµi ii∈I can be identified with the measure algebra of λ, while µ is a Radon measure (417Q). So µ witnesses that κ0 ∈ MahR (X); by 531Ef, κ ∈ MahR (X). (ii) Suppose that ω ≤ κ ≤ #(I 0 ) where I 0 = {i : i ∈ I, #(Xi ) ≥ 2}. For i ∈ I 0 , let xi , yi be distinct points of Xi and µi the point-supported probability measure on Xi such that µi {xi } = µi {yi } = 21 ; for i ∈ I \ I 0 , let µi be the unique Radon probability measure on Xi . As in (i) above, the Radon measure product of hµi ii∈I is Maharam-typehomogeneous, with Maharam type #(I 0 ), so #(I 0 ) ∈ MahR (X); by 531Ef again, κ ∈ MahR (X). (b) Now suppose that ω ≤ κ ∈ MahR (X) and that κ > #(I 0 ). For each i ∈ I, let θi be the least cardinal greater than every member of MahR (Xi ). Note that κ0 ∈ MahR (Xi ) whenever ω ≤ κ0 < θi . Set Q I1 = {i : i ∈ I, κ < θi }, Z1 = i∈I1 Xi ,

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531G

I2 = {i : i ∈ I, θi ≤ κ, cf θi > ω},

Z2 =

Q

Xi ,

I3 = {i : i ∈ I, θi = κ, cf θi = ω},

Z3 =

Q

Xi ,

Z4 =

Q

Xi ,

I4 = {i : i ∈ I, θi < κ, cf θi = ω},

i∈I2 i∈I3 i∈I4

I5 = {i : i ∈ I, θi = 1, #(Xi ) > 1},

Z5 =

I6 = {i : i ∈ I, #(Xi ) = 1},

Q

Z6 =

Q

i∈I6

i∈I5

Xi ,

Xi .

Q

Then X can be identified with 1≤k≤6 Zk , so 531Eg tells us that κ ∈ MahR (Zk ) for some k. As Z6 is a singleton, we actually have κ ∈ MahR (Zk ) for some k ≤ 5. case 1 Suppose κ ∈ MahR (Z1 ). Then, in particular, I1 6= ∅ and thereQis a j ∈ I such that κ < θj . In this case, κ ∈ MahR (Xj ), and we can set κj = κ, κi = 0 for i 6= j to find a family in i∈I MahR (Xi ) with supremum κ. case 2 Suppose that κ ∈ MahR (Z2 ). Let µ be a Radon probability measure on Z2 with Maharam type κ. P For each i ∈ Z2 , let µ0i be the marginal measure on Xi , and κ0i its Maharam type. By 531C, κ ≤ i∈I2 κ0i ; since I2 ⊆ I 0 , κ > #(I2 ); since κ is infinite, it must be less than or equal to supi∈I2 max(ω, κ0i ). On the other hand, by 531F, each κ0i is either finite or the supremum of some countable subset of MahR (Xi ); because cf θi > ω, κ0i < θi and max(ω, κ0i ) ∈ MahR (Xi ). Setting κi = med(κ0i , ω, κ) for i ∈ I2 , = 0 for i ∈ I \ I2 , we have κi ∈ MahR (Xi ) for every i ∈ I and κ = supi∈I κi . case 3 Suppose that κ ∈ MahR (Z3 ). Because κ = θi ∈ / MahR (Xi ) for i ∈ I3 , 531Eg tells us that I3 must be infinite. Let hin in∈N be a sequence of distinct elements of I3 . Of course κ itself is uncountable and has countable cofinality, so we can find a sequence κ0n of infinite cardinals less than κ with supremum κ. Setting κin = κ0n , κi = 0 for i ∈ I \ {in : n ∈ N}, we have κi ∈ MahR (Xi ) for every i and κ = supi∈I κi . case 4 Suppose that κ ∈ MahR (Z4 ). Following the scheme of case 2 above, let µ be a Radon probability measure on Z4 with Maharam type κ, and for each i ∈ I4 let µ0i be the marginal measure on Xi and κ0i its Maharam type. Then, as before, κ ≤ supi∈I4 max(ω, κ0i ). At the same time, κ0i ≤ θi < κ for every i, so we must have κ = supi∈I4 θi . Set δ = cf κ. Then we can choose hiξ iξ<δ inductively in I4 so that θiη < θiξ whenever η < ξ < δ and supξ<δ θiξ = κ. Now define hκi ii∈I by saying κiξ+1 = θiξ whenever ξ < δ, κi = 0 if i ∈ I \ {iξ+1 : ξ < δ}. This gives κi ∈ MahR (Xi ) for every i and κ = supi∈I κi . case 5 ?? Suppose, if possible, that κ ∈ MahR (Z5 ). Once again, we can find a Radon probability measure µ on Z5 with Maharam type κ, and look at its marginal measures µ0i for i ∈ I5 . This every µ0i must be purely Ptime, however, 0 0 atomic and has Maharam S type κi ≤ ω; also #(I5 ) < κ. So our formula κ ≤ i∈I5 κi becomes κ = ω. In this case I5 must be finite and κ ∈ i∈I5 MahR (Xi ) = {0}, which is absurd. X X Thus this case evaporates and the proof is complete. 531H Remarks The results above already enable us to calculate MahR (X) for many spaces. Of course we begin with compact spaces (531Ec). If X is compact and Hausdorff, and [0, 1]κ is a continuous image of X, where κ is an infinite cardinal, then κ ∈ MahR (X) (531Ed); so if [0, 1]w(X) is a continuous image of X, then MahR (X) is completely specified, being {0} ∪ {κ : ω ≤ κ ≤ w(X)} (531Ea, 531Ef). Of course it is not generally true that w(X) ∈ MahR (X) (531Xc). But it is quite often the case that [0, 1]κ is a continuous image of X for every κ ∈ MahR (X), and I will now investigate this phenomenon. 531I Notation For the rest of the section, I will use the following notation, mostly familiar from earlier chapters of this volume. For any ordinal ζ, let νζ be the usual measure on {0, 1}ζ , Σζ its domain, Nζ its null ideal and (Bζ , ν¯ζ ) its measure algebra. In this context, I will write eξ ∈ Bζ for Eξ• , where Eξ = {x : x ∈ {0, 1}ζ , x(ξ) = 1}. For I ⊆ ζ let CI be the closed subalgebra of Bζ generated by {eξ : ξ ∈ I}. For each ξ < ζ, let φξ : Bζ → Bζ be the measure-preserving involution corresponding to reversal of the ξth coordinate, that is, φξ (eξ ) = 1 \ eξ and φξ (eη ) = eη for η 6= ξ.

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531J S Lemma Let κ be a cardinal, and take Bκ , CI , for I ⊆ κ, and φξ , for ξ < κ, as in 531I. (a) {CI : I ∈ [κ]<ω } is dense in Bκ for the measure-algebra topology of Bκ . (b) For every a ∈ Bκ , there is a (unique) countable J ∗ (a) ⊆ κ such that, for I ⊆ κ, a ∈ CI iff I ⊇ J ∗ (a). (c) φξ φη = φη φξ for all ξ, η < κ. (d) If I ⊆ κ, a ∈ CI and ξ < κ, then a ∩ φξ a ∈ CI\{ξ} . (e) For a ∈ Bκ and ξ < κ we have φξ a = a iff ξ ∈ / J ∗ (a). (f) φξ a ∈ CI whenever I ⊆ κ, ξ < κ and a ∈ CI . proof (a) See 254Fe. (b) See 254Rd or 325Mb. (c) Because {eζ : ζ < κ} τ -generates Bκ , it is enough to check that φξ φη eζ = φη φξ eζ for all ξ, η, ζ < κ, and this is easy. (d) The subalgebra {(c ∩ eξ ) ∪ (c0 \ eξ ) : c, c0 ∈ CI\{ξ} } generated by CI\{ξ} ∪ {eξ } is closed (323K), so is the whole of CI and contains a. If c, c0 ∈ CI\{ξ} are such that a = (c ∩ eξ ) ∪ (c0 \ eξ ), then φξ a = (c \ eξ ) ∪ (c0 ∩ eξ ) and a ∩ φξ a = c ∩ c0 ∈ CI\{ξ} . (e) If ξ ∈ / J ∗ (a) then φξ a = a because φξ (eη ) = eη for every η 6= ξ. If φξ a = a then a = a ∩ φξ a ∈ Cκ\{ξ} , by (d), and J ∗ (a) ⊆ κ \ {ξ}, that is, ξ ∈ / J ∗ (a). (f ) CI is the closed subalgebra of A generated by {eη : η ∈ I}, so φξ [CI ] is the closed subalgebra generated by {φξ eη : η ∈ I} ⊆ CI (324L). 531K Lemma Let κ ≥ ω2 be a cardinal. Suppose that we are given a family haξ iξ<κ in Bκ . Then there are a set Γ ∈ [κ]κ and a family hcξ iξ<κ in Bκ such that cξ ⊆ aξ ,

ν¯κ cξ ≥ 2¯ νκ aξ − 1

for every ξ, and, in the notation of 531I, ν¯κ (inf ξ∈I cξ ∩ eξ ∩ inf η∈J cη \ eη ) =

1 ν¯κ (inf ξ∈I∪J cξ ) 2#(I∪J)

whenever I, J ⊆ Γ are disjoint finite sets. proof Let eξ , φξ , for ξ < κ, CL , for L ⊆ κ, and J ∗ (a), for a ∈ Bκ , be as in 531I-531J. Set Lξ = J ∗ (aξ ) and cξ = aξ ∩ φξ aξ for each ξ; then ν¯κ cξ = ν¯κ aξ + ν¯κ (φξ aξ ) − ν¯κ (aξ ∪ φξ aξ ) ≥ 2¯ νκ aξ − 1 and cξ ∈ CLξ \{ξ} (531Jd). By Hajnal’s Free Set Theorem (5A1Hc), there is a set Γ ∈ [κ]κ such that ξ ∈ / Lη whenever ξ, η are distinct members of Γ. (This is where we use the hypothesis that κ ≥ ω2 .) Now suppose that I, J ⊆ Γ are finite and disjoint. Then (Lξ \ {ξ}) ∩ (I ∪ J) = ∅, so cξ ∈ Cκ\(I∪J) , for every ξ ∈ I ∪ J. Accordingly c = inf ξ∈I∪J cξ belongs to Cκ\(I∪J) . This means that c and the eξ , for ξ ∈ I ∪ J, are stochastically independent, and ν¯κ (c ∩ inf ξ∈I eξ ∩ inf η∈J (1 \ eη )) = ν¯κ c ·

Q

ξ∈I

ν¯κ eξ ·

Q

η∈J

ν¯κ (1 \ eη ) =

1 ν¯κ c, 2#(I∪J)

as claimed. 531L Theorem Let X be a normal Hausdorff space. (a) (Haydon 77) If ω ∈ MahR (X) then [0, 1]ω is a continuous image of X. (b) (Haydon 77, Plebanek 97) If κ ≥ ω2 belongs to MahR (X) and λ ≤ κ is an infinite cardinal such that (κ, λ) is a measure-precaliber pair of every probability algebra, then [0, 1]λ is a continuous image of X. proof (a) If ω ∈ MahR (X) then X has a compact subset K which is not scattered (531Ee) and there is a continuous surjection from K onto [0, 1] (4A2G(j-iv)). By Tietze’s theorem (4A2Fd), this has an extension to a continuous map from X to [0, 1] which is still a surjection. As there is a continuous surjection from [0, 1] onto [0, 1]ω (5A4J(b-ii)), there is a continuous surjection from X onto [0, 1]ω . (b) Let µ be a Maharam-type-homogeneous Radon probability measure on X with Maharam type κ, Σ its domain, and (A, µ ¯) its measure algebra, so that (A, µ ¯) is isomorphic to the measure algebra (Bκ , νκ ) as discussed in 531I-531K. Let he0ξ iξ<κ be a stochastically independent generating set of elements of measure 21 in A, so that (A, he0ξ iξ<κ ) is isomorphic to (B, heξ iξ<κ ). For each ξ < κ, let Eξ ∈ Σ be such that Eξ• = e0ξ in A. Let Kξ0 ⊆ Eξ , Kξ00 ⊆ X \ Eξ be compact sets of measure at least 13 , and set Kξ = Kξ0 ∪ Kξ00 , aξ = Kξ• for ξ < κ. By 531K, copied into A, there are hcξ iξ<κ and Γ0 ∈ [κ]κ such that cξ ⊆ aξ and µ ¯cξ ≥ 13 for each ξ, and

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ν¯κ (inf ξ∈I cξ ∩ e0ξ ∩ inf η∈J cη \ e0η ) =

531L

1 ν¯κ (inf ξ∈I∪J cξ ) 2#(I∪J)

whenever I, J ⊆ Γ0 are disjoint finite sets. At this point, recall that (κ, λ) is supposed to be a measure-precaliber pair of every probability algebra. So there is a Γ ∈ [Γ0 ]λ such that inf ξ∈I cξ 6= 0 for every finite I ⊆ Γ. It follows at once that inf ξ∈I aξ ∩ e0ξ ∩ inf η∈J aη \ e0η is T T non-zero for all disjoint finite sets I, J ⊆ Γ. But this means that X ∩ ξ∈I Kξ0 ∩ η∈J Kη00 is non-negligible, therefore non-empty, for T all disjoint finite I, J ⊆ Γ. Set K = ξ∈Γ Kξ , so that K ⊆ X is compact. Then we have a continuous function f : K → {0, 1}Γ defined by setting f (x)(ξ) = 1 if x ∈ K ∩ Eξ = K ∩ Kξ0 , = 0 if x ∈ K \ Eξ = K ∩ Kξ00 . Now f is surjective. P P If w ∈ {0, 1}Γ and L ⊆ Γ is finite, then FL = {x : x ∈ X, x ∈ Kξ0 whenever ξ ∈ L and w(ξ) = 1, x ∈ Kξ00 whenever ξ ∈ L and w(ξ) = 0} is a non-empty closed set. The family {FL : L ∈ [Γ]<ω } is downwards-directed, so has non-empty intersection; and if x is any point of the intersection, x ∈ K and f (x) = w. Q Q As #(Γ) = λ, {0, 1}λ is a continuous image of a closed subset of X and [0, 1]λ is a continuous image of X (5A4Fa). 531M Proposition (Plebanek 97) If κ is an infinite cardinal which is not a measure-precaliber of every probability algebra, then there is a compact Hausdorff space X such that κ ∈ MahR (X) but [0, 1]κ is not a continuous image of X. proof Recall that since there is some probability algebra of which κ is not a measure-precaliber, it cannot be a measure-precaliber of (Bκ , ν¯κ ) (525J(a-i)). Let haξ iξ<κ be a family in Bκ , with no centered subfamily of size κ, such that inf ξ<κ ν¯κ aξ = α > 0. Choose hbξ iξ<κ in Bκ inductively, as follows. Given hbη iη<ξ , let Dξ be the closed subalgebra of Bκ generated by {bη : η < ξ} ∪ {aξ }. Because Bκ is homogeneous with Maharam type κ > τ (Dξ ), it is relatively atomless over Dξ , and there is a b ∈ Bκ such that ν¯κ (b ∩ c) = 12 ν¯κ c for every c ∈ Dξ (331B). Set bξ = b ∩ aξ ; then for any η < ξ we have 1 2

1 2

α 2

ν¯κ (bξ 4 bη ) = ν¯κ bξ + ν¯κ bη − 2¯ νκ (bξ ∩ bη ) = ν¯κ aξ + ν¯κ bη − ν¯κ (aξ ∩ bη ) ≥ ν¯κ aξ ≥ . Continue. Let C be the subalgebra of Bκ generated by {bξ : ξ < κ}, and X its Stone space. Then C is isomorphic to the algebra of open-and-closed subsets of X, so we have a Radon measure µ on X defined by saying that µb c = ν¯κ c for every c ∈ C, writing b c for the open-and-closed subset of X corresponding to c (416Qa). Now µ is strictly positive and we can identify C with a topologically dense subalgebra of the measure algebra of µ. It follows that µ has a Maharam-typehomogeneous component of type at least κ. P P?? Otherwise, there would be a set E ⊆ X, of measure at least 1 − 41 α, such that the Maharam type of the subspace measure µE was less than κ. But α α µ(E ∩ bbξ 4bbη ) ≥ ν¯κ (bξ 4 bη ) − ≥ 4

4

whenever η < ξ < κ, so the topological density of the measure algebra of µE is at least κ (5A4B(h-ii)) and the Maharam type of µE is at least κ (521O). X XQ Q It follows that κ ∈ MahR (X). ?? Suppose, if possible, that [0, 1]κ is a continuous image of X. By 5A4C(d-iii), there is a non-empty closed subset F of X such that χ(x, F ) ≥ κ for every x ∈ F . Let D ⊆ κ be a maximal set such that {F } ∪ {bbξ : ξ ∈ D} has the finite T intersection property. Set Z = F ∩ ξ∈D bbξ ; then Z contains a point z say. Because {bξ : ξ ∈ D} is centered, so is T b {aξ : ξ ∈ D}, and #(D) < κ. Accordingly {F ∩ bξ : I ∈ [D]<ω } is too small to be a base of neighbourhoods of z in ξ∈I

F ; by 4A2Gd, Z 6= {z}. Let x ∈ Z \ {z}. Because {b c : c ∈ C} is a base for the topology of X there is a ζ < κ such that one of x, z belongs to bbζ and the other does not. But in this case ζ ∈ / D and Z ∩ bbζ 6= ∅, so {bbξ : ξ ∈ D ∪ {ζ}} ∪ {F } is centered, contrary to the choice of D. X X Thus X has the required properties. 531N From 531L-531M we see that if κ is any infinite cardinal other than ω1 , then the following are equiveridical: (i) κ is a measure-precaliber of every probability algebra; (ii) whenever X is a compact Hausdorff space and κ ∈ MahR (X), then [0, 1]κ is a continuous image of X. The following variation on the construction in 531M shows that ω1 really is a special case.

531O

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191

Proposition S (Plebanek 97) Suppose that there is a family hWξ iξ<ω1 in Nω1 such that every closed subset of {0, 1}ω1 \ ξ<ω1 Wξ is scattered. Then there is a compact Hausdorff space X such that ω1 ∈ MahR (X) but [0, 1]ω1 is not a continuous image of X. proof (a) in {0, 1}ω1 for each j 1 − 2−j−4 . Set

Set Eξ = {z : z ∈ {0, 1}ω1 , z(ξ) = 1} for each ξ < ω1 . Choose a family hKξn iξ<ω1 ,n∈N of compact sets S as follows. Given hKηn iη<ξ,n∈N , where ξ < ω1 , such that n∈NT Kηn S is conegligible for every η < ξ, then ∈ N we can find a family hn(ξ, η, j)iη<ξ in N such that Lξj = η<ξ i≤n(ξ,η,j) Kηi has measure at least S S 0 0 0 For j ∈ N choose a compact set Kξj ⊆ Lξj \ (Wξ ∪ i<j Kξi ) of measure at least 1 − 2−j−3 − νω1 ( i<j Kξi ). 0 Kξ,2i = Kξi ∩ Eξ ,

0 Kξ,2i+1 = Kξi \ Eξ

for each i ∈ N, and continue. (b) At the end of the induction, let C be the algebra of subsets of {0, 1}ω1 generated by {Kξi : ξ < ω1 , i ∈ N}, and b = νω C for every C ∈ C, X its Stone space. Then we have a Radon probability measure µ on X defined by setting µC 1 b where C is the open-and-closed subset of X corresponding to C. For η < ξ < ω1 , we have b η0 4K b ξ0 ) = νω (Kη0 4Kξ0 ) µ(K 1 0 0 = νω1 ((Eη ∩ Kη0 )4(Eξ ∩ Kξ0 )) 0 0 ≥ νω1 (Eη 4Eξ ) − νω1 (Eη \ Kη0 ) − νω1 (Eξ \ Kξ0 )

≥

1 2

−

1 8

−

1 8

1 4

= ,

so the Maharam type of µ is at least ω1 and ω1 ∈ MahR (X). (c) Let F ⊆ X be a non-scattered closed set. Then there is a ζ < ω1 such that F 6⊆ T S b ξi ) × Kξi ⊆ X × {0, 1}ω1 . R = ξ<ω1 i∈N (F ∩ K

S

i∈N

b ζi . P K P?? Otherwise, set

b ξi are open-and-closed sets covering F , so {i : F ∩ K b ξi 6= ∅} is finite and S (F ∩ Note that for each ξ < ω1 the K i∈N b ξi ) × Kξi is compact; so R is compact. If (x, z) and (x0 , z 0 ) ∈ R and x 6= x0 , there must be some C ∈ C such that K b and x0 ∈ b so there must be some ξ < ω1 and i ∈ N such that just one of x, x0 belongs to K b ξi ; in this case, x∈C / C, only the corresponding one of z, z 0 can belong to Kξi , and z 6= z 0 . Conversely, if (x, z) and (x0 , z 0 ) ∈ R and z 6= z 0 , there is some ξ such that z(ξ) 6= z 0 (ξ). In this case, if i, j ∈ N are b ξi × Kξi and (x0 , z 0 ) ∈ K b ξj × Kξj , i 6= j and x 6= x0 . such that (x, z) ∈ K This shows that R is the graph of a bijection from F to R[F ]. Because R isSa compact subset of F × R[F ], it is a homeomorphism, and R[F ] Sis not scattered. But, for each ξ < ω1 , R[F ] ⊆ i∈N Kξi is disjoint from Wξ ; and all compact subsets of {0, 1}ω1 \ ξ<ω1 Wξ are supposed to be scattered. X XQ Q S b ζi . Then χ(x, X) ≤ ω. P (d) Take x ∈ F \ i∈N K P Consider the set T b ηi ⇐⇒ x ∈ K b ηi }. V = η≤ζ,i∈N {x0 : x0 ∈ X, x0 ∈ K This is a Gδ set containing x. ?? If there is an x0 ∈ V \ {x}, there must be some ξ < ω1 and j ∈ N such that just one S b ξj . In this case, ξ > ζ, so Kξj ⊆ S b b of x, x0 belongs to K i≤k Kζi and Kξj ⊆ i≤k Kζi for some k ∈ N. But neither x nor S 0 b x belongs to i≤k Kζi . X X Thus V = {x}; by 4A2Gd, χ(x, X) ≤ ω. Q Q (e) Thus we see that whenever F ⊆ X is a non-scattered closed set, there is an x ∈ F such that χ(x, X) is countable. By 5A4C(d-iii), [0, 1]ω1 is not a continuous image of X. 531O In 531M we have a space X from which there is no surjection onto [0, 1]κ because every non-empty closed set has a point of character less than κ. From stronger properties of κ we can get compact spaces with stronger topological properties, as in the next two results. Proposition Let κ, κ0 and λ be infinite cardinals such that (κ, κ0 ) is not a measure-precaliber pair of (Bλ , ν¯λ ). Then there is a compact Hausdorff space X such that κ ∈ MahR (X) and χ(x, X) < max(κ0 , λ+ ) for every x ∈ X. proof Let haξ iξ<κ be a family in Bλ , with no centered subfamily of cardinal κ0 , such that inf ξ<κ µ ¯aξ = α > 0. Let 1 ψ : Bλ → Σλ be a lifting; for each ξ < κ, let Kξ ⊆ ψaξ be a compact set of measure at least T T 2 α. If D ⊆ κ and #(D) = κ0 , then there is a finite set I ⊆ D such that inf ξ∈I aξ = 0, in which case ξ∈I Kξ ⊆ ξ∈I ψaξ = ∅. Thus {ξ : x ∈ Kξ } has cardinal less than κ0 for every x ∈ {0, 1}λ . Set

192

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X=

T

ξ<κ0 {(x, y)

531O

: x ∈ {0, 1}λ , y ∈ {0, 1}κ , x ∈ Kξ or y(ξ) = 0},

so that X is a compact subset of {0, 1}λ × {0, 1}κ . Now χ((x, y), X) < max(κ0 , λ+ ) for every (x, y) ∈ X. P P Set D = {ξ : ξ < κ, x ∈ Kξ }, so that #(D) < κ0 . For I ∈ [λ]<ω and J ∈ [D]<ω set VIJ = {(x0 , y 0 ) : (x0 , y 0 ) ∈ X, x0 I = xI, y 0 J = yJ}, so that V T = {VIJ : I ∈ [λ]<ω , J ∈ [D]<ω } is a downwards-directed family of closed neighbourhoods of (x, y). If 0 0 0 (x , y ) ∈ V, then xT = x, so x0 ∈ / Kξ for ξ ∈ κ \ D, and y 0 (ξ) = y(ξ) = 0 for ξ ∈ / D; also y 0 D = yD, so 0 0 (x , y ) = (x, y). Thus V = {(x, y)}; by 4A2Gd, V is a base of neighbourhoods of (x, y), and χ((x, y), X) ≤ #(V) ≤ max(#(D), λ) < max(κ0 , λ+ ). Q Q Define g : {0, 1}λ × {0, 1}κ → {0, 1}κ and h : {0, 1}λ × {0, 1}κ → X by setting g(x, y)(ξ) = y(ξ) if x ∈ Kξ , = 0 otherwise, h(x, y) = (x, g(x, y)), for ξ < κ, x ∈ {0, 1}λ and y ∈ {0, 1}κ . Write Σ for the domain of the product measure ν = νλ × νκ on {0, 1}λ × {0, 1}κ . Then the σ-algebra {F : F ⊆ X, h−1 [F ] ∈ Σ} contains all sets of the form {(x, y) : x(η) = 1} and {(x, y) : y(ξ) = 1}, so includes a base for the topology of X and therefore contains every open-and-closed set. Accordingly we have an additive functional U 7→ νh−1 [U ] on the algebra of open-and-closed subsets of X, which extends to a Radon probability measure µ on X (416Qa). Set Fξ = {(x, y) : (x, y) ∈ X, y(ξ) = 1} for each ξ < κ; then for any η < ξ < κ, µ(Fξ \ Fη ) = νh−1 [Fξ \ Fη ] 1 4

1 8

≥ ν{(x, y) : x ∈ Kξ , y(ξ) = 1, y(η) = 0} = νλ Kξ ≥ α. Once again, this shows that the measure algebra of µ must have a homogeneous principal ideal with Maharam type at least κ, and κ ∈ Mah(X). 531P Putting these ideas together with 531L, we have the following. Proposition (see Plebanek 95) Let κ be a regular uncountable cardinal. Then the following are equiveridical: (i) κ is a precaliber of every measurable algebra; (ii) if X is a compact Hausdorff space such that κ ∈ MahR (X), then χ(x, X) ≥ κ for some x ∈ X. proof (i)⇒(ii) If κ ≥ ω2 , then 531Lb tells us that [0, 1]κ is a continuous image of X. By 5A4C(d-iii) and 5A4Bb, it follows at once that χ(x, X) ≥ κ for many points x ∈ X. ?? Suppose, if possible, that ω1 is a precaliber of every probability algebra, but that there is a first-countable compact Hausdorff space X with ω1 ∈ MahR (X). Let µ be a Maharam-type-homogeneous Radon probability measure on X with Maharam type ω1 , and (A, µ ¯) its measure algebra; let heξ iξ<ω1 be a τ -generating stochastically independent family of elements of measure 21 in A. As in 531J, there is for each a ∈ A a countable J ∗ (a) ⊆ ω1 such that a belongs to the closed subalgebra of A generated by {eξ : ξ ∈ J ∗ (a)}. x ∈ X, let Ux be a countable base of open neighbourhoods of x, andSset Ax = {U • : U ∈ Ux }, J † (x) = S For each ∗ † † a∈Ax J (a). Then J (x) is countable. For ξ < ω1 , set Cξ = {x : J (x) ⊆ ξ}; then ξ<ω1 Cx = X. Now ω1 is supposed to be a precaliber of A, so there must be a ξ < ω1 such that Cξ has full outer measure (525Dc). Let G ⊆ X be open. Then G• belongs to the closed subalgebraSCξ of A generated by {eη : η < ξ}. P P For each x ∈ G ∩ Cξ , there is a Ux ∈ Ux such that Ux ⊆ G. Set H = {Ux : x ∈ G ∩ Cξ }, so that H ⊆ G is open and G ∩ Cξ = H ∩ Cξ ; as Cξ has full outer measure, G \ H is negligible and H • = G• . But 414Aa tells us that H • = supx∈Cξ Ux• , and this belongs to Cξ , because J † (x) ⊆ ξ for every x ∈ Cξ . Q Q It follows at once that F • ∈ Cξ for every closed F ⊆ X. Because µ is inner regular with respect to the closed sets, Cξ is order-dense in A and A = Cξ has Maharam type #(ξ) < ω1 . X X Thus (i)⇒(ii) also when κ = ω1 . not-(i)⇒not-(ii) Again, the case κ ≥ ω2 is covered by earlier results. P P If κ ≥ ω2 is a regular cardinal which is not a precaliber of every probability algebra, then by 525Jb there is a λ < κ such that κ is not a precaliber of Bλ . Now 531O tells us that there is a compact Hausdorff space X such that κ ∈ MahR (X) and χ(x, X) < max(κ, λ+ ) = κ for every x ∈ X. Q Q So once again we are led to the case κ = ω1 , and we are supposing that ω1 is not a precaliber of every probability algebra, and therefore not of Bω1 . By 525Hc, cov Nω1 = ω1 and there is a non-decreasing family hEξ iξ<ω1 of negligible subsets of {0, 1}ω1 covering {0, 1}ω1 . Let Eξ0 ⊇ Eξ be a negligible set determined by coordinates in a countable set

531Q

Maharam types of Radon measures

193

˜ξ = S{Eη0 : η < ξ, Jη ⊆ ξ}; then each E ˜ξ is determined by coordinates Jξ ⊆ ω1 , for each ξ < ω1 . For each ξ < ω1 , let E ˜ less than ξ, hEξ iξ<ω1 is non-decreasing, and S S 0 ω1 ˜ ξ<ω1 Eξ = ξ<ω1 Eξ = {0, 1} . ˜ξ is a non-negligible measurable set determined by Let ζ < ω1 be such that J ⊆ ζ. For ζ ≤ ξ < ω1 , {0, 1}ω1 \ E ˜ξ , of measure at least 3 , also determined by coordinates in ξ. There is therefore a compact set Kξ ⊆ {0, 1}ω1 \ E 4 coordinates in ξ. Set T X = ζ≤ξ<ω1 {x : x ∈ {0, 1}ω1 , x ∈ Kξ or x(ξ) = 0}, ˜ξ . so that X ⊆ {0, 1}ω1 is compact. Then X is first-countable. P P If x ∈ X, then there is some ξ ≥ ζ such that x ∈ E In this case, if yξ = xξ, neither x nor y can belong to Kη for any η ≥ ξ, so y = x and {x} is a Gδ set in X and χ(x, X) ≤ ω (4A2Gd). Q Q ω1 Define f : {0, 1} → {0, 1}ω1 by setting f (x)(ξ) = x(ξ) if x ∈ Kξ , = 0 otherwise. Then f (x) ∈ X for every x and we can define a Radon measure µ on X by setting µV = νω1 f −1 [V ] for every open-and-closed set V ⊆ X (416Qa). Consider Gξ = {x : x ∈ X, x(ξ) = 1} for ξ < ω1 . If η < ξ < ω1 , then µ(Gξ \ Gη ) ≥ νω1 {x : x ∈ Kξ , x(ξ) = 1, x(η) = 0} 1 2

1 2

1 8

≥ (νω1 Kξ − ) ≥ . So the Maharam type of µ is at least ω1 and ω1 ∈ MahR (X) (531F). Thus in this case too we have the required example. 531Q In 531P we see that if ω1 is not a precaliber of every measurable algebra then there is a first-countable compact Hausdorff space with a Radon measure with Maharam type ω1 . With a sharper hypothesis, and rather more work, we can get a substantially stronger version, as follows. Proposition Suppose that cf Nω = ω1 . Then there is a hereditarily separable perfectly normal compact Hausdorff space X such that ω1 ∈ MahR (X). proof For η ≤ ξ ≤ ω1 and x ∈ {0, 1}ξ , set πξη (x) = xη; write πη for πω1 η : {0, 1}ω1 → {0, 1}η . (a) It will be helpful to have the following fact out in the open. Let Y be a zero-dimensional metrizable compact Hausdorff space, µ an atomless Radon probability measure on Y , A ⊆ Y a µ-negligible set and Q a countable family of closed subsets of Y . Then there are closed sets K, L ⊆ Y , with union Y , such that K ∪ L = Y , K ∩ L ∩ A = ∅, µ(K ∩ L) ≥ 12 , K ∩ Q = Q \ L and L ∩ Q = Q \ K for every Q ∈ Q. P P We can of S course suppose that Q is non-empty. For each Q ∈ Q let DQ be a countable dense subset of Q; let S ⊆ Y \ (A ∪ Q∈Q DQ ) be a closed set of measure at least 12 . (This is where we need to know that µ is atomless, so that every DQ is negligible.) Let U be a countable base for the topology of Y consisting of open-and-closed sets and let h(Un , Qn )in∈N run over U × Q. Choose inductively sequences hGn in∈N , hHn in∈N of open-and-closed subsets of Y \ S, as follows. Start with G0 = H0 = ∅. Given that Gn and Hn are disjoint from each other and from S, then if Un ∩ Qn ∩ S = ∅, take Gn+1 = Gn ∪ (Un \ Hn ) and Hn+1 = Hn . Otherwise, S ∩ Qn is nowhere dense in Qn (because it does not meet DQn ), while (Gn ∪ Hn ) ∩ Qn is a relatively closed set in Qn not meeting S, therefore not including Qn ∩ Un ; so Qn ∩ Un \ (S ∪ Gn ∪ Hn ) must have at least two points y, y 0 say, and we can enlarge Gn and Hn to disjoint 0 open-and-closed subsets Gn+1 , Hn+1 of YS\ S containing y, yS respectively, and therefore both meeting Un ∩ Qn . At the end of the induction, set G = n∈N Gn and H = n∈N Hn , so that G and H are disjoint open subsets of Y \ S. Now if y is any point of Y \ S, there must be some n such that y ∈ Un ⊆ Y \ S, so that y ∈ Gn+1 ∪ Hn ; thus Y = G ∪ H ∪ S. Set K = G ∪ S = Y \ H, L = H ∪ S = Y \ G; then K and L are closed sets with union Y , and K ∩ L = S has measure at least 21 and is disjoint from A. If Q ∈ Q, y ∈ S ∩ Q and U is any neighbourhood of y, there is an n ∈ N such that Qn = Q and y ∈ Un ⊆ U . In this case, Gn+1 and Hn+1 meet Un ∩ Q. As U is arbitrary, y ∈ Q ∩ G ∩ Q ∩ H = Q \ L ∩ Q \ K. Thus K ∩ L ∩ Q ⊆ Q \ L ∩ Q \ K, so K ∩ Q ⊆ Q \ L and L ∩ Q ⊆ Q \ K. On the other hand, K ⊇ Q \ L and L ⊇ Q \ K, so K ∩ Q = Q \ L and L ∩ Q = Q \ K. Thus K and L fulfil all the specifications. Q Q

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531Q

(b) Choose hfξ iω≤ξ≤ω1 , hµξ iω≤ξ<ω1 , hKξ iω≤ξ<ω1 , hLξ iω≤ξ<ω1 , hXξ iω≤ξ≤ω1 , hQ0ζξ iω≤ξ≤ζ<ω1 , hQξδ iω≤δ≤ξ<ω1 , hQξδη iω≤η≤δ≤ξ<ω1 , hAζξ iω≤ξ≤ζ<ω1 , hAξ iω≤ξ<ω1 as follows. fω (x) = x for every x ∈ {0, 1}ω = Xω . Given that ω ≤ ξ < ω1 and fξ is a Borel measurable surjection from {0, 1}ξ onto a compact subset Xξ of {0, 1}ξ , let µξ be the image measure νξ fξ−1 , so that µξ is a Radon measure on {0, 1}ξ (433E, 418I); of course µξ Xξ = 1. Because cf Nω = ω1 , µξ is inner regular with respect to a family of size at most ω1 (524Pb), which we may suppose to consist of closed sets; let hQ0ζξ iξ≤ζ<ω1 run over such a family. Similarly, there −1 is a family hAζξ iξ≤ζ<ω1 running over a cofinal subset of the null ideal of µξ . Next, for ω ≤ δ ≤ ξ, let Qξδ ⊆ πξδ [Q0ξδ ] −1 be a compact µξ -self-supporting set of the same µξ -measure as πξδ [Q0ξδ ]. Note that, as µξ Xξ = 1, every Qξδ must be −1 included in Xξ . Set Qξδη = Xξ ∩ πξδ [Qδη ] for ω ≤ η ≤ δ ≤ ξ. Set S −1 Aξ = {πξη [Aδη ] : ω ≤ η ≤ δ ≤ ξ}, Aξ = {A : A ∈ Aξ , µξ A = 0}; because Aξ is countable, Aξ is µξ -negligible. By (a) above, we can find closed sets Kξ , Lξ covering {0, 1}ξ such that µξ (Kξ ∩ Lξ ) ≥ 21 , Kξ ∩ Lξ ∩ Aξ = ∅, and Kξ ∩ Qξδη = Qξδη \ Lξ , Lξ ∩ Qξδη = Qξδη \ Kξ whenever ω ≤ η ≤ δ ≤ ξ. Given that ω < ξ ≤ ω1 and that Kη , Lη are closed subsets of {0, 1}η covering {0, 1}η for ω ≤ η < ξ, define fξ (x)(η) inductively, for x ∈ {0, 1}ξ and η < ξ, by setting fξ (x)(η) = x(η) if η < ω or η ≥ ω and xη ∈ Kη ∩ Lη , = 1 if η ≥ ω and xη ∈ / Lη , = 0 if η ≥ ω and xη ∈ / Kη . ξ

ξ

Then fξ : {0, 1} → {0, 1} is Baire measurable. Set T Xξ = ω≤η<ξ {x : x ∈ {0, 1}ξ , x(η) = 1 or xη ∈ Lη , x(η) = 0 or xη ∈ Kη }; then Xξ ⊆ {0, 1}ξ is compact, fξ (x) ∈ Xξ for every x ∈ {0, 1}ξ , and fξ (x) = x for every x ∈ Xξ . So fξ [{0, 1}ξ ] = Xξ and (if ξ < ω1 ) the process continues. (c) At the end of the induction, write f for fω1 and X for Xω1 . If z ∈ {0, 1}ω1 and ξ < ω1 , an easy induction on η shows that f (z)(η) = fξ (zξ)(η) for every η < ξ, that is, that f (z)ξ = fξ (zξ). Next, because f is Baire measurable, we have a Radon measure µ on {0, 1}ω1 defined by saying that µV = νω1 f −1 [V ] for every Baire set V ⊆ {0, 1}ω1 (432F); of course µV = 0 for every open-and-closed set V disjoint from X, so µX = 1. (d) A couple of simple facts. I have already observed that πξ f = fξ πξ for ω ≤ ξ < ω1 ; consequently Xξ = fξ [{0, 1}ξ ] = fξ [πξ [{0, 1}ω1 ]] = πξ [f [{0, 1}ω1 ]] = πξ [X] and µξ [V ] = νξ fξ−1 [V ] = (νω1 πξ−1 )fξ−1 [V ] = νω1 (fξ πξ )−1 [V ] = νω1 (πξ f )−1 [V ] = (νω1 f −1 )πξ−1 [V ] = µπξ−1 [V ] for every open-and-closed set V ⊆ {0, 1}ξ . Thus the Radon measures µπξ−1 and µξ are identical. Equally, if ω ≤ η ≤ ξ < ω1 , Xη = πη [X] = πξη [πξ [X]] = πξη [Xξ ] and −1 µη = µπη−1 = µ(πξη πξ )−1 = µξ πξη .

Accordingly, if ω ≤ η ≤ δ ≤ ξ < ω1 , −1 µξ πξη [Aδη ] = µη Aδη = 0.

Thus in fact Aξ =

S

−1 Aξ and Kξ ∩ Lξ is disjoint from πξη [Aδη ] whenever ω ≤ η ≤ δ ≤ ξ.

(e) We come now to the first key idea of this construction. If ω ≤ η ≤ δ ≤ ξ < ω1 , then gξδη = πξδ Qξδη is an irreducible surjection onto Qδη . P P First, let us confirm that if δ ≤ ζ ≤ ξ then −1 −1 −1 πξζ [Qξδη ] = πξζ [Xξ ∩ πξδ [Qδη ]] = πξζ [Xξ ∩ πξζ [πζδ [Qδη ]]] −1 −1 = πξζ [Xξ ] ∩ πζδ [Qδη ] = Xζ ∩ πζδ [Qδη ] = Qζδη .

(∗)

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195

In particular, −1 πξδ [Qξδη ] = Qδδη = Xδ ∩ πδδ [Qδη ] = Qδη .

To see that gξδη is irreducible, induce on ξ. At the start, gδδη is an identity function, so is certainly irreducible. For the inductive step to ξ + 1, given that δ ≤ ξ < ω1 and gξδη is irreducible, consider h = πξ+1,ξ Qξ+1,δ,η . By (*), h[Qξ+1,δ,η ] = Qξδη . Note that Xξ+1 can be identified with {(x, 1) : x ∈ Xξ ∩ Kξ } ∪ {(x, 0) : x ∈ Xξ ∩ Lξ }. ξ+1

If V ⊆ {0, 1} Because

is a cylinder set meeting Qξ+1,δ,η , then V 0 = πξ+1,ξ [V ] is a cylinder set in {0, 1}ξ meeting Qξδη . Kξ ∩ Qξδη = Qξδη \ Lξ ,

Lξ ∩ Qξδη = Qξδη \ Kξ ,

Kξ ∪ Lξ ⊇ Qξδη ,

V 0 must meet both Qξδη \ Lξ and Qξδη \ Kξ . Now V must cover at least one of {(x, 1) : x ∈ V 0 ∩ Qξδη \ Lξ },

{(x, 0) : x ∈ V 0 ∩ Qξδη \ Kξ },

and h[Qξ+1,δ,η \ V ] is disjoint from at least one of the non-empty sets V 0 ∩ Qξδη \ Lξ , V 0 ∩ Qξδη \ Kξ . As V is arbitrary, h is irreducible, and the composition gξδη h is irreducible (5A4C(d-iv)); but this is gξ+1,δ,η . For the inductive step to a limit ordinal ξ such that δ < ξ < ω1 , again take a cylinder set V ⊆ {0, 1}ξ meeting Qξδη . This time, because ξ is a limit ordinal, there is a ζ such that δ ≤ ζ < ξ and V is determined by coordinates less than ζ. Set V 0 = πξζ [V ]; then V 0 is a cylinder set in {0, 1}ζ meeting πξζ [Qξδη ] = Qζδη . Now −1 πξδ [Qξδη \ V ] = πζδ [πξζ [Qξδη \ πξζ [V 0 ]]]

= πζδ [πξζ [Qξδη ] \ V 0 ] = πζδ [Qζδη \ V 0 ] ⊂ Qδη because gζδη is irreducible. Thus the induction continues. Q Q (f ) Repeating the last step of the argument in (e) with ξ = ω1 , we see that in fact πδ X ∩ πδ−1 [Qδη ] is an irreducible surjection onto Qδη whenever ω ≤ η ≤ δ < ω1 . The next point to confirm is that if z, z 0 ∈ X, ω ≤ η ≤ δ < ω1 , zδ = z 0 δ and zη ∈ Aδη , then z 0 = z. P P Suppose that δ ≤ ξ < ω1 and zξ = z 0 ξ. Then Kξ ∩ Lξ does not meet −1 πξη [Aδη ], so does not contain zξ. Accordingly z(ξ) = 1 =⇒ zξ ∈ Kξ =⇒ zξ ∈ / Lξ =⇒ z 0 ξ ∈ / Lξ =⇒ z 0 (ξ) = 1, and similarly z(ξ) = 0 ⇒ z 0 (ξ) = 0. So an easy induction on ξ shows that z(ξ) = z 0 (ξ) whenever δ ≤ ξ < ω1 , and z = z0. Q Q (g) It follows that if H ⊆ X is closed, there is a ξ < ω1 such that H = X ∩ πξ−1 [πξ [H]] and πξ H is irreducible. P P For ω ≤ η ≤ ξ < ω1 , −1 −1 µη πη [H] = µξ πξη [πη [H]] = µξ πξη [πξη [πξ [H]]] ≥ µξ πξ [H].

So we have an η < ω1 such that µη πη [H] = µξ πξ [H] whenever η ≤ ξ < ω1 . Now recall that µη is inner regular with respect to {Q0δη : η ≤ δ < ω1 }. So there is a countable set I ⊆ ω1 \ η such that hQ0δη iδ∈I is disjoint, Q0δη ⊆ πη [H] for P every δ ∈ I and δ∈I µη Q0δη = µπη [H]. For each δ ∈ I, −1 µδ (Qδη \ πδ [H]) ≤ µδ (πδη [Q0δη ] \ πδ [H]) −1 ≤ µδ (πδη [πη [H]] \ πδ [H]) −1 = µδ (πδη [πη [H]]) − µδ πδ [H] −1 (because surely πδ [H] ⊆ πδη [πη [H]])

= µη πη [H] − µδ πδ [H] = 0 by the choice of η. Because Qδη was µδ -self-supporting, and πδ [H] is closed, Qδη ⊆ πδ [H]. Because πδ X ∩ πδ−1 [Qδη ] is irreducible, X ∩ πδ−1 [Qδη ] ⊆ H. −1 −1 Set ζ = sup(I ∪ {η}) < ω1 . Since Qδη ⊆ πδη [Q0δη ], πζδ [Qδη ] ⊆ πζη [Q0δη ] for each δ ∈ I; as hQ0δη iδ∈I is disjoint, so is −1 hπζδ [Qδη ]iδ∈I ; and X X X −1 −1 µδ πδη [Q0δη ] µζ πζδ [Qδη ] = µδ Qδη = δ∈I

δ∈I

=

X δ∈I

δ∈I

µη Q0δη

= µη πη [H] = µζ πζ [H].

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S −1 −1 Because X ∩ πδ−1 [Qδη ] ⊆ H, πζ [H] ⊇ Xζ ∩ πζδ [Qδη ] for every δ ∈ I. So πζ [H] \ δ∈I πζδ [Qδη ] is µζ -negligible and is included in Aξζ for some ξ ≥ ζ. Repeating the arguments of the last two sentences at the new level, we see that S S −1 −1 −1 Xξ ∩ δ∈I πξδ [Qδη ] ⊆ πξ [H] ⊆ δ∈I πξδ [Qδη ] ∪ πξζ Aξζ . Now suppose that V ⊆ {0, 1}ω1 is an open set meeting H. If there is a z belonging to V ∩ H ∩ πζ−1 [Aξζ ], then zξ 6= z 0 ξ for any other z 0 ∈ X, by (f); so zξ ∈ / πξ [H \ V ] and πξ [H \ V ] 6= πξ [H]. Otherwise, there is a δ ∈ I such that V ∩ πδ−1 [Qδη ] is not empty. Because πδ X ∩ πδ−1 [Qδη ] is irreducible, πδ [H \ V ] cannot cover Qδη ⊆ πδ [H]. Thus πδ [H \ V ] 6= πδ [H]; it follows at once that πξ [H \ V ] 6= πξ [H]. As V is arbitrary, πξ H is irreducible. I have still to check that H = X ∩ πξ−1 [πξ [H]]. If z, z 0 ∈ X, z ∈ H and z 0 ξ = zξ, then if z ∈ πζ−1 [Aξζ ] we have z 0 = z ∈ H. Otherwise, there is some δ ∈ I such that z ∈ πδ−1 [Qδη ]. In this case, z 0 δ = zδ ∈ Qδη ; but X ∩ πδ−1 [Qδη ] ⊆ H, so again z 0 ∈ H. So we have the result. Q Q (h) We are within sight of the end. From (g) we see, first, that if H ⊆ X is closed then it is of the form X ∩πη−1 [πη [H]] for some η, so is a zero set in X; accordingly X is perfectly normal, therefore first-countable (5A4Cb). Second, for any closed H ⊆ X, there is an irreducible continuous surjection from H onto a compact metrizable space πξ [H]; because πξ [H] is separable, so is H (5A4C(d-i)). It follows that X is hereditarily separable. P P If A ⊆ X, then A is separable; let D ⊆ A be a countable dense set. Because X is first-countable, each member of A is in the closure of a countable Q subset of A, and there is a countable C ⊆ A such that D ⊆ C. Now C is a countable dense subset of A. Q (i) Finally, we need to check that ω1 ∈ MahR (X). For ω ≤ ξ < ω1 , set Uξ = {z : z ∈ {0, 1}ω1 , zξ ∈ Kξ ∩ Lξ , z(ξ) = 1}. Then µ(Uξ 4E) ≥ 14 whenever E ⊆ {0, 1}ω1 is a Baire set determined by coordinates less than ξ. P P Set E 0 = πξ [E], so that E = πξ−1 [E 0 ] and E 0 is a Baire set. Then µ(Uξ \ E) = νω1 f −1 [Uξ \ E] = νω1 {z : f (z)ξ ∈ Kξ ∩ Lξ \ E 0 , f (z)(ξ) = 1} = νω1 {z : fξ (zξ) ∈ Kξ ∩ Lξ \ E 0 , z(ξ) = 1} 1 2

1 2

= νω1 {z : fξ (zξ) ∈ Kξ ∩ Lξ \ E 0 } = µξ (Kξ ∩ Lξ \ E 0 ), while µ(E \ Uξ ) = νω1 f −1 [E \ Uξ ] ≥ νω1 f −1 [E ∩ πξ−1 [Kξ ∩ Lξ ] \ Uξ ] = νω1 {z : f (z)ξ ∈ Kξ ∩ Lξ ∩ E 0 , f (z)(ξ) = 0} = νω1 {z : fξ (zξ) ∈ Kξ ∩ Lξ ∩ E 0 , z(ξ) = 0} 1 2

1 2

= νω1 {z : fξ (zξ) ∈ Kξ ∩ Lξ ∩ E 0 } = µξ (Kξ ∩ Lξ ∩ E 0 ). Putting these together, 1 2

1 4

µ(E4Uξ ) ≥ µξ (Kξ ∩ Lξ ) ≥ . Q Q In particular, µ(Uη 4Uξ ) ≥ 14 whenever ω ≤ η < ξ < ω1 . So µ has uncountable Maharam type. As µX = 1, so has the subspace measure on X, and ω1 ∈ MahR (X). Thus X has all the properties announced. 531R Returning to the ideas of 531J, we have the following construction. Lemma Let κ be a cardinal, and let Bκ , heξ iξ<κ , hφξ iξ<κ , J ∗ and hCI iI⊆κ be as in 531I-531J. For a ∈ Bκ and I ⊆ κ, set SI (a) = sup{c : c ∈ Cκ\I , c ⊆ a}. (a) For all a ∈ Bκ , ξ < κ and I, J ⊆ κ, (i) S∅ (a) = a, (ii) SJ (a) ⊆ SI (a) if I ⊆ J, (iii) J ∗ (SI (a)) ⊆ J ∗ (a) \ I, (iv) S{ξ} (a) = a ∩ φξ a, (v) SI (SJ (a)) = SI∪J (a). (b) Whenever a ∈ Bκ ,  > 0 and m ∈ N, there is a finite K ⊆ κ such that ν¯κ (a \ SI (a)) ≤  whenever I ⊆ κ \ K and #(I) ≤ m. proof (a)(i) Cκ = Bκ . (ii) If I ⊆ J then Cκ\J ⊆ Cκ\I .

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(iii) Set C = {c : c ∈ Cκ\I , c ⊆ a}. If ξ ∈ κ \ (J ∗ (a) \ I) and c ∈ C, then φξ c ∈ C. P P By 531Jf, φξ c ∈ Cκ\I . If ξ ∈ I then φξ c = c ⊆ a. If ξ ∈ κ \ J ∗ (a) then φξ c ⊆ φξ a = a. Q Q Because φξ is an order-continuous involution, φξ (SI (a)) = SI (a) and ξ ∈ / J ∗ (S(a)), by 531Je. (iv) If c ∈ Cκ\{ξ} and c ⊆ a, then c = φξ c ⊆ φξ a; so S{ξ} (a) ⊆ a ∩ φξ a. On the other hand, by 531Jd, a ∩ φξ belongs to Cκ\{ξ} , so is included in S{ξ} (a). (v) By (iii), J ∗ (SI (SJ (a))) ⊆ J ∗ (SJ (a)) \ I ⊆ J ∗ (a) \ (I ∪ J), and SI (SJ (a)) ∈ Cκ\(I∪J) ; since also SI (SJ (a)) ⊆ SJ (a) ⊆ a, SI (SJ (a)) ⊆ SI∪J (a). On the other hand, SI∪J (a) ∈ Cκ\I and is included in SJ (a), so is included in SI (SJ (a)). (b) Induce on m. For m = 0 the result is trivial. For the inductive step to m + 1, take K0 ∈ [κ]<ω such that ν¯κ (a \ SJ (a)) ≤ 14  whenever J ∈ [κ \ K0 ]≤m . By 531Ja, there are a finite set K1 ⊆ κ and a b ∈ CK1 such that ν¯κ (a 4 b) ≤ 14 ; set K = K0 ∪ K1 . Suppose I ∈ [κ \ K]m+1 ; take ξ ∈ I and set J = I \ {ξ}. Then SI (a) = S{ξ} (SJ (a)) = SJ (a) ∩ φξ (SJ (a)) by (a-iv) and (a-v). So

ν¯κ (a \ SI (a)) ≤ ν¯κ (a \ SJ (a)) + ν¯κ (a \ φξ SJ (a)) ≤

 4

+ ν¯κ (a \ φξ a) + ν¯κ φξ (a \ SJ (a))

≤

 4

+ ν¯κ (a \ b) + ν¯κ (b \ φξ b) + ν¯κ φξ (b \ a) + ν¯κ (a \ SJ (a)) ≤ 

because φξ is a measure-preserving Boolean homomorphism and φξ b = b. Thus the induction continues. 531S Moving on from hypotheses expressible as statements about measure algebras, we have a further result which can be used when Martin’s axiom is true. Lemma Suppose that ω1 < mK (definition: 517O). Let haξ iξ<ω1 be a family of elements of Bω1 of measure greater than 12 . Then there is an uncountable set Γ ⊆ ω1 such that inf ξ∈I (aξ ∩ eξ ) ∩ inf η∈J (aη \ eη ) is non-zero whenever I, J ⊆ Γ are finite and disjoint. proof (a) Define J ∗ (a), for a ∈ Bω1 , and SI (a), for a ∈ Bω1 and I ⊆ ω1 , as in 531J and 531R. Let P be the set of pairs (c, I) where I ⊆ C is finite, 0 6= c ⊆ inf ξ∈I aξ and I ∩ J ∗ (c) = ∅. Order P by saying that (c, I) ≤ (c0 , I 0 ) if I ⊆ I 0 and c0 ⊆ c. Then P is a partially ordered set. For each ξ ∈ C, S{ξ} (aξ ) 6= 0, so pξ = (S{ξ} (aξ ), {ξ}) belongs to P . The point of the proof is the following fact. (b) P satisfies Knaster’s condition upwards. P P Let h(cξ , Iξ )iξ<ω1 be a family in P . Then there are an α > 0 and an uncountable A0 ⊆ ω1 such that ν¯ω1 (cξ ∩ cη ) ≥ α for all ξ, η ∈ A0 (525Uc). Next, there is an uncountable A1 ⊆ A0 such that hIξ iξ∈A1 is a ∆-system with root I say (4A1Db); let m ∈ N be such that A2 = {ξ : ξ ∈ A1 , #(Iξ \ I) = m} is uncountable. Finally, because J ∗ (cη ) is countable for each η, and hIξ \ Iiξ∈A2 is disjoint, we can find an uncountable A3 ⊆ A2 such that J ∗ (cη ) ∩ Iξ \ I = ∅ whenever η, ξ ∈ A3 and η < ξ. Take a strictly increasing sequence hηk ik∈N in A3 and a ζ ∈ A3 greater than every ηk . By 531Rb, there is a finite set K ⊆ ω1 such that ν¯ω1 (cζ \ SJ (cζ )) < α whenever J ⊆ ω1 \ K and #(J) = m. Let k ∈ N be such that Iηk \ I does not meet K. Set c0ζ = SIηk \I (cζ ). Then ν¯ω1 (cζ \ c0ζ ) < ν¯ω1 (cζ ∩ cηk ), so c = cηk ∩ c0ζ 6= 0. Set L = Iηk ∪ Iζ . Then J ∗ (cηk ) is disjoint from Iηk and from Iζ \ I, by the choice of A3 , while J ∗ (c0ζ ) ⊆ J ∗ (cζ ) \ Iηk (531R(a-iii)) is also disjoint from L; so J ∗ (c) ⊆ J ∗ (cηk ) ∪ J ∗ (c0ζ ) is disjoint from L. Finally, c ⊆ cηk ∩ cζ ⊆ inf ξ∈Iηk aξ ∩ inf ξ∈Iζ aξ = inf ξ∈L aξ , so (c, L) ∈ P . Now (c, L) dominates both (cηk , Iηk ) and (cζ , Iζ ). What this shows is that if we write Q for {{η, ζ} : η, ζ ∈ A3 , (cη , Iη ) and (cζ , Iζ ) are compatible upwards in P }, then whenever ζ ∈ A3 and M ⊆ A3 ∩ ζ is infinite there is an η ∈ M such that {η, ζ} ∈ M . By 5A1Gb, there is an uncountable A4 ⊆ A3 such that [A4 ]2 ⊆ Q, that is, h(cξ , Iξ )iξ∈A4 is upwards-linked. As h(cξ , Iξ )iξ<ω1 is arbitrary, P satisfies Knaster’s condition upwards. Q Q (c) By 517S, there is a sequence hRn in∈N of upwards-directed subsets of P covering {pξ : ξ ∈ C}; as C is uncountable, there must be some n such that Γ = {ξ : pξ ∈ Rn } has cardinal ω1 . In this case, {pξ : ξ ∈ Γ} is upwards-centered in

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P . If I, J ⊆ Γ are finite and disjoint, then there must be a (c, K) ∈ P which is an upper bound for {pξ : ξ ∈ I ∪ J}; now I ∪ J does not meet K, while c ⊆ inf ξ∈I∪J aξ . But this means that ν¯ω1 (c ∩ inf ξ∈I eξ ∩ inf η∈J (1 \ eη )) = 2−#(I∪J) ν¯ω1 c > 0, and 0 6= c ∩ inf ξ∈I eξ ∩ inf η∈J (1 \ eη ) ⊆ inf ξ∈I (aξ ∩ eξ ) ∩ inf η∈J (aη \ eη ). So we have a set Γ of the kind required. 531T Theorem (Fremlin 97) Suppose that ω ≤ κ < mK . If X is a normal Hausdorff space and κ ∈ MahR (X), then [0, 1]κ is a continuous image of X. proof For κ = 0 this is trivial. If κ ≥ ω2 , then κ is a measure-precaliber of all probability algebras (525Gb), so 531Lb gives the result. If κ = ω1 , let µ be a Maharam-type-homogeneous Radon probability measure on X with Maharam type ω1 , and (A, µ ¯) its measure algebra. Let hdξ iξ<ω1 be a τ -generating stochastically independent family of elements of A of measure 1 . For ξ < ω1 let Eξ ∈ dom µ be such that Eξ• = dξ , and Kξ0 ⊆ Eξ , Kξ00 ⊆ X \ Eξ compact sets of measure greater than 2 1 0 00 • ¯, hdξ iξ<ω1 ) is isomorphic to (Bω1 , ν¯ω1 , heξ iξ<ω1 ), 531S tells us 4 ; set Kξ = Kξ ∪ Kξ and aξ = Kξ in A. Because (A, µ that there is an uncountable set Γ ⊆ ω1 such that T T 0 6= inf ξ∈I (aξ ∩ dξ ) ∩ inf η∈J (aη \ dη ) = (X ∩ ξ∈I Kξ0 ∩ η∈J Kη00 )• whenever T I, J ⊆ Γ are finite. Just as in part (b) of the proof of 531L, it follows that there is a continuous surjection from ξ∈Γ Kξ onto {0, 1}Γ ≡ {0, 1}Γ , and therefore a continuous surjection from X onto [0, 1]ω1 . 531X Basic exercises (a) Show that there is a Hausdorff completely regular quasi-Radon probability space (X, T, Σ, µ) with Maharam type greater than #(X). (Hint: 523Id.) (b) Give an example of a separable Radon measure space with magnitude 2c . (c) Let I k be the split interval (343J, 419L). Show that MahR (I ) = {0, ω}. ˇ (d) Let I be an infinite set, and βI the Stone-Cech compactification of the discrete space I. Show that 2#(I) is the greatest member of MahR (βI). (Hint: 5A4Ja or 515H.) (e) For a topological space X, write MahqR (X) for the set of Maharam types of Maharam-type-homogeneous quasiRadon probability measures on X. (i) Show that κ ≤ w(X) for every κ ∈ MahqR (X). (ii) Show that MahqR (Y ) ⊆ MahqR (X) for every Y ⊆ X. (iii) Show that if Y is another topological space, and neither X nor Y is empty, then MahqR (X × Y ) = MahqR (X) ∪ MahqR (Y ). >(f ) Let X be a Hausdorff topological group carrying Haar measures, and A its Haar measure algebra (442H). Show that w(X) = max(c(A), τ (A)). (Hint: 443Ge, 529Ba.) Show that if X is σ-compact, locally compact, Hausdorff and not discrete then w(X) ∈ MahR (X). (g) Let X be a normal Hausdorff space and κ an infinite cardinal. Suppose that whenever Y is a Hausdorff continuous image of X of weight κ then MahR (Y ) ⊆ κ. Show that MahR (X) ⊆ κ. (h) Let X be a Hausdorff space, and hEiS ii∈I a family S of universally Radon-measurable subsets of X such that #(I) < cov Nκ for every κ. Show that MahR ( i∈I Ei ) = i∈I MahR (Ei ). (i) Let K be an Eberlein compactum. Show that MahR (K) ⊆ {0, ω}. (Hint: 467Xj.) S (j) Let W ⊆ Nω be such that every compact subset of {0, 1}ω \ W isSscattered. Show that there is a family W 0 ⊆ Nω1 such that #(W 0 ) = #(W) and every compact subset of {0, 1}ω1 \ W 0 is scattered. (k) Let (X, T, Σ, µ) be a Hausdorff quasi-Radon probability space. Show that the Maharam type of µ is at most 2χ(X) . (Hint: 5A4Ba, 5A4Bg.) (l) In the language of 531R, show that if a, b ∈ Bκ and I ⊆ κ \ J ∗ (b) is finite, then ν¯κ (a 4 SI (a)) ≤ 2#(I) ν¯κ (a 4 b). (m) Show that if mK > ω1 and X is a countably tight compact Hausdorff space, then ω1 ∈ / MahR (X).

§532 intro.

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(n) Let X be an infinite compact Hausdorff space with a strictly positive Radon measure µ, and P the set of Radon probability measures on X with its narrow topology. Show that the topological density of P is at most the Maharam type of µ. (Hint: the indefinite-integral measures over µ are dense in P .) 531Y Further exercises (a) Let κ be an infinite cardinal such that κ = κc . Show that there is a set X ⊆ {0, 1}κ , of full outer measure for νκ , such that every subset of X of cardinal c is discrete. Show that MahqR (X) (531Xe) contains κ but not ω. (b) Let X and Y be infinite compact Hausdorff spaces, and suppose that the dual spaces C(X)∗ and C(Y )∗ are isomorphic as normed spaces. Show that MahR (X) = MahR (Y ). (c) Let µ be a τ -additive Borel probability measure on a topological space X, and κ an cardinal of uncountable cofinality such that (i) χ(x, X) < cf κ for every x ∈ X (ii) no non-negligible measurable set can be covered by cf κ negligible sets. Show that the Maharam type of µ cannot be κ. 531Z Problem Can there be a countably tight compact Hausdorff space X such that ω2 ∈ MahR (X)? What if X is perfectly normal? What if X is hereditarily separable? 531 Notes and comments This section is directed to Radon measures, studying MahR (X); of course we can look at Maharam types of quasi-Radon measures (531Xe, 531Ya), or Borel or Baire measures for that matter. In the next section I shall have something to say about completion regular measures. The function X 7→ MahR (X) has a much more satisfying list of basic properties (531E, 531G) than the others. From 531L and 531T we see that there are many cardinals κ such that whenever X is a compact Hausdorff space and κ ∈ MahR (X), then there is a continuous function from X onto [0, 1]κ . Such cardinals are said to have Haydon’s property. From 531L, 531M and 531T we see that ω has Haydon’s property (531La); if κ ≥ ω2 and κ is a measure-precaliber of Bκ then κ has Haydon’s property (531Lb); if κ ≥ ω is not a measure-precaliber of Bκ then κ does not have Haydon’s property (531M); if ω1 < mK then ω1 has Haydon’s property (531T). Thus if mK > ω1 , an infinite cardinal κ has Haydon’s property iff it is a measure-precaliber of every probability algebra. ω1 really is different; it is possible that ω1 is a precaliber of every probability algebra but does not have Haydon’s property. To check this, it is enough to find a model of set theory in which cov Nω1 > ω1 but there is a family hWξ iξ<ω1 as in 531N; one is described in 553F. You will observe that all the key arguments of this section depend on analysis of the measure algebras Bκ . We have already seen in §524 that many properties of a Radon measure can be determined from its measure algebra. Here we find that some important topological properties of compact Hausdorff spaces can be determined by the measure algebras of the Radon measures they carry. The results here largely depend for their applications on knowing enough about precalibers; I remind you that it seems to be still unknown whether it is possible that every infinite cardinal is a measure-precaliber of every probability algebra. The remarks above have concerned the existence of continuous surjections onto [0, 1]κ ; a natural place to start, because measures of Maharam type κ arise immediately from such surjections. In 531O-531Q I look at different measures of the richness of a compact space X. Concerning characters, 531O-531P give us quite a lot of information, slightly irregular at the edges. I ought to offer a remark on the context of 531Q. In some set theories (for instance, when m > ω1 ), we find not only that ω1 is a precaliber of every measurable algebra, but also that a compact Hausdorff space is hereditarily separable iff it is hereditarily Lindel¨of (Fremlin 84a, 44H); so that, for instance, a hereditarily separable compact Hausdorff space cannot carry a Radon measure of uncountable Maharam type. Typically, the situation is very different if the continuum hypothesis or ♦ is true, and 531Q is a descendant of the construction in Kunen 81 of a non-separable hereditarily Lindel¨ of compact Hausdorff space. See Dˇ zamonja & Kunen 93 and Kunen & Mill 95 for further exploration of these questions.

532 Completion regular measures on {0, 1}I As I remarked in the introduction to §434, the trouble with topological measure theory is that there are too many questions to ask. In §531 I looked at the problem of determining the possible Maharam types of Radon measures on a Hausdorff space X. But we can ask the same question for any of the other classes of topological measures listed in §411. It turns out that the very narrowly focused topic of completion regular Radon measures on powers of {0, 1} already leads us to some interesting arguments.

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I define the classes MahcrR (X), corresponding to the MahR (X) examined in §531, in 532A. They are less accessible, and I almost immediately specialize to the relation λ ∈ MahcrR ({0, 1}κ ). This at least is more or less convex (532G, 532K), and can be characterized in terms of the measure algebras Bλ (532I). On the way it is helpful to extend the treatment of completion regular measures given in §434 (532D, 532E, 532H). For fixed infinite λ, there is a critical cardinal κ0 ≤ (2λ )+ such that λ ∈ MahcrR ({0, 1}κ ) iff λ ≤ κ < κ0 ; under certain conditions, when λ = ω, we can locate κ0 in terms of the cardinals of Cicho´ n’s diagram (532P, 532Q). This depends on facts about the Lebesgue measure algebra (532M, 532O) which are of independent interest. Finally, for other λ of countable cofinality, the square principle and Chang’s transfer principle are relevant (532R-532S). 532A Definition If X is a topological space, I write Mahcr (X) for the set of Maharam types of Maharam-typehomogeneous completion regular topological probability measures on X. If X is a Hausdorff space, I write MahcrR (X) for the set of Maharam types of Maharam-type-homogeneous completion regular Radon probability measures on X. 532B Proposition Let X be a Hausdorff space. Then a probability algebra (A, µ ¯) is isomorphic to the measure algebra of a completion regular Radon probability measure on X iff (α) τ (Aa ) ∈ MahcrR (X) whenever Aa is a non-zero homogeneous principal ideal of A (β) the number of atoms of A is not greater than the number of points x ∈ X such that {x} is a zero set. proof (a) Suppose that µ is a completion regular Radon probability measure on X and Aa is a non-zero homogeneous principal ideal of its measure algebra A. Let F be such that F • = a and ν the indefinite-integral measure over µ defined by the function

1 χF . µF

Then ν is a Radon measure (416S), inner regular with respect to the zero sets (412Q); and

its measure algebra is isomorphic, up to a scalar multiple, to Aa , so is homogeneous with Maharam type τ (Aa ). So ν witnesses that τ (Aa ) ∈ MahcrR (X). This shows that A satisfies condition (α). As for condition (β), each atom of A is of the form {x}• for some x ∈ X such that µ{x} > 0 (414G, or otherwise). In this case, because µ is completion regular, {x} must be a zero set. So we have at least as many singleton zero sets as we have atoms in A. (b) Now suppose that (A, µ ¯) satisfies the conditions. I copy the argment of 531F. Express (A, µ ¯) as the simple product of a countable family h(Ai , µ ¯0i )ii∈I of non-zero homogeneous measure algebras. For i ∈ I, set κi = τ (Ai ) and γi = µ ¯0i 1Ai . Set J = {i : i ∈ I, κi ≥ ω}. (β) tells us that #(I \ J) is less than or equal to the number of singleton zero sets in X; let hxi ii∈I\J be a family of distinct elements of X such that every {xi } is a zero set. For each i ∈ J, (α) tells us that there is a completion regular Maharam-type-homogeneous Radon probability measure µi on X with Maharam type κi . Now there is a disjoint family hEi ii∈J of Baire subsets of X such that µi Ei > 0 for every i ∈ J. P P We may suppose that J ⊆ N. Choose hEi ii∈N , hFi ii∈N inductively, as follows. F0 = X \ {xi : i ∈ I \ J}. Given that Fi is a Baire set and µj Fi > 0 for every j ∈ J \ i, then if i ∈ / J set Ei = ∅ and Fi+1 = Fi ; otherwise, because µi is atomless and completion regular, we can find, for each j ∈ J such that j > i, a Baire set Gij ⊆ Fi such that S µi Gij < 2−j µi Fi and µj Gij > 0; set Fi+1 = j∈J,j>i Gij and Ei = Fi \ Fi+1 ; continue. Q Q Now set P P µE = i∈I\J,xi ∈E γi + i∈J (µi Ei )−1 γi µi (E ∩ Ei ) whenever E ⊆ X is such that µi measures E ∩ Ei for every i ∈ J. Of course µ is a probability measure. Because every µi is a topological measure, so is µ; because every µi is inner regular with respect to the compact sets, so is µ; because every µi is complete, so is µ; so µ is a Radon measure. Because every subspace measure (µi )Ei is Maharamtype-homogeneous with Maharam type κi , the measure algebra of µ is isomorphic to (A, µ ¯). Because all the {xi } are zero sets and all the µi are completion regular, µ is completion regular. 532C Remarks Nearly the whole of this section will be devoted to the usual measures on powers of {0, 1}. Accordingly the following notation will be useful, as previously in this volume. If I is any set, νI will be the usual measure on {0, 1}I , BI its measure algebra and NI its null ideal. If X is a topological space, B(X) will be its Borel σ-algebra. Let κ be an infinite cardinal. Then νκ is a completion regular Radon probability measure (416U), and Bκ is homogeneous with Maharam type κ. So κ ∈ MahcrR ({0, 1}κ ). Next, any Radon measure on {0, 1}κ can have Maharam type at most w({0, 1}κ ) (531Aa), so λ ≤ κ for every λ ∈ MahcrR ({0, 1}κ ). At the bottom end, 0 ∈ MahcrR ({0, 1}κ ) iff {0, 1}κ has a singleton Gδ set, that is, iff κ = ω. From this we see already that we do not have direct equivalents of any of the results 531Eb-531Ef. However the class {(λ, κ) : λ ∈ MahcrR ({0, 1}κ )} is convex in two senses (532G, 532K). For the first of these, it will be useful to have a result left over from §434. 532D Theorem (Fremlin & Grekas 95) Let (X, µ1 ) and (Y, µ2 ) be effectively locally finite topological measure spaces of which X is quasi-dyadic (definition: 434O), µ1 is completion regular and µ2 is τ -additive. Let µ be the c.l.d. product measure on X × Y as defined in §251. Then µ is a τ -additive topological measure.

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201

proof (a) To begin with (down to the end of (e)) let us suppose that µ1 and µ2 are complete and totally finite and inner regular with respect to the Q Borel sets. Let hXi ii∈I be a family of separable metrizable spaces such that there is a continuous surjection f : i∈I Xi → X. For each i ∈ I, let Ui be a countable base for the Q topology of Xi not containing ∅; for J ⊆ I, let CJ be the family of cylinder sets expressible in the form {z : z ∈ i∈I Xi , z(i) ∈ Ui for every i ∈ K} where K ⊆ J is finite and Ui ∈ Ui for each i ∈ K. (b) ?? Suppose, if possible, that µ is not a topological measure. Let W ⊆ X × Y be a closed set which is not measured by µ. By 434Q, µ1 is τ -additive; by 417C, there is a τ -additive topological measure µ ˜ extending µ, and µ∗ W = µ ˜W (apply 417C(iv) to the complement of W ). (c) If J ⊆ I is countable, there are sets H, V , V 0 such that H ⊆ Y is open, V ∈ CJ , V 0 ∈ CI\J , f [V ∩ V 0 ] × H is disjoint from W , and µ∗ (W ∩ (f [V ] × H)) > 0. P P For V ∈ CJ , set S HV = V 0 ∈CI\J {H : H ⊆ Y is open, W ∩ (f [V ∩ V 0 ] × H) = ∅}, HV =

S

HV ,

and choose a measurable envelope FV of f [V ]. As CJ is countable, S W1 = (X × Y ) \ V ∈CJ FV × HV is measured by µ; also W1 ⊆ W because {f [V ∩ V 0 ] × H : V ∈ CJ , V 0 ∈ CI\J , H ⊆ Y is open} is a network for the topology of X × Y . So µ ˜W1 = µW1 ≤ µ∗ W < µ∗ W = µ ˜W and µ ˜(W \W1 ) > 0. There must therefore be a V ∈ S CJ such that µ ˜(W ∩(FV ×HV )) > 0.SNext, because µ2 is τ -additive, there is a countable H ⊆ HV such that µ2 (HV \ H) = 0, and now µ ˜(W ∩ (FV × H)) = µ ˜(W ∩ (FV × HV )) is non-zero. Accordingly there is an H ∈ H such that µ ˜(W ∩ (FV × H)) > 0. By 417H,

R FV

µ2 (W [{x}] ∩ H)µ1 (dx) = µ ˜(W ∩ (FV × H))

is greater than 0. But this means that µ1 {x : x ∈ FV , µ2 (W [{x}] ∩ H) > 0} > 0. (Recall that we are supposing that µ1 is complete.) So {x : x ∈ f [V ], µ2 (W [{x}] ∩ H) > 0} is not µ1 -negligible, and W ∩ (f [V ] × H) is not µ-negligible. Finally, because H ∈ HV , there is a V 0 ∈ CI\J such that W ∩ (f [V ∩ V 0 ] × H) = ∅. Q Q (d) We may therefore choose inductively families hJξ iξ<ω1 , hHξ iξ<ω1 , hVξ iξ<ω1 , hVξ0 iξ<ω1 in such a way that, for every ξ < ω1 , Jξ is a countable subset of I, Hξ is an open subset of Y , Vξ ∈ CJξ , Vξ0 ∈ CI\Jξ , W ∩ (f [Vξ ∩ Vξ0 ] × Hξ ) = ∅, µ∗ (W ∩ (f [Vξ ] × Hξ )) > 0, S η<ξ Jη ⊆ Jξ , Vξ , Vξ0 ∈ CJξ+1 . For each ξ < ω1 , let Kξ be a finite subset of Jξ+1 such that Vξ and Vξ0 are determined by coordinates in Kξ . By the ∆-system Lemma (4A1Db), there is an uncountable set A ⊆ ω1 such that hKξ iξ∈A is a ∆-system with root K say. Set ζ0 = min A. Express each Vξ as V˜ξ ∩ Vˆξ where V˜ξ ∈ CK and Vˆξ ∈ CKξ \K ; because CK is countable, there is a V˜ such that B = {ξ : ξ ∈ A, ξ > ζ0 , V˜ξ = V˜ } is uncountable. Note that µ∗1 f [V˜ ] > 0, because µ∗1 f [V˜ ] ≥ µ∗ (W ∩ (f [Vξ ] × Hξ )) for any ξ ∈ B. Also K ⊆ Kζ0 ⊆ Jζ0 +1 ⊆ Jξ , Vξ0

is determined by coordinates in Kξ \ Jξ ⊆ Kξ \ K, for every ξ ∈ B. S (e) Set Hξ0 = η∈B\ξ Hη for each ξ < ω1 . Then hHξ0 iξ<ω1 is non-increasing, so there is a ζ < ω1 such that µ2 Hξ0 = µ2 Hζ0 whenever ξ ≥ ζ. Now consider F = {x : µ2 (W [{x}] ∩ Hζ0 ) > 0}. Applying 417H to the characteristic function of W ∩(X ×Hζ0 ), and recalling once more that µ1 is complete, we see that µ1 measures F . Also µ∗1 (F ∩f [V˜ ]) > 0. P P Take any ξ ∈ B \ ζ. Then so

F ∩ f [V˜ ] ⊇ {x : x ∈ f [V˜ξ ], µ2 (W [{x}] ∩ Hξ ) > 0} must be non-µ1 -negligible because W ∩ (f [V˜ξ ] × Hξ ) is not µ-negligible. Q Q

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At this point, recall that we are supposing that µ1 is completion regular. So there is a zero set Z ⊆ F such that µ1 Z > µ1 F − µ∗1 (F ∩ f [V˜ ]), and Z ∩ f [V˜ ] 6= ∅, that is, V˜ ∩ f −1 [Z] is not empty. f −1 [Z] is a zero set (4A2C(b-iv)), so there is a countable set J ⊆ I such that f −1 [Z] is determined by coordinates in J (4A3Nc); we may suppose that K ⊆ J. Because hKη \ Kiη∈A is disjoint, there is a ξ ≥ ζ such that J ∩ Kη = K for every η ∈ A \ ξ. Q Take any w ∈ V˜ ∩ f −1 [Z] and modify it to produce w0 ∈ i∈I Xi such that w0 J = wJ and w0 ∈ Vˆη ∩ Vη0 for every η ∈ B \ ξ; this is possible because Vˆη ∩ Vη0 is determined by coordinates in Kη \ K for each η, and J and the Kη \ K are disjoint. Set x = f (w0 ); then x ∈ Z ⊆ F , so µ2 (W [{x}] ∩ Hζ0 ) > 0. If η ∈ B \ξ, then w0 ∈ V˜ , because w ∈ V˜ and V˜ is determined by coordinates in K ⊆ J; so w0 ∈ V˜ ∩ Vˆη ∩Vη0 = Vη ∩Vη0 . Accordingly x ∈ f [Vη ∩ Vη0 ]; as W ∩ (f [Vη ∩ Vη0 ] × Hη ) = ∅, W [{x}] does not meet Hη . As η is arbitrary, W [{x}] does not meet Hξ0 and W [{x}] ∩ Hζ0 is µ2 -negligible. But this is impossible. X X (f ) This contradiction shows that µ will be a topological measure, at least if µ1 and µ2 are complete, totally finite and inner regular with respect to the Borel sets. Now suppose just that µ1 and µ2 are totally finite. Let µ01 and µ02 be the completions of the Borel measures µ1 B(X) and µ2 B(Y ), and µ0 their c.l.d. product. Then µ1 B(X) and µ01 are completion regular topological measures, while µ2 B(Y ) and µ02 are τ -additive. So (a)-(e) tell us that µ0 measures every open set. Now the completions µ ˆ1 , µ ˆ2 extend µ01 and µ02 , and µ is the c.l.d. product of µ ˆ1 and µ ˆ2 (251T), so µ 0 1 extends µ (251L ). Thus we again have a topological product measure µ. (g) In the general case, let W ⊆ X × Y be an open set, E ⊆ X a zero set of finite measure, and F ⊆ Y any set of finite measure. Then µ measures W ∩(E ×F ). P P Let (µ1 )E and (µ2 )F be the subspace measures. Then both are totally finite topological measures, (µ1 )E is inner regular with respect to the zero sets (412Pd), E is quasi-dyadic (434Pc), and (µ2 )F is τ -additive (414K). So the product (µ1 )E × (µ2 )F is a topological measure and measures W ∩ (E × F ). By 251Q, µ measures W ∩ (E × F ). Q Q Let K be the family of zero sets of finite measure in X, L the family of Borel sets of finite measure in Y , and M the family of sets M ⊆ X × Y such that µ measures W ∩ M . Because µ1 is inner regular with respect to K, µ2 is inner regular with respect to L, E × F ∈ M for every E ∈ K and F ∈ M, and M is a σ-algebra of sets, 412R tells us that µ is inner regular with respect to M. As µ is complete and locally determined, it must measure W (412Ja). As W is arbitrary, µ is a topological measure. (h) Finally, as noted in (b), µ1 is τ -additive and there is a τ -additive topological measure µ ˜ on X × Y extending µ. (434Q and 417C still apply.) So µ too must be τ -additive. 532E Corollary Let hXi ii∈I be a family of regular spaces with countable networks, and Y any topological space. Suppose that we are given a strictly positive topological probability measure µiQon each Xi , and a τ -additive topological probability measure ν on Y . Let µ be the ordinary product measure on Z = i∈I Xi × Y . (a) µ is a topological measure. (b) µ is τ -additive. (c) If ν is completion regular, so is µ. proof (a) For each i, let µ0i be the completion of the Borel measure µi B(Xi ). Then µ0i is a completion regular quasiQ Radon measure (415C). By 434Pb, i∈I Xi is quasi-dyadic. The product ν1 of the µ0i is a topological measure (453I) and inner regular with respect to the zero sets (412Ub); so the product µ0 of ν1 and ν is a topological measure, by 532D. Now µ0 is also the product of the measures µi B(Xi ) and ν, so µ extends µ0 and µ also is a topological measure. (b) Because every µi is τ -additive, as is ν, 417E tells us that there is a τ -additive measure extending µ, so µ itself must be τ -additive. (c) By 412Ub again, µ is inner regular with respect to the zero sets, so is completion regular. 532F Corollary Let h(Xi , µi )ii∈I be a family of quasi-dyadic compact Hausdorff spaces with strictly positive Q completion regular Radon measures. Then the ordinary product measure λ on i∈I Xi is a completion regular Radon measure. Q proof By 532D, the ordinary productQmeasure on i∈J Xi is a topological measure, for every finite J ⊆ I. By 417T, λ is the τ -additive product measure on i∈I Xi , which by 417Q is a Radon measure. By 412Ub, λ is completion regular. 532G Proposition Suppose that λ, λ0 and κ are cardinals such that max(ω, λ) ≤ λ0 ≤ κ and λ ∈ MahcrR ({0, 1}κ ). Then λ0 ∈ MahcrR ({0, 1}κ ). 1 Later

editions only.

532I

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203

proof Let ν be a completion regular Maharam-type-homogeneous Radon probability measure on {0, 1}κ with Maharam 0 type λ, and consider the ordinary product measure µ of νλ0 and ν on X = {0, 1}λ × {0, 1}κ . Applying 532E with κ 0 Y = {0, 1} and Xξ = {0, 1} for ξ < λ , we see that µ is a completion regular topological measure, therefore a Radon measure. By 334A, the Maharam type of µ is at most max(ω, λ0 , λ) = λ0 , so the measure algebra (A, µ ¯) of µ can λ0 0 be embedded in Bλ . At the same time, the inverse-measure-preserving projection from X onto {0, 1} induces a measure-preserving embedding of Bλ0 into A. By 332Q, (A, µ ¯) and (Bλ0 , ν¯λ0 ) are isomorphic, that is, µ is Maharam0 type-homogeneous with Maharam type λ . So µ witnesses that λ0 ∈ MahcrR (X) = MahcrR ({0, 1}κ ). 532H Lemma Let hXi ii∈I Q be a family of separable metrizable spaces, and µ a totally finite completion regular topological measure on X = i∈I Xi . Then (a) the support of µ is a zero set; (b) µ is inner regular with respect to the self-supporting zero sets. proof (a) Recall from 434Q that µ is τ -additive, so has a support Z. Let hKn in∈N be a sequence of zero sets such that Kn ⊆ Z and µKn ≥S µX − 2−n for each n. S Then there is a countable set J ⊆ I such that every Kn is determined by coordinates in J. So n∈N Kn and Z 0 = n∈N Kn are determined by coordinates in J, and Z 0 is a zero set. But Z 0 ⊆ Z and µZ 0 = µZ so Z = Z 0 is a zero set. (b) If µE > γ then there is a zero set K ⊆ E such that µK ≥ γ. Now µ K is a totally finite completion regular topological measure on X so its support Z is a zero set, by (a); and Z ⊆ E is self-supporting for µ with µZ ≥ γ. 532I There is a useful general characterization of the sets MahcrR ({0, 1}κ ) in terms of the measure algebras Bλ . At the same time, we can check that other products of separable metrizable spaces follow powers of {0, 1}, as follows. Theorem (Choksi & Fremlin 79) Let λ ≤ κ be infinite cardinals. Then the following are equiveridical: (i) λ ∈ MahcrR ({0, 1}κ ); Q (ii) there is a family hXξ iξ<κ of non-singleton separable metrizable spaces such that λ ∈ Mahcr ( ξ<κ Xξ ); (iii) there is a Boolean-independent family hbξ iξ<κ in Bλ with the following property: for every a ∈ Bλ there is a countable set J ⊆ κ such that the subalgebras generated by {a} ∪ {bξ : ξ ∈ J} and {bη : η ∈ κ \ J} are Booleanindependent. proof If κ = ω then λ = ω and (i)-(iii) are all true. So we may assume that κ is uncountable. (i)⇒(ii) is trivial. Q (ii)⇒(iii) If λ ∈ Mahcr (X), where every Xξ is a non-trivial separable metrizable space and X = ξ<κ Xξ , let µ be a Maharam-type-homogeneous completion regular topological probability measure on X with Maharam type λ. By 532Ha and 4A3Nc, the support Z of µ is determined by coordinates in a countable subset L of κ. Let A be the measure algebra of µ. For each ξ < κ, let fξ : Xξ → [0, 1] be a continuous function taking both values 0 and 1; let tξ ∈ ]0, 1[ be such that µ{x : x ∈ X, fξ (x(ξ)) = tξ } = 0. Set Uξ = {x : fξ (x(ξ)) < tξ }, Vξ = {x : fξ (x) > tξ }; then Uξ and Vξ are disjoint non-empty open sets in X, both determined by coordinates in {ξ}, and µ(Uξ ∪Vξ ) = 1. Set bξ = Uξ• in A. Then hbξ iξ<κ\L is Boolean-independent. P P If I, I 0 ⊆ κ\L are disjoint finite sets, T T then H = X ∩ ξ∈I Uξ ∩ ξ∈I 0 Vξ is a non-empty open set in X. As H is determined by coordinates in I ∪ I 0 and Z is determined by coordinates in L, Z ∩ H is non-empty and therefore non-negligible; so µH > 0 and inf ξ∈I bξ \ supξ∈I 0 bξ is non-zero in A. Q Q If a ∈ A let E be such that E • = a. By 532Hb, we can choose for each n ∈ N self-supporting zero sets Kn ⊆ E, ˜ n ⊆ X \ E such that µKn + µK ˜ n ≥ 1 − 2−n . Let J ⊆ κ \ L be a countable set such that every Kn and every K ˜ n is K determined by coordinates in J ∪L. Now the subalgebras D1 , D2 generated by {a}∪{bξ : ξ ∈ J} and {bξ : ξ ∈ (κ\L)\J} are Boolean-independent. P P Take non-zero d1 ∈ D1 and d2 ∈ D2 . Suppose for the moment that d1 ∩ a 6= 0. Because 1 \ bξ = Vξ• for each ξ, there is an open set G, determined by coordinates in J, such that 0 6= a ∩ G• ⊆ d1 . There is also an open set H, determined by coordinates in κ \ (J ∪ L), such that 0 6= H • ⊆ d2 . Next, as a = supn∈N Kn• , there is an n ∈ N such that 0 6= Kn• ∩ G• , that is, Kn ∩ G 6= ∅. As Kn ∩ G is determined by coordinates in J ∪ L and H is determined by coordinates in κ \ (J ∪ L), Kn ∩ G ∩ V 6= ∅; as Kn is self-supporting, 0 6= (Kn ∩ G ∩ H)• ⊆ d1 ∩ d2 . In the same way, using Kn0 in place of Kn , we see that d1 ∩ d2 6= 0 if d1 \ a 6= 0. As d1 and d2 are arbitrary, D1 and D2 are Boolean-independent. Q Q As #(κ \ L) = κ and A ∼ = Bλ , hbξ iξ∈κ\L , suitably reinterpreted, witnesses that (iii) is satisfied. (iii)⇒(i) Now suppose that the conditions of (iii) are satisfied. Let (Z, ν) be the Stone space of (Bλ , ν¯λ ). (See 411P for a summary of the properties of these spaces.) For b ∈ Bλ write bb for the corresponding open-and-closed subset of Z. Define φ : Z → {0, 1}κ by setting φ(z) = hχbbξ (z)iξ<κ for z ∈ Z. Then φ is continuous; let µ = νφ−1 be the image Radon

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measure on {0, 1}κ (418I). Now µ is completion regular. P P Suppose that K ⊆ {0, 1}κ is compact and self-supporting. Identifying Bλ with the measure algebra of ν, we have a Boolean homomorphism ψ : dom µ → Bλ defined by setting ψE = (φ−1 [E])• whenever µ measures E, and ν¯λ ψE = νφ−1 [E] = µE for every E; setting Eξ = {x : x ∈ {0, 1}κ , x(ξ) = 1}, ψEξ = bξ . Set a = ψK. Let J ⊆ κ be a countable set such that the subalgebras D1 , D2 generated by {a} ∪ {bξ : ξ ∈ J} and {bη : η ∈ κ \ J} are Boolean-independent. ?? If x ∈ K, y ∈ {0, 1}κ \ K and xJ = yJ, let U be an open cylinder containing y and disjoint from K. Express U as U 0 ∩ U 00 where U 0 is determined by coordinates in J and U 00 by coordinates in κ \ J. Then ψU 0 ∈ D1 and ψU 00 ∈ D2 . As hbξ iξ<κ is Boolean-independent, ψU 00 6= 0. Now K is self-supporting and x ∈ K ∩ U 0 , so µ(K ∩ U 0 ) > 0 and ψ(K ∩ U 0 ) = a ∩ ψU 0 is non-zero; also a ∩ ψU 0 ∈ D1 ; because D1 and D2 are Boolean-independent, ψ(K ∩ U ) = a ∩ ψU 0 ∩ ψU 00 6= 0 and K ∩ U cannot be empty, contrary to the choice of U . X X This shows that K is determined by coordinates in J and is a zero set (4A3Nc, in the other direction). As K is arbitrary, we see that all self-supporting compact sets are zero sets. But as µ is a Radon measure, it is inner regular with respect to the self-supporting compact sets, therefore with respect to the zero sets, and is completion regular. Q Q The inverse-measure-preserving function φ (and, of course, the Boolean homomorphism ψ) correspond to an embedding of the measure algebra of µ into Bλ . So the Maharam type of µ is at most λ. There is therefore a λ0 ∈ MahcrR ({0, 1}κ ) such that λ0 ≤ λ (532B). By 532G, λ ∈ MahcrR ({0, 1}κ ). 532J Corollary (a) Suppose that λ, κ are infinite cardinals and λ ∈ MahcrR ({0, 1}κ ). Then κ is at most the cardinal power λω . (b) If κ is an infinite cardinal such that λω < κ for every λ < κ (e.g., κ = c+ ), then MahcrR ({0, 1}κ ) = {κ}. proof (a) By 532I, κ ≤ #(Bλ ); by 524Ma, #(Bλ ) ≤ λω . (b) By (a), no infinite cardinal less than κ can belong to MahcrR ({0, 1}κ ). Also κ is uncountable, so the remarks in 532C tell us the rest of what we need. 0

532K Corollary If ω ≤ λ ≤ κ0 ≤ κ and λ ∈ MahcrR ({0, 1}κ ) then λ ∈ MahcrR ({0, 1}κ ). proof If hbξ iξ<κ witnesses the truth of 532I(iii) for λ and κ, then its subfamily hbξ iξ<κ0 witnesses the truth of 532I(iii) for λ and κ0 . 532L Corollary If ω ≤ λ ≤ λ0 and cf[λ0 ]≤λ < cf κ and λ0 ∈ MahcrR ({0, 1}κ ), then λ ∈ MahcrR ({0, 1}κ ). proof Let hbξ iξ<κ be a family in Bλ0 satisfying (iii) of 532I. Let heη iη<λ0 be the standard generating family in Bλ0 (525A), and let J be a cofinal subset of [λ0 ]λ of cardinal less than cf κ. For each ξ < κ, there are a countable set L ⊆ λ0 such that bξ belongs to the closed subalgebra of Bλ0 generated by {eη : η ∈ L}, and a Jξ ∈ J such that L ⊆ Jξ . Because #(J) < cf κ, there is a J ∈ J such that A = {ξ : ξ < κ, Jξ = J} has cardinal κ. Now the closed subalgebra B of Bλ0 generated by {eη : η ∈ J} is isomorphic to Bλ , and hbξ iξ∈A witnesses that 532I(iii) is true of λ and κ. 532M I turn now to the question of identifying those κ for which ω ∈ MahcrR ({0, 1}κ ). We know from 532C and 532Ja that they all lie between ω and c. To go farther we need to look at some of the cardinals from §522. Proposition If A ⊆ Bω \ {0} and #(A) < d = cf(N N ), then there is a c ∈ Bω such that neither c nor 1 \ c includes any member of A. proof Let hen in∈N be the standard generating family in Bω . For each a ∈ A, n ∈ N let fa (n) ∈ N be such that there is a b in the subalgebra generated by {bi : i < fa (n)2 } such that ν¯ω (b 4 a) < 2−n−3 µ ¯a. Because #(A) < d, there is an f ∈ N N such that f 6≤ fa for every a ∈ A; we may suppose that f is strictly increasing and f (0) > 0. Note that f (n)2 + n + 1 < f (n)2 + f (n) + 1 ≤ (f (n) + 1)2 ≤ f (n + 1)2 for every n. For each n ∈ N, set cn = inf f (n)2 ≤i≤f (n)2 +n+1 bi ; then hcn in∈N is stochastically independent and ν¯ω cn = 2−n−2 for each n. Define c0n , for n ∈ N, by setting c00 = 0 and c0n+1 = c0n 4 cn for each n. Then ν¯N (c0n+1 4 c0n ) = 2−n−2 for every n, so hc0n in∈N is a Cauchy sequence for the measure metric on Bω , and has a limit c. Note that Pm−1 −i−3 Pm−1 ≤ ν¯ω (c0m 4 c0n ) ≤ i=n 2−i−2 i=n 2 Pn−1 whenever m ≥ n. P P Induce on m. For m = n the result is trivial (interpreting i=n as zero). For the inductive step to m + 1, cm = c0m+1 4 c0m is stochastically independent of c0m 4 c0n , so

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ν¯ω (c0m+1 4 c0n ) = ν¯ω (cm 4 c0m 4 c0n ) = ν¯ω cm + ν¯ω (c0m 4 c0n ) − 2¯ νω (cm ∩ (c0m 4 c0n )) = 2−m−2 + (1 − 2−m−1 )¯ νω (c0m 4 c0n ) ≥ 2−m−2 + (1 − 2−m−1 )

m−1 X

2−i−3

i=n

=

m−1 X

2−i−3 + 2−m−3 (2 − 4

i=n

m−1 X

2−i−3 ) ≥

i=n

m X

2−i−3 ;

i=n

on the other hand, ν¯ω (c0m+1 4 c0n ) ≤ 2−m−2 + ν¯ω (c0m 4 c0n ) ≤

Pm

i=n

2−i−2 .

So the induction proceeds. Q Q Taking the limit, we see that 2−n−2 ≤ ν¯ω (c 4 c0n ) ≤ 2−n−1 for every n ∈ N. Take any a ∈ A. Let n ∈ N be such that fa (n) < f (n). Then there is a b in the algebra generated by {bi : i < f (n)2 } such that ν¯ω (a 4 b) < 2−n−3 µ ¯a. Now c 4 c0n is stochastically independent of both b \ c0n and b ∩ c0n , so ν¯ω (b \ c) = ν¯ω (b \ c0n )(1 − ν¯ω (c 4 c0n )) + ν¯ω (b ∩ c0n ) · ν¯ω (c 4 c0n ) ≥ ν¯ω (b \ c0n )(1 − 2−n−1 ) + 2−n−2 ν¯ω (b ∩ c0n ) ≥ 2−n−2 ν¯ω b ≥ 2−n−3 ν¯ω a. So ν¯ω (a \ c) ≥ 2n−3 ν¯ω a − ν¯ω (b \ a) > 0, and a 6⊆ c. Similarly, ν¯ω (b ∩ c) = ν¯ω (b ∩ c0n )(1 − ν¯ω (c 4 c0n )) + ν¯ω (b \ c0n ) · ν¯ω (c 4 c0n ) ≥ ν¯ω (b ∩ c0n )(1 − 2−n−1 ) + 2−n−2 ν¯ω (b \ c0n ) ≥ 2−n−2 ν¯ω b, and ν¯ω (a ∩ c) > 0. As a is arbitrary, we have found an appropriate c. 532N It will be useful to have a classic example relevant to a question already examined in 325F. Lemma There is a Borel set W ⊆ {0, 1}N × {0, 1}N such that whenever E, F ⊆ {0, 1}N have positive measure for νω then neither (E × F ) ∩ W nor (E × F ) \ W is negligible for the product measure νω2 on {0, 1}N × {0, 1}N . proof (a) (Cf. 134Jb.) There is a Borel set H ⊆ {0, 1}N such that both H and its complement meet every non-empty set open in a set of non-zero measure. P P For x ∈ {0, 1}N set Ix = {n : x(i) = 0 for n ≤ i < 2n}. Set H = {x : Ix is finite and not empty and max Ix is even}. Q Q (b) Let + be the usual group operation on {0, 1}N ≡ ZN 2 . In this group, addition and subtraction are identical, as x + x = 0 for every x; but the formulae may be easier to read if I use the symbol − when it seems appropriate. Set W = {(x, y) : x, y ∈ {0, 1}N , x − y ∈ H}. Let E, F ⊆ {0, 1}N be sets of positive measure. Then {z : z ∈ {0, 1}N , νω (E ∩ (F + z)) > 0} is open (443C) and not empty (443Da), so meets H in a set of positive measure. Now νω2 ((E × F ) ∩ W ) = νω2 {(x, y) : x ∈ E, y ∈ F, x − y ∈ H} = νω2 {(x, z) : x ∈ E, x − z ∈ F, z ∈ H} (because (x, y) 7→ (x, x − y) is a measure space automorphism for νω2 , as in 255Ae or 443Xa) = νω2 {(x, z) : x ∈ E, x ∈ F + z, z ∈ H} Z = νω (E ∩ (F + z))νω (dz) > 0. H

Applying the same argument with {0, 1}N \ H in the place of H, we see that the same is true of (E × F ) \ W . 532O Proposition If A ⊆ Bω \ {0} and #(A) < cov Nω , then there is a c ∈ Bω such that neither c nor 1 \ c includes any member of A.

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proof Take W ⊆ {0, 1}N × {0, 1}N as in 532N. For x ∈ {0, 1}N , set cx = W [{x}]• in Bω . If a ∈ A, then {x : a ⊆ cx } is negligible. P P Let F ⊆ {0, 1}N be such that F • = a, and set E = {x : a ⊆ cx }. Because x 7→ cx is measurable when Bω is given its measure-algebra topology (418Ta), E is measurable. For every x ∈ E, F \ W [{x}] is negligible, so (E × F ) \ W is negligible. But this means that at least one of E, F must be negligible; since F • = a 6= 0, νω E = 0, as required. Q Q Similarly, {x : a ∩ cx = 0} is negligible. Since {0, 1}N cannot be covered by #(A) negligible sets, there is an x ∈ {0, 1}N such that cx neither includes, nor is disjoint from, any member of A. 532P Proposition Set κ = max(d, cov Nω ). If FN(PN) = ω1 , then ω ∈ MahcrR ({0, 1}κ ). In particular, if c = ω1 then ω ∈ MahcrR ({0, 1}ω1 ). proof (a) By 524O(b-ii), FN(Bω ) = ω1 ; let f : Bω → [Bω ]≤ω be a Freese-Nation function. By 532M (if κ = d) or 532O (if κ = cov Nω ), we can choose inductively a family hbξ iξ<κ in Bω such that neither bξ nor 1 \ bξ includes any nonzero member of Cξ , where Cξ is the smallest subalgebra of Bω including {bη : η < ξ} and closed under f . Of course this implies that hbξ iξ<κ is Boolean-independent. (b) For K, L ⊆ κ set dKL = inf ξ∈K bξ \ supξ∈L bξ . For a ∈ Bω , set Qa = {(K, L) : K, L ∈ [κ]<ω are disjoint, dKL ⊆ a}, and let Q0a be the set of minimal members of Qa , taking (K, L) ≤ (K 0 , L0 ) if K ⊆ K 0 and L ⊆ L0 . Of course Qa is well-founded so Q0a is coinitial with Qa . Now Ran = {(K, L) : (K, L) ∈ Q0a , #(K ∪ L) = n} is countable for every n ∈ N and a ∈ Bω . P P Induce on n. If n = 0 this is trivial. For the inductive step to n + 1, set Rζ0 = {(K, L) : K ∪ L ⊆ ζ, (K ∪ {ζ}, L) ∈ Ra,n+1 } for each ζ < κ. For (K, L) ∈ Rζ0 , bζ ∩ dKL = dK∪{ζ},L is included in a, so there is a cKLζ ∈ f (dK∪{ζ},L ) ∩ f (a) such that dK∪{ζ},L ⊆ cKLζ ⊆ a, in which case bζ ⊆ cKLζ ∪ (1 \ dKL ). If ζ < ζ 0 < κ, (K, L) ∈ Rζ0 and (K 0 , L0 ) ∈ Rζ0 0 , then dK 0 L0 6⊆ a (because (K 0 ∪ {ζ 0 }, L0 ) is a minimal member of Qa ), so cKLζ ∪ (1 \ dK 0 L0 ) 6= 1; as cKLζ and dK 0 L0 both belong to Cζ 0 , bζ 0 ⊆ 6 cKLζ ∪ (1 \ dK 0 L0 ) and cKLζ 6= cK 0 L0 ζ 0 . As f (a) is countable, A = {ζ : Rζ0 6= ∅} is countable. Next, for any ζ ∈ A and (K, L) ∈ Rζ0 , we see that dKL ⊆ a ∪ (1 \ bζ ), and indeed that (K, L) ∈ Q0a ∪ (1 \ bζ ) , so that (K, L) ∈ Ra ∪ (1 \ bζ ),n . By the inductive hypothesis, Rζ0 is countable. This shows that {(K, L, ζ) : K ∪ L ⊆ ζ, (K ∪ {ζ}, L) ∈ Ra,n+1 } is countable. In the same way, applying the ideas above to 1 \ bζ in place of bζ , {(K, L, ζ) : K ∪ L ⊆ ζ, (K, L ∪ {ζ}) ∈ Ra,n+1 } is countable; so Ra,n+1 is countable and the induction proceeds. Q Q It follows that Q0a is countable for every a ∈ Bω . (c) Now take any a ∈ Bω and let J ⊆ κ be a countable set such that K ∪ L ⊆ J whenever (K, L) ∈ Q0a ∪ Q01 \ a . ?? Suppose, if possible, that the algebras D1 , D2 generated by {a} ∪ {bξ : ξ ∈ J} and {bη : η ∈ κ \ J} are not Booleanindependent. Then there must be finite subsets K, L, K 0 and L0 of κ such that K ∪ L ⊆ J, K 0 ∪ L0 ⊆ κ \ J, dK 0 L0 6= 0, and either dKL ∩ a 6= 0, dK 0 L0 ∩ dKL ∩ a = 0 or dKL \ a 6= 0, dK 0 L0 ∩ dKL \ a = 0. 0

Suppose the former. Then (K ∪ K , L ∪ L0 ) ∈ Q1 \ a so there is a (K 00 , L00 ) ∈ Q01 \ a such that K 00 ⊆ K ∪ K 0 and L00 ⊆ L ∪ L0 ; in which case K 00 ∪ L00 ⊆ J so in fact K 00 ⊆ K, L00 ⊆ L and dKL ∩ a ⊆ dK 00 L00 ∩ a = 0, which is impossible. Replacing a by 1 \ a we get a similar contradiction in the second case. X X So D1 and D2 are Boolean-independent. (d) As a is arbitrary, (c) shows that hbξ iξ<κ satisfies the conditions of 532I(iii), so that ω belongs to MahcrR ({0, 1}κ ), as claimed. 532Q Proposition Suppose that add Nω > ω1 . (a) λ ∈ / MahcrR ({0, 1}κ ) whenever λ ≥ ω and cf[λ]≤ω < κ. (b) If ω1 ≤ κ ≤ ωω then MahcrR ({0, 1}κ ) = {κ}. 0

proof (a) ?? If ω ≤ λ, cf[λ]≤ω < κ and λ ∈ MahcrR ({0, 1}κ ), set κ0 = (cf[λ]≤ω )+ ; then λ ∈ MahcrR ({0, 1}κ ) (532K). + As cf[λ]≤ω < cf κ0 , ω belongs to MahcrR ({0, 1}λ ) (532L) and therefore to MahcrR ({0, 1}ω1 ) (532K again). Let hbξ iξ<ω1 be a family in Bω satisfying the conditions of 532I(iii). By 524Mb, ω1 < wdistr(Bω ); by 514K, there is a countable C ⊆ Bω \ {0} such that for every ξ < ω1 there is a c ∈ C such that c ⊆ bξ . Let a ∈ C be such that {ξ : ξ < ω1 , c ⊆ bξ } is uncountable. There is supposed to be a countable J ⊆ ω1 such that the subalgebras generated by {a} and {bξ : ξ ∈ ω1 \ J} are Boolean-independent; but there must be a ξ ∈ C \ J, in which case 0 6= a ⊆ bξ , which is impossible. X X This shows that (a) is true. (b) If ω ≤ λ < κ ≤ ωω , then cf[λ]≤ω ≤ λ < κ (5A1E(e-iv)), so (a) tells us that λ ∈ / MahcrR ({0, 1}κ ). From 532C we see that MahcrR ({0, 1}κ ) must be {κ} exactly.

532Xd

Completion regular measures on {0, 1}I

207

532R Two combinatorial principles already used in 524O are relevant to the questions treated here. Proposition Suppose that λ is an uncountable cardinal with countable cofinality such that λ (definition: 5A1S) is true. Set κ = λ+ . Then λ ∈ MahcrR ({0, 1}κ ). proof (a) Let hIξ iξ<κ be a family of countably infinite subsets of λ as in 5A1T. For each ξ < κ, let hIξn in∈N , hαξn in∈N be such that hIξn in∈N S is a disjoint sequence of subsets of Iξ with #(Iξn ) = n for each n and hαξn in∈N is a sequence of distinct points in Iξ \ n∈N Iξn . Set Uξn = {x : x ∈ {0, 1}λ , x(η) = 0 for every η ∈ Iξn }, S Vξn = {x : x ∈ Uξn \ m>n Uξm , x(αξn ) = 1}, S V˜ξn = {x : x ∈ Uξn \ m>n Uξm , x(αξn ) = 0} for n ∈ N. Note that S as νκ Uξm = 2−m for each n, Vξn and V˜ξn are non-negligible, while both are determined by coordinates in {αξn } ∪ m≥n Iξm ⊆ Iξ . Set S Fξ = n∈N Vξn , bξ = Fξ• ∈ Bλ . Note that Fξ ∩ V˜ξn = ∅. (b) Take any a ∈ Bλ . Then we can express a as E • where E ⊆ {0, 1}λ is a Baire set; let I ⊆ λ be a countable set such that E is determined by coordinates in I. By the choice of hIξ iξ<κ there is a countable set J ⊆ κ such that I ∩ Iξ is finite for every ξ ∈ κ \ J. Let D1 , D2 be the subalgebras of Bλ generated by {a} ∪ {bξ : ξ ∈ J} and {bξ : ξ ∈ κ \ J} respectively. Then D1 and D2 are Boolean-independent. P P If d1 ∈ D1 and d2S∈ D2 are non-zero, we can express d1 as H1• where H1 ⊆ {0, 1}λ is a Baire set determined by coordinates in L = I ∪ ξ∈K Iξ for some finite K ⊆ J. Next, we can find disjoint finite sets K 0 , K 00 ⊆ κ \ J such that d2 ⊇ inf ξ∈K 0 bξ \ supξ∈K 00 bξ . Because all the sets Iξ ∩ Iη , for distinct ξ, η < κ, and also the sets I ∩ Iξ , for ξ ∈ κ \ J, are finite, there is an m ∈ N such that all the S sets Jξ = {αξm } ∪ n≥m Iξn , for ξ ∈ K 0 ∪ K 00 , are disjoint from each other and from I. Look at the sets sets Vξm , for T T ξ ∈ K 0 , and V˜ξm , for ξ ∈ K 00 . Set H2 = {0, 1}λ ∩ ξ∈K 0 Vξm ∩ ξ∈K 00 V˜ξm . Then H2• ⊆ d2 . But observe now that all the Vξm and V˜ξm are non-negligible and that Vξm , V˜ξm are determined by coordinates in Jξ for each ξ ∈ K 0 ∪ K 00 . So the sets H1 , Vξm (for ξ ∈ K 0 ) and V˜ξm (for ξ ∈ K 00 ) are stochastically independent, and Q Q ν¯λ (d1 ∩ d2 ) ≥ νλ (H1 ∩ H2 ) = νλ H1 · ξ∈K 0 νλ Vξm · ξ∈K 00 νλ V˜ξm > 0. Thus d1 ∩ d2 6= 0; as d1 and d2 are arbitrary, D1 and D2 are stochastically independent. Q Q (c) The argument of (b) works equally well with J an arbitrary finite subset of κ and a any member of the subalgebra generated by {bξ : ξ ∈ J} to show that hbξ iξ<κ is Boolean-independent. So the conditions of 532I(iii) are satisfied and κ ∈ MahcrR (λ), as claimed. 532S Proposition Suppose that add Nω > ω1 and that λ is an infinite cardinal such that CTP(λ+ , λ) (definition: 5A1U) is true. Then λ ∈ / MahcrR ({0, 1}κ ) for any κ > λ. proof By 532K, it is enough to consider the case κ = λ+ . ?? Suppose, if possible, that there is a family hbξ iξ<κ in Bλ satisfying the conditions of 532I(iii). Let heη iη<λ be the standard generating family in Bλ . Then for each ξ < κ we have a countable set Iξ ⊆ λ such that bξ belongs to the closed subalgebra of Bλ generated by {eη : η ∈ Iξ }. Because S CTP(κ, λ) is true, there is an uncountable set A ⊆ κ such that J = ξ∈A Iξ is countable (5A1U(b-ii)). Now the closed subalgebra CJ generated by {eη : η ∈ J} is isomorphic to Bω , so hbξ iξ∈A witnesses that ω ∈ MahcrR ({0, 1}ω1 ); but this contradicts 532Qa. X X 532X Basic exercises (a) Let X be a normal Hausdorff space and Y ⊆ X a zero set. Show that MahcrR (Y ) ⊆ MahcrR (X). ˇ (b) Let βN be the Stone-Cech compactification of N. (i) Show that MahcrR (βN) = {0}. (ii) Give an example of a non-empty compact Hausdorff space X such that MahcrR (X) = ∅. (c) Let X and Y be compact Hausdorff spaces. Show that MahcrR (X × Y ) ⊆ MahcrR (X) ∪ MahcrR (Y ). (Hint: 454T.) (d) Let λ and κ be infinite cardinals such that λ ∈ MahcrR ({0, 1}κ ). (i) Show that there is a strictly positive Maharam-type-homogeneous completion regular Radon probability measure on {0, 1}κ with Maharam type λ. (ii) Suppose that λ is uncountable and that H ⊆ {0, 1}κ is a non-empty Gδ set. Show that λ ∈ MahcrR (H).

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(e) Find a proof of 532E which does not rely on 532D. (Hint: 415E.) (f ) Let h(Xi , µi )ii∈I be a family of quasi-dyadic spaces with Q strictly positive completion regular topological probability measures. Show that the ordinary product measure on i∈I Xi is a strictly positive completion regular τ -additive topological probability measure. 532Y Further exercises (a) Let Z be the Stone space of Bλ , where λ ≥ ω. (i) Show that if F ⊆ Z is a non-empty nowhere dense zero set then it is not ccc. (ii) Show that MahcrR (Z) = {λ}. (iii) Show that MahcrR (Z × Z) = ∅. (b) Let hXi ii∈I be a family of topological spaces with countable networks, and Y any topological space. Suppose that we are given a strictly positive topological probability measureQ µi on each Xi , and a τ -additive topological probability measure ν on Y . Show that the ordinary product measure on i∈I Xi × Y is a topological measure. (c) Suppose that FN(PN) = ω1 . Show that there are a Hausdorff space X and a completion regular Radon measure µ on X such that the Maharam type of µ is ω, but the Maharam type of µB(X) is ω1 . (Hint: 532P, 419C.) 532Z Problems (a) In 532P, can we take κ = cf N ? (b) We have ω ∈ MahcrR ({0, 1}ω1 ) if FN(PN) = ω1 (532P, 532K) and not if add Nω > ω1 (532Q). Can we narrow the gap? (c) For a Hausdorff space X let MahspcrR (X) be the set of Maharam types of strictly positive Maharam homogeneous completion regular Radon measures on X. Describe the sets Γ of cardinals for which there are a compact Hausdorff spaces X such that MahspcrR (X) = Γ. 532 Notes and comments I have spent a good many pages on a rather specialized topic. But I think the patterns here are instructive. When looking at MahR (X), as in §531, we quickly come to feel that it is a measure of a certain kind of complexity; the richer the space X, the larger MahR (X) will be. 531Eb and 531Ed are direct manifestations of this, and 531G develops the theme. MahcrR (X) can sometimes tell us more about X; knowing MahcrR (X) we may have a lower bound on the complexity of X as well as an upper bound. (On the other hand, MahcrR (X) can evaporate for non-trivial reasons, as in 532Xb and 532Ya, and leave us with very little idea of what X might be like.) In place of the straightforward facts in 531E, we have the relatively complex and partial results in 532G and 532K. As soon as we leave the constrained context of powers of {0, 1}, the most natural questions seem to be obscure (532Zc). However, if we follow the paths which are open, rather than those we might otherwise have chosen, we come to some interesting ideas, starting with 532I. Here, as happened in §531, we see that a proper understanding of the measure algebras Bλ will take us a long way; and once again we find that this understanding has to be conditional on the model of set theory we are working in. Even to decide which powers of {0, 1} carry completion regular Radon measures with countable Maharam type we need to examine some new aspects of the Lebesgue measure algebra (532M-532O). Moreover, as well as the familiar cardinals of Cicho´ n’s diagram, we have to look at the Freese-Nation number of PN (532P). For larger Maharam types, in a way that we are becoming accustomed to, other combinatorial principles become relevant (532R, 532S).

533 Special topics I present notes on certain questions which can be answered if we make particular assumptions concerning values of the cardinals considered in §§523-524. The first cluster (533A-533E) looks at Radon and quasi-Radon measures in contexts in which the additivity of Lebesgue measure is large compared with other cardinals of the structures considered. Developing ideas which arose in the course of §531, I discuss ‘uniform regularity’ in perfectly normal and first-countable spaces (533H). We also have a complete description of the cardinals κ for which R κ is measure-compact (533J). As previously, I write N (µ) for the null ideal of a measure µ; νκ will be the usual measure on {0, 1}κ and Nκ = N (νκ ) its null ideal. 533A Lemma Let (X, Σ, µ) be a semi-finite measure space with measure algebra (A, µ ¯). If hKξ iξ<κ is a family of ideals in Σ such that µ T is inner regular with respect to every Kξ amd κ < min(add N (µ), wdistr(A)), then µ is inner regular with respect to ξ<κ Kξ . proof Take E ∈ Σ and γ < µE. Then there is an E1 ∈ Σ such that E1 ⊆ E and γ < µE1 < ∞. For ξ < κ, Dξ = {K • : K ∈ Kξ } is closed under finite unions and is order-dense in A, so includes a partition of unity Aξ . Now

533D

Special topics

209

there is a partition B of unity in A such that {a : a ∈ Aξ , a ∩ b 6= 0} is finite for every b ∈ B and ξ < κ. Let B 0 ⊆ B be a finite set such that µ ¯(E1• ∩ sup B 0 ) ≥ γ, and let E2 ⊆ E1 be such that E2• = E1• ∩ sup B 0 . For any ξ < κ, S A0ξ = {a : a ∈ Aξ , a ∩ E2• 6= 0} ⊆ b∈B 0 {a : a ∈ Aξ , a ∩ b 6= 0} is finite, so sup A0ξ belongs to Dξ and can be expressed as Kξ• for some Kξ ∈ Kξ . Now E2• ⊆ sup A0ξ so E2 \ Kξ is S negligible. As κ < add N (µ), we have a negligible H ∈ Σ including ξ<κ E2 \ Kξ ; now E 0 = E2 \ H ⊆ E, µE 0 ≥ γ and T T E 0 ∈ ξ<κ Kξ . As E and γ are arbitrary, µ is inner regular with respect to ξ<κ Kξ . Remark Of course this result is covered by 412Ac unless wdistr(A) > ω1 , which nearly forces A to have countable Maharam type (524Mb). 533B Corollary Let (X, Σ, µ) be a totally finite measure space with countable Maharam type. If E ⊆ Σ, #(E) < min(add Nω , add N (µ)) and  > 0, there is a set F ∈ Σ such that µ(X \ F ) ≤  and {E ∩ F : E ∈ E} is countable. proof Let (A, µ ¯) be the measure algebra of µ. Then A is separable in its measure-algebra topology (521Oa). Let H ⊆ Σ be such that {H • : H ∈ H} is dense in A. For E ∈ E and n ∈ N choose HEn ∈ H such that µ(E4HEn ) ≤ 2−n ; let S KE be the family of measurable sets K such that K is disjoint from i≥n E4HEi for some n. Then µ is inner regular T with respect to KE . BecauseT#(E) < min(wdistr(A), add N (µ)) (524Mb), µ is inner regular with respect to E∈E KE (533A) and there is an F ∈ KE such that µF ≥ µX − . If E ∈ E, there is an n ∈ N such that F ∩ (E4HEn ) = ∅, that is, F ∩ E = F ∩ HEn ; so {F ∩ E : E ∈ E} ⊆ {F ∩ H : H ∈ H} is countable. 533C Proposition Let (X, T, Σ, µ) be a quasi-Radon measure space with countable Maharam type. (a) If w(X) < add Nω , then µ is inner regular with respect to the second-countable subsets of X; if moreover T is regular and Hausdorff, then µ is inner regular with respect to the metrizable subsets of X. (b) If Y is a topological space of weight less than add Nω , then any measurable function f : X → Y is almost continuous. (c) If hYi ii∈I is a family of topological Q spaces, with #(I) < add Nω , and fi : X → Yi is almost continuous for every i, then x 7→ f (x) =hfi (x)ii∈I : X → i∈I Yi is almost continuous. proof Note first that add N (µ) ≥ add Nω , by 524Ta. (a) Let U be a base for T with #(U) < add Nω . Set F = {F : F ⊆ X, {F ∩ U : U ∈ U} is countable}. Then µ is inner regular with respect to F. P P If E ∈ Σ and γ < µE, let H ∈ Σ be such that H ⊆ E and γ < µH < ∞. Then the subspace measure µH still has countable Maharam type (use 322I and 514Ed) and add N (µH ) ≥ add N (µ) ≥ add Nω > #({H ∩ U : U ∈ U}). By 533B, there is an F ∈ dom µH such that µH F ≥ γ and {F ∩ H ∩ U : U ∈ U} is countable; now F ∈ F, F ⊆ E and µF ≥ γ. Q Q But every member of F is second-countable (use 4A2B(a-iv)). If T is regular and Hausdorff, then every member of F is separable and metrizable (4A2Pb). (b) If f : X → Y is measurable, let V be a base for the topology of Y with #(V) < add Nω . Suppose that E ∈ Σ and γ < µE. By 533B, there is an F ∈ Σ such that F ⊆ E, γ < µF < ∞ and {F ∩ f −1 [V ] : V ∈ V} is countable. It follows that {f [F ] ∩ V : V ∈ V} is countable, so that the subspace topology on f [F ] is second-countable (4A2B(a-iv)). Giving F its subspace topology TF and measure µF , µF is inner regular with respect to the closed sets (412Pc). If H ⊆ f [F ] is relatively open in f [F ], it is of the form G ∩ f [F ] where G is an open subset of Y , so that (f F )−1 [H] = F ∩ f −1 [G] is measured by µF ; thus f F : F → f [F ] is measurable. By 418J, f F is almost continuous, and there is a K ∈ Σ such that K ⊆ F , µK ≥ γ and f K is continuous. As E and γ are arbitrary, f is almost continuous. (c) For each i ∈ I, set Ki = {K : K ∈ Σ, fi K is continuous}. Then Ki is an ideal in Σ and µ is inner regular with respect T to Ki . Also, as in 533B, wdistr(A) > κ, where A is the measure algebra of µ. So µ is inner regular with respect to K = i∈I Ki , by 533A. But f K is continuous for every K ∈ K, so f is almost continuous. 533D Proposition Let (X, T) be a first-countable compact Hausdorff space such that cf[w(X)]≤ω < add Nω , and µ a Radon measure on X with countable Maharam type. Then µ is inner regular with respect to the metrizable zero sets. proof Set κ = w(X). Then there is an injective continuous function f : X → [0, 1]κ (5A4Cc). Let I be a cofinal subset of [κ]≤ω with #(I) < add Nω . By 524Pa, add µ ≥ add Nω .

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For I ∈ I and x ∈ X set fI (x) = f (x)I. We need to know that for every x ∈ X there is an I ∈ I such that {x} = fI−1 [fI [{x}]]. P P Set S FI = fI−1T [fI [{x}]] for each I. Because I is upwards-directed, hFI iI∈I is downwards-directed. Because f is injective and I = κ, I∈I FI = {x}. Let V be a countable base of S open neighbourhoods of x. For each V ∈ V there is an IV ∈ I such that FIV ∩ (X \ V ) = ∅. Let I ∈ I be such that V ∈V IV ⊆ I; then FI = {x}. Q Q −1 I For I ∈ I, let λI be the image measure µfI on [0, 1] ; note that λI is a Radon measure (418I). Of course add λI is also at least add Nω , and in particular is greater than κ. If G ⊆ X is open, then G and fI [G] are expressible as unions of at most κ compact sets, so λI measures fI [G]. There is an I ∈ I such that µfI−1 [fI [G]] = µG for every open set G ⊆ X. P P?? Suppose, if possible, otherwise. For each I ∈ I choose an open S set GI ⊆ X such that EI = fI−1 [fI [GI ]] \ GI is non-negligible; because λI measures fI [GI ], µ measures EI . Set EI0 = J∈I,J⊇I EJ for each I ∈ I; because #(I) < add µ, µ measures EI0 . Note that EI0 ⊆ EJ0 whenever J ⊆ I in I; moreover, any sequence in I has an upper bound T in I. There is therefore an M ∈ I such that 0 0 0 EM \ EI0 is negligible for every I ∈ I. Again because #(I) < add µ, EM \ I∈I EI0 is negligible; as EM is not negligible, T −1 0 there is an x ∈ I∈I EI . But there is an I ∈ I such that {x} = fI [fI [{x}]], so x ∈ / EJ for any J ⊇ I. X XQ Q S −1 Let U be a base for the topology of X with #(U) = κ. Then U ∈U fI [fI [U ]] \ U is µ-negligible; let Y be its complement. If x ∈ X and y ∈ Y and x 6= y, there is a U ∈ U containing x but not y, so fI−1 [fI [U ]] contains x and not y and f (x) 6= f (y). If F ⊆ Y is compact, then F is homeomorphic to the metrizable fI [F ], so is metrizable, and F = fI−1 [fI [F ]] is a zero set. As µ is surely inner regular with respect to the compact subsets of the conegligible set Y , it is inner regular with respect to the metrizable zero sets. 533E Corollary Suppose that cov Nω1 > ω1 . Let (X, T) be a first-countable K-analytic Hausdorff space such that cf[w(X)]≤ω < add Nω . Then X is a Radon space. proof Let µ be a totally finite Borel measure on X, E ⊆ X a Borel set and γ < µE. Because X is K-analytic, there Ris a compact R set K ⊆ X such that µ(E ∩ K) > γ (432B). Let λ be the Radon measure on K defined by saying that f dλ = K f dµ for every f ∈ C(K) (using the Riesz Representation Theorem, 436J/436K). Because cov Nω1 > ω1 , ω1 is a precaliber of every measurable algebra (525K); as K is first-countable, ω1 ∈ / MahR (K) (531P) and λ must have countable Maharam type (531Ef). By 533D, λ is completion regular. But if F ⊆ K is a zero set (for the subspace topology of K), there is a non-increasing sequence hfn in∈N in C(K) with infimum χF , so λF = limn→∞

R

fn dλ = limn→∞

R K

fn dµ = µF .

Accordingly λH = sup{λF : F ⊆ H is a zero set} = sup{µF : F ⊆ H is a zero set} ≤ µH for every Borel set H ⊆ K. As λK = µK, λ agrees with µ on the Borel subsets of K. In particular, λ(E ∩ K) > γ; now there is a compact set L ⊆ E ∩ K such that γ ≤ λL = µL. As E and γ are arbitrary, µ is tight; as µ is arbitrary, X is a Radon space. 533F Definition Let X be a topological space and µ a topological measure on X. I will say that µ is uniformly S regular if there is a countable family V of open sets in X such that G \ {V : V ∈ V, V ⊆ G} is negligible for every open set G ⊆ X. 533G Lemma Let (X, T, Σ, µ) be a compact Radon measure space. (a) The following are equiveridical: (i) µ is uniformly regular; (ii) there are a metrizable space Z and a continuous function f : X → Z such that µf −1 [f [F ]] = µF for every closed F ⊆ X; (iii) there is a countable family H of cozero sets in X such that µG = sup{µH : H ∈ H, H ⊆ G} for every open set G ⊆ X; (iv) there is a countable family E of zero sets in X such that µG = sup{µE : E ∈ E, E ⊆ G} for every open set G ⊆ X. (b) If T is perfectly normal, the following are equiveridical: (i) µ is uniformly regular; (ii) there are a metrizable space Z and a continuous function f : X → Z such that µf −1 [f [E]] = µE for every E ∈ Σ; (iii) there are a metrizable space Z and a continuous function f : X → Z such that f [G] 6= f [X] whenever G ⊆ X is open and µG < µX; (iv) there is a countable family E of closed sets in X such that µG = sup{µE : E ∈ E, E ⊆ G} for every open set G ⊆ X.

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proof (a)(i)⇒(iii) Given V as in 533F, then for each V ∈ V there is a cozero set HV ⊆ V of the same measure. P P T is completely regular, so H = {H : H ⊆ V is a cozero set} has union V ; µ is τ -additive, so there is a sequence hH i n n∈N in SV S HV such that µV = µ( n∈N Hn ); set HV = n∈N Hn ; by 4A2C(b-iii), HV is a cozero set. Q Q Now H = {HV : V ∈ V} witnesses that (iii) is true. (iii)⇒(iv) Given H as in (iii), then for each H ∈ H let hFn (H)in∈N be a non-decreasing sequence of zero sets with union H (4A2C(b-vi)). Set E = {Fn (H) : H ∈ H, n ∈ N}, so that E is a countable family of zero sets. If G ⊆ X is open, µG = supH∈H,H⊆G µH = supH∈H,H⊆G,n∈N µFn (H) ≤ supE∈E,E⊆G µE ≤ µG, so E witnesses that (iv) is true. (iv)⇒(ii) Given E as in (iv), then for each E ∈ E choose a continuous fE : X → R such that E = fE−1 [{0}], and −1 set f (x) = hfE (x)iE∈E for x ∈ X. Then f : X → Z = R E is continuous and S Z is metrizable and f [f [E]] = E for every E ∈ E. If F ⊆ X is closed, set E0 = {E : E ∈ E, E ∩ F = ∅}. Then E0 has the same measure as X \ F and does not meet f −1 [f [F ]], so µf −1 [f [F ]] = µF . As F is arbitrary, f and Z witness that µ is uniformly regular. (ii)⇒(i) Take Z and f : X → Z as in (ii). Replacing Z by f [X] if necessary, we may suppose that f is surjective, so that Z is compact, therefore second-countable (4A2P(a-ii)). Let U be a countable base for the topology of Z closed under finite unions, and set V = {f −1 [U ] : U ∈ U}, so that V is a countable family of open sets in X. If S G ⊆ X is open, set F =SX \ G, U0 = {U : U ∈ U, U ∩ f [F ] = ∅}, V0 = {f −1 [U ] : U ∈ U0 }. Then Z \ f [F ] = U0 so X \ f −1 [f [F ]] = V0 and (because U0 and V0 are closed under finite unions) sup{µV : V ∈ V, V ⊆ G} ≥ sup µV = µ(X \ f −1 [f [F ]]) V ∈V0

= µX − µf −1 [f [F ]] = µX − µF = µG. Thus V witnesses that µ is uniformly regular. (b)(i)⇒(iii) If µ is uniformly regular, let Z be a metrizable space and f : X → Z a continuous function such that µf −1 [f [F ]] = µF for every closed F ⊆ X. If now G ⊆ X is open and µG < S µX, there is a sequence hFn in∈N of closed sets with union G, because T is perfectly normal. In this case f −1 [f [G]] = n∈N f −1 [f [Fn ]] has the same measure as G, so is not the whole of X, and f [G] 6= f [X]. Thus f and Z witness that (iii) is true. (iii)⇒(ii) Take Z and f from (iii). Let ν be the image measure µf −1 on Z; then µ is a Radon measure (418I). ?? If E ∈ Σ and µ∗ f −1 [f [E]] > µE, let E 0 ⊇ E be a Borel set such that µE 0 = µE. Because X is perfectly normal, E 0 belongs to the Baire σ-algebra of X (4A3Kb), so is Souslin-F (421L), therefore K-analytic (422Hb); consequently f [E 0 ] is K-analytic (422Gd) therefore measured by ν (432A). This means that f −1 [f [E 0 ]] ∈ Σ, and of course µf −1 [f [E 0 ]] ≥ µ∗ f −1 [f [E]] > µE = µE 0 . We can therefore find open sets G ⊇ E 0 and G0 ⊇ X \ f −1 [f [E 0 ]] such that µG + µG0 < µX. But now G ∪ G0 is an open set of measure less than µX and f [G ∪ G0 ] = f [X], which is supposed to be impossible. X X Thus, for any E ∈ Σ, we have µ∗ f −1 [f [E]] = µE; of course it follows at once that f −1 [f [E]] is measurable, with the same measure as E, as required by (ii). (ii)⇒(i)⇔(iv) (ii)⇒(i) is trivial, and (i)⇔(iv) is immediate from (a), because all closed sets in X are zero sets. 533H Theorem (a) Suppose that cov Nω1 > ω1 . Let X be a perfectly normal compact Hausdorff space. Then every Radon measure on X is uniformly regular. (b) (Plebanek 00) Suppose that cov Nω1 > ω1 = non Nω . Let X be a first-countable compact Hausdorff space. Then every Radon measure on X is uniformly regular. proof (a) Let µ be a Radon measure on X. ?? If µ is not uniformly regular, then we can choose hgξ iξ<ω1 and hGξ iξ<ω1 inductively, as follows. Given that gη : X → R is continuous for every η < ξ, set fξ (x) = hgη (x)iη<ξ for x ∈ X, so that fξ : X → R ξ is continuous. By 533G(b-iii), there is an open set Gξ such that µGξ < µX and fξ [Gξ ] = fξ [X]; now Gξ is a cozero set and there is a continuous function gξ : X → R such that Gξ = {x : gξ (x) 6= 0}. Continue. At the end of the induction, we have a continuous function fω1 : X → R ω1 , setting fω1 (x) = hgξ (x)iξ<ω1 for each x. Now ω1 is a precaliber of every measurable algebra (525K), and µ(X \ Gξ ) > 0 for each ξ, so there is an x ∈ X such that A =S{ξ : x ∈ / Gξ } is uncountable (525Da). Set H = {y : fω1 (y) 6= fω1 (x)}; then H is an open set, so expressible as n∈N Kn where each Kn is compact. For each ξ ∈ A there is an xξ ∈ Gξ such that fξ (xξ ) = fξ (x). As gξ (xξ ) 6= 0 = gξ (x), xξ ∈ H. Let n ∈ N be such that A0 = {ξ : ξ ∈ A, xξ ∈ Kn } is uncountable. Then fω1 (x) ∈ {fω1 (xξ ) : ξ ∈ A0 } ⊆ fω1 [Kn ];

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but this is impossible, because Kn ⊆ H. X X So µ must be uniformly regular, as required. (b) Let µ be a Radon measure on X. If µX = 0 then of course µ is uniformly regular; suppose µX > 0. As in (a) and 533E, ω1 is a precaliber of measure algebras, so ω1 ∈ / MahR (X) and the Maharam type of µ is countable. Let A be the measure algebra of µ; then d(A) ≤ non Nω (524Me), so there is a set A ⊆ X, of full outer measure, with #(A) ≤ ω1 (521Jc). For each x ∈ X, let hVxn in∈N run over a base of neighbourhoods of x. Let H be the family of sets expressible as finite unions of Vxn for x ∈ A and n ∈ N, so that H is a family of openSsets in X and #(H) ≤ ω1 . For any open G ⊆ X, µG = sup{µH : H ∈ H, H ⊆ G}. P P Set H ∗ = {H : H ∈ H, H ⊆ G}. For any x ∈ A ∩ G, there is an n ∈ N such that Vxn ⊆ G, and now Vxn ∈ H, so x ∈ H ∗ . Thus G \ H ∗ does not meet A; as A has full outer measure, µG = µH ∗ = sup{µH : H ∈ H, H ⊆ G} because {H : H ∈ H, H ⊆ G} is closed under finite unions. Q Q So there is a countable H0 ⊆ {H : H ∈ H, H ⊆ G} such that µG = supH∈H0 µH. Let hHξ iξ<ω1 run over H. For ξ < ω1 , set Gξ = {G : G ⊆ X is open, µG = sup{µHη : η ≤ ξ, Hη ⊆ G}}. Then

S

ξ<ω1

Gξ = T. For each ξ < ω1 , set Yξ = {y : y ∈ X, Vyn ∈ Gξ for every n ∈ N};

then X = ξ<ω1 Yξ . Now there is a ξ < ω1 such that Yξ has full outer measure. P P Let ξ be such that µ∗ Yξ = µ∗ Yη for ∗ every η ≥ ξ. ?? If µ Yξ < µX, let K ⊆ X \ Yξ be a non-negligible measurable set. Then the subspace measure µK is a Radon measure with countable Maharam type, so S

cov N (µK ) ≥ cov Nω ≥ cov Nω1 > ω1 . Since K ⊆

S

η<ω1

Yη , there must be some η < ω1 such that µ∗K (K ∩ Yη ) > 0; but now µ∗ (K ∩ Yη ) > 0 and η > ξ and µ∗ Yη = µ∗ (Yη \ K) + µ∗ (Yη ∩ K) > µ∗ Yξ . X X

So Yξ has Set Hξ V = {Vyn countable

full outer measure. Q Q S = {Hη : η ≤ ξ}. If G ⊆ X is open,Sand H ∗ = {H : H ∈ Hξ , H ⊆ G}, then G \ H ∗ is negligible. P P Set : y ∈ Yξ , n ∈ N,SVyn ⊆ G}, H1∗ = V. Then Yξ does not meet G \ H1∗ , so µH1∗ = µG. Let V0 ⊆ V be a set such that µ( V0 ) = µG. If V ∈ V0 , then V ∈ Gξ and V ⊆ G so V \ H ∗ is negligible. Accordingly S S G \ H ∗ ⊆ (G \ V0 ) ∪ V ∈V0 (V \ H ∗ )

is negligible. Q Q So if we take H0 to be the set of finite unions of members of Hξ , H0 will be a countable family of open sets and µG = sup{µH : H ∈ H0 , H ⊆ G} for every open G ⊆ X. Thus µ is uniformly regular. 533I We know from 435Fb/435H and 439P that R N is measure-compact and R c is not. It turns out that we already have a language in which to express a necessary and sufficient condition for R κ to be measure-compact. To give the result in its full strength I repeat a definition from 435Xk. Definition A completely regular space X is strongly measure-compact if µX = sup{µ∗ K : K ⊆ X is compact} for every totally finite Baire measure µ on X. Remark For the elementary properties of these spaces, see 435Xk. I repeat one here: a completely regular space X is strongly measure-compact iff it is measure-compact and pre-Radon. P P(i) Suppose that X is measure-compact and pre-Radon and that µ is a totally finite Baire measure on X. Because X is measure-compact, µ has an extension to a quasi-Radon measure µ ˜ (435D); because X is pre-Radon, µ ˜ is Radon (434Jb) and µX = µ ˜X =

sup

µ ˜K

K⊆X is compact

=

sup K⊆X is compact

µ ˜∗ K ≤

sup

µ∗ K ≤ µX.

K⊆X is compact

As µ is arbitrary, X is strongly measure-compact. (ii) Suppose that X is strongly measure-compact. (α) Let µ be a Baire probability measure on X. Then there is a non-negligible compact set, so X cannot be covered by the negligible open sets; by 435Fa, this is enough to ensure that X is measure-compact. (β) Now let µ be a totally finite τ -additive Borel measure on X. Write ν for the restriction of µ to the Baire σ-algebra of X. Then there is a compact set K ⊆ X which is not ν-negligible. ?? If µ(X \ K) = µX, then, because µ is τ -additive and X is regular, there is a closed set F ⊆ X \ K such that µF + ν ∗ K > µX. Because X is completely regular, there is a zero set G including K and disjoint from F , in which case ν ∗ K > µG = νG, which is impossible. X X So µK > 0; by 434J(a-iii), this tells us that X is pre-Radon. Q Q

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533J Theorem (see Fremlin 77) Let κ be a cardinal. Then the following are equiveridical: (i) R κ is measure-compact; Q (ii) if hXξ iξ<κ is a family of strongly measure-compact completely regular Hausdorff spaces then ξ<κ Xξ is measurecompact; T (iii) whenever X is a compact Hausdorff space and hGξ iξ<κ is a family of cozero sets in X, then X ∩ ξ<κ Gξ is measure-compact; (iv) for any Radon measure, the union of κ or fewer closed negligible sets has inner measure zero; (v) for any Radon measure, the union of κ or fewer negligible sets has inner measure zero; (vi) κ < cov N (µ) for any Radon measure µ; (vii) κ < cov Nκ ; (viii) κ < m(A) for every measurable algebra A. proof not-(iv)⇒not-(i) Suppose that X is a Hausdorff space, µ is a Radon measure on X and hF S Sξ iξ<κ is a family of closed µ-negligible subsets of X such that µ∗ ( ξ<κ Fξ ) > 0. Then there is a compact set K ⊆ ξ<κ Fξ such that µK > 0. For each ξ < κ, there is a continuous gξ : K → [0, 1[ such that gξ (z) = 0 for z ∈ K ∩ Fξ and gξ−1 [{0}] is negligible. P P −n For each n ∈ N, there is a compact set Ln ⊆ K \ Fξ such that µL ≥ µK − 2 ; there is a continuous f : K → [0, 1] n n P∞ such that fn (z) = 0 for z ∈ K ∩ Fξ , 1 for z ∈ Ln ; set gξ = n=0 2−n−2 fn . Q Q Set g(z) = hgξ (z)iξ<κ for z ∈ K, so that κ g : K → [0, 1[ is continuous. κ Let ν be the Baire measure on [0, 1]κ defined by setting νH = µg −1 [H] for every Baire set H ⊆ [0, 1]κ . Then ]0, 1[ κ κ has full outer measure for ν. P P If H ⊆ [0, 1] is a Baire set including ]0, 1[ , then H is determined by coordinates in I some countable subset I of κ (4A3Mb). If z ∈ K and gξ (z) > 0 for every ξ ∈ I, then g(z)I ∈ ]0, 1[ is equal to wI −1 for some w ∈ H, so g(z) ∈ H. Thus g [H] includes {z : z ∈ K, gξ (z) > 0 for every ξ ∈ I} and νH = µg −1 [H] ≥ µ{z : gξ (z) > 0 for every ξ ∈ I} = µK = ν[0, 1]κ . Q Q κ

On the other hand, every point y of ]0, 1[ belongs to a ν-negligible cozero set. P P g[K] is a compact set not containing y, so there is a cozero set W containing y and disjoint from g[K], and now νW = 0. Q Q κ κ κ Let ν0 be the subspace measure on ]0, 1[ . By 4A3Nd, ν0 is a Baire measure on ]0, 1[ . If y ∈ ]0, 1[ it belongs to κ κ κ a ν-negligible cozero set W ⊆ [0, 1] , and now W ∩ ]0, 1[ is a ν0 -negligible cozero set in ]0, 1[ containing y. At the same time, κ

ν0 ]0, 1[ = ν[0, 1]κ = µK > 0. κ

κ

So ν0 witnesses that ]0, 1[ is not measure-compact; as R κ is homeomorphic to ]0, 1[ , it also is not measure-compact. (iv)⇒(iii) Suppose that (iv) is true and that we have X and hGξ iξ<κ , as in (iii), with a Baire measure R probability R T µ on Y = X ∩ ξ<κ Gξ . Let ν be the Radon probability measure on X defined by saying that f dν = (f Y )dµ for every f ∈ C(X) (see 436J/436K). Then νGξ = 1 for each ξ < κ. P P Let f : X → R be a continuous function such that Gξ = {x : x ∈ X, f (x) 6= 0}. Set fn = n|f | ∧ χX for each n. Then limn→∞ fn = χGξ , so νGξ = limn→∞

R

fn dν = limn→∞

R

(fn Y )dµ = µY = 1. Q Q S By (iv), ν∗ ( ξ<κ (X \ Gξ )) = 0, that is, Y has full outer measure. In particular, Y must meet the support of ν; take any z in the intersection. If U is a cozero set in Y containing z, there is an open set G ⊆ X such that U = G ∩ Y ; now there is a continuous f : X → [0, 1] such that f (z) = 1 and f (x) = 0 for x ∈ X \ G; in this case µU ≥

R

(f Y )dµ =

R

f dν > 0

because {x : f (x) > 0} is an open set meeting the support of ν. This shows that Y is not covered by the µ-negligible relatively cozero sets; as µ is arbitrary, Y is measure-compact (435Fa). (iii)⇒(i) We can express R κ in the form of (iii) by taking X = [−∞, ∞]κ and Gξ = {x : x(ξ) is finite} for each ξ. (iv)⇒(vii) Let Z be the Stone space of the measure algebra of νκ , and λ its usual measure. If hEξ iξ<κ is a family of λ-negligible sets, then, because λ is inner regular with respect to the open-and-closed sets, we can find negligible zero sets Fξ ⊇ Eξ for each ξ. By (iv), {Fξ : ξ < κ} cannot cover Z, so the same is true of {Eξ : ξ < κ}. Thus cov N (λ) > κ. By 524Jb, cov Nκ > κ. (vii)⇒(vi) Let θ be min{cov N (ν) : ν is a non-zero Radon measure}. By 524Pc, there is an infinite cardinal κ0 such that θ = cov Nκ0 ; by 523F, θ = κ0 . ?? If θ ≤ κ, then 523B tells us that κ < cov Nκ ≤ cov Nθ = θ. X X So θ > κ, as required.

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S (vi)⇒(v) If (vi) is true, (X, µ) is a Radon measure space, hFξ iξ<κ is a family of negligible sets, and E ⊆ ξ<κ Fξ is a measurable set, then the subspace measure S µE is a Radon measure (416Rb), while E can be covered by κ negligible sets; by (vi), µE = 0; as E is arbitrary, µ∗ ( ξ<κ Fξ ) = 0. (v)⇒(ii) Suppose that (v) is true, that hXξ iξ<κ is a family of strongly measure-compact completely regular Hausdorff ˇ spaces with product X, and that Q µ is a Baire probability measure on X. For each ξ < κ let Zξ be the Stone-Cech compactification of Xξ ; set Z = ξ<κ Zξ , and πξ (z) = z(ξ) for z ∈ Z, ξ < κ. Then we have a Radon probability R R measure λ on Z defined by saying that g dλ = X (gX)dµ for every g ∈ C(Z). Note that if W ⊆ Z is a zero set, there is a non-increasing sequence hgn in∈N in C(Z) with infimum χW , so that λW = inf n∈N Now λπξ−1 [Xξ ] = 1 for µξ E = µ(X ∩ πξ−1 [E]) for measure-compact, there is subset of Zξ , so there is a Xξ including K, so

R

gn dλ = inf n∈N

R X

(gn X)dµ = µ(W ∩ X).

each ξ. P P Let  > 0. We have a Baire probability measure µξ on Xξ defined by setting every Baire set E ⊆ Xξ , and a Radon measure λξ = λπξ−1 on Zξ . Because Xξ is strongly a compact set K ⊆ Xξ such that µ∗ξ K ≥ 1 − . Now K is still compact when regarded as a zero set F ⊆ Zξ , including K, such that λξ F = λξ K. In this case, F ∩ Xξ is a zero set in λ∗ πξ−1 [Xξ ] ≥ λπξ−1 [K] = λξ K = λξ F = λπξ−1 [F ] = µ(X ∩ πξ−1 [F ]) = µξ (F ∩ Xξ ) ≥ µ∗ξ K ≥ 1 − .

As  is arbitrary, we T have the result. Q Q By (v), X = Z ∩ ξ<κ πξ−1 [Xξ ] has full outer measure for λ. Let G be the family of µ-negligible cozero sets in X and H the family of λ-negligible open sets in Z. If x ∈ G ∈ G, then there is a Rcontinuous function g : Z → [0, 1] such R that g(x) = 1 and H = {y : y ∈ X, g(y) > 0} is included in G; now g dλ = (gX)dµ = 0, so λH = 0. This shows S S that G ⊆ H is λ-negligible, and, in particular, is not the whole of X. By 435Fa as usual, this is enough to show that X is measure-compact, as required. (ii)⇒(i) is elementary, because R is certainly strongly measure-compact. (vi)⇒(viii)⇒(vii) are immediate from 524Md. 533X Basic exercises (a) Exhibit a family hKt iT t∈R such that every Kt consists of compact sets, Lebesgue measure on R is inner regular with respect to every Kt , but t∈R Kt = ∅. (b) Let µ be a uniformly regular topological measure on a topological space X. (i) Show that if A ⊆ X then the subspace measure on A is uniformly regular. (ii) Show that any indefinite-integral measure over µ is uniformly regular. (iii) Show that if Y is another topological space and f : X → Y is a continuous open map, then the image measure µf −1 is uniformly regular. (c) Show that any Radon measure on the split interval is uniformly regular. (Hint: 419L.) (d) (Babiker 76) Let X and Y be compact Hausdorff spaces, µ a Radon measure on X, f : X → Y a continuous −1 the image surjection and ν = µf R R measure on Y . Show that the following are equiveridical: (i) νf [F ] = µF for every closed F ⊆ X; (ii) g dµ = inf{ h dν : h ∈ C(Y ), hf ≥ g} for every g ∈ C(X); (iii) for every g ∈ C(X), {y : g is constant on f −1 [{y}]} is ν-conegligible. (e) Show that any uniformly regular Borel measure has countable Maharam type. (f ) Let hXi ii∈I be a countable family of topological spaces with product X, and µ a τ -additive topological measure on X. Suppose that the marginal measure of µ on Xi is uniformly regular for every i ∈ I. Show that µ is uniformly regular. (g) Let X be [0, 1] × {0, 1} with the topology generated by {G × {0, 1} : G ⊆ [0, 1] is relatively open for the usual topology} ∪ {{(t, 1)} : t ∈ [0, 1]} ∪ {X \ {(t, 1)} : t ∈ [0, 1]}. Show that X is compact and Hausdorff. Let µ be the Radon measure on X which is the image of Lebesgue measure on [0, 1] under the map t 7→ (t, 0). Show that µ is uniformly regular but not completion regular. (h) Let X be a topological space and µ a uniformly regular topological probability measure on X. Show that there is an equidistributed sequence in X.

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(i) Show that there is a first-countable compact Hausdorff space with a uniformly regular topological probability measure, inner regular with respect to the closed sets, which is not τ -additive. (Hint: 439K.) 533Y Further exercises (a) (Pol 82) Let X be a compact Hausdorff space and µ a uniformly regular Radon measure on X. Show that if we give the space MR+ of Radon measures on X its narrow topology (437J) then χ(µ, MR+ ) ≤ ω. (b) For a topological S measure µ on a space X, write ureg(µ) for the smallest size of any family V of open subsets of X such that G \ {V : V ∈ V, V ⊆ G} is negligible for every open G ⊆ X. (i) Show that the Maharam type τ (µ) of µ is at most ureg(µ). (ii) Show that if X is compact and Hausdorff and µ is a Radon measure, then ureg(µ) ≤ max(non Nτ (µ) , χ(X)). (iii) Show that if X is compact and Hausdorff, µ is a Radon probability measure and cov Nτ (µ) > ureg(µ), then µ has an equidistributed sequence. (c) (Plebanek 00) Suppose that κ is a regular infinite cardinal such that non Nκ < cov Nκ = κ. Let (X, µ) be a Radon probability space such that χ(X) < κ. Show that µ has an equidistributed sequence. 533Z Problem For which cardinals κ is R κ Borel-measure-compact? 533 Notes and comments I suppose that from the standpoint of measure theory the most fundamental of all the properties of ω is the fact that the union of countably many Lebesgue negligible sets is again Lebesgue negligible; this is of course shared by every κ < add Nω (which is in effect the definition of add Nω ). In 533A-533E and 533J we have results showing that uncountable cardinals can be ‘almost countable’ in other ways. In each case the fact that ω has the property examined is either trivial (as in 533B) or a basic result from Volume 4 (as in 533Cb, 533Cc and 533E). Similarly, the fact that c does not have any of these properties is attested by classical examples. If you are familiar with Martin’s axiom you will not be surprised to observe that everything here is sorted out if we assume that m = c. 533H does not quite fit this pattern, and the hypothesis in 533Hb definitely contradicts Martin’s axiom. ‘Uniformly regular’ measures got squeezed out of §434 by shortage of space; in the exercises 533Xb-533Xi I sketch some of what was missed. Here I mention them just to show that there is more to say on the subject of first-countable and perfectly normal spaces than I put into 531P and 531Q. Another phenomenon of interest is the occurrence of measures which are inner regular with respect to a family of compact metrizable sets (462J, 533Ca, 533D).

534 Hausdorff measures and strong measure zero In this section I look at constructions which are primarily metric rather than topological. I start with a note on Hausdorff measures, spelling out connexions between Hausdorff r-dimensional measure on a separable metric space and the basic σ-ideal N (534B). The main part of the section section is a brief introduction to a class of ideals which are of great interest in settheoretic analysis. While the most important ones are based on separable metric spaces, some of the ideas can be expressed in more general contexts, and I give the definition and most elementary properties in terms of uniformities (534C-534D). An associated topological notion is what I call ‘Rothberger’s property’ (534E-534G). A famous characterization of sets of strong measure zero in R in terms of translations of meager sets can also be represented as a theorem about σ-compact groups (534H). There are few elementary results describing the cardinal functions of strong measure zero ideals, but I give some information on their uniformities (534I) and additivities (534K). There seem to be some interesting questions concerning spaces with isomorphic strong measure zero ideals, which I consider in 534L-534N. A particularly important question, from the very beginning of the topic in Borel 1919, concerns the possible cardinals of sets of strong measure zero; in 534O I give some sample facts. 534A An elementary lemma will be useful. Lemma Let (X, ρ) be a separable metric space. Then there is a countable family C of subsets of X such that whenever A ⊆ X has finite diameter and η > 0 then there is a C ∈ C such that A ⊆ C and diam C ≤ η + 2 diam A. proof Let D be a countable dense subset of X and set C = {∅} ∪ {B(x, q) : x ∈ D, q ∈ Q, q ≥ 0}. If A ⊆ X has finite diameter and η > 0, then if A = ∅ we can take C = ∅. Otherwise, take y ∈ A and q ∈ Q such that diam A + 41 η ≤ q ≤ diam A + 12 η. Let x ∈ D be such that ρ(x, y) ≤ 14 η; then C = B(x, q) ∈ C, A ⊆ B(y, diam A) ⊆ C and diam C ≤ 2q ≤ η + 2 diam A. 534B Hausdorff measures There are difficult questions concerning the cardinal functions of even the most familiar Hausdorff measures. However we do have some easy results.

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Topology and measure III

534B

Theorem Let (X, ρ) be a metric space and r > 0. Write µHr for r-dimensional Hausdorff measure on X, N for the null ideal of Lebesgue measure on R and M for the ideal of meager subsets of R. (a) add µHr = add N (µHr ). (b) If X is separable, N (µHr ) 4T N , so that add µHr ≥ add N and cf N (µHr ) ≤ cf N . (c) If X is separable, (X, ∈, N (µHr )) 4GT (M, 63, R), so that cov N (µHr ) ≤ non M and mcountable ≤ non N (µHr ). (d) If X is analytic and µHr X > 0, then add µHr = add N , cf N (µHr ) = cf N , non N (µHr ) ≤ non N and cov N (µHr ) ≥ cov N . proof (a) 521Ac. (b)(i) Let C be a countable family of subsets of X such that whenever A ⊆ X has finite diameter and η > 0 there is a C ∈ C such that A ⊆ C and diam C ≤ η + 2 diam A (534A). S If A ⊆ X, then A ∈ N (µHr ) iff for every  > 0 there is a sequence hCn in∈N in C such that A ⊆ n∈N Cn and P∞ r P If A is negligible 0, then (by the definition in 471A) there must be a sequence n=0 (diam Cn ) ≤ . P S and  > P ∞ hAn in∈N of subsets of X such that A ⊆ n∈N An and n=0 (diam An )r < 2−r . Let hηn in∈N be a sequence of strictly P∞ r positive real numbers such that n=0 (ηn + 2 diam An ) ≤ . For each P∞ S n we can find Cn ∈ Cn such that An ⊆ Cn and diam Cn ≤ ηn + 2 diam An , so that n=0 (diam Cn )r ≤ , while A ⊆ n∈N Cn . On the otherShand, if A satisfies P∞ the condition, then for every , δ > 0 there is a sequence hCn in∈N of subsets of X such that A ⊆ n∈N Cn and n=0 (diam Cn )r ≤ min(, δ r ). In this case, diam Cn ≤ δ for every n, so θrδ A, as defined in 471A, is at most . As  is arbitrary, θrδ A = 0; as δ is arbitrary, A is µHr -negligible. Q Q (ii) It follows that (N (µHr ), ⊆, N (µHr )) 4GT (N N , ⊆∗ , S), where (N N , ⊆∗ , S)P is the N-localization relation (522K). α) For each n ∈ N, let In be the family of finite subsets I of C such that C∈I (diam C)r ≤ 4−n . Let hInj ij∈N P P(α be a sequence running In . Now, given A ∈ N (µHr ), then for each n ∈ N let hCni ii∈N be a sequence in C, Pover ∞ covering A, such thatP i=0 (diam Cni )r ≤ 2−n−1 . LetThCi ii∈N S be a re-indexing of the family hCni in,i∈N , so that hCi ii∈N ∞ is a sequence in C, i=0 (diam Ci )r ≤ 1, and A ⊆ m≥n i≥m Ci . Let hk(n)in∈N be a strictly increasing sequence P∞ r −n in N such that k(0) = 0 and for every n. Now, for n ∈ N, let φ(A)(n) be such that i=k(n) (diam Ci ) ≤ 4 {Ci : k(n) ≤ i < k(n + 1)} = In,φ(A)(n) . This process defines a function φ : N (µHr ) → N N such that T S S A ⊆ m∈N n≥m In,φ(A)(n) for every A ∈ N (µHr ). β ) For S ∈ S, set (β ψ(S) =

T

m∈N

S

n≥m

S

i∈S[{n}]

S

Ini ⊆ X.

If n ∈ N, then P because #(S[{n}]) ≤ 2n and

{(diam C)r : C ∈

S

i∈S[{n}] Ini }

≤ 2n · 4−n = 2−n ,

P {(diam C)r : C ∈ Ini } ≤ 4−n for every i. But this means that, for any m ∈ N, P S S {(diam C)r : C ∈ n≥m i∈S[{n}] Ini } ≤ 2−m+1 ,

while ψ(S) ⊆

SS S ( n≥m i∈S[{n}] Ini ).

So ψ(S) ∈ N (µHr ) for every S ∈ S. (γγ ) Suppose that A ∈ N (µHr ) and φ(A) ⊆∗ S ∈ S. Then there is some m0 ∈ N such that (n, φ(A)(n)) ∈ S for every n ≥ m0 . Now, for any m ∈ N, we have A⊆

[

[

In,φ(A)(n)

n≥max(m,m0 )

⊆

[

[

n≥max(m,m0 ) i∈S[{n}]

[

Ini ⊆

[

[

[

Ini ,

n≥m i∈S[{n}]

so A ⊆ ψ(S). This shows that (φ, ψ) is a Galois-Tukey connection from (N (µHr ), ⊆, N (µHr )) to (N N , ⊆∗ , S), and (N (µHr ), ⊆, N (µHr )) 4GT (N N , ⊆∗ , S). Q Q (iii) Since (N , ⊆, N ) ≡GT (N N , ⊆∗ , S) (522M), (N (µHr ), ⊆, N (µHr )) 4GT (N , ⊆, N ), that is, N (µHr ) 4T N . (iv) By 513Ee, as usual, we can conclude that add N (µHr ) ≥ add N and cf N (µHr ) ≤ cf N .

534C

Hausdorff measures and strong measure zero

217

(c)(i) If µHr X = 0, the result is trivial. P P Set φ(x) = ∅ for x ∈ X, ψ(t) = X for t ∈ R; then (φ, ψ) is a Galois-Tukey connection from (X, ∈, N (µHr )) to (M, 63, R). Q Q So let us suppose that X is infinite. (ii) Let F be the set of 1-Lipschitz functions f : X → [0, 1]. Define T : X → `∞ (F ) by setting (T x)(f ) = f (x) for f ∈ F and x ∈ X. Then kT x − T yk∞ = supf ∈F |f (x) − f (y)| = min(1, ρ(x, y)) for all x, y ∈ X. P P Of course supf ∈F |f (x) − f (y)| ≤ min(1, ρ(x, y)), by the definition of F . On the other hand, given x ∈ X, we can set f (z) = min(1, ρ(z, x)) for every z ∈ X; then f ∈ F and |f (x) − f (y)| = min(1, ρ(x, y)). So we have equality. Q Q Thus T is 1-Lipschitz for ρ and the usual metric on `∞ (F ), and T [X] is a separable subset of `∞ (F ) (4A2B(e-iii)). Let V be the closed linear subspace of `∞ (F ) generated by T [X]; then V is separable (4A4Bi). Being a closed subset of the complete metric space `∞ (F ), V is a Polish space. Since X has more than one point, and T is injective, V is non-empty and has no isolated points. Let hvn in∈N enumerate a dense subset of V . Set T S E = n∈N i≥n U (vi , 2−i−1 ) where U (v, δ) = {u : u ∈ V , ku − vk∞ < δ} for v ∈ V , δ > 0. Then E is the intersection of a sequence of dense open sets in V , so is comeager, and M = V \ E belongs to the ideal M(V ) of meager subsets of V . For any v ∈ V , the map u 7→ u − v : V → V is a homeomorphism, so M − v ∈ M(V ). Define φ : X → M(V ) by setting φ(x) = M − T x for x ∈ X. In the other direction, define ψ : V → PX by setting ψ(v) = T −1 [E − v] for v ∈ V . Then ψ(v) ∈ N (µHr ) for every v ∈ V. P P If v ∈ V and δ ≤ 21 , then ku − u0 k∞ < 1 for all u, u0 ∈ U (v, δ), so ρ(x, x0 ) ≤ kT x − T x0 k∞ whenever x, 0 −1 x ∈ T [U (v, δ)]. Accordingly diam T −1 [U (vi − v, 2−i−1 )] ≤ 2−i for every i ∈ N. This means that \ [ µ∗Hr T −1 [E + v] = µ∗Hr ( T −1 [U (vi − v, 2−i−1 )] n∈N i≥n ∞ X (2−i )r = 0. Q Q ≤ inf n∈N

i=n

So ψ is a function from V to N (µHr ). We now see that φ(x) 63 v =⇒ v ∈ / M − T x =⇒ T x ∈ / M −v =⇒ T x ∈ E − v =⇒ x ∈ ψ(v). Thus (φ, ψ) is a Galois-Tukey connection from (X, ∈, N (µHr )) to (M(V ), 63, V ) and (X, ∈, N (µHr )) 4GT (M(V ), 63, V ) ∼ = (M, 63, R) (522Vb). (iii) Now cov N (µHr ) = cov(X, ∈, N (µHr )) ≤ cov(M, 63, R) = non M, non N (µHr ) = add(X, ∈, N (µHr )) ≥ add(M, 63, R) = cov M = mcountable (512D, 512Ed, 522Ra). (d) If X is analytic and µHr X > 0, then by Howroyd’s theorem (471S) there is a compact set K ⊆ X such that (K) 0 < µHr K < ∞. Now the subspace measure µHr on K is an atomless Radon measure (471E, 471Dg, 471F) on a compact metric space, so (K)

add N ≤ add N (µHr ) ≤ add N (µHr ) = add N , (K)

cf N ≥ cf(X, N (µHr )) ≥ cf(K, N (µHr )) = cf N , (K)

non N (µHr ) ≤ non N (µHr ) = non N , (K)

cov(X, N (µHr )) ≥ cov(K, N (µHr )) = cov N by (b) above, 521D and 522Va. 534C Strong measure zero Let (X, W) be a uniform space. I say that X has strong measure zero or property C if for any sequence hWn in∈N in W there is a cover hAn in∈N of X such that An × An ⊆ Wn for every n ∈ N. A subset

218

Topology and measure III

534C

A of X has strong measure zero if it has strong measure zero in its subspace uniformity, that is, if for any sequence hWn in∈N in W there is a cover hAn in∈N of A such that An × An ⊆ Wn for every n ∈ N. If (X, ρ) is a metric space, it has strong measure zero if it has strong measure zero for the associated uniformity (3A4B), that is, for any sequence hn in∈N of strictly positive real numbers there is a cover hAn in∈N of X such that diam An ≤ n for every n ∈ N. I will write Smz(X, W) or Smz(X, ρ) for the family of sets of strong measure zero in a uniform space (X, W) or a metric space (X, ρ). 534D Proposition (a) If (X, W) is a uniform space, then Smz(X, W) is a σ-ideal containing all the countable subsets of X. (b) If (X, W) is a uniform space with strong measure zero, (Y, V) is a uniform space, and f : X → Y is uniformly continuous, then f [X] ∈ Smz(Y, V). (c) Let (X, W) be a uniform space and A ⊆ X. Then A ∈ Smz(X, W) iff f [A] ∈ Smz(Y, ρ) whenever (Y, ρ) is a metric space and f : X → Y is uniformly continuous. (d) Let (X, W) be a uniform space and A ∈ Smz(X, W). If B ⊆ X is such that B \ G ∈ Smz(X, W) whenever G is an open set including A, then B ∈ Smz(X, W). (e) If (X, ρ) is a metric space with strong measure zero, then X is separable and universally negligible. proof (a) It is immediate from the definition that any subset of a set in Smz(X, W) belongs to Smz(X, W), and so does any countable set. Now suppose that hAn in∈N is a sequence in Smz(X, W). Let hWn in∈N be any sequence in W. For each k ∈ N, hW2k (2i+1) ii∈N is a sequence in W, so there is a sequence hAki ii∈N , covering Ak , such that S Aki × Aki ⊆ W2k (2i+1) for every i. Set B0 = ∅, Bn = Aki if n = 2k (2i + 1) where k, i ∈ N; then A ⊆ n∈N Bn and Bn × Bn ⊆ Wn for every n. As hWn in∈N is arbitrary, A has strong measure zero; as hAn in∈N is arbitrary, Smz(X, W) is a σ-ideal. (b) Let hVn in∈N be a sequence in V. For each n ∈ N, there is a Wn ∈ W such that (f (x), f (x0 )) ∈ Vn whenever (x, x0 ) ∈ Wn . Because X ∈ Smz(X,SW), there is a cover hAn in∈N of X such that An × An ⊆ Wn for every n; now f [An ] × f [An ] ⊆ Vn for every n and n∈N f [An ] = f [X]. As hVn in∈N is arbitrary, f [X] ∈ Smz(Y, V). (c) If A has strong measure zero, then of course f [A] has strong measure zero for any uniformly continuous function f from X to a metric space, by (b) applied to f A. Now suppose that A satisfies the condition, and that hWn in∈N is a sequence in W. Then there is a pseudometric ρ on X, compatible with the uniformity in the sense that {(x, y) : ρ(x, y) ≤ } ∈ W for every  > 0, such that {(x, y) : ρ(x, y) < 2−n } ⊆ Wn for every n (4A2Ja). Set ∼ = {(x, y) : ρ(x, y) = 0}. Then ∼ is an equivalence relation on X. If Y is the set of equivalence classes, we have a metric ρ˜ on Y defined by setting ρ˜(x• , y • ) = ρ(x, y) for all x, y ∈ X. Setting f (x) = x• for x ∈ X, f : X → Y is uniformly continuous. So f [A] ∈ Smz(Y, ρ˜). Let hBn in∈N be a cover of f [A] such that diam Bn ≤ 2−n−1 for every n, and set An = f −1 [Bn ] for each n. Then hAn in∈N is a cover of A. If n ∈ N and x, y ∈ An , then ρ(x, y) = ρ˜(f (x), f (y)) ≤ 2−n−1 , so (x, y) ∈ Wn . Thus An × An ⊆ Wn . As hWn in∈N is arbitrary, A ∈ Smz(X, W). (d) Let hWn in∈N be any sequence in W. For each n ∈ N, let Vn ∈ W be such that Vn ◦ Vn ◦ Vn−1 ⊆ W2n . Then thereSis a sequence hAn in∈N , covering A, such that An × An ⊆ Vn for every n. Set B2n = int Vn [An ] for each n, and G = n∈N B2n ; then B2n × B2n ⊆ W2n for every n and G is an open set including A. Accordingly B \ G ∈ Smz(X, W) and there is a sequence hB2n+1 in∈N , covering B \ G, such that B2n+1 × B2n+1 ⊆ W2n+1 for every n. Now hBn in∈N covers B and Bn × Bn ⊆ Wn for every n. As hWn in∈N is arbitrary, B has strong measure zero. (e) ?? If X is not separable, there is an uncountable A ⊆ X such that  = inf x,y∈A,x6=y ρ(x, y) is greater than 0 (5A4B(h-iii)). Now there can be no cover hAn in∈N of X by sets of diameter less than . X X Thus X is separable. Now suppose that µ is a Borel probability measure on X. Then there is a δ > 0 such that for every n ∈ N there is a Borel set En with diam En ≤ 2−n and µEn ≥ δ. P P?? Otherwise, we can find for each n ∈ N an n > 0 such that µE ≤ 2−n−2 whenever E is a Borel set and diam E ≤ n . Let hAn in∈N be a cover of X such that diam An ≤ n for every n; then diam An ≤ n , so µAn ≤ 2−n−2 for every n, and P∞ µX ≤ n=0 µAn < 1. X XQ Q T S Now consider E = n∈N m≥n Em . Since µE ≥ δ > 0, there is an x ∈ E. For any n ∈ N, there is an m ≥ n such that x ∈ Em ⊆ B(x, 2−m ) ⊆ B(x, 2−n ), so µ{x} = inf n∈N µB(x, 2−n ) ≥ δ > 0. As µ is arbitrary, this shows that X is universally negligible.

534H

Hausdorff measures and strong measure zero

219

534E Rothberger’s property ‘Strong measure zero’ is a uniform-space property. The topological notion closest to it seems to be the following. If X is a topological space and A is a subset of X, I will say that A has Rothberger’s property in X if for every sequence S hGn in∈N of non-empty open covers of X there is a sequence hGn in∈N such that Gn ∈ Gn for every n ∈ N and A ⊆ n∈N Gn . Note that this is a relative property of the pair (A, X), not an intrinsic property of the set A (see 534Xf, 534Xr). I will write Rbg(X) for the family of subsets of X with Rothberger’s property. 534F Proposition Let X be a topological space. (a) Rbg(X) is a σ-ideal containing all the countable subsets of X. (b) If Y is another topological space, f : X → Y is continuous and A ∈ Rbg(X), then f [A] ∈ Rbg(Y ). (c) If X is completely regular, and W is any uniformity on X compatible with its topology, then Rbg(X) ⊆ Smz(X, W). (d) If X is σ-compact and completely regular, and W is any uniformity on X compatible with its topology, then Rbg(X) = Smz(X, W). proof (a) It is immediate from the definition that any subset of a set in Rbg(X), and any countable subset of X, belong to Rbg(X). Now suppose that hAn in∈N is a sequence in Rbg(X), with union A. Let hGn in∈N be any sequence of non-empty open covers of X. For each k ∈ N, hG2k (2i+1) ii∈N is a sequence of open covers of X, so there is a sequence hGki ii∈N , covering Ak , such that Gki ∈ G2k (2i+1) for every i. Take H0 to be any member of G0 , and set Hn = Gki if S n = 2k (2i+1) where k, i ∈ N; then A ⊆ n∈N Hn and Hn ∈ Gn for every n. As hGn in∈N is arbitrary, A has Rothberger’s property in X. (b) Let hHn in∈N be a sequence of non-empty open covers of Y . For each n ∈ N, set Gn = {f −1 [H] : H ∈ Hn }; then Gn is a non-empty open cover of X. Because A ∈ Rbg(X), there is a cover hGS n in∈N of A such that Gn ∈ Gn for every n ∈ N. Express Gn as f −1 [Hn ], where Hn ∈ Hn , for each n ∈ N; then f [A] ⊆ n∈N Hn . As hHn in∈N is arbitrary, f [A] has Rothberger’s property in Y . (c) Suppose that A ∈ Rbg(X), and that hWn in∈N is any sequence in W. For each n ∈ N, set Gn = {G : G ⊆ X is open, G × G ⊆ Wn }; then Gn is a non-empty open cover of X. So we can find a cover hGn in∈N of A such that Gn ∈ Gn , that is, Gn × Gn ⊆ Wn , for each n. As hWn in∈N is arbitrary, A ∈ Smz(X, W). (d)(i) Let K ⊆ X be compact and G an open cover of X. Then there is a W ∈ W such that whenever x ∈ K there is a G ∈ G such that W [{x}] ⊆ G. P P Set Q = {(x, V ) : x ∈ X, V ∈ W, V [V [{x}]] ⊆ G for some G ∈ G}. Then for every x ∈ X there are a G ∈ G such that x ∈ G and a V ∈ W such that V [V [{x}]] ⊆ G, and in this case (x, V ) ∈ QSand x ∈ int V [{x}]. So {int V [{x}] : (x, V ) ∈ Q} is an open cover of X and there is a finite set Q0 ⊆ Q such that K ⊆ {int V [{x}] : (x, V ) ∈ Q0 }. Let W ∈ W be such that W ⊆ V whenever (x, V ) ∈ Q0 . If x ∈ K, there is an (x0 , V ) ∈ Q0 such that x ∈ V [{x0 }]; and now there is a G ∈ G including V [V [{x0 }]] ⊇ W [{x}]. Q Q (ii) Let K ⊆ X be compact and A ∈ Smz(X, W). Then A ∩ K ∈ Rbg(X). P P Let hGn in∈N be a sequence of non-empty open covers of X. For each n ∈ N let Wn ∈ W be such that {W [{x}] : x ∈ K} refines Gn . Let hAn in∈N be a cover of A such that An × An ⊆ Wn for every n. If n ∈ N and An ∩ K = ∅, take any Gn ∈ Gn . Otherwise, take xn ∈ An ∩ K and Gn ∈ Gn such that Wn [{xn }] ⊆ Gn . If x ∈ A ∩ K, there isSan n ∈ N such that x ∈ An ; now (xn , x) ∈ An × An ⊆ Wn and x ∈ Wn [{xn }] ⊆ Gn . As x is arbitrary, A ∩ K ⊆ n∈N Gn . As hGn in∈N is arbitrary, A ∩ K has Rothberger’s property in X. Q Q (iii) Smz(X, W) ⊆ Rbg(X). P P If A ∈ Smz(X, W), let hKn in∈N be a sequence of compact subsets of X covering X. By (iii), A ∩ Kn ∈ Rbg(X) for each n; by (a) above, A ∈ Rbg(X). Q Q By (c), we have equality. 534G Corollary If X is a σ-compact completely regular topological space, then any uniformity on X compatible with its topology gives the same sets of strong measure zero. proof Immediate from 534Fd. 534H Theorem Let X be a σ-compact locally compact Hausdorff topological group and A a subset of X. Then the following are equiveridical: (i) A has Rothberger’s property in X; (ii)Sfor any sequence hUn in∈N of neighbourhoods of the identity e of X, there is a sequence hxn in∈N in X such that A ⊆ n∈N Un xn ; (iii) F A 6= X for any nowhere dense set F ⊆ X; (iv) EA 6= X for any meager set E ⊆ X; (v) AF 6= X for any nowhere dense set F ⊆ X;

220

Topology and measure III

534H

(vi) AE 6= X for any meager set E ⊆ X. Remark For the general theory of topological groups see §4A5 and Chapter 44. Readers unfamiliar with this theory, or impatient with the extra discipline needed to deal with non-commutative groups, may prefer to start by assuming that X = R 2 , so that every xU becomes x + U , and every V −1 V becomes V − V . proof (i)⇒(ii) Suppose that (i) is true, and that hUn in∈N is any sequence of neighbourhoods of e. Then Gn = {int Un x : S x ∈ X} is an open cover of X, so there is a sequence hxn in∈N such that A ⊆ n∈N Un xn . (ii)⇒(i) Suppose that (ii) is true, and that hWn in∈N is any sequence in the right uniformity of X. Then for each n ∈ N there is a neighbourhood Un of e such that Wn ⊇ {(x, y) : xy −1S∈ Un }; let Vn be a neighbourhood of e such that Vn Vn−1 ⊆ Un . By (ii), there is a sequence hxn in∈N such that A ⊆ n∈N Vn xn . Set An = Vn xn for each n. Then −1 An A−1 ⊆ Un , so An × An ⊆ Wn , for each n, while hAn in∈N covers A. As hWn in∈N is arbitrary, A has strong n = Vn Vn measure zero. By 534Fd, A ∈ Rbg(X). (ii)⇒(iv) Suppose that A satisfies (ii), and that E ⊆ X is meager. α) If K ⊆ X is compact and nowhere dense, then there is a sequence hUn in∈N of neighbourhoods of e such that (α T K 0 = n∈N Un K is still compactTand nowhere dense. P P By 443N(ii), there is a nowhere dense zero set F ⊇ K. Now F is a Gδ set; suppose that F = n∈N Gn where Gn is open for each n. As K ⊆ Gn , the open set Un0 = {x : xK ⊆ G} (4A5Ei) of e included in Un0 . Then Un K ⊆ Gn for every n, so T contains e; let Un be a compact neighbourhood 0 0 K = n∈N Un K ⊆ F is nowhere dense, while K is compact (use 4A5Ef). Q Q β ) Let K ⊆ X be compact and nowhere dense and U a neighbourhood of e. Then there is a neighbourhood V of e (β such that for every x T ∈ X there is an x0 ∈ U x such that V x0 ∩ K = ∅. P P Let hUn in∈N be a sequence of neighbourhoods 0 of e such that K = n∈N Un K is compact and nowhere dense ((α) above). Choose a sequence hVn in∈N of compact T −1 neighbourhoods of e such that V0 ⊆ U and Vn+1 Vn+1 T ⊆ Un ∩ Vn for each n ∈ N. Then Y = n∈N Vn is a compact subgroup of X (see the proof of 4A5S), and Y K = n∈N Vn K (4A5Eh). ?? If for every n ∈ N there is an xn ∈ X such that Vn−1 x0 ∩ K 6= ∅ for every x0 ∈ U xn , then, in particular, Vn−1 xn ∩ K 6= ∅, so xn ∈ Vn K. Since hVn Kin∈N is a non-decreasing sequence of compact sets, hxn in∈N has a cluster point T x∗ ∈ n∈N Vn K = Y K ⊆ K 0 . Because K 0 is nowhere dense, V1 x∗ 6⊆ K 0 ; take x ∈ V1 x∗ \ K 0 . Let W be an open neighbourhood of e such that W x ∩TK 0 = ∅. Then W x is disjoint from Y K = Y −1 Y K so Y W x ∩ Y K = ∅. Now Y W is an open set including Y = n∈N Vn , and all the Vn are compact, so there is an m ≥ 1 such that Vm ⊆ Y W and Vm x ∩ Y K = ∅. Now observe that there is an n > m such that xn ∈ V1 x∗ , so that x ∈ V1 V1−1 xn ⊆ V0 xn ⊆ U xn , X while Vn−1 x ∩ K ⊆ Vm x ∩ Y K is empty. X Thus we can take V = Vn−1 for some n. Q Q (γγ ) Because X is σ-compact, any Fσ set in X is actually Kσ , and there is a sequence hKn in∈N of nowhere dense compact sets covering E; we can suppose that hKn in∈N is non-decreasing. Choose inductively sequences hUn in∈N , hVn in∈N , hVn0 in∈N and hVn00 in∈N of neighbourhoods of e such that U0 is any compact neighbourhood of e, given Un , Vn is to be a neighbourhood of e such that Vn Vn ⊆ Un , given Vn , Vn0 is to be a neighbourhood of e such that for every y ∈ X there is a z ∈ Vn y such that 0 Vn z ∩ Kn+1 = ∅ (using (β)), given Vn0 , Vn00 is to be an open neighbourhood of e such that (Vn00 )−1 Vn00 ⊆ Vn0 , given Vn00 , Un+1 is to be a compact neighbourhood of e, included in Vn ∩ Vn00 , such that Kn+1 Un+1 ⊆ Vn00 Kn+1 . (This last is possible by 4A5Ei, because Vn00 Kn+1 is an open set including Kn+1 , so {x : Kn+1 x ⊆ Vn00 Kn+1 } is an open set containing e.) (k) (δδ ) For each k ∈ N, hU2k (2i+1) ii∈N is a sequence of neighbourhoods of e, so there must be a sequence hxi ii∈N S (k) (k) such that A ⊆ i∈N U2k (2i+1) xi . Set x0 = e and xn = xi if n = 2k (2i + 1). For any k ∈ N, S S S S (k) A ⊆ i∈N Aki ⊆ i∈N U2k (2i+1) xi ⊆ n≥2k Un xn ⊆ n≥k Un xn . S This means that EA ⊆ n≥1 Kn Un xn . P P If z ∈ EA, we can express it as xy where x ∈ E and y ∈ A. There are a k ≥ 1 such that x ∈ Kk and an n ≥ k such that y ∈ Un xn , in which case z ∈ Kn Un xn . Q Q

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() Now choose hyn in∈N , hzn in∈N as follows. Start from y0 = e. Given yn , let zn ∈ Vn yn x−1 n+1 be such that ∩ Kn+1 = ∅; this is possible by the choice of Vn0 . Now set yn+1 = zn xn+1 , and continue. For each n, Un+1 yn+1 ⊆ Un yn . P P

Vn0 zn

Un+1 yn+1 ⊆ Vn yn+1 (by the choice of Un+1 ) = Vn zn xn+1 ⊆ Vn Vn yn x−1 n+1 xn+1 (by the choice of zn ) ⊆ Un yn by the choice of Vn . Q Q For each n ∈ N, Un+1 yn+1 ∩ Kn+1 Un+1 xn+1 = ∅. P P We chose zn such that Vn0 zn ∩ Kn+1 = ∅. Because (Vn00 )−1 Vn00 ⊆ 0 00 00 Vn , Vn zn ∩ Vn Kn+1 = ∅. Because Kn+1 Un+1 ⊆ Vn00 Kn+1 and Un+1 ⊆ Vn00 , Un+1 zn ∩ Kn+1 Un+1 = ∅, that is, Un+1 yn+1 ∩ Kn+1 Un+1 xn+1 = ∅. Q Q (ζζ ) FromT() we see that hUn yn in∈N is a non-increasing sequence of S compact sets, so has non-empty intersection. Take any x ∈ n∈N Un yn . Then x ∈ / Kn+1 Un+1 xn+1 for any n, so x ∈ / n≥1 Kn Un xn ⊇ EA. Thus EA 6= X. As E is arbitrary, (iv) is true. (iv)⇒(iii) is trivial. (iii)⇒(ii) Suppose that (iii) is true. S Let hUn in∈N be any sequence of open neighbourhoods of e. Then there is a sequence hxn in∈N in X such that n∈N xn Un−1 is dense. P P Let hVn in∈N be a sequence of neighbourhoods of e −1 −1 such that T Vn+1 Vn+1 ⊆ Vn ∩ Un for every n ∈ N. Then there is a compact normal subgroup Y of X such that Y ⊆ n∈N Vn and X/Y is metrizable (4A5S). The canonical map x 7→ x• : X → X/Y is continuous, so X/Y is σ-compact, in X such that {x•n : n ∈ N} is dense in X/Y . S therefore separable (4A2P(a-ii)). Let hxn in∈N be a sequence • Set G0 = n∈N xn Vn+1 Y . ?? If H = X \ G0 is non-empty, then {x : x ∈ H} is open (4A5Ja) so contains x•n for some X Thus G0 is dense. But, for any n ∈ N, n. But xn Y ⊆ xn Vn+1 Y ⊆ G0 , so there x ∈ H such that x• = x•n . X S can be no −1 −1 −1 Q Y ⊆ Vn+1 so Vn+1 Y ⊆ Un and G = n∈N xn Un includes G0 . Thus G is dense, as required. Q Accordingly F = X \ G is nowhere dense, and F A = 6 X; suppose x ∈ X \ F A. Then F ∩ xA−1 = ∅, that is, S S S −1 −1 −1 −1 −1 −1 xA ⊆ n∈N xn Un , that is, A ⊆ n∈N x xn Un , that is, A ⊆ n∈N Un xn x. As hUn in∈N is arbitrary, (ii) is true. (i)⇔(v)⇔(vi) Because x 7→ x−1 is a homeomorphism, A ∈ Rbg(X) ⇐⇒ A−1 ∈ Rbg(X) =⇒ EA−1 6= X whenever E ⊆ X is meager =⇒ E −1 A−1 6= X whenever E ⊆ X is meager (because E −1 is meager if E is) ⇐⇒ AE 6= X whenever E ⊆ X is meager =⇒ AF −1 6= X whenever F ⊆ X is nowhere dense (because F −1 is nowhere dense if F is) =⇒ F A−1 6= X whenever F ⊆ X is nowhere dense =⇒ A−1 ∈ Rbg(X). Remark The case X = R is due to Galvin Mycielski & Solovay 79. 534I Proposition (a) Let X be a Lindel¨ of space. Then non Rbg(X) ≥ mcountable . (b) (Fremlin & Miller 88) Give N N the metric ρ defined by setting ρ(x, y) = inf{2−n : n ∈ N, xn = yn}. Then non Smz(N N , ρ) = mcountable . proof (a) Suppose that A ⊆ X and #(A) < mcountable . Let hGn in∈N be a sequence of non-empty open covers of X. Because X is Lindel¨ of, we can choose for each n a non-empty countable Gn0 ⊆ Gn covering X. Let P be the set of finite sequences p = hp(i)ii
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partially ordered set. For each x ∈ A, the set Qx = {p : x ∈ p(i) for some i < #(p)} is cofinal with P . P P Given p ∈ P , set n = #(p); let G ∈ Gn0 be such that x ∈ G; set q = p ∪ {(n, G)}; then p ≤ q ∈ Qx . Q Q ↑ Because #(A) < mS every Qx countable ≤ m (P ) (517Pc), there is an upwards-directed family R ⊆ P meeting S ∗ (517B(iv)). Now p = R is a function; its domain I is either N or a natural number; in either case, A ⊆ i∈I p∗ (i) and p∗ (i) ∈ Gi for every i ∈ I. As hGn in∈N is arbitrary, A has Rothberger’s property in X; as A is arbitrary, non Rbg(X) ≥ mcountable . (b) By (a) and 534Fc, non Smz(N N , ρ) ≥ mcountable . By 522Rb, there is a set A ⊆ N N , of cardinal mcountable , such that for every y ∈ N N there is S an x ∈ A such that x(n) 6= y(n) for every n. ?? If A ∈ Smz(N N , ρ), take a sequence hyn in∈N in N N such that A ⊆ n∈N B(yn , 2−n−1 ). Set y(n) = yn (n) for every n. Then there is an x ∈ A such that x(n) 6= y(n) for every n. But in this case x(n) 6= yn (n) and xn + 1 6= yn n + 1 and x ∈ / B(yn , 2−n−1 ) for every n. X X N Thus A witnesses that non Smz(N , ρ) ≤ mcountable . Remark Observe that (N N , ρ), as described in (b) above, is isometric with Z N with the metric defined by the same formula, and that this metric is translation-invariant, so induces the topological group uniformity of Z N . 534J Proposition (Fremlin 91a) Let (X, ρ) be a separable metric space. Then Smz(X, ρ) 4T N d , where N is the null ideal of Lebesgue measure on R and d is the dominating number (522A). proof (a) By 534A, there is a countable family C of subsets of X such that whenever A ⊆ X has finite diameter and η > 0, there is a C ∈ C such that A ⊆ C and diam C ≤ η + 2 diam A. For each i ∈ N, let hCij ij∈N be a sequence running over {C : C ∈ C, diam C ≤ 2−i }. Let (N N , ⊆∗ , S) be the N-localization relation. (b) Let D ⊆ T N N beSa cofinal set with cardinal d. For each d ∈ D we can find a function φd : Smz(X, ρ) → N N such that A ⊆ n∈N i≥n Cd(i),φd (A)(i) for every A ∈ Smz(X, ρ). P P For A ∈ Smz(X, ρ) and k ∈ N, choose a k

sequence hAki ii∈N of sets covering A such that 2 diam Aki < 2−d(2 (2i+1)) for every i ∈ N. For n = 2k (2i + 1), let An ∈ C be such that Aki ⊆ An and diam An ≤ 2−d(n) ; choose φd (A)(n) such that An = Cd(n),φd (A)(n) . Q Q Define φ : Smz(X, ρ) → (N N )D by setting φ(A) = hφd (A)id∈D for A ∈ Smz(X, ρ). (c) For S ∈ S and d ∈ D, define T S S ψd (S) = n∈N i≥n j∈S[{i}] Cd(i),j ⊆ X. T For hSd id∈D ∈ S D set ψ(hSd id∈D ) = d∈D ψd (Sd ). Then A = ψ(hSd id∈D ) has strong measure zero. P P Let hi ii∈N be any family of strictly positive real numbers. Let d ∈ D be such that 2−d(k) ≤ i whenever k ∈ N and i < 2k+1 . For each k ∈ N, #(Sd [{k}]) ≤ 2k , so we can find a sequence hAi ii∈N such that hAi i2k ≤i<2k+1 is a re-enumeration of hCd(k),j ij∈S[{k}] supplemented by empty sets if necessary. This will ensure that if 2k ≤ i < 2k+1 then diam Ai ≤ 2−d(k) ≤ i , while S S A ⊆ ψd (Sd ) ⊆ (k,j)∈Sd Cd(k),j = i∈N Ai . As hi ii∈N is arbitrary, A ∈ Smz(X, ρ). Q Q (d) Now (φ, ψ) is a Galois-Tukey connection from (Smz(X, ρ), ⊆, Smz(X, ρ)) to (N N , ⊆∗ , S)D , that is, ((N N )D , T, S D ), where T is the simple product relation as defined in 512H. P P φ : Smz(X, ρ) → (N N )D and ψ : S D → Smz(X, ρ) are functions. Suppose that A ∈ Smz(X, ρ) and hSd id∈D are such that (φ(A), hSd id∈D ) ∈ T , that is, φd (A) ⊆∗ Sd for every d. Fix d ∈ D for the moment. Then there is an n ∈ N such that (i, φd (A)(i)) ∈ Sd for every i ≥ n. Now, for any m ≥ n, S S S A ⊆ i≥m Cd(i),φd (A)(i) ⊆ i≥m j∈Sd [{i}] Cd(i),j . Thus A⊆

T

m∈N

S

i≥m

S

j∈Sd [{i}]

Cd(i),j) = ψd (Sd ).

This is true for every d, so A ⊆ ψ(hSd id∈D ). As A and hSd id∈D are arbitrary, (φ, ψ) is a Galois-Tukey connection. Q Q (e) Thus (Smz(X, ρ), ⊆, Smz(X, ρ)) 4GT (N N , ⊆∗ , S)D . But (N N , ⊆∗ , S) ≡GT (N , ⊆, N ) (522M), so (N N , ⊆∗ , S) ≡GT (N , ⊆, N )D (512Hb) and D

(Smz(X, ρ), ⊆, Smz(X, ρ)) 4GT (N , ⊆, N )D = (N D , ≤, N D ) where ≤ is the natural partial order of the product partially ordered set N D . Accordingly Smz(X, ρ) 4T N D ∼ = N d, as claimed. 534K Corollary Let (X, W) be a Lindel¨ of uniform space. Then add Smz(X, W) ≥ add N , where N is the null ideal of Lebesgue measure on R.

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proof (a) Suppose to begin with that W is metrizable. Then X is separable (4A2Pd), so Smz(X, ρ) 4T N d , and add N = add N d ≤ add Smz(X, ρ) by 513E(e-ii). S (b) For the general case, suppose that A ⊆ Smz(X, ρ) and #(A) < add N , and set A∗ = A. Let f be a uniformly continuous function from X to a metric space S Y . Then f [X] is Lindel¨of (5A4Bc), and f [A] has strong measure zero in f [X] for every A ∈ A (534Db), so f [A∗ ] = A∈A f [A] has strong measure zero, by (a). As f is arbitrary, A∗ has strong measure zero, by 534Dc; as A is arbitrary, add Smz(X, ρ) ≥ add N . 534L Smz-equivalence (a) If (X, V) and (Y, W) are uniform spaces, I say that they are Smz-equivalent if there is a bijection f : X → Y such that a set A ⊆ X has strong measure zero in X iff f [A] has strong measure zero in Y . Of course this is an equivalence relation on the class of uniform spaces. (b) If (X, V) and (Y, W) are uniform spaces, I say that X is Smz-embeddable in Y if it is Smz-equivalent to a subspace of Y (with the subspace uniformity, of course). Evidently this is transitive in the sense that if X is Smz-embeddable in Y and Y is Smz-embeddable in Z then X is Smz-embeddable in Z. (c) When X and Y are topological spaces, I will say that they are Rbg-equivalent if there is a bijection f : X → Y such that A ⊆ X has Rothberger’s property in X iff f [A] has Rothberger’s property in Y . 534M Lemma (a) Suppose that (X, V) and (Y, W) are uniform spaces, and that hXn in∈N , hYn in∈N are partitions of X, Y respectively such that Xn is Smz-equivalent to Yn for every n. Then X is Smz-equivalent to Y . (b) Suppose that (X, V) and (Y, W) are uniform spaces, and that X is Smz-embeddable in Y and Y is Smzembeddable in X. Then X and Y are Smz-equivalent. proof (a) SFor each n ∈ N, let fn : Xn → Yn be a bijection identifying the ideals of sets with strong measure zero. Then f = n∈N fn is a bijection identifying Smz(X, V) and Smz(Y, W). (b) (Compare 344D.) Let X1 ⊆ X and Y1 ⊆ Y be Smz-equivalent to Y , X respectively; let f : X → Y1 and g : Y → X1 be bijections identifying the ideals of strong measure zero in each pair. Set X0 = X, Y0 = Y , Xn+1 = g[Yn ] and Yn+1 = f [Xn ] for each T n ∈ N; then hX Tn in∈N is a non-increasing sequence in Σ and hYn in∈N is a non-increasing sequence in T. Set X∞ = n∈N Xn , Y∞ = n∈N Yn . Then f X2k \ X2k+1 is an Smz-equivalence between X2k \ X2k+1 and Y2k+1 \ Y2k+2 , while g Y2k \ Y2k+1 is an Smz-equivalence between Y2k \ Y2k+1 and X2k+1 \ X2k+2 ; and g Y∞ is an Smz-equivalence between Y∞ and X∞ . So (a) gives the required Smz-equivalence between X and Y . 534N Proposition R r , [0, 1]r and {0, 1}N are Rbg-equivalent for every integer r ≥ 1. proof (a) Give R its usual uniformity. Of course the identity map is an Smz-embedding of [0, 1] in R. In the other direction, any homeomorphism from R to ]0, 1[ is an Rbg-equivalence between R and ]0, 1[ and therefore, because R and ]0, 1[ are σ-compact, an Smz-embedding of R in [0, 1] (534Fd). By 534Mb, R and [0, 1] and ]0, 1[ are Smz-equivalent. (b) Give {0, 1}N the metric ρ defined by saying that ρ(x, y) = inf{2−n : n ∈ N, xn = yn} P∞ for x, y ∈ {0, 1}N . Define f : {0, 1}N → [0, 1] by setting f (x) = n=0 2−n−1 x(n) for x ∈ {0, 1}N . Then f is continuous, therefore uniformly continuous, so f [A] has strong measure zero in [0, 1] whenever A ⊆ {0, 1}N has strong measure zero in {0, 1}N . It is also the case that f −1 [B] has strong measure zero whenever B ⊆ [0, 1] does. P P Let hn in∈N be any sequence of strictly positive numbers. Then there is a sequence hBn in∈N , covering B, such that diam Bn < 12 min(1, 2n , 2n+1 ) for every n. Fix n for the moment and consider f −1 [Bn ]. If k is such that 2−k−1 ≤ diam Bn < 2−k , then Bn can meet at most two intervals of the type Iki = [2−k i, 2−k (i + 1)]. So f −1 [Bn ] can meet at most two sets of the type {x : xk = z}, and we can express it as A2n ∪ A2n+1 where max(diam A2n , diam A2n+1 ) ≤ 2−k ≤ 2 diam Bn ≤ min(2n , 2n+1 ). S Putting these together, we have a cover hAn in∈N of n∈N f −1 [Bn ] ⊇ f −1 [B] such that diam An ≤ n for every n; as hn in∈N is arbitrary, f −1 [B] has strong measure zero. Q Q Of course f is not a bijection, so it is not in itself an Smz-equivalence. But if we set D1 = {x : x ∈ {0, 1}N , x is eventually constant}, D2 = {2−k i : k ∈ N, i ≤ 2k }, then D1 ⊆ {0, 1}N and D2 ⊆ [0, 1] are countably infinite, and f {0, 1}N \ D1 is an Smz-equivalence between {0, 1}N \ D1 and [0, 1] \ D2 . Putting this together with any bijection between D1 and D2 , we have an Smz-equivalence between {0, 1}N and [0, 1].

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(c)(i) I show by induction on r that [0, 1]r is Smz-equivalent to R and therefore to [0, 1]. The case r = 1 is covered by (a). For the inductive step to r ≥ 2, I adapt the method of (b). Give {0, 1}N×r the metric ρ defined by setting ρ(x, y) = inf{2−n : n ∈ N, x(n × r) = y(n × r)} for x, y ∈ {0, 1}N×r . Define f : {0, 1}N×r → [0, 1]r by setting P∞ f (x) = h i=0 2−i−1 x(i, j)ij
S

j
: z ∈ [0, 1]r , z(j) ∈ D2 },

where D1 ⊆ {0, 1}N , D2 ⊆ [0, 1] are defined as in the proof of (b), then f is a bijection between {0, 1}N×r \ D1∗ and [0, 1]r \ D2∗ , so is an Smz-equivalence between these. Accordingly [0, 1]r \ D2∗ is Smz-embeddable in {0, 1}N×r , which is homeomorphic, therefore uniformly equivalent, to {0, 1}N , which is in turn Smz-equivalent to ]0, 1[; so [0, 1]r \ D2∗ is Smz-embeddable in ]0, 1[. r−1 Now consider D2∗ . This is a countable union of sets which are isometric, therefore and S Smz-equivalent, to [0, 1] ∗ therefore to ]0, 1[, by the inductive hypothesis. We can therefore express D2 as n∈N Xn where hXn in∈N is disjoint and every Xn is Smz-embeddable in ]0, 1[ and therefore in ]n + 1, n + 2[. Assembling these with the Smz-equivalence between [0, 1]r \D2∗ and ]0, 1[ we have already found, we have a Smz-embedding from [0, 1]r to R. In the other direction, we certainly have an isometric embedding of [0, 1] in [0, 1]r and therefore a Smz-embedding of R in [0, 1]r ; so R and [0, 1]r are Smz-equivalent. Thus the induction proceeds. r

(ii) As for R r , we have a homeomorphism between R r and ]0, 1[ , which (because these again are σ-compact) is a Smz-equivalence and therefore a Smz-embedding of R r in [0, 1]r . So 534Mb, once more, tells us that R r and [0, 1]r and [0, 1] are Smz-equivalent. (d) So R r , [0, 1]r and {0, 1}N are Smz-equivalent, for their usual uniformities. Now 534Fd tells us that they are also Rbg-equivalent. 534O Large sets with strong measure zero It is a remarkable fact that it is relatively consistent with ZFC to suppose that the only subsets of R with strong measure zero are the countable sets (Laver 76, Ihoda 88 or ´ ski & Judah 95, §8.3). We therefore find ourselves investigating constructions of non-trivial sets with Bartoszyn strong measure zero under special axioms. I start by clearing the ground a little. Proposition (a) If (X, ρ) is a separable metric space and A ⊆ X has cardinal less than c, there is a Lipschitz function f : X → R such that f A is injective. (b) (Carlson 93) For any cardinal κ < c, if there is any separable metric space with a set of size κ which is of strong measure zero, then there is a subset of R, of size κ, which has strong measure zero. (c) (Rothberger 41) If b = ω1 there is a subset of R, of size ω1 , which has strong measure zero. proof (a) If X = ∅ this is trivial. Otherwise, let hxn in∈N run over a dense sequence in X, and for x ∈ X define gx : R → R by setting P∞ min(1,ρ(x,xn )) n t gx (t) = n=0 n!

for t ∈ R. Then gx is a real-entire function (5A5A). If x, y ∈ X are distinct, then there must be some n such that min(1, ρ(x, xn )) 6= min(1, ρ(y, xn )), so that one of the coefficients of gx − gy is non-zero, and {t : gx (t) = gy (t)} is countable (5A5A). So if A ⊆ X and #(A) < c, we can find a t ≥ 0 such that gx (t) 6= gy (t) for all distinct x, y ∈ A. Set f (x) = gx (t) for x ∈ X; then f : X → R is a function such that f A is injective. If x, y ∈ X then

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Hausdorff measures and strong measure zero

|f (x) − f (y)| = |

∞ X

225

tn n!

(min(1, ρ(x, xn )) − min(1, ρ(y, yn )) |

n=0

≤ exp(t) sup |ρ(x, xn ) − ρ(y, yn )| ≤ exp(t)ρ(x, y), n∈N

so that f is Lipschitz. (b) Let (X, ρ) be a separable metric space with a set A ∈ [X]κ of strong measure zero. Then (a) tells us that we have a uniformly continuous function f : X → R which is injective on A, so that f [A] ∈ [R]κ has strong measure zero (534Db). (c) Now suppose that b = ω1 . Then there is a family hxξ iξ<ω1 in N N which is increasing and unbounded for the pre-order ≤∗ of 522C(i). In this case {ξ : xξ ≤∗ x} is countable for every x ∈ N N , so that {ξ : xξ ∈ K} is countable for every compact K ⊆ N N . Let f : N N → [0, 1] \ Q be any homeomorphism (4A2Ub), and consider A = {f (xξ ) : ξ < ω1 }. Then #(A) = ω1 . Also A has strong measure zero. P P Of course Q, being countable, has strong measure zero. Let G ⊆ R be an open set including Q. Set K = [0, 1] \ G, so that K ⊆ [0, 1] \ Q is compact and f −1 [K] is compact. As {ξ : xξ ∈ f −1 [K]} is countable, so is A ∩ K = A \ G, and A \ G has strong measure zero. By 534Dd, this is enough to show that A has strong measure zero. Q Q 534P Subject to the continuum hypothesis we have many ways of building sets with strong measure zero. I give one example to suggest what can be done. Example Suppose that mcountable = c. Then there is a set A ⊆ R \ Q, with Rothberger’s property in R, such that (i) A + A = R, (ii) A × A ∈ / Rbg(R 2 ), (iii) there is a continuous surjection from A onto R, (iv) A ∈ / Rbg(R \ Q). proof (a) For x ∈ N N , define ψ(x) ∈ {0, 1}N by setting ψ(x)(n) = 0 if x(n) is even, 1 if x(n) is odd. Then ψ : N N → {0, 1}N is a continuous surjection. Let φ : N N → [0, 1] \ Q be a homeomorphism (4A2Ub). Enumerate N N as hxξ iξ
A = A0 + Z.

Then A ∩ [0, 1] has strong measure zero. P P Let G ⊆ R be any open set including Q. Then [0, 1] \ G is a compact subset of [0, 1]\Q, and K = φ−1 [ [0, 1]\G] is a compact subset of N N . There is therefore some ξ < c such that K ⊆ {x : x ≤ xξ } and [0, 1] \ G ⊆ φ[Kξ ]. Now if η ≥ ξ, neither bη nor cη belongs to φ[Kξ ] + Z; and also aη ∈ ]0, 1[ \ φ[Kξ ], so aη also does not belong to φ[Kξ ] + Z. What this means is that A ∩ [0, 1] \ G ⊆ {aη + n : η ≤ ξ, n ∈ Z} ∪ {bη + n : η ≤ ξ, n ∈ Z} ∪ {cη + n : η ≤ ξ, n ∈ Z} has cardinal less than c = mcountable . By 534Ia and 534Fc, it has strong measure zero. As G is arbitrary, A ∩ [0, 1] has strong measure zero, by 534Dd. Q Q Because A + Z = A, S A = n∈Z (A ∩ [0, 1]) + n also has strong measure zero, therefore has Rothberger’s property (534Fd).

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(c) Let us consider the properties (i)-(iv). Because no aξ , bξ or cξ belongs to Q, A ∩ Q = ∅. For every ξ < c, tξ = bξ + cξ ∈ A + A; as htξ iξ 0. Let I be the σ-ideal of subsets of X generated by {A : µ∗Hr A < ∞}. Show that non N (µHr ) = min(non N , non I) = non N if µHr is σ-finite, = non I otherwise. (Hint: if µHr is not σ-finite, then by 471S there is an uncountable disjoint family of Borel sets of non-zero finite measure.) (b) (i) Let (X, ρ) be a metric space, r > 0 and A ⊆ X a set with strong measure zero. Show that A has zero Hausdorff r-dimensional measure. (ii) Find a subset of R 2 which is universally negligible but does not have strong measure zero (for the usual metric on R 2 ). (Hint: 439H.) (iii) Find a subset of {0, 1}N which is universally negligible but does not have strong measure zero for the metric of 534Ib. (c) Let r, s ≥ 1 be integers. Let A ⊆ R r be a set with strong measure zero, and f : A → R s a function which is differentiable relative to its domain at every point of A. Show that f [A] has strong measure zero. (Hint: 262N.) (d) Let (X, W) be a Hausdorff uniform space with strong measure zero. Show that X is universally negligible iff it is a Radon space. (e) Give ω1 + 1 its order topology and the corresponding uniformity (4A2Jf). Show that it has strong measure zero but is not universally negligible. (f )(i) Show that a topological space which has Rothberger’s property in itself must be Lindel¨of. (ii) Give X = ω1 + 1 its order topology. Show that ω1 has Rothberger’s property in ω1 + 1 but not in itself. (g) Let X be a topological space and A ⊆ X. Show that A ∈ Rbg(X) iff A ∈ Rbg(A). (h) Let X be a σ-compact topological space which is either Hausdorff or regular, and A ⊆ X. Show that A ∈ Rbg(X) iff for every sequence hGn in∈N of finite open covers of X, there is a sequence hGn in∈N , covering A, such that Gn ∈ Gn for every n. (i) Let X be a topological space and A ∈ Rbg(X). Suppose that B ⊆ X is such that B \ G ∈ Rbg(X) for every open subset G of X including A. Show that B ∈ Rbg(X). (j) Let X be a paracompact Hausdorff space, and A a subset of X. Show that the following are equiveridical: (i) A ∈ Rbg(X); (ii) f [A] ∈ Rbg(Y ) whenever Y is a metrizable space and f : X → Y is continuous; (iii) f [A] ∈ Smz(Y, ρ) whenever (Y, ρ) is a metric space and f : X → Y is continuous; (iv) A ∈ Smz(X, W) whenever W is a uniformity on X inducing the topology of X. (Hint: 5A4Fb.) (k)(i) Let X be a Hausdorff topological space. Show that if A ∈ Rbg(X) then A is universally τ -negligible (definition: 439Xh). (ii) Let X be a Hausdorff uniform space. Show that if X has strong measure zero then it is universally τ -negligible. (l) Let X be a locally compact Hausdorff topological group. Show that a subset of X has Rothberger’s property in X iff it has strong measure zero for the right uniformity of X iff it has strong measure zero for the bilateral uniformity of X.

534Zd

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227

(m) Let X be a Hausdorff space. Show that a compact subset of X has Rothberger’s property in X iff it is scattered. (n) Show that non Rbg(N N ) = mcountable . (o) Let X be a Lindel¨ of topological space. Show that add Rbg(X) ≥ add N , where N is the null ideal of Lebesgue measure on R. (p) Let (X, W) be a Lindel¨ of uniform space. Show that there is some κ such that Smz(X, W) 4T N κ , where N is the null ideal of Lebesgue measure on R. (q) Show that every separable metric space (X, ρ) is uniformly equivalent to a subspace of [0, 1]N and is therefore Smz-embeddable in [0, 1]N . (r) Suppose that b = ω1 . Show that there is a set A ⊆ R \ Q such that A has Rothberger’s property in R but not in R \ Q. (s) Let A be the set constructed in 534P on the assumption that mcountable = c. Show that A ∪ Q has Rothberger’s property in itself, but A does not have Rothberger’s property in itself. 534Y Further exercises (a) (i) Set I = {[4−m i, 4−m (i + 1)[ : m ∈ N, i ∈ Z}. For A ⊆ R set θ(A) = √ P (1) (1) inf{ I∈I 0 diam I : I 0 ⊆ I covers A}. Show that if µH,1/2 is Hausdorff 12 -dimensional measure on R, then µH,1/2 (A) = 0 iff θ(A) = 0. (ii) Set J = {[2−m i, 2−m (i + 1)[ × [2−m j, 2−m (j + 1)[ : m ∈ N, i, j ∈ Z}, and for A ⊆ R 2 set P (2) θ0 (A) = inf{ J∈J 0 diam J : J 0 ⊆ J covers A}. Show that if µH1 is Hausdorff 1-dimensional measure on R 2 , then (2)

(1)

(2)

µH1 (A) = 0 iff θ0 (A) = 0. (iii) Show that the null ideals N (µH,1/2 ) and N (µH1 ) are isomorphic. (b) Show that if either non N = cf N or non N < cov N , where N is the null ideal of Lebesgue measure on R, then Hausdorff one-dimensional measure on R 2 does not have the measurable envelope property. (c) Let G be a collection of families of sets. Let us say that a set A has the G-Rothberger property if for every sequence hGn in∈N in G there is a cover hGn in∈N of A such that Gn ∈ Gn for every n ∈ N. (i) Show T that S the family I of sets with the G-Rothberger property is a σ-ideal of sets containing every countable subset of G∈G G. (ii) Show that if H is another collection of families of sets, and f is a function such that for every H ∈ H there is a member of G refining {f −1 [H] : H ∈ H}, then f [A] has the H-Rothberger property whenever A ∈ I. (iii) Suppose that G is a collection of families of open subsets of a topological space X. Suppose that A ∈ I has the G-Rothberger property, and S that B ⊆ X is such that B \ G ∈ I for every open set G ⊇ A. Show that B ∈ I. (iv) Suppose that X = G for every G ∈ G, and that every member of G is countable. Show that non(I, X) ≥ mcountable . (v) Suppose that every member of G is countable, and that for any two members of G there is another refining them both. Show that add I ≥ add N , where N is the null ideal of Lebesgue measure on R. (d) Suppose that mcountable = c. Let X be the group of all bijections of N, regarded as the isometry group of N with its {0, 1}-valued metric, so that X is a Polish group (441Xp-441Xq). Show that there is a subset A of X such that A has strong measure zero for the right uniformity of X but A−1 does not. (e) Suppose that mcountable = d. Show that there is a subset of R, of cardinal mcountable , with strong measure zero. (2)

(2)

534Z Problems (a) Let µH1 be one-dimensional Hausdorff measure on R 2 . Is the covering number cov N (µH1 ) (2) necessarily equal to cov N ? As observed in 534Bc-534Bd, we have cov N ≤ cov N (µH1 ) ≤ non M. We can ask the same n question for r-dimensional Hausdorff measure on R whenever 0 < r < n; in particular, for r-dimensional Hausdorff ¯ ns 05 measure on [0, 1], where 0 < r < 1, and these questions are strongly connected (534Ya). Shelah & Stepra (2) (2) show that non N (µH1 ) can be less than non N ; of course this is possible only because µH1 is not semi-finite (439H, 521Xb). (b) Can cf Rbg(R) be ω1 ? (c) How many types of Polish spaces under Smz-equivalence can there be? If we give N N the metric of 534Ib, can it fail to be Smz-equivalent to [0, 1]N ? (d) Suppose that there is a separable metric space (X, ρ) of cardinal c with strong measure zero. Must there be a subset of R, of cardinal c, with Rothberger’s property in R?

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534 Notes and comments I have very little to say about Hausdorff measures, and 534B is here only because it would seem even lonelier in a section by itself. All I have tried to do is to run through the obvious questions connecting §471 with Chapter 52. But at the next level there is surely much more to be done (534Za). ‘Strong measure zero’ has attracted a great deal of attention, starting with the work of E.Borel, who suggested that every subset of R with strong measure zero must be countable; this is the Borel conjecture. It turns out that this is undecidable in ZFC, and that if the Borel conjecture is true then there are no uncountable sets of strong measure zero in any separable metric space (534O). So we have some questions of a new kind: in the ideals Smz(X, W) of sets of strong measure zero, in addition to the standard cardinals add, non, cov and cf, we find ourselves asking for the possible cardinals of sets belonging to the ideal. The next point is that strong measure zero is not (or rather, not always) either a topological property or a metric property; it is a property of uniform spaces. We must therefore be prepared to examine uniformities, even if we are happy to stay with metrizable ones. In 534P(iv), using an axiom which is a consequence of the continuum hypothesis, I show that we can have a set which has strong measure zero for one of two equivalent metrics and not for the other. Goldstern Judah & Shelah 93 describe a model in which the ideal Rbg(R) has add Rbg(R) = non Rbg(R) = ω2 while mcountable = ω1 . So in this case 534Ib tells us that N N , with the metric described there, is not even Smzembeddable in R. Of course in models of set theory in which the Borel conjecture is true we do have topologically determined structures, for trivial reasons. Note that for any uncountable complete separable metric space X, there is a subset of X homeomorphic to {0, 1}N (423J), and the homeomorphism must be a uniform equivalence; so that {0, 1}N and its companions [0, 1]r , R r must be Smz-embeddable in X. In this sense they are the ‘simplest’ uncountable complete metric spaces. In the same sense, [0, 1]N is the most complex (534Xq). For σ-compact spaces, strong measure zero becomes a topological property (534G), corresponding to what I call ‘Rothberger’s property’ (534E). Rothberger 38b investigated subsets of R which have Rothberger’s property in themselves, under the name ‘property C0 ’. The ideas of 534D and 534J-534K can all be re-presented as theorems about Rothberger’s property (534F, 534Xi, 534Xn, 534Xo); the machinery of 534Yc is supposed to suggest a reason for this. It is natural to be attracted to a topological concept, but there is a difficulty in that Rothberger’s property is not hereditary in the usual way (534Xf, 534Xr, 534Xs). I note that while 534N is stated in terms of Rbg-equivalence, isomorphism of the ideals of sets with the appropriate Rothberger’s property, the concept of strong measure zero seems to be necessary in the Schr¨ oder-Bernstein arguments based on 534M. ´ ski & Judah 95, chap. 8, from which much of For a fuller discussion of strong measure zero in R, see Bartoszyn the material of this section is taken.

535 Liftings I introduced the Lifting Theorem (§341) as one of the fundamental facts about complete strictly localizable measure spaces. Of course we can always complete a measure space and thereby in effect obtain a lifting of any σ-finite measure. For the applications of the Lifting Theorem in §§452-453 this procedure is natural and effective; and generally in this treatise I have taken the view that one should work with completed measures unless there is some strong reason not to. But I have also embraced the principle of maximal convenient generality, seeking formulations which will exhibit the full force of each idea in the context appropriate to that idea, uncluttered by the special features of intended applications. So the question of when, and why, liftings of incomplete measures can be found is one which automatically arises. It turns out to be a fruitful question, in the sense that it leads us to new arguments, even though the answers so far available are unsatisfying. As usual, much of what we want to know depends on the behaviour of the usual measures on powers of {0, 1} (535B). An old argument relying on the continuum hypothesis shows that Lebesgue measure can have a Borel lifting; this has been usefully refined, and I give a strong version in 535D-535E. We know that we cannot expect to have translation-invariant Borel liftings (345F), but strong Borel liftings are possible (535H-535I), and in some cases can be built from Borel liftings (535J-535N). For certain applications in functional analysis, we are more interested in liftings of L∞ spaces than in liftings of measure algebras; and it is sometimes sufficient to have a ‘linear lifting’, not necessarily corresponding to a lifting in the strict sense (535O, 535P). I give a couple of paragraphs to linear liftings because in some ways they are easier to handle and it is conceivable that they are relevant to the main outstanding problem (535Zf). 535A Notation (a) The most interesting questions to be examined in this section can be phrased in the following language. If (X, Σ, µ) is a measure space and T a topology on X, I will say that a Borel lifting of µ is a lifting which takes values in the Borel σ-algebra B(X) of X. (As usual, I will use the word ‘lifting’ indifferently for homomorphisms from Σ to itself, or from A to Σ, where A is the measure algebra of µ. Of course a homomorphism θ : A → Σ is a Borel lifting iff the corresponding homomorphism E 7→ θE • : Σ → Σ is a Borel lifting.) Similarly, a Baire lifting of µ is a lifting which takes values in the Baire σ-algebra Ba(X) of X.

535C

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229

(b) I remark at once that if (X, T, Σ, µ) is a topological measure space and φ : Σ → B(X) is a Borel lifting of µ, then φB(X) is a lifting of the Borel measure µB(X). Conversely, if φ0 : B(X) → B(X) is a lifting of µB(X), and if for every E ∈ Σ there is a Borel set E 0 such that E4E 0 is negligible, then φ0 extends uniquely to a Borel lifting φ of µ. In the same way, any Baire lifting of a measure µ which measures every zero set will give us a lifting of µBa(X); and a lifting of µBa(X) will correspond to a Baire lifting of µ if, for instance, µ is completion regular, as in 535B below. (c) As in Chapter 52, I will say that, for any set I, νI is the usual measure on {0, 1}I and BI its measure algebra. 535B Proposition Let (X, Σ, µ) be a strictly localizable measure space with non-zero measure. Suppose that νκ has a Baire lifting (that is, νκ Ba({0, 1}κ ) has a lifting) for every infinite cardinal κ such that the Maharam-type-κ component of the measure algebra of µ is non-zero. Then µ has a lifting. proof Write (A, µ ¯) for the measure algebra of µ. (a) Suppose first that µ is a Maharam-type-homogeneous probability measure. In this case A is either {0, 1} or isomorphic to Bκ for some infinite κ. The case A = {0, 1} is trivial, as we can set φE = ∅ if E ∈ Σ is negligible, φE = X if E ∈ Σ is conegligible. Otherwise, A is τ -generated by a stochastically independent family heξ iξ<κ of elements of measure 21 . For each ξ < κ, choose Eξ ∈ Σ such that Eξ• = eξ , and define f : X → {0, 1}κ by setting f (x)(ξ) = χEξ (x) for x ∈ X and ξ < κ. Then {F : F ⊆ {0, 1}κ , νF and µf −1 [F ] are defined and equal} is a Dynkin class containing all the measurable cylinders in {0, 1}κ , so includes Baκ = Ba({0, 1}κ ), and f is inverse-measure-preserving for µ and νκ0 = νκ Baκ . Note that Bκ can be identified with the measure algebra of νκ0 (put 415E and 322Da together, or see 415Xp). So we have an induced measure-preserving Boolean homomorphism π : Bκ → A defined by setting πF • = f −1 [F ]• for every F ∈ Baκ . Since π[Bκ ] is an order-closed subalgebra of A (324Kb) containing every eξ , it is the whole of A. We are supposing that there is a lifting θ : Bκ → Baκ of νκ . Define θ1 : A → Σ by setting θ1 a = f −1 [θπ −1 a] for every a ∈ A; then θ1 is a Boolean homomorphism because θ and π −1 are, and (θ1 a)• = π((θπ −1 a)• ) = ππ −1 a = a for every a ∈ A, so θ1 is a lifting of µ. (b) It follows at once that if µ is any non-zero totally finite Maharam-type-homogeneous measure, then it will have a lifting, as we can apply (a) to a scalar multiple of µ. Now consider the general case. Let K be the family of measurable subsets K of X such that the subspace measure µK is non-zero, totally finite and Maharam-type-homogeneous. Then µ is inner regular with respect to K, by Maharam’s theorem (332B). By 412Ia, there is a decomposition hXi ii∈I of X such that at most one Xi does not belong to K, and that exceptional one, if any, is negligible; adding a trivial element Xk = ∅ if necessary, we may suppose that there is exactly one k ∈ I such that µXk = ∅. For each i ∈ I \ {k}, let µi be the subspace measure on Xi , and Σi its domain; then µi has a lifting φi : Σi → Σi . (The point is that if the Maharam type κ of µi is infinite, then the Maharam-type-κ component of A includes Xi• and is non-zero, so our hypothesis tells us that νκ has a Baire lifting.) At this point, recall that we are also supposing that µX > 0, so there is some j ∈ I \ {k}; fix z ∈ Xj , and define φ : Σ → PX by setting [ φE = φi (E ∩ Xi ) if z ∈ / φj (E ∩ Xj ), i∈I\{k}

= Xk ∪

[

φi (E ∩ Xi ) if z ∈ φj (E ∩ Xj ).

i∈I\{k}

Then φ is a lifting of µ. P P It is a Boolean homomorphism because every φi is. If E ∈ Σ, then Xi ∩ φE = φi (E ∩ Xi ) if i ∈ I \ {k}, and is either Xk or ∅ if i = k; in any case, it belongs to Σi ; as hXi ii∈I is a decomposition for µ, φE ∈ Σ. Also P µ(E4φE) ≤ µXk + i∈I\{k} µi ((E ∩ Xi )4φi (E ∩ Xi )) = 0. Finally, if µE = 0, then µi (E ∩ Xi ) = 0 and φi (E ∩ Xi ) = ∅ for every i ∈ I \ {k}, so φE = ∅. Q Q 535C Proposition If λ and κ are cardinals with λ = λω ≤ κ, and νκ has a Baire lifting, then νλ has a Baire lifting. proof If λ is finite, the result is trivial, so we may suppose that λ ≥ ω (and therefore that λ ≥ c). For I ⊆ κ, write BaI for the Baire σ-algebra of {0, 1}I and TI for the family of those E ∈ Baκ which are determined by coordinates in I. Set πI (x) = xI for every x ∈ {0, 1}κ ; then H 7→ πI−1 [H] is a Boolean isomorphism between BaI and TI , with inverse E 7→ πI [E]. P P Because πI is continuous, πI−1 [H] ∈ Baκ for every H ∈ BaI . Of course H 7→ πI−1 [H] is a Boolean homomorphism, and it is injective because πI is surjective. Identifying {0, 1}κ with {0, 1}I × {0, 1}κ\I , we

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535C

have a function h : {0, 1}I → {0, 1}κ defined by setting h(v) = (v, 0) for v ∈ {0, 1}I . This is continuous, therefore (BaI , Baκ )-measurable. If E ∈ TI , then E = πI−1 [πI [E]] = πI−1 [h−1 [E]]; so H 7→ πI−1 [H] is surjective and is an isomorphism. Q Q Consequently #(TI ) ≤Sc for every countable I ⊆ κ (4A1O, because BaI is σ-generated by the cylinder sets, by 4A3Na). For any I, TI = J∈[I]≤ω TJ , because every member of BaI is determined by coordinates in a countable set (4A3Nb). So #(TI ) ≤ max(c, #(I)ω ) = λ whenever I ⊆ κ and #(I) = λ. λ Let φ be a Baire lifting of νκ . ChooseSa non-decreasing family S hJξ iξ<ω1 in [κ] such that J0 = λ and φE ∈ TJξ+1 whenever ξ < ω1 and E ∈ TJξ . Set J = ξ<ω1 Jξ ; then TJ = ξ<ω1 TJξ , so φE ∈ TJ for every E ∈ TJ . We therefore have a Boolean homomorphism φ1 : BaJ → BaJ defined by setting φ1 H = πJ [φ(πJ−1 [H])] for every H ∈ BaJ . If νJ H = 0, then νκ πJ−1 [H] = 0 and φ1 H = φ(πJ−1 [H]) = 0. For any H ∈ BaJ , πJ−1 [H4φ1 H] = πJ−1 [H]4φ(πJ−1 [H]) is νκ -negligible, so H4φ1 H is νJ -negligible. Thus φ1 is a lifting of νJ BaJ . As νJ BaJ is isomorphic to νλ Baλ , the latter also has a lifting. As νλ is completion regular (416U), the measure algebra of νλ Baλ can be identified with Bλ , and we can interpret a lifting of νλ Baλ as a Baire lifting of its completion νλ . 535D The following result covers most of the cases in which non-complete probability measures are known to have liftings. Theorem Let (X, Σ, µ) be a measure space such that µX > 0, and suppose that its measure algebra is tightly ω1 -filtered (definition: 511Di). Then µ has a lifting. proof This is a special case of 518L. 535E Proposition Suppose that c ≤ ω2 and the Freese-Nation number FN(PN) is ω1 . (a) If A is a measurable algebra of cardinal at most ω2 , it is tightly ω1 -filtered. (b) (Mokobodzki 7?) Let (X, Σ, µ) be a σ-finite measure space with Maharam type at most ω2 . (i) µ has a lifting. (ii) If T is a topology on X such that µ is inner regular with respect to the Borel sets, then µ has a Borel lifting. (iii) If T is a topology on X such that µ is inner regular with respect to the zero sets, then µ has a Baire lifting. proof (a) By 524O(b-iii), FN(A) ≤ ω1 , so 518M gives the result. (b)(i) By 514De once more, the measure algebra of µ has cardinal at most ω2ω = max(c, ω2 ) ≤ ω2 (5A1E(e-iv)). So we can put (a) and 535D together. (ii) Because µ is σ-finite and inner regular with respect to the Borel sets, every measurable set can be expressed as the union of a Borel set and a negligible set. By (i), µB(X) has a lifting, which can be interpreted as a Borel lifting of µ, as in 535Ab. (iii) As (ii), but with Ba(X) in place of B(X). 535F With the continuum hypothesis, we can go a little farther. Proposition Let (X, Σ, µ) be a measure space such that µX > 0 and #(A) ≤ ω1 , where A is the measure algebra of µ, and suppose that φ : A → Σ is such that φ0 = ∅,

φ(a ∩ b) = φa ∩ φb for all a, b ∈ A,

(φa)• ⊆ a for every a ∈ A.

Then µ has a lifting θ : A → Σ such that θE • ⊇ φE for every E ∈ Σ. proof (a) Adjusting φ1 if necessary, we can suppose that φ1 = X. Note that φa ⊆ φb whenever a ⊆ b in A. Let haξ iξ<ω1 be a family running over A, and for α ≤ ω1 let Cα be the subalgebra of A generated by {aξ : ξ < α}. Define Boolean homomorphisms θα : Cα → Σ inductively, as follows. The inductive hypothesis will be that (θα c)• = c and θα c ⊇ φc for every c ∈ Cα , while θα extends θβ for every β ≤ α. Start with θ0 0 = ∅, θ0 1 = X. (b) Given θα , where α < ω1 , set S F = {φ(c ∪ aα ) \ θα c : c ∈ Cα }, S G = {φ(c ∪ (1 \ aα )) \ θα c : c ∈ Cα }. Because Cα is countable, F and G belong to Σ. If c ∈ Cα , then

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(φ(c ∪ aα ) \ θα c)• = φ(c ∪ aα )• \ c ⊆ (c ∪ aα ) \ c ⊆ aα , so F • ⊆ aα ; similarly, G• ⊆ 1 \ aα . Next, F ∩ G = ∅. P P If b, c ∈ Cα , then (φ(b ∪ aα ) \ θα b) ∩ (φ(c ∪ (1 \ aα )) \ θα c) = φ((b ∪ aα ) ∩ (c ∪ (1 \ aα ))) \ (θα b ∪ θα c) ⊆ φ(b ∪ c) \ θα (b ∪ c) = ∅. Q Q Choose any E ∈ Σ such that E • = aα and set Eα = (E ∪ F ) \ G; then Eα• = aα , F ⊆ Eα and G ∩ Eα = ∅. If c ∈ Cα and c ⊆ aα , then φ((1 \ c) ∪ aα ) = φ1 = X, so θα c = φ((1 \ c) ∪ aα ) \ θα (1 \ c) ⊆ F ⊆ Eα . Similarly, if c ∈ Cα and c ∩ aα = 0, then θα c = φ((1 \ c) ∪ (1 \ aα )) \ θα (1 \ c) ⊆ G is disjoint from Eα . We can therefore define a Boolean homomorphism θα+1 : Cα+1 → Σ by setting θα+1 ((b ∩ aα ) ∪ (c \ aα )) = (θα b ∩ Eα ) ∪ (θα c \ Eα ) for all b, c ∈ Cα (312O again), and θα+1 will extend θβ for every β ≤ α + 1. Because (θα+1 aα )• = aα and θα+1 c = θα c for every c ∈ Cα , (θα+1 a)• = a for every a ∈ Cα+1 . I have still to check the other part of the inductive hypothesis. If b, c ∈ Cα , then φ((b ∩ aα ) ∪ (c \ aα )) = φ((b ∪ c) ∩ (c ∪ aα ) ∩ (b ∪ (1 \ aα ))) = φ(b ∪ c) ∩ φ(c ∪ aα ) ∩ φ(b ∪ (1 \ aα )) ⊆ θα (b ∪ c) ∩ (F ∪ θα c) ∩ (G ∪ θα b) ⊆ θα+1 (b ∪ c) ∩ (θα+1 aα ∪ θα+1 c) ∩ (θα+1 (1 \ aα ) ∪ θα+1 b) = θα+1 ((b ∩ aα ) ∪ (c \ aα )), which is what we need to know. S S (c) For non-zero limit ordinals α ≤ ω1 , we have Cα = β<α Cβ so we can, and must, take θα = β<α θβ . At the end of the induction, θω1 : A → Σ is an appropriate lifting. 535G Corollary (see Neumann 31) Suppose that c = ω1 . Then for any integer r ≥ 1 there is a Borel lifting θ of Lebesgue measure on Rr such that x ∈ θE • whenever E ⊆ Rr is a Borel set and x is a density point of E. proof In 535F, let φ be lower Lebesgue density (341E), interpreted as a function from the Lebesgue measure algebra to the Borel σ-algebra. We need to check that φE • is indeed always a Borel set; this is because φE • = {x : limn→∞

µ(E∩B(x,2−n )) µB(x,2−n )

=1

and the functions x 7→ µ(E ∩ B(x, 2−n )) are all continuous (use 443A/443B). 535H Again using the continuum hypothesis, we have some results on ‘strong’ liftings, as described in §453. Theorem Let (X, T, Σ, µ) be an effectively locally finite τ -additive topological measure space with measure algebra A. If #(A) ≤ add µ and µ is strictly positive, then µ has a strong lifting. proof (a) For each a ∈ A, set a=

T

{F : F ⊆ X is closed, F • ⊇ a}.

Then a is closed and a• ⊇ a (414Ac). If a, b ∈ A, then a ∪ b = a ∪ b. P P Of course a ∪ b ⊇ a ∪ b, because the operation is order-preserving. On the other hand, a ∪ b is a closed set and (a ∪ b)• ⊇ a ∪ b, so a ∪ b ⊇ a ∪ b. Q Q For a subalgebra B of A, say that a function θ : B → Σ is ‘potentially a strong lifting’ if it is a Boolean homomorphism and (θb)• = b and θb ⊆ b for every b ∈ B. (b) (The key.) Suppose that B is a subalgebra of A, of cardinal less than add µ, and c ∈ A; let B1 be the subalgebra of A generated by B ∪ {c}. If θ : B → Σ is potentially a strong lifting, then it has an extension θ1 : B1 → Σ which is also potentially a strong lifting. P P Set S C0 = {θa : a ∈ B, a ⊆ c}, T D0 = {θb : b ∈ B, c ⊆ b},

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S C1 = {θa \ a \ c : a ∈ B}, T D1 = {(X \ θb) ∪ b ∩ c : b ∈ B}. Fix E0 ∈ Σ such that E0• = c. If a, a0 , b, b0 ∈ B and a0 ⊆ c ⊆ b0 , then a0 ⊆ b0 , so θa0 ⊆ θb0 ; θa0 ∩ θb = θ(a0 ∩ b) ⊆ a0 ∩ b ⊆ b ∩ c, so θa0 ⊆ (X \ θb) ∪ b ∩ c; θa \ θb0 = θ(a \ b0 ) ⊆ a \ b0 ⊆ a \ c, so θa \ a \ c ⊆ θb0 ; θa ∩ θb = θ(a ∩ b) ⊆ a ∩ b = a ∩ b ∩ c ∪ a ∩ b \ c ⊆ a \ c ∪ b ∩ c, so θa \ a \ c ⊆ (X \ θb) ∪ b ∩ c. This shows that C0 ∪ C1 ⊆ D0 ∩ D1 . At the same time, E0• = c ⊇ a0 , so θa0 \ E0 is negligible; E0• = c ⊆ b0 , so E0 \ θb0 is negligible; (E0 ∪ a \ c)• ⊇ c ∪ (a \ c) ⊇ a = (θa)• so (θa \ a ∩ c) \ E0 is negligible; •

E0• = c ⊆ (1 \ b) ∪ (b ∩ c) ⊆ (X \ θb)• ∪ b ∩ c , so E0 \ ((X \ θb) ∪ b ∩ c) is negligible. Because #(B) < add µ, (C0 ∪ C1 ) \ E0 and E0 \ (D0 ∩ D1 ) are measurable and negligible. If we set E = (E0 ∪ C0 ∪ C1 ) ∩ (D0 ∩ D1 ), then E ∈ Σ, E = c and C0 ∪ C1 ⊆ E0 ⊆ D0 ∩ D1 . So we can set θ1 c = E to define a homomorphism from B1 to Σ (312O), and we shall have (θ1 d)• = d for every d ∈ B1 . We must check that θ1 d ⊆ d for every d ∈ B1 . Now d is expressible as (b ∩ c) ∪ (a \ c) for some a, b ∈ B, and in this case •

θb ∩ E ⊆ θb ∩ ((X \ θb) ∪ b ∩ c) ⊆ b ∩ c, θa \ E ⊆ θa \ (θa \ a \ c) ⊆ a \ c, so θ1 d = (θb ∩ E) ∪ (θa \ E) ⊆ b ∩ c ∪ a \ c = d. So θ1 is a potential strong lifting, as required. Q Q (c) Enumerate A as haξ iξ∈κ where κ ≤ add µ, and for α ≤ κ let Bα be the subalgebra of A generated by {aξ : ξ < α}. Then (b) tells us that we can choose inductively a family hθα iα<κ such that S θα : Bα → Σ is a potential strong lifting and θα+1 extends θα for each α < κ. (At non-zero limit ordinals α, Bα = ξ<α Bξ so we can take θα to be the common S extension of ξ<α θξ . We need to know that µ is strictly positive in order to be sure that 1 = X, so that we can take θ0 1 = X.) In this way we obtain a lifting θ = θκ of µ. Also θa ⊆ a for every a ∈ A. Looking at this from the other side, if F ⊆ X is closed then F • ⊆ F so θ(F • ) ⊆ F , and θ is a strong lifting. 535I Corollary (see Mokobodzki 75) Suppose that c = ω1 . Let (X, T, Σ, µ) be a strictly positive σ-finite quasiRadon measure space with Maharam type at most ω1 = c. Then µ has a strong Borel lifting. proof Because µ is σ-finite, its measure algebra A is ccc, and has size at most cω = ω1 ; so we can apply 535H. 535J Under certain conditions, we can deduce the existence of a strong lifting from the existence of a lifting. The basic case is the following. Lemma Let (X, T, Σ, µ) be a completely regular totally finite topological measure space with a Borel lifting φ. Suppose that K ⊆ X is a self-supporting set of non-zero measure, homeomorphic to {0, 1}N , such that K ∩ G ⊆ φG for every open set G ⊆ X. Then the subspace measure µK has a strong Borel lifting.

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proof (a) Taking E to be the algebra of relatively open-and-closed subsets of K, we have a Boolean homomorphism ψ0 : E → B(X) such that E ⊆ int ψ0 E for every E ∈ E. P P We have a Boolean-independent sequence hEn in∈N in E which generates E and separates the points of K. Because every member of E is compact, we can choose for each n ∈ N an open Hn ⊆ X such that En = K ∩ Hn = K ∩ H n . Define h : X → K by saying that, for every n ∈ N and x ∈ X, h(x) ∈ En iff x ∈ Hn . Define ψ0 : E → B(X) by setting ψ0 E = h−1 [E] for E ∈ E. Then ψ0 is a Boolean homomorphism. The set {E : E ∈ E, E ⊆ int ψ0 E, K \ E ⊆ int ψ0 (K \ E)} is a subalgebra of E containing every En , so is the whole of E, and ψ0 has the required property. Q Q (b) Let A be the measure algebra of µ, and θ : A → B(X) the lifting corresponding to φ. Set ψ1 E = (ψ0 E)• for E ∈ E, so that ψ1 : E → A is a Boolean homomorphism. Let I be the null ideal of µK . Because K is self-supporting, E ∩ I = {∅}. Taking E 0 = {E4F : E ∈ E, F ∈ I}, E 0 is a subalgebra of PK, and we have a Boolean homomorphism ψ 0 : E 0 → E defined by setting ψ 0 (E4F ) = E whenever E ∈ E and F ∈ I; set ψ10 = ψ1 ψ 0 , so that ψ10 : E 0 → A is a Boolean homomorphism extending ψ1 , and ψ10 F = 0 whenever F ∈ I. Because µ is totally finite, A is Dedekind complete, and there is a Boolean homomorphism ψ˜1 : PK → A extending ψ10 (314K). Now set φ1 E = K ∩ (φE ∪ (θψ˜1 E \ φK)) for every measurable E ⊆ K. Then φ1 is a strong lifting of µK . P P φΣK is a Boolean homomorphism from the domain ΣK of µK to B(φK), while E 7→ θψ˜1 E \ φK is a Boolean homomorphism from ΣK to B(X \ φK); putting these together, φ1 is a Boolean homomorphism from ΣK to B(K). If E ∈ ΣK , then E4(K ∩ φE) and K \ φK are negligible, so E4φ1 E is negligible. If E ∈ ΣK is negligible, then φE = ψ10 E = ∅ and φ1 E is empty. Thus φ1 is a lifting of µK . Morover, if E ∈ E, set G = int ψ0 E, so that E ⊆ K ∩ G. In this case, E ⊆ φG = θG• ⊆ θ(ψ0 E)• = θψ1 E = θψ˜1 E, while E ∩ φK ⊆ φG ∩ φK = φE; so E ⊆ φ1 E. So if V ⊆ K is relatively open, S S V = {E : E ∈ E, E ⊆ V } ⊆ {φ1 E : E ∈ E, E ⊆ V } ⊆ φ1 V . Thus φ1 is strong. Q Q 535K Lemma Let X be a metrizable space, µ an atomless Radon measure on X and ν an atomless strictly positive Radon measure on {0, 1}N . Let K be the family of those subsets K of X such that K, with the subspace topology and measure, is isomorphic to {0, 1}N with its usual topology and a scalar multiple of ν. Then µ is inner regular with respect to K. proof (a) It will be helpful to note that if E ∈ dom µ and γ < µE there is a compact set K ⊆ E such that µK = γ. P P Let hγn in∈N be a strictly decreasing sequence with γ0 < µE and inf n∈N γn = γ. Choose hKn in∈N , hEn in∈N inductively as follows. E0 = E. Given that µEn > γn , let Kn ⊆ En be a compact set such that µKn ≥ γn ; now T let En+1 be a measurable set with measure γn (215D, because µ is atomless). At the end of the induction, set K = n∈N Kn . Q Q (b) Now for the main argument. Suppose that 0 ≤ γ < µE. Let S hγn in∈N be a strictly decreasing sequence with γ0 < µE and inf n∈N γn = γ. Set γn0 = 21 (γn +γn+1 ) for each n. For σ ∈ n∈N {0, 1}n , set Iσ = {z : σ ⊆ z ∈ {0, 1}N }. Let K0 be a compact subset of E of measure γ0 ; because X is metrizable, K0 is second-countable; let hVn in∈N run over a base for the topology of K0 . Choose hm(n)in∈N and Lσ , for σ ∈ {0, 1}m(n) , as follows. Start with m(0) = 0 and L∅ = K0 . Given that hLσ iσ∈{0,1}m(n) is a disjoint family of compact subsets of X with µLσ = γn νIσ for every σ ∈ {0, 1}m(n) , let m(n + 1) > m(n) be so large that γn+1 νIτ < (γn − γn+1 )νIσ whenever σ ∈ {0, 1}m(n) and τ ∈ {0, 1}m(n+1) . (This is where we need to know that ν is atomless and strictly positive.) Now, for each σ ∈ {0, 1}m(n) , enumerate {τ : σ ⊆ τ ∈ {0, 1}m(n+1) } as hτ (σ, i)ii<2m(n+1)−m(n) . Choose inductively disjoint compact sets Lτ (σ,i) ⊆ Lσ , for i < 2m(n+1)−m(n) , in such a way that µLτ (σ,i) = γn+1 νIτ (σ,i) and Lτ (σ,i) is always either included S in Vn or disjoint from it; this will be possible because when we come to choose Lτ (σ,i) , the measure of the set F = Lσ \ j
X

γn+1 νIτ (σ,j) ≥ (γn − γn+1 )νIσ + γn+1 νIτ (σ,i)

j
> 2γn+1 νIτ (σ,i) , so at least one of F ∩ Vn , F \ Vn will be of measure greater than γn+1 νIτ (σ,i) . Continue.

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S T Set Kn = {Lσ : σ ∈ {0, 1}m(n) } for each n ∈ N, and K = n∈N Kn . The construction ensures that whenever n ≤ k, σ ∈ {0, 1}m(n) , τ ∈ {0, 1}m(k) and σ ⊆ τ , then Lτ ⊆ Lσ . We therefore have a function f : K → {0, 1}N defined by saying that f (x)m(n) = σ whenever n ∈ N, σ ∈ {0, 1}m(n) and x ∈ K ∩ Lσ . Because all the Lσ are compact, f is continuous. But it is also injective. P P If x, y ∈ K are different, there is an n ∈ N such that x ∈ Vn and y ∈ / Vn ; now x and y must belong to different members of Ln+1 . Q Q For any n ∈ N, σ ∈ {0, 1}m(n) and k ≥ n, S P µ( {Lτ : σ ⊆ τ ∈ {0, 1}m(k) }) = σ⊆τ ∈{0,1}m(k) γk νIτ = γk νIσ . So µ(f −1 [Iσ ]) = inf k≥n γk νIσ = γνIσ . S Thus the Radon measure µf −1 on {0, 1}N agrees with the Radon measure γν on {Iσ : σ ∈ n∈N {0, 1}m(n) }; as this is a base for the topology of {0, 1}N closed under finite intersections, µf −1 and γν are identical (415H(v)). Once again because ν is strictly positive, f is surjective and is a homeomorphism. So f witnesses that K ∈ K. As E and γ are arbitrary, µ is inner regular with respect to K. 535L Lemma (a) If (X, T) is a separable metrizable space, there is a zero-dimensional separable metrizable topology S on X, finer than T, with the same Borel sets as T, such that T is a π-base for S. (b) If X is a non-empty zero-dimensional separable metrizable space without isolated points, it is homeomorphic to a dense subset of {0, 1}N . (c) Any completely regular space of size less than c is zero-dimensional. proof (a) Enumerate a countable base for T as hUn in∈N . Define a sequence hSn in∈N of topologies on X by saying that S0 = T and that Sn+1 is the topology on X generated by Sn ∪ {Vn }, where Vn is the closure of Un for Sn . Inducing on n, we see that S Sn is second-countable and has the same Borel sets as T, for every n. So taking S to be the topology generated by n∈N Sn (that is, the topology generated by {Un : n ∈ N} ∪ {Vn : n ∈ N}), this also is second-countable and has the same Borel sets as T. Each Vn is open for Sn+1 and closed for Sn , so is open-and-closed for S. Moreover, since S S T Un = {Um : m ∈ N, U m ⊆ Un } = {Vm : m ∈ N, Vm ⊆ Un } for each n, {Vn : n ∈ N} is a base for S consisting of open-and-closed sets for S, and S0 is zero-dimensional. Finally, observe that if Vn is not empty, then Vn ⊇ Un 6= ∅, so T ⊇ {Un : n ∈ N} is a π-base for S. (b) The family E0 of open-and-closed subsets of X is a base for the topology of X, so includes a countable base U (4A2P(a-iii)). Because X has no isolated points, the subalgebra E1 of E0 generated by U is countable, atomless and non-trivial, and must be isomorphic to the algebra E of open-and-closed subsets of {0, 1}N (316M2 ). Let π : E → E1 be an isomorphism. Then we have a function f : X → {0, 1}N defined by saying that, for E ∈ E, f (x) ∈ E iff x ∈ πE. Because πE 6= ∅ for every non-empty E ∈ E, f [X] is dense in {0, 1}N . Because {f −1 [E] : E ∈ E} = E1 ⊇ U is a base for the topology of X, f is a homeomorphism between X and f [X]. (c) If X is a completely regular space and #(X) < c, G ⊆ X is open and x ∈ G, let f : X → [0, 1] be a continuous function such that f (x) = 1 and f (y) = 0 for y ∈ X \ G. Because #(X) < c, there is an α ∈ [0, 1] \ f [X], and now {y : f (x) > α} = {y : f (x) ≥ α} is an open-and-closed set containing x and included in G. As x and G are arbitrary, X is zero-dimensional. 535M Lemma Suppose that there is a Borel probability measure on {0, 1}N with a strong lifting. Then whenever X is a separable metrizable space and D ⊆ X is a dense set, there is a Boolean homomorphism φ from PD to the Borel σ-algebra B(X) of X such that φA ⊆ A for every A ⊆ D. proof case 1 Suppose that X is countable. Then it is zero-dimensional (535Lc), so has a base U consisting of open-andclosed sets; let E be the algebra of sets generated by U. For E ∈ E set πE = E ∩ D; then π is an isomorphism between E and a subalgebra E 0 of PD. Because B(X) = PX is Dedekind complete, the Boolean homomorphism π −1 : E 0 → E extends to a Boolean homomorphism φ : PD → PX = B(X) (314K). If A ⊆ D and x ∈ X \ A, then there is a U ∈ U such that x ∈ U and A ∩ U = ∅, in which case φA ⊆ π −1 (D \ U ) = X \ U does not contain x. As x is arbitrary, φA ⊆ A; as A is arbitrary, φ has the required property. case 2 Suppose that X is zero-dimensional and has no isolated points. If X is empty the result is trivial; otherwise, by 535Lb, we may suppose that X is a dense subset of {0, 1}N . This time, let E be the algebra of open-and-closed 2 Formerly

393F.

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subsets of {0, 1}N . For E ∈ E, set πE = E ∩ D. Because D is dense in X and therefore in {0, 1}N , π is an isomorphism between E and a subalgebra E 0 of PD. Fix a Borel probability measure µ on {0, 1}N with a strong lifting θ, and let A be the measure algebra of µ. Then A 7→ (π −1 A)• is a Boolean homomorphism from E 0 to A; because A is Dedekind complete, it extends to a Boolean homomorphism ψ : PD → A. For E ⊆ {0, 1}N , set π ˜ E = E ∩ X. Then φ = π ˜ θψ is a Boolean homomorphism from PD to B(X). If A ⊆ D and x ∈ X \ A, then there is an E ∈ E such that x ∈ E and A ∩ E = ∅, in which case φA ⊆ θψA ⊆ θψ(D \ E) = θ({0, 1}N \ E)• = {0, 1}N \ E, and x ∈ / φA. As x and A are arbitrary, φ is a suitable homomorphism. case 3 Suppose that X has no isolated points. Write T for the given topology on X. By 535La, there is a finer zero-dimensional separable metrizable topology S on X, with the same Borel sets, such that T is a π-base for S. If V ∈ S is non-empty, there is a non-empty U ∈ T such that U ⊆ V , and D ∩ V ⊇ D ∩ U is non-empty; so D is S S-dense. By case 2, there is a Boolean homomorphism φ : PD → B(X, S) such that φA ⊆ A for every A ⊆ D. As S T B(X, S) = B(X, T), and A ⊆ A for every A ⊆ X, this φ satisfies the conditions required. S general case In general, let G be the family of countable open subsets of X, and G0 = G; because X is separable and metrizable, therefore hereditarily Lindel¨of, G0 is countable. Set Z = X \ G0 , and let D0 be a countable dense subset of Z; set Y = D ∪ G0 ∪ D0 . By case 1, there is a Boolean homomorphism φ0 : PD → PY such that φ0 A ⊆ A for every A ⊆ D. By case 3, there is a Boolean homomorphism φ1 : P(Y ∩ Z) → B(Z) such that φ1 B ⊆ B for every B ⊆ Y ∩ Z. Now set φA = (φ0 A \ Z) ∪ φ1 (Z ∩ φ0 A) for every A ⊆ D. Then φ is a Boolean homomorphism from PD to B(X); and if A ⊆ D, then φA ⊆ φ0 A ∪ φ1 (Z ∩ φ0 A) ⊆ A ∪ Z ∩ A = A, so in this case also we have a homomorphism of the kind we need. 535N Theorem Suppose there is a metrizable space X with a non-zero atomless semi-finite tight Borel measure µ which has a lifting. Then whenever Y is a metrizable space and ν is a strictly positive σ-finite Borel measure on Y , ν has a strong lifting. proof (a) Let φ be a lifting of µ. Then there is a Borel set E ⊆ X, of non-zero finite measure, such that E ∩ G ⊆ φG for every open GS⊆ X. P P Let L0 ⊆ X be a compact set of non-zero measure; then L0 has a countable base U; set E = L0 ∩ φL0 \ U ∈U (U 4φU ), so that µE = µL0 ∈ ]0, ∞[. If G ⊆ X is open and x ∈ E ∩ G, then there is a U ∈ U such that x ∈ U ⊆ G. Since x ∈ E ∩ U , x ∈ φU ⊆ φG. As x and G are arbitrary, we have an appropriate E. Q Q (b) Let λ be any strictly positive atomless Radon measure on {0, 1}N . There is a compact set K ⊆ E such that K, with its induced topology and measure, is isomorphic to {0, 1}N with its usual topology and a non-zero multiple of λ, by 535K. In particular, K is self-supporting. By 535J, the subspace measure on K has a strong Borel lifting. It follows at once that λ has a strong Borel lifting. (c) Refining (b) slightly, we see that if Y ⊆ {0, 1}N is a dense set and λ is a strictly positive atomless totally finite Borel measure on Y , then λ has a strong lifting. P P There is a Radon measure ν on {0, 1}N such that νE = λ(Y ∩ E) N for every Borel set E ⊆ {0, 1} ; because λ is atomless, so is ν; because λ is strictly positive and Y is dense, ν is strictly positive. So ν has a strong Borel lifting ψ0 say. If E, F ∈ B({0, 1}N ) and E ∩ Y = F ∩ Y , then ν(E4F ) = 0 and ψ0 E = ψ0 F ; we therefore have a Boolean homomorphism ψ : B(Y ) → B(Y ) defined by setting ψ(E ∩ Y ) = Y ∩ ψ0 E for every Borel set E ⊆ {0, 1}N . It is easy to check that ψ is a lifting of λ, and it is strong because if G ⊆ {0, 1}N is open then ψ(Y ∩ G) = Y ∩ ψ0 G ⊆ Y ∩ G. Q Q (d) If (Y, S) is a separable metrizable space with a strictly positive atomless totally finite Borel measure ν, then ν has a strong lifting. P P If Y = ∅ the result is trivial. Otherwise, by 535La, there is a finer separable metrizable topology S0 on Y with the same Borel sets such that S is a π-base for S0 . Because S and S0 have the same Borel sets, ν is a Borel measure for S0 ; because every non-empty S0 -open set includes a non-empty S-open set, ν is strictly positive for S0 ; because ν is atomless, Y has no S0 -isolated points. By 535Lb, (Y, S0 ) is homeomorphic to a dense subset of {0, 1}N ; by (c) above, ν has a lifting φ which is strong with respect to the topology S0 . But now φ is still strong with respect to the coarser topology S. Q Q (e) Now suppose that Y is a separable metrizable space with a strictly positive totally finite Borel measure ν. Then ν has a strong lifting. P P The set D = {y : ν{y} > 0} is countable. If D is empty, then the result is immediate from (d) applied to a scalar multiple of ν. (If νY = 0 then Y = ∅ and the result is trivial.) Otherwise, let νY \D be the subspace measure; then νY \D is a totally finite Borel measure on Y \ D, and is zero on singletons, so must be atomless. Because

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Y \ D is hereditarily Lindel¨ of, νY \D is τ -additive; let Z be its support, and νZ the subspace measure on Z. Then νZ has a strong Borel lifting ψ0 , by (d) again. Next, Z is relatively closed in Y \ D, so is expressible as F \ D for some closed set F ⊆ Y . If x ∈ Y \ F and G is an open set containing x, then G0 = G \ F is a non-empty open set, so has non-zero measure, while νY \D (G0 \ D) = 0; accordingly G0 ∩ D 6= ∅. This shows that Y \ F ⊆ D so D is dense in Y \ Z. Now 535M (with (b) above) tells us that there is a Boolean homomorphism ψ1 : PD → B(Y \ Z) such that ψ1 A ⊆ A for every A ⊆ D. Define ψ : B(Y ) → B(Y ) by setting ψE = ψ0 (E ∩ Z) ∪ (E ∩ D) ∪ (ψ1 (E ∩ D) \ D) for every Borel set E ⊆ Y . ψ is a Boolean homomorphism because ψ0 and ψ1 are. If νE = 0, then νZ (E ∩ Z) = 0 and E ∩ D = ∅, so ψE = 0. For any E ∈ B(Y ), ψ0 (E ∩ Z)4(E ∩ Z) and Y \ (D ∪ Z) are negligible, so E4ψE is negligible. Thus ψ is a lifting of ν. Finally, for any E, ψE ⊆ ψ0 (E ∩ Z) ∪ (E ∩ D) ∪ ψ1 (E ∩ D) ⊆ E, so ψ is a strong lifting. Q Q (f ) Finally, if Y is a metrizable space and ν is a strictly positive σ-finite Borel measure on Y , then Y must be ccc, therefore separable; and there is a totally finite Borel measure ν 0 with the same null ideal as ν, so that ν 0 has a strong lifting, by (e), which is also a strong lifting of ν. 535O Linear liftings Let (X, Σ, µ) be a measure space, with measure algebra A. Write L∞ (Σ) for the space of bounded Σ-measurable real-valued functions on X. A linear lifting of µ is either a positive linear operator T : L∞ (µ) → L∞ (Σ) such that T (χX • ) = χX and (T u)• = u for every u ∈ L∞ (µ) or a positive linear operator S : L∞ (Σ) → L∞ (Σ) such that S(χX) = χX, Sf = 0 whenever f =a.e. 0 and Sf =a.e. f for every f ∈ L∞ (Σ). As with liftings (see 341A-341B) we have a direct correspondence between the two kinds of linear operator; given T as in the first formulation, we can set Sf = T (f • ) for every f ∈ L∞ (Σ); given S as in the second formulation, we can set T (f • ) = Sf for every f ∈ L∞ (Σ). If θ : A → Σ is a lifting of µ, then we have a corresponding Riesz homomorphism T : L∞ (A) → L∞ (Σ) such that T (χa) = χ(θa) for every a ∈ A, by 363F. Identifying L∞ (A) with L∞ (µ) as in 363I, we see that T can be regarded as a linear lifting. Of course the associated linear operator from L∞ (Σ) to itself is the operator derived by the process of 363F from the Boolean homomorphism E 7→ θE • : Σ → Σ. As in 535Aa, I will say that a Borel linear lifting is a linear lifting such that all its values are Borel measurable functions; similarly, a Baire linear lifting is a linear lifting such that all its values are Baire measurable functions. 535P I give a sample result to show that for some purposes linear liftings are adequate. Proposition Let (X, Σ, µ) be a countably compact measure space such that Σ is countably generated, (Y, T, ν) a σ-finite measure space with a linear lifting, and f : X → Y an inverse-measure-preserving function. Then there is a disintegration hµy iy∈Y of µ over ν, consistent with f , such that y 7→ µy E is a T-measurable function for every E ∈ Σ. proof I use the method of 452H-452I. ∞ (a) Suppose first that µ and ν are probability measures. Let S : L∞ (ν) ν. Let R → L R (T) be a linear lifting of ∞ ∞ T : L (µ) → L (ν) be the positive linear operator defined by saying that F T u = f −1 [F ] u whenever u ∈ L∞ (µ) and F ∈ T (as in part (a) of the proof of 452I). For y ∈ Y and E ∈ Σ, set

ψy E = (ST (χE • ))(y) as in part (b) of the proof of 452H. Because µ is countably compact, we can use the argument of 452H to see that we have a family hµ0y iy∈Y of totally finite measures on X such that, for any E ∈ Σ, µ0y E = ψy E for almost every y ∈ Y . Let H be a countable subalgebra of Σ such that Σ is the σ-algebra of sets generated by H. Set Y0 = {y : µ0y H = ψy H for every H ∈ H}, so that Y0 is conegligible; let Y1 ⊆ Y0 be a measurable conegligible set; set µy = µ0y for y ∈ Y1 , and take µy to be the zero measure on X for y ∈ Y \ Y1 . If H ∈ H, then µy H = ψy H = ST (χH • )(y) for every y ∈ Y1 , so y 7→ µy H is T-measurable; also, of course,

R

R

R

R

χH • = µ(H ∩ f −1 [F ]). R Now consider the family E of those E ∈ Σ such that y 7→ µy E is T-measurable and F µy Eν(dy) = µ(E ∩ f −1 [F ]) for every F ∈ T. This is a Dynkin class including H, so is the whole of Σ; which is what we need to know. F

µy Hν(dy) =

F

ST (χH • )dν =

F

T (χH • ) =

f −1 [F ]

535Xd

Liftings

237

(b) In general, if νY = 0, the result is trivial. Otherwise, apply (a) to a suitable pair of indefinite-integral measures over µ and ν, as in part (c) of the proof of 452I. 535Q Proposition Let (X, Σ, µ) and (Y, T, ν) be probability spaces, and λ the c.l.d. product measure on X × Y . b has a linear lifting. Then µ has a linear lifting. Suppose that λΣ⊗T R b → L∞ (Σ⊗T) b b b proof Let S : L∞ (Σ⊗T) be a linear lifting of λΣ⊗T. For h ∈ L∞ (Σ⊗T), set (U h)(x) = h(x, y)ν(dy) b for every x ∈ X; by 252P, U h is well-defined and is Σ-measurable. Now U is a positive linear operator from L∞ (Σ⊗T) to L∞ (Σ), and U (χ(X × Y )) = χX, because νY = 1. Note that

R

|U h|dµ ≤

R

U |h|dµ =

RR

|h(x, y)|ν(dy)µ(dx) =

R

|h|dλ

b for every ∈ L∞ (Σ⊗T) (252P again). Next, for f ∈ L∞ (Σ) set (V f )(x, y) = f (x) for every x ∈ X and y ∈ Y , so that b V is a positive linear operator from L∞ (Σ) to L∞ (Σ⊗T). ∞ ∞ Consider S1 = U SV : L (Σ) → L (Σ). This is a positive linear operator and S1 (χX) = χX. If f ∈ L∞ (Σ) and f = 0 µ-a.e., then V f = 0 λ-a.e. and SV f = 0, so S1 f = 0. For any f ∈ L∞ (Σ),

R

|f − S1 f |dµ =

R

|f − U SV f |dµ =

R

|U V f − U SV f |dµ ≤

R

|V f − SV f |dλ = 0,

so f =a.e. S1 f ; thus S1 is a linear lifting of µ. 535R Proposition Write νω2 for the usual measure on ({0, 1}ω )2 , and T(2) ω for its domain. Suppose that νκ has a Baire linear lifting for some κ ≥ c++ . Then there is a Borel linear lifting S of νω2 which respects coordinates in the sense that if f ∈ L∞ (T(2) ω ) is determined by a single coordinate, then Sf is determined by the same coordinate. proof Because ({0, 1}κ , νκ ) is isomorphic, as topological measure space, to ({0, 1}κ×ω , νκ×ω ), the latter has a Baire linear lifting S0 say. For I ⊆ κ, let TI be the σ-algebra of Baire S subsets of {0, 1}κ×ω determined by coordinates in κ×ω I × ω. Then #(TI ) ≤ c whenever #(I) ≤ c. Also Ba({0, 1} ) = {TI : I ∈ [κ]≤ω } (4A3N). It follows that for every ξ < κ there is a set Iξ ⊆ κ, of size at most c, such that ξ ∈ Iξ and S0 (χE) is TIξ -measurable whenever E ∈ TIξ ; so that S0 f is TIξ -measurable whenever f : {0, 1}κ×ω → R is bounded and TIξ -measurable. Because κ ≥ c++ , there are ξ, η < κ such that ξ ∈ / Iη and η ∈ / Iξ (5A1I(a-iii)). Set J = {ξ} × ω, K = {η} × ω and L = (κ × ω) \ (J ∪ K), so that {0, 1}κ×ω can be identified with {0, 1}J∪K × {0, 1}L and Ba({0, 1}κ×ω ) with b Ba({0, 1}J∪K )⊗Ba({0, 1}L ). RSet (V f )(w, z) = f (w) when f : {0, 1}J∪K → R is a function, w ∈ {0, 1}J∪K and L z ∈ {0, 1} ; and (U h)(w) = h(w, z)νL (dz) when h : {0, 1}κ×ω → R is a bounded Baire measurable function and w ∈ {0, 1}J∪K . Then S1 = U S0 V is a Baire linear lifting of νJ∪K , just as in 535Q. Moreover, if f : {0, 1}J∪K → R is a bounded Baire measurable function determined by coordinates in J, in the sense that f (x, y) = f (x, y 0 ) whenever x ∈ {0, 1}J and y, y 0 ∈ {0, 1}K , then S1 f is determined by coordinates in J. P P V f is determined by coordinates in J, so S0 V f is determined by coordinates in Iξ × ω; since K ∩ (Iξ × ω) is empty, S0 V f (x, y, z) = S0 V f (x, y 0 , z) for all x ∈ {0, 1}J , z ∈ {0, 1}L and y, y 0 ∈ {0, 1}K . It follows at once that S1 f (x, y) =

R

S0 V f (x, y 0 , z)νL (dz) =

R

S0 V f (x, y 0 , z)νL (dz) = S1 f (x, y 0 )

whenever x ∈ {0, 1}J and y, y 0 ∈ {0, 1}K . Q Q Similarly, if f : {0, 1}J∪K → R is a bounded Baire measurable function determined by coordinates in K, then S1 f is determined by coordinates in K. Now we can transfer S1 from {0, 1}J∪K ∼ = {0, 1}J × {0, 1}K to ({0, 1}ω )2 , and we shall obtain a Baire (or Borel) 2 linear lifting S of νω which respects coordinates. 535X Basic exercises (a) Let (X, Σ, µ) be a measure space with a lifting, and A any subset of X. Show that if A has a measurable envelope then the subspace measure µA has a lifting. (Hint: 322I.) (b) Let h(Xi , Σi , µi )ii∈I be a family of measure spaces, with µi Xi > 0 for every i ∈ I, and (X, Σ, µ) their direct sum. Show that µ has a lifting iff every µi has a lifting. (c) Let A be a Boolean algebra and I a proper ideal of A. Suppose that sup A is defined in A and belongs to I whenever A ⊆ I and #(A) < #(A). Show that there is a Boolean homomorphism θ : A/I → A such that (θb)• = b for every b ∈ A/I. (Hint: enumerate A as {aξ : ξ < κ}; let Cξ be the subalgebra of A/I generated by {a•η : η < ξ}; construct θ Cξ inductively by choosing θa•ξ appropriately.) (d) Let A be a Dedekind σ-complete Boolean algebra and I a proper ideal of A. Show that if the quotient Boolean algebra A/I is tightly ω1 -filtered, then there is a Boolean homomorphism θ : A/I → A such that (θb)• = b for every b ∈ A/I.

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(e) Let A be a tightly ω1 -filtered Boolean algebra, B a Dedekind σ-complete Boolean algebra and A0 a countable subalgebra of A. Show that every Boolean homomorphism from A0 to B extends to a Boolean homomorphism from A to B. (f ) Let A, B be Dedekind σ-complete Boolean algebras and π : A → B a surjective sequentially order-continuous Boolean homomorphism. Suppose that φ : B → A is such that φ0 = 0, πφb ⊆ b for every b ∈ B and φ(b ∩ c) = φb ∩ φc for all b, c ∈ B. Show that if #(B) ≤ ω1 , there is a Boolean homomorphism θ : B → A such that φb ⊆ θb and πθb = b for every b ∈ B. (g) Suppose that c ≤ ω2 and FN(PN) = ω1 . Show that νκ has a strong Baire lifting whenever κ ≤ ω2 . (Hint: let heξ iξ<κ be the standard generating family for Bκ . Show that there is a tight ω1 -filtration haη iη<ζ of Bκ such that for every ξ < κ there is an η < ζ such that the closed subalgebras generated by {eδ : δ < ξ} and {aδ : δ < η} are the same and eξ = aη .) (h) Suppose that c ≤ ω2 and FN(PN) = ω1 . Show that whenever X is a separable metrizable space and D ⊆ X is a dense set, there is a Boolean homomorphism φ : PD → B(X) such that φA ⊆ A for every A ⊆ D. (i) Let (X, Σ, µ) be a measure space. Show that a linear lifting S : L∞ (Σ) → L∞ (Σ) of µ corresponds to a lifting iff it is ‘multiplicative’, that is, S(f × g) = Sf × Sg for all f , g ∈ L∞ (Σ). (j) Let (X, Σ, µ) be a strictly localizable measure space with non-zero measure. Suppose that νκ has a Baire linear lifting for every infinite cardinal κ such that the Maharam-type-κ component of the measure algebra of µ is non-zero. Show that µ has a linear lifting. S S (k) Let (X, Σ, µ) be a probability space such that whenever E ⊆ Σ, #(E) ≤ c and E is negligible, then E ∈ Σ. Show that µ has a linear lifting. (Hint: 363Yg.) (l) Let (Y, T, ν) be a σ-finite measure space with a linear lifting, Z a set, Υ a countably generated σ-algebra of b such that ν is the marginal measure of µ on Y . Show that there subsets of Z, and µ a measure with domain T⊗Υ is a family hµ i of measures with domain Υ such that y 7→ µy H is a T-measurable function for every H ∈ Υ and y y∈Y R b µW = µy W [{y}]ν(dy) for every W ∈ T⊗Υ. (m) Let (X, T, Σ, µ) and (Y, S, T, ν) be τ -additive topological probability spaces, and λ the τ -additive product measure on X × Y (417G). Suppose that λ has a Borel linear lifting and that µ is inner regular with respect to the Borel sets. Show that µ has a Borel linear lifting. 535Y Further exercises (a) Suppose that we are provided with a bijection between B(R) and ω1 , but are otherwise not permitted to use the axiom of choice. Show that we can construct a Borel lifting of Lebesgue measure. (b) Suppose that for every cardinal κ there is a Baire linear lifting of νκ . Show that for every n ∈ N there is a Borel linear lifting S of Lebesgue measure on [0, 1]n which (i) respects coordinates in the sense that if f : [0, 1]n → R is a bounded measurable function determined by coordinates in I ⊆ n, then Sf also is determined by coordinates in I (ii) is symmetric in the sense that if ρ : n → n is any permutation and (ˆ ρf )(x) = f (xρ) for x ∈ [0, 1]n and f : [0, 1]n → R, then S commutes with ρˆ. (Hint: 5A1Ib.) (c) Let (X, Σ, µ) be a countably compact measure space, (Y, T, ν) a σ-finite measure space with a linear lifting, and f : X → Y an inverse-measure-preserving function. Suppose there is a family H ⊆ Σ such that Σ is the σ-algebra of sets generated by H and #(H) < add ν. Show that there is a disintegration hµy iy∈Y of µ over ν, consistent with f , such that y 7→ µy E is a T-measurable function for every E ∈ Σ. 535Z Problems (a) Can it be that every probability space has a lifting? By 535B, it is enough to consider ({0, 1}κ , Ba({0, 1}κ ), νκ Ba({0, 1}κ )) where κ is a cardinal. Since Mokobodzki’s theorem (535Eb) deals with κ ≤ ω2 when c = ω1 , the key case to consider seems to be κ = ω3 . (b) Suppose that c ≥ ω3 . Does νω have a Borel lifting? It is known to be relatively consistent with ZFC to suppose that c = ω2 and that FN(PN) = ω1 (554G-554H). In this case νω B({0, 1}ω ) will have a lifting, by 535Eb, so νω has a Borel lifting. But if c ≥ ω3 then Bω is not tightly ω1 -filtered (518S). (c) (A.H.Stone) Can there be a countable ordinal ζ and a lifting φ of νω such that φE is a Borel set, with Baire class at most ζ, for every Borel set E ⊆ {0, 1}ω ? The point of this question is that while, subject to the continuum hypothesis, we can almost write down a formula for a Borel lifting of Lebesgue measure (535Ya), the method gives no control over the Baire classes of the sets constructed.

536B

Alexandra Bellow’s problem

239

(d) Can there be a strictly positive Radon probability measure of countable Maharam type which does not have a strong lifting? (See 453G, 535I, 535Xg.) (e) Is there a probability space which has a linear lifting but no lifting? (f ) Can there be a Borel linear lifting of the usual measure on ({0, 1}ω )2 which respects coordinates in the sense of 535R? It seems possible that there is a proof in ZFC that there is no such lifting; in which case 535R shows that we should have a negative answer to (a). 535 Notes and comments For a fuller account of this topic, see Burke 93. Neumann & Stone 35 used a direct construction along the lines of 535Xc to show that if the continuum hypothesis is true then Lebesgue measure has a Borel lifting. The method works equally well for νω1 , but for νω2 we need a further idea from Mokobodzki 7?; the version I give here is based on Geschke 02, itself derived at some remove from an unpublished result of T.J.Carlson, who showed that we could have a Borel lifting of Lebesgue measure in a model in which the continuum hypothesis is false (554I). It is not a surprise that there should be a model of set theory in which Lebesgue measure has no Borel lifting. Nor is it a surprise that the first such model should have been found by S.Shelah (Shelah 83). What does remain surprising is that in most of the vast number of models of set theory which have been studied, we do not know whether there is such a lifting. Only in the familiar case c = ω1 , the special combination c = ω2 = FN(PN)+ (535E), and in variations of Shelah’s model, do we have definite information. It remains possible that in any model in which m > ω1 or c = ω3 there is no Borel lifting of Lebesgue measure. When we leave the real line, the position is even more open; conceivably it is relatively consistent with ZFC to suppose that every probability space has a lifting, and at least equally believably it is a theorem of ZFC that νω3 does not have a Baire lifting. From 535I we see that ω2 appears in Losert’s example (453N) for a good reason. Once again, it seems to be unknown whether it is consistent to suppose that there is a (completed) strictly positive Radon probability measure with countable Maharam type which has no strong lifting (535Zd). When we come to look for strong Borel liftings, we have some useful information in the separable metrizable case (535N). The result is natural enough. We are used to supposing that Polish spaces are all very much the same, and that point-supported measures are trivial. But because the concept of ‘strong’ lifting is topological, and cannot easily be reduced to the Borel structure, we have to work a bit; and it seems also that point-supported measures need care (535M). ‘Linear liftings’ (535O-535R) remain poor relations. I give them house room here partly for completeness and partly because of a slender hope that they will lead us to a solution of 535Za. Of course the match between ω3 in 535Za and c++ in 535R may show only a temporarily coincidental frontier of ignorance. Burke & Shelah 92 have shown that it is relatively consistent with ZFC to suppose that νω has no Borel linear lifting.

536 Alexandra Bellow’s problem In 463Za I mentioned a curious problem concerning pointwise compact sets of continuous functions. This problem is known to be soluble if we are allowed to assume the continuum hypothesis, for instance. Here I present the relevant arguments. 536A The problem I recall some ideas from §463. Let (X, Σ, µ) be a measure space, and L0 = L0 (Σ) the space of all Σ-measurable functions from X to R, so that L0 is a linear subspace of R X . On L0 we have the linear space topologies Tp and Tm of pointwise convergence and convergence in measure (462Ab, 245Ab). Tp is Hausdorff and locally convex; if µ is σ-finite, Tm is pseudometrizable. The problem, already asked in 463Za, is this: suppose that K ⊆ L0 is compact for Tp , and that Tm is Hausdorff on K. Does it follow that Tp and Tm agree on K? 536B Known cases Let (X, Σ, µ) be a σ-finite measure space. Given that K ⊆ L0 is compact for Tp , and Tm is Hausdorff on K, and either K is sequentially compact for Tp or K is countably tight for Tp or K is convex or X has a topology for which K ⊆ C(X), µ is a strictly positive topological measure, and every function h ∈ R X which is continuous on every relatively countably compact set is continuous or µ is perfect, then K is metrizable for Tp , and Tp and Tm agree on K (463Cd, 463F, 463G, 463H, 463Lc, 463Cc). Now for the new results.

240

Topologies and measures III

536C

536C Theorem Let (X, Σ, µ) be a probability space, and L0 the space of Σ-measurable real-valued functions on X. Write Tp , Tm for the topologies of pointwise convergence and convergence in measure on L0 . Suppose that K ⊆ L0 is Tp -compact and that µ{x : f (x) 6= g(x)} > 0 for any distinct f , g ∈ K, but that K is not Tp -metrizable. (a) Every infinite Hausdorff space which is a continuous image of a closed subset of K has a non-trivial convergent sequence. (b) There is a continuous surjection from a closed subset of K onto {0, 1}ω1 . (c) Every infinite compact Hausdorff space of weight at most ω1 has a non-trivial convergent sequence. (d) c > ω1 . (e) The Maharam type of µ is at least 2ω1 . (f) There is a non-negligible measurable set in Σ which can be covered by ω1 negligible sets. (g) mcountable = ω1 . R proof For f , g ∈ L0 set ρ(f, g) = min(1, |f − g|); then ρ is a pseudometric on L0 defining Tm , and ρK × K is a metric on K. Set ∆(∅) = 0, and for non-empty A ⊆ L0 set ∆(A) = sup{ρ(inf L, sup L) : ∅ 6= L ∈ [A]<ω }. Note that if A ⊆ K has more than one member then ∆(A) > 0, and that ∆(A) ≤ ∆(B) whenever A ⊆ B. (a)(i) ?? Suppose, if possible, that Z is an infinite Hausdorff space, K0 ⊆ K is closed, φ : K0 → Z is a continuous surjection and there is no non-trivial convergent sequence in Z. Write L for the family of closed subsets L of K0 such T that φ[L] is infinite. Then L = n∈N Ln belongs to L for every non-increasing sequence hLn in∈N in L. P P hφ[Ln ]in∈N is a non-increasing sequence of infinite closed subsets of Z; because Z is supposed to have no non-trivial convergent T sequence, M = n∈N φ[Ln ] is infinite (4A2G(h-i)). Since φ[L] = M (5A4Cf), L ∈ L. Q Q By 513O, inverted, there is a K1 ∈ L such that ∆(L) = ∆(K1 ) for every L ∈ L such that L ⊆ K1 . (ii) Now there is no non-trivial convergent sequence in φ[K1 ], so φ[K1 ] cannot be scattered (4A2G(h-ii)), and it must have an infinite closed subset M1 without isolated points which has a countable π-base (4A2G(j-iii)). Let K2 ⊆ K1 be a closed set such that φ[K2 ] = M1 and φK2 is irreducible (4A2G(i-i)). Let V be a countable π-base for the topology of M1 , not containing ∅. For each V ∈ V, choose hV ∈ K2 ∩ φ−1 [V ]. Set g0 = inf V ∈V hV , g1 = supV ∈V hV in R X . Then g0 and g1 are measurable. Setting L = {f : f ∈ K2 , g0 ≤ f ≤ g1 }, φ[L] ⊇ {hV : V ∈ V} is infinite, so L ∈ L and

R

g1 − g0 ≥ ∆(L) = ∆(K1 ) > 0.

Set g(x) = max( 21 (g0 (x) + g1 (x)), g1 (x) − 12 ) for x ∈ X, and E = {x : g0 (x) < g1 (x)} = {x : g(x) < g1 (x)} = {x : g0 (x) < g(x)}, so that µE > 0. For x ∈ E, the set Fx = {f : f ∈ K2 , f (x) ≤ g(x)} is a proper closed subset of K2 , so φ[Fx ] 6= M1 and there is some Vx ∈ V such that Vx ∩ φ[Fx ] = ∅. Because V is countable, there is a V ∈ V such that D = {x : x ∈ E, Vx = V } is non-negligible. But now observe that f (x) > g(x) whenever f ∈ K2 ∩ φ−1 [V ] and x ∈ D, so hU (x) > g(x) whenever U ∈ V, U ⊆ V and x ∈ D. Set V 0 = {U : U ∈ V, U ⊆ V }, g00 = inf U ∈V 0 hU and L0 = {f : f ∈ K2 , g00 ≤ f ≤ g1 }. Then g ≤ g00 and {x : x ∈ X, g1 (x) − g00 (x) < min(1, g1 (x) − g0 (x))} ⊇ D is non-negligible, so ∆(L0 ) ≤

R

min(1, g1 − g00 ) <

R

min(1, g1 − g0 ) = ∆(K1 ).

On the other hand, φ[L0 ] meets every member of V 0 , so is dense in V and (because M1 has no isolated points) must be infinite. And this contradicts the choice of K1 . X X Thus (a) is true. (b) If hfn in∈N is a sequence in K which converges at almost every point of X, then any two Tp -cluster points of hfn in∈N must be equal a.e. and therefore equal, so hfn in∈N is Tp -convergent (5A4Ce). ?? Suppose, if possible, that there is no continuous surjection from a closed subset of K onto {0, 1}ω1 . Then 463D tells us that every sequence in K has a subsequence which is convergent almost everywhere, therefore convergent. So K is sequentially compact, which is impossible, as noted in 536B. X X (c) Since [0, 1] is a continuous image of {0, 1}N , [0, 1]ω1 is a continuous image of {0, 1}ω1 ×N ∼ = {0, 1}ω1 and therefore of a closed subset of K. If Z is an infinite compact Hausdorff space of weight at most ω1 , it is homeomorphic to a closed subset of [0, 1]ω1 (5A4Cc) and therefore to a continuous image of a closed subset of K. By (a), Z must have a non-trivial convergent sequence. (d) Since βN has weight c (5A4Ja), is infinite, but has no non-trivial convergent subsequence (4A2I(b-v)), we must have ω1 < c.

536C

Alexandra Bellow’s problem

241

(e)(i) If F1 , F2 are disjoint non-empty Tp -closed subsets of K, then ρ(F1 , F2 ) > 0. P P?? Otherwise, there are sequences hfn in∈N in F1 , hgn in∈N in F2 such that ρ(fn , gn ) ≤ 2−n for every n ∈ N. Let F be any non-principal ultrafilter on N and set f = limn→F fn , g = limn→F gn , taking the limits in K for the topology Tp . Then, for any n ∈ N, S {x : |f (x) − g(x)| > 2−n } ⊆ i≥2n {x : |fi (x) − gi (x)| > 2−n } P∞ has measure at most i=2n 2−i+n = 2−n+1 , so f =a.e. g and f = g; but f ∈ F1 and g ∈ F2 , so this is impossible. X XQ Q (ii) By (b), there are a closed subset K0 of K and a continuous surjection ψ : K0 → {0, 1}ω1 . For ξ < ω1 , set Fξ = {f : f ∈ K0 , ψ(f )(ξ) = 0}, Fξ0 = {f : f ∈ K0 , ψ(f )(ξ) = 1}; then ρ(Fξ , Fξ0 ) > 0. There must therefore be a δ > 0 such that C = {ξ : ρ(Fξ , Fξ0 ) ≥ δ} is uncountable. For each D ⊆ C, choose hD ∈ K0 such that ψ(hD )C = χD. Then ρ(hD , hD0 ) ≥ δ for all distinct D, D0 ⊆ C. Thus A = {h•D : D ⊆ C} is a subset of L0 = L0 (µ), of cardinal 2ω1 , such that any two members of A are distance at least δ apart for the metric on L0 corresponding to ρ. Accordingly the cellularity and topological density of L0 are at least 2ω1 ; by 529Bb, the Maharam type of µ is at least 2ω1 . (f )(i) By (b), there is a continuous surjection ψ : K0 → {0, 1}ω1 where K0 ⊆ K is closed. Let Q be the set of pairs (F, C) such that F ⊆ K0 is closed, C ⊆ ω1 is closed and cofinal and {ψ(f )C :Tf ∈ F } = {0, 1}C .TIf h(Fn , Cn )in∈N is a non-increasing sequence in Q, then it has a lower bound in Q. P P Set F = n∈N Fn and C = n∈N Cn . Then for any z ∈ {0, 1}C and n ∈ N there is an fn ∈ Fn such that ψ(fn )C = z; now take a Tp -cluster point f of hfn in∈N , and see that f ∈ F and that ψ(f )C = z. As z is arbitrary, (F, C) ∈ Q. Q Q By 513O again, there is a member (K1 , C ∗ ) of Q such that ∆(F ) = ∆(K1 ) whenever (F, C) ∈ Q, F ⊆ K1 and C ⊆ C ∗ . Now C ∗ is order-isomorphic to ω1 and its order topology agrees with the subspace topology induced by the order topology of ω1 (4A2Rm). Let θ : ω1 → C ∗ be an order-isomorphism and set ψ1 (f ) = ψ(f )θ for f ∈ K1 . Then ψ1 : K1 → {0, 1}ω1 is a continuous surjection, and if F ⊆ K1 is closed, C ⊆ ω1 is closed and cofinal and {ψ1 (f )C : f ∈ F } = {0, 1}C , then (F, θ[C]) ∈ Q so ∆(F ) = ∆(K1 ). (ii) Let K2 ⊆ K1 be a compact set such that ψ1 K2 is an irreducible surjection onto {0, 1}ω1 (4A2G(i-i) again). Because {0, 1}ω1 is separable (4A2B(e-ii)), so is K2 (5A4C(d-i)). Let hfn in∈N enumerate a dense subset of K2 . Because K2 is compact in RX , h1 = supn∈N fn and h0 = inf n∈N fn are defined in R X , and of course they belong to L0 . If f ∈ K2 , then f (x) ∈ {fn (x) : n ∈ N} ⊆ [h0 (x), h1 (x)] for every x, and h0 ≤ f ≤ h1 . Accordingly we have ∆(K2 ) ≤ ρ(h0 , h1 ) = supn∈N ρ(inf i≤n fi , supi≤n fi ) ≤ ∆(K2 ). Let U be the family of non-empty cylinder sets in {0, 1}ω1 . For U ∈ U set IU = {n : n ∈ N, ψ1 (fn ) ∈ U } and gU = inf{fn : n ∈ IU }. Observe that FU = {f : f ∈ K2 , gU ≤ f ≤ h1 } is a closed subset of K1 and that FU ∩ ψ1−1 [U ] is dense in ψ1−1 [U ], so U ∩ψ1 [FU ] must be dense in U and U ⊆ ψ1 [FU ]. There is a finite set I ⊆ ω1 such that U is determined by coordinates in I; in this case, C = ω1 \ I is closed and cofinal in ω1 , and {zC : z ∈ U } = {0, 1}C . By the choice of K1 , ∆(FU ) = ∆(K1 ). As FU ⊆ [gU , h1 ] in L0 , ρ(gU , h1 ) = ∆(K1 ) = ρ(h0 , h1 ), and min(1, h1 −gU ) =a.e. min(1, h1 −h0 ). Set h(x) = max( 21 (h0 (x) + h1 (x)), h0 (x) − 12 ) for x ∈ X, and E = {x : h0 (x) < h1 (x)} = {x : h(x) < h1 (x)}, so that E is measurable and not negligible. If U ∈ U, then EU = {x : x ∈ E, h(x) ≤ gU (x)} ⊆ {x : x ∈ E, h1 (x) − gU (x) < min(1, h1 (x) − h0 (x))} is negligible. For every x ∈ E, Fx0 = {f : f ∈ K2 , f (x) ≤ h(x)} is a proper closed subset of K2 , so ψ1 [Fx0 ] 6= {0, 1}ω1 and there is 0 some U ∈ U / Fx0 , that is, fn (x) > h(x), for every n ∈ IU , so gU (x) ≥ h(x). S such that U ∩ ψ1 [Fx ] = ∅. In this case fn ∈ Thus E = U ∈U EU is a non-negligible measurable set covered by ω1 negligible sets. (g) Continuing the argument from (f), define φ : X → R N by setting φ(x) = hfn (x)in∈N for x ∈ X. Then φ is measurable (418Bd), so we have a non-zero totally finite Borel measure on R N defined by setting νH = µ(E ∩φ−1 [H]) S νT N ∞ ∞ for every Borel set H ⊆ R . Note that φ[X] ⊆ ` and that ` = n∈N i∈N {w : |w(i)| ≤ n} is an Fσ set in R N . Now set h01 (w) = supn∈N w(n),

h00 (w) = inf n∈N w(n),

h0 (w) = max( 21 (h00 (w) + h01 (w)), h01 (w) − 21 ) for w ∈ `∞ , so that h1 = h01 φ, h0 = h00 φ and h = h0 φ; for U ∈ U, set EU0 = {w : w ∈ `∞ , h0 (w) ≤ inf n∈IU w(n)} so that EU0 is an Fσ set and EU = E ∩ φ−1 [EU0 ]; accordingly νEU0 = 0. Because E ⊆

S

U ∈U

EU , φ[E] ⊆

S

U ∈U

EU0 .

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Topologies and measures III

536C

Thus we have a non-negligible subset of R N which is covered by ω1 negligible Fσ sets and therefore by ω1 closed negligible sets. By 526L, mcountable = ω1 . 536X Basic exercises (a) Let (X, Σ, µ) be a complete measure space, with null ideal N (µ). Suppose that add N (µ) = cov N (µ). Show that there is a Tp -compact K ⊆ L0 (Σ) such that the identity map on K is not (Tp , Tm )continuous. (b) Let (X, Σ, µ) be a perfect measure space. For E ⊆ X, write N (µE ) for the null ideal of the subspace measure on E. Suppose that non(E, N (µE )) < cov(E, N (µE )) for every non-negligible measurable set E of finite measure. Show that if K ⊆ L0 (Σ) is Tp -compact, then the identity map on K is (Tp , Tm )-continuous. 536Y Further exercises (a) Assume that the continuum hypothesis is true. Find a strictly localizable perfect measure space (X, Σ, µ) and a Tp -compact K ⊆ L0 (Σ) such that Tm is Hausdorff on K but K is not Tm -compact. 536 Notes and comments The methods here are derived from ideas of M.Talagrand. They seem frustratingly close to delivering an answer to the original question. But it seems clear that even if a positive answer – every Tp -compact Tm -separated set is metrizable – is true in ZFC, some further idea will be needed in the proof. On the other side, while it may well be that in some familiar model of set theory there is a negative answer, parts (c), (d) and (g) of 536C give simple tests to rule out many candidates.

537 Sierpi´ nski sets, shrinking numbers and strong Fubini theorems W.Sierpi´ nski observed that if the continuum hypothesis is true then there are uncountable subsets of R which have no uncountable negligible subsets, and that such sets lead to curious phenomena; he also observed that, again assuming the continuum hypothesis, there would be a (non-measurable) function f : [0, 1]2 → {0, 1} for which Fubini’s theorem failed radically, in the sense that

RR

f (x, y)dxdy = 0,

RR

f (x, y)dydx = 1.

In this section I set out to explore these two insights in the light of the concepts introduced in Chapter 52. I start with definitions of ‘Sierpi´ nski’ and ‘strongly Sierpi´ nski’ set (537A), with elementary facts and an excursion into the theory of ‘entangled’ sets (537C-537G). Turning to repeated integration, I look at three interesting cases in which, for different reasons, some form of separate measurability is enough to ensure equality of repeated integrals (537I, 537L, RR RR dxdy ≤ dydx 537S). Working a bit harder, we find that there can be valid non-trivial inequalities of the form (537N-537Q). As elsewhere, I will write N (µ) for the null ideal of a measure µ. 537A Definitions (a) If (X, Σ, µ) is a measure space, a subset of X is a Sierpi´ nski set if it is uncountable but meets every negligible set in a countable set. (b) If (X, Σ, µ) is a measure space, a subset A of X is a strongly Sierpi´ nski set if it is uncountable and for every n ≥ 1 and for every set W ⊆ X n which is negligible for the (c.l.d.) product measure on X n , the set {u : u ∈ An ∩ W , u(i) 6= u(j) for i < j < n} is countable. 537B Proposition (a) Let (X, Σ, µ) be a measure space and A ⊆ X a Sierpi´ nski set. (i) add N (µ) = non N (µ) = ω1 and cov N (µ) ≥ #(A). (ii) If {x} is negligible for every x ∈ A, then cf N (µ) ≥ cf([#(A)]≤ω ). (b) Suppose that (X, Σ, µ) and (Y, T, ν) are measure spaces such that singleton subsets of Y are negligible. Let f : X → Y be an inverse-measure-preserving function. (i) If A ⊆ X is a Sierpi´ nski set, then f [A] is a Sierpi´ nski set in Y and #(f [A]) = #(A). (ii) Now suppose that ν is σ-finite. If A ⊆ X is a strongly Sierpi´ nski set, then f [A] is a strongly Sierpi´ nski set in Y. (c) Suppose that λ and κ are infinite cardinals and that (X, Σ, µ) is a locally compact semi-finite measure space of Maharam type at most λ in which singletons are negligible and µX > 0. Give {0, 1}λ its usual measure. (i) If {0, 1}λ has a Sierpi´ nski subset of size κ, then X has a Sierpi´ nski subset of size κ. (ii) If {0, 1}λ has a strongly Sierpi´ nski subset of size κ, then X has a strongly Sierpi´ nski subset of size κ. proof (a)(i) We are told that A is uncountable; now any subset of A with ω1 members witnesses that non N (ν) ≤ ω1 . On the other hand, if E is a family of negligible sets covering X, then #(A) ≤ max(ω, #(E)), so #(E) ≥ #(A); as E is arbitrary, cov N (µ) ≥ #(A).

537D

Sierpi´ nski sets, shrinking numbers and strong Fubini theorems

243

(ii) If {x} is negligible for every x ∈ A, then [A]≤ω ⊆ N (µ), and the identity function is a Tukey function from [A] to N (µ); so cf[A]≤ω ≤ cf N (µ). ≤ω

(b)(i) If y ∈ Y , then {y} and f −1 [{y}] are negligible, so A ∩ f −1 [{y}] is countable; consequently #(A) ≤ max(ω, #(f [A])) and #(f [A]) = #(A). If F ⊆ Y is negligible, then f −1 [F ] is negligible so A ∩ f −1 [F ] and f [A] ∩ F are countable. So f [A] is a Sierpi´ nski set. (ii) Let W ⊆ Y n be a negligible set for the product measure λ0 on Y n , where n ≥ 1. Define f : X n → Y n by saying f that (x0 , . . . , xn−1 ) = (f (x0 ), . . . , f (xn−1 )) for x0 , . . . , xn−1 ∈ X. Then f −1 [W ] is negligible for the product measure 0 ii∈N of cylinder sets of finite measure in Y n such λ on X n . P P we can find, for each m ∈ N, a sequence hCmi S By 251Wg, P∞ 0 0 −m 0 0 ] that W ⊆ i∈N Cmi and i=0 λ Cmi ≤ 2 . (This is where we need to know that ν is σ-finite.) Set Cmi = f −1 [Cmi T S n 0 0 −1 for m, i ∈ N; then Cmi ⊆ X is a cylinder set and λCmi = λ Cmi for all m and i, and f [W ] ⊆ m∈N i∈N Cmi is negligible. Q Q Accordingly B = {u : u ∈ An ∩ f −1 [W ], u(i) 6= u(j) for i < j < n} is countable, so {v : v ∈ f [A]n ∩ W, v(i) 6= v(j) for i < j < n} ⊆ f [B] is countable. (c) Take any set E ⊆ X of non-zero finite measure, and give E its normalized subspace measure µ0E = (µE)−1 µE . Then there is an f : {0, 1}λ → E which is inverse-measure-preserving for νλ and µ0E (343Cd3 ). So (b) tells us that E has a subset A of size κ which is Sierpi´ nski or strongly Sierpi´ nski for µ0E . But now A is still Sierpi´ nski or strongly Sierpi´ nski for µ. 537C Entangled sets (a) Definition If X is a totally ordered set, then X is ω1 -entangled if whenever n ≥ 1, I ⊆ n and hxξi iξ<ω1 ,i
xξ1 > xη1 ,

xξ2 ≤ xη2 .

So hxξi iξ<ω1 ,i<3 witnesses that X is not ω1 -entangled. X XQ Q (ii) Set A = {(x, y) : x < y, [x, y] ∩ D0 = ∅}. Note that if (x, y), (x0 , y 0 ) ∈ A are distinct, then [x, y] ∩ [x0 , y 0 ] = ∅, since otherwise [min(x, x0 ), max(y, y 0 )] would be an interval disjoint from D0 with at least three elements. It follows that A is countable. P P?? Otherwise, let h(xξ0 , xξ1 )iξ<ω1 be a family of distinct elements of A. Then all the xξi are distinct. But if ξ, η < ω1 are different, we cannot have xξ0 ≤ xη0 , xξ1 > xη1 . So hxξi iξ<ω1 ,i<2 witnesses that X is not entangled. X XQ Q (iii) So if we set D = D0 ∪ {x : (x, y) ∈ A} we shall have a suitable countable set. (b) For i < n write ≤i = ≤ if i ∈ I, ≤i = ≥ if i ∈ n \ I; we are seeking ξ < η such that xξi ≤i xηi for every i < n. For each family d = hdi ii
editions only.

244

Topologies and measures III

537E

537E Lemma Suppose that n ≥ 1, I ⊆ n and that A ⊆ ({0, 1}N )n is non-negligible for the usual product measure on ({0, 1}N )n . Let ≤ be the lexicographic ordering of {0, 1}N . Then there are v, w ∈ A such that v(i) 6= w(i) for every i < n and {i : i < n, v(i) ≤ w(i)} = I. νNn

proof For each k ∈ N let Σk be the algebra of subsets of X = ({0, 1}N )n generated by sets of the form {v : v ∈ X, v(i)(j) = 1} for S i < n and j < k. Then hΣk ik∈N is a non-decreasing sequence of finite algebras and the σ-algebra generated by k∈N Σk is the Borel σ-algebra B(X) of X. Let E ∈ B(X) be a measurable envelope of A for νNn . For each k ∈ N, let fk be the conditional expectation of χE on Σk , that is, fk (u) = 2kn νNn {v : v ∈ E, v(i)k = u(i)k for every i < n} for u ∈ X. By L´evy’s martingale theorem (275I), χE =a.e. limk→∞ fk . In particular, there are a u ∈ A and a k ∈ N such that fk (u) > 1 − 2−n . But this means that F = {v : v ∈ E, v(i)k = u(i)k for every i < n} has measure greater than 2−kn (1 − 2−n ), and both the sets F 0 = {v : v ∈ F , v(i)(k) = 0 for i ∈ I, v(i)(k) = 1 for i ∈ n \ I}, F 00 = {v : v ∈ F , v(i)(k) = 1 for i ∈ I, v(i)(k) = 0 for i ∈ n \ I}, must have positive measure. Accordingly we can find v ∈ A ∩ F 0 and w ∈ A ∩ F 00 , and these will serve. 537F Corollary Suppose that A ⊆ {0, 1}N is strongly Sierpi´ nski for the usual measure on {0, 1}N . Then A is N ω1 -entangled for the lexicographic ordering of {0, 1} . proof Let hxξi iξ<ω1 ,i
537I

Sierpi´ nski sets, shrinking numbers and strong Fubini theorems

245

(c) Similarly, Q is upwards-ccc. P P The principal changes needed in the argument above are 0 —– in the choice of the dαij , we need to write ‘f (xαi ) ≥ d0αij ≥ f (xαj )’; —– in the choice of particular α and β in the set C, we need to write ‘yαi ≤ yβi for i < n and yαi ≥ yβi for n ≤ i < 2n’. Q Q So P and Q satisfy our requirements. 537H Scalarly measurable functions (a) Definition Let X be a set, Σ a σ-algebra of subsets of X and U a linear topological space. A function φ : X → U is scalarly (Σ-)measurable if f φ : X → R is (Σ-)measurable for every f ∈ U ∗ . (b) If φ : X → U is scalarly measurable, V is another linear topological space and T : U → V is a continuous linear operator, then T φ : X → V is scalarly measurable, because hT ∈ U ∗ for every h ∈ V ∗ . (c) If U is a separable metrizable locally convex space and φ : X → U is scalarly measurable, then it is measurable. P P T = {F : F ⊆ U , φ−1 [F ] ∈ Σ} includes the cylindrical σ-algebra of U (4A3T), which is the Borel σ-algebra (4A3V). Q Q 537I Proposition Let (X, Σ, µ) and (Y, T, ν) be probability spaces and U a reflexive Banach RR space. Suppose that x 7→ ux : X → U and y 7→ fy : Y → U ∗ are bounded scalarly measurable functions. Then fy (ux )µ(dx)ν(dy) and fy (ux )ν(dy)µ(dx) are defined and equal. proof (a)(i) Recall from 467Hc that if V ⊆ U and W ⊆ U ∗ are closed linear subspaces, they are a ‘projection pair’ if U = V ⊕ W ◦ and kv + v 0 k ≥ kvk for all v ∈ V and v 0 ∈ W ◦ . We need to know that this is symmetric; that is, that in this case U ∗ = W ⊕ V ◦,

kg + g 0 k ≥ kgk for all g ∈ W , g 0 ∈ V ◦ .

P P Note first that if g ∈ W ∩ V ◦ , then g(u) = 0 for every u ∈ W ◦ + V , that is, g = 0. Now take any f ∈ U ∗ . Define g : U → R by saying that g(v + v 0 ) = f (v) for v ∈ V , v 0 ∈ W ◦ . Then g is linear and continuous and kgk ≤ kf k. Now g(v 0 ) = 0 for every v 0 ∈ W ◦ , that is, g ∈ W ◦◦ , which is the weak*-closure of W (4A4Eg); but as U and U ∗ are reflexive, this is just the norm-closure of W , which is equal to W . Set g 0 = f − g. Then g 0 ∈ V ◦ . This shows that f ∈ W + V ◦ ; as f is arbitrary, U ∗ = W ⊕ V ◦ . Finally, I remarked in the course of the argument that kgk ≤ kf k, which is what we need to know to check that kgk ≤ kg + g 0 k whenever g ∈ W and g 0 ∈ V ◦ . Q Q (ii) Because U is reflexive, its unit ball is weakly compact, so U is surely weakly compactly generated, therefore weakly K-countably determined (467L). Now turn to Lemma 467G. This tells us that there is a family M of subsets of U ∪ U ∗ such that for every B ⊆ X ∪ X ∗ there is an M ∈ M such S that B ⊆ M and #(M ) ≤ max(ω, #(B)); whenever M0 ⊆ M is upwards-directed, then M0 ∈ M; whenever M ∈ M then (VM , WM ) is a projection pair of subspaces of U and U ∗ , where I write VM = M ∩ U and WM = M ∩ U ∗ . For M ∈ M, ◦ U = VM ⊕ W M ,

◦ U ∗ = WM ⊕ VM ;

let PM : U → VM and QM : U ∗ → WM be the corresponding projections. Since kvk ≤ kv + v 0 k whenever v ∈ VM and ◦ v 0 ∈ WM , kPM k ≤ 1; similarly, kQM k ≤ 1. If u ∈ U , f ∈ U ∗ and M ∈ M, then f (PM u) = (Qm f )(u) = (QM f )(PM u). 0

0

◦ ◦ P P Express u as v + v and f as g + g , where v ∈ VM , v 0 ∈ WM , g ∈ WM and g 0 ∈ VM . Then

f (v) = g(v) = g(u), that is, f (PM u) = (QM f )(PM u) = (Qm f )(u). Q Q (iii) If M0 , M1 ∈ M and M0 ⊆ M1 then PM0 = PM0 PM1 = PM1 PM0 . P P If u ∈ U , express it as v0 + v00 where 0 ◦ 0 0 0 ◦ v0 ∈ VM0 and v0 ∈ WM0 ; now express v0 as v1 + v1 where v1 ∈ VM1 and v1 ∈ WM . Then 1 PM0 u = v0 ∈ VM1 , ◦ so PM1 PM0 u = PM0 u. On the other hand, u = v0 + v1 + v10 where v0 + v1 ∈ VM1 and v10 ∈ WM , so PM1 u = v0 + v1 ; 1 ◦ and as v1 = v00 − v10 belongs to WM , P P u = v = P u. Q Q M0 M1 0 M0 0

246

S

Topologies and measures III

537I

(iv) If hMζ iξ<ζ is a non-decreasing family in M, where ζ is a non-zero limit ordinal, then we know that M = ξ<ζ Mξ belongs to M. Now PM u = limξ↑ζ PMξ u

for every u ∈ U , the limit being for the norm topology on U . P P Let  > 0. We know that PM u ∈ VM = M ∩ U , so there is a u0 ∈ M ∩ U such that ku0 − PM uk ≤ 12 . Let ξ < ζ be such that u0 ∈ Mξ . If ξ ≤ η < ζ, then kPMη u − PM uk = kPMη (PM u − u0 ) + PM (u0 − PM u)k (because u0 ∈ VMη , so PM u0 = PMη u0 = u0 ) ≤ 2kPM u − u0 k ≤ . Q Q (v) Similarly, QM0 = QM0 QM1 = QM1 QM0 whenever M0 , M1 ∈ M and M0 ⊆ M1 , and QM f = limξ↑ζ QMξ f whenever f ∈ U ∗ and ζ is a non-zero limit ordinal and hMζ iξ<ζ is a non-decreasing family in M with union M . (b) Now let M0 be {M : M ∈ M, #(M ) ≤ ω}. Then there is an M0 ∈ M0 such that PM0 (ux ) = PM (ux ) µ-a.e.(x) whenever M0 ⊆ M ∈ M0 . P P?? Suppose, if possible, otherwise. Then we can choose inductively an increasing family hMξ iξ<ω1 in M0 such that µ{x : PMξ+1 (ux ) 6= PMξ (ux )} > 0 for every ξ < ω1 , Mξ =

S

η<ξ

Mη whenever ξ < ω1 is a non-zero countable limit ordinal.

(The set of x for which PMξ+1 (ux ) 6= PMξ (ux ) is necessarily measurable because x 7→ PMξ+1 ux − PMξ ux is scalarly measurable, by 537Hb, therefore measurable for the norm topology, by 537Hc, since VMξ+1 is separable.) Now there must be a δ > 0 such that A = {ξ : ξ < ω1 , µEξ ≥ δ} is infinite, where Eξ = {x : kPMξ+1 (ux ) − PMξ (ux )k ≥ δ} for each ξ < ω1 . But in this case there must be an x ∈ X such that is infinite. (Take a sequence hξn in∈N A0 in ω1 . Then

A0 = {ξ : ξ ∈ A, x ∈ Eξ } T S of distinct points in A, and x ∈ n∈N m≥n Eξm .) Let ζ be any cluster point of PMζ (ux ) = limξ↑ζ PMξ (ux )

((a-iv) above), which is impossible. X XQ Q (c) Similarly, there is an M1 ∈ M0 such that M1 ⊇ M0 and PM1 (fy ) = PM (fy ) ν-a.e.(y) whenever M1 ⊆ M ∈ M0 . Because x 7→ PM1 (ux ) and y → 7 QM1 (fy ) are scalarly measurable maps to norm-separable spaces, they are norm-measurable; again because VM1 and WM1 are separable, (x, y) → (PM1 ux , QM1 fy ) : X × b Y → VM1 × WM1 is Σ⊗T-measurable (418B). Because (f, x) → 7 f (x) : U ∗ × U → R is norm-continuous, (x, y) 7→ b (QM1 fy )(PM1 ux ) is Σ⊗T-measurable, and

RR

(QM1 fy )(PM1 ux )µ(dx)ν(dy) =

RR

(QM1 fy )(PM1 ux )ν(dy)µ(dx)

by Fubini’s theorem (252C). Now observe that if y ∈ Y there is an M ∈ M0 such that M1 ⊆ M and fy ∈ M . So Z Z Z fy (ux )µ(dx) = (QM fy )(ux )µ(dx) = fy (PM ux )µ(dx) Z Z = fy (PM1 ux )µ(dx) = (QM1 fy )(PM1 ux )µ(dx).

537K

Sierpi´ nski sets, shrinking numbers and strong Fubini theorems

This is true for every y. So

RR

fy (ux )µ(dx)ν(dy) is defined and equal to

RR

fy (ux )ν(dy)µ(dx) =

RR

RR

247

(QM1 fy )(PM1 ux )µ(dx)ν(dy). Similarly,

(QM1 fy )(PM1 ux )ν(dy)µ(dx).

Putting these together, we have the result. 537J Corollary Let (X, Σ, µ), (Y, T, ν) and (Z, Λ, σ) be probability spaces. Let x 7→ Ux : X → Λ and y 7→ Vy : Y → Λ be functions such that x 7→ σ(Ux ∩ W ), y 7→ σ(Vy ∩ W ) RR RR are measurable for every W ∈ Λ. Then σ(Ux ∩ Vy )µ(dx)ν(dy) and σ(Ux ∩ Vy )ν(dy)µ(dx) are defined and equal. proof (a) For x ∈RX set ux = (χUx )• in L2 (σ). Then x 7→ ux is scalarly measurable. P P If f ∈ U ∗ , there is a v ∈ L2 (σ) such that f (u) P = u × v for every u ∈ L2 (σ) (244K). Let  > 0. Then there are W0 , . . . , Wn ∈ Λ and α0 , . . . , αn ∈ R n such that kv − i=0 αi (χWi )• k2 ≤  (244Ha), so that R R Pn Pn |f (ux ) − i=0 αi σ(Ux ∩ Wi )| = | ux × v − ux × i=0 αi (χWi )• | ≤ kux k2 ≤  Pn for every x ∈ X. Now the function x 7→ i=0 αi σ(Ux ∩ Wi ) is Σ-measurable. So we see that the function x 7→ f (ux ) is uniformly approximated by Σ-measurable functions and is itself Σ-measurable. As f is arbitrary, x 7→ ux is scalarly measurable. Q Q (b) Similarly, setting vy = (χVy )• for y ∈ Y , y 7→ vy : Y → L2 (σ) is scalarly measurable. Identifying L2 (σ) with its dual, 537I tells us that

RR

(ux |vy )µ(dx)ν(dy) =

RR

(ux |vy )ν(dy)µ(dx),

σ(Ux ∩ Vy )µ(dx)ν(dy) =

RR

σ(Ux ∩ Vy )ν(dy)µ(dx).

that is, that

RR

537K The next few paragraphs will be concerned with upper and lower integrals. For the basic theory of these, see §133 and 214N4 . Theorem (Freiling 86, Shipman 90) Let h(Xj , Σj , µj )ij≤m be a finite sequence of probability spaces and hκj ij≤m a sequence of cardinals such that XjN , with its product measure µN has a set of cardinal κj which is not covered by κj−1 j,Q negligible sets (if j ≥ 1) and is not negligible (if j = 0). Let f : j≤m Xj → R be a bounded function, and suppose that σ : m + 1 → m + 1 and τ : m + 1 → m + 1 are permutations. Set

R

I=

R

. . . f (x0 , . . . , xm )dxσ(m) . . . dxσ(0) ,

I0 =

R

. . . f (x0 , . . . , xm )dxτ (m) . . . dxτ (0) .

R

Then I ≤ I 0 . proof Let M ≥ 0 be such that |f (x0 , . . . , xm )| ≤ M for all x0 , . . . , xm . Q Q u) ⊆ XkN , for k ∈ N and u ∈ j≤m,j6=k XjN , (a) Set Z = j≤m XjN . The key fact is that we can find negligible sets W (u such that 1 Pn I ≤ lim inf n→∞ i=0 f (t0i , . . . , tmi ) n+1

whenever htj ij≤m = hhtji ii∈N ij≤m is such that tk ∈ / W (t0 , . . . , tk−1 , tk+1 , . . . , tm ) for every k. P P Because the formula 1 Pn lim inf n→∞ i=0 f (t0i , . . . , tmi ) n+1

is tolerant of permutations in the coordinates 0, . . . , m, it is enough to consider the case σ(j) = j for j ≤ m, so that I=

R

R

. . . f (x0 , . . . , xm )dxm . . . dx0 .

(i) Define D0 , . . . , Dm+1 as follows. D0 = {∅} = Q (t0 , . . . , tk−1 ) ∈ j
editions only.

1 Pn R n+1 i=0

R

Q

j<0

XjN . For 0 < k ≤ m let Dk be the set of those

. . . f (t0i , . . . , tk−1,i , xk , . . . , xm )dxm . . . dxk ,

248

Topologies and measures III

537K

where tj = htji ii∈N for j < k. For k < m and u = (u0 , . . . , uk−1 , uk+1 , . . . , um ) in

Q

j≤m,j6=k

XjN , set

u) = ∅ if (u0 , . . . , uk−1 ) ∈ W (u / Dk , = {t : t ∈ XkN , (u0 , . . . , uk−1 , t) ∈ / Dk+1 } otherwise. u) ⊆ XkN is negligible. To see this, we need consider only the case in which (u0 , . . . , uk−1 ) belongs to Dk . (ii) W (u Express uj as huji ii∈N for j < k, and for i ∈ N define hi : Xk → R by setting hi (x) =

R

R

. . . f (u0i , . . . , uk−1,i , x, xk+1 , . . . , xm )dxm . . . dxk+1

for x ∈ Xk . Now the definition of Dk tells us just that R 1 Pn R . . . f (u0i , . . . , uk−1,i , xk , . . . , xm )dxm . . . dxk , I ≤ lim inf n→∞ i=0 n+1

that is, that I ≤ lim inf n→∞

1 Pn R hi (x)dx. n+1 i=0

R For R each i ∈ N let gi : Xk → [−M, M ] be aNmeasurable function such that gi (x) ≤ hi (x) for every x andN gi dµk = h dµ . Now consider the functions g˜i : Xk → R defined by setting g˜i (t) = gi (ti ) for t = hti ii∈N ∈ Xk . We have R i kN R g˜i dµk = hi dµk for each i, while h˜ gi ii∈N is a uniformly bounded independent sequence of random variables. By the strong law of large numbers in the form 273H, R 1 Pn limn→∞ gi (t) − g˜i dµN i=0 (˜ k) = 0 n+1

for almost every t ∈

XkN .

Since lim inf n→∞

1 Pn R n+1 i=0

g˜i dµN k = lim inf n→∞

1 Pn R hi dµk n+1 i=0

≥ I,

we have 1 n→∞ n+1

I ≤ lim inf =

1 lim inf n→∞ n+1

n X

1 n→∞ n+1

g˜i (t) ≤ lim inf

i=0 n Z X

n X

hi (ti )

i=0

Z ...

f (u0i , . . . , uk−1,i , ti , xk+1 , . . . , xm )dxm . . . dxk+1

i=0

u) is negligible, for almost every t = hti ii∈N ∈ XkN , that is, (u0 , . . . , uk−1 , t) ∈ Dk+1 for almost every t ∈ XkN , that is, W (u as required. (iii) Suppose that t = (t0 , . . . , tm ) ∈ Z is such that tk ∈ / W (t0 , . . . , tk−1 , tk+1 , . . . , tm ) for every k < m. Then (t0 , . . . , tk ) ∈ Dk+1 for every k; in particular, t ∈ Dm+1 and, writing tj = htji ii∈N for each j, 1 Pn Q I ≤ lim inf n→∞ i=0 f (t0i , . . . , tmi ). Q n+1

u) ⊆ XkN , for k ∈ N and u ∈ or applying the argument above to −f , we have negligible sets W 0 (u Q (b) Similarly, N j≤m,j6=k Xj , such that I 0 ≥ lim supn→∞

1 Pn f (t0i , . . . n+1 i=0

, tmi )

u) if whenever htj ij≤m = hhtji ii∈N ij≤m is such that tk ∈ / W 0 (t0 , . . . , tk−1 , tk+1 , . . . , tm ) for every k. Enlarging the W 0 (u 0 u u) for every u . necessary, we may suppose that W (u ) ⊇ W (u (c) Now the point of the construction is that we can find a t = (t0 , . . . , tm ) ∈ Z such that tk ∈ / W 0 (t0 , . . . , tk−1 , tk+1 , . . . , tm ) for every k. P P For each k ≤ m let Ak ⊆ XkN be a non-negligible set of size κk which (if k ≥ 1) cannot be covered by κk−1 negligible sets. Choose tm , tm−1 , . . . , t0 in such a way that Q Q u) whenever u ∈ j
537N

Sierpi´ nski sets, shrinking numbers and strong Fubini theorems

I≤

1 lim inf n→∞ n+1

n X

f (t0i , . . . , tmi )

i=0 n X

1 n+1 n→∞ i=0

≤ lim sup

249

f (t0i , . . . , tmi ) ≤ I 0 ,

as claimed. 537L Corollary Let h(Xj , Σj , µj )ij≤m be a finite sequence of probability spaces such that XjN , with its product Q measure µN nski set of cardinal ωj+1 for each j ≤ m. Let f : j≤m Xj → R be a bounded function, and j , has a Sierpi´ suppose that σ : m + 1 → m + 1 and τ : m + 1 → m + 1 are permutations such that the two repeated integrals I=

R

I0 =

R

...

R

f (x0 , . . . , xm )dxσ(m) . . . dxσ(0) ,

R

f (x0 , . . . , xm )dxτ (m) . . . dxτ (0) ,

...

0

are both defined. Then I = I . proof Apply 537K in both directions. 537M A simple fact which I never got round to spelling out will be useful below. Lemma Suppose that (X, Σ, µ) is a totally finite measure space and f is a [0, ∞]-valued function defined almost everywhere inR X. R (a) If γ < f , then there is a measurable integrable function g : X → [0, ∞[ such that g ≥ γ and {x : x ∈ dom f , g(x) ≤ f (x)} has full outer measure in X. R R (b) If f < γ, then there is a measurable integrable function g : X → [0, ∞[ such that g ≤ γ and {x : x ∈ dom f , f (x) ≤ g(x)} has full outer measure in X. proof (a) By 135H(b-i),

R

f = supk∈N

R

min(f (x), k)µ(dx);

R

R let k ∈ N be such that fk > γ, where fk (x) = min(f (x), k) for x ∈ dom f . Because µX < ∞, fk is finite. By R R 133J(a-i), there is an integrable h such that h = fk and fk ≤a.e. h; adjusting h on a negligible set if necessary, R we can arrange that h is defined (and finite) everywhere on X and is measurable. Set  = ( h − γ)/(1 + µX), and g = h − χX; then by the last part of 133J(a-i)5 , {x : x ∈ dom f , g(x) ≤ f (x)} has full outer measure in X, while inf g ≥ γ. R R (b) By 133J(a-ii)/135Ha, there is a measurable h : X → [0, ∞[ such that h ≤a.e. f and h = f . Set  = R R (γ − h)/(1 + µX), and g = h + χX; then g ≤ γ and {x : f (x) ≤ g(x)} has full outer measure. 537N For ordinary two-variable repeated integrals we can squeeze a little bit more out than is given by 537K. Proposition Let (X, Σ, µ) be a non-empty semi-finite measure space, (Y, T, ν) a probability space, and ν N the product measure on Y N . If non(E, N (µ)) < cov N (ν N ) for every E ∈ Σ \ N (µ), then

RR

f (x, y)ν(dy)µ(dx) ≤

RR

f (x, y)µ(dx)ν(dy)

for every function f : X × Y → [0, ∞]. proof (a) To begin with, suppose that µX < ∞ and #(X) < cov N (ν N ). For each y ∈ Y , let hy : X → [0, ∞] be a R R measurable function such that f (x, y) ≤ hy (x) for every x ∈ X and hy dµ = f (x, y)µ(dx); let v : Y → [0, ∞] be a R R RR measurable function such that hy dµ ≤ v(y) for every y ∈ Y and v dν = f (x, y)µ(dx)ν(dy). If this is infinite, we can stop. Otherwise, for each x ∈ X let g : Y → [0, ∞] be a measurable function such that gx (y) ≤ Rf (x, y) for every x R R y ∈ Y and gx dν = f (x, y)ν(dy), and let u : X → [0, ∞] be a measurable function such that u(x) ≤ gx dν for every R RR x and u dµ = f (x, y)ν(dy)µ(dx). As #(X) < cov N (ν N , we can find a sequence hyi ii∈N in Y such that R 1 Pn v dν = limn→∞ i=0 v(yi ) n+1

5 Later

editions only.

250

Topologies and measures III

537N

and

R

gx dν ≤ limn→∞

1 Pn gx (yi ) n+1 i=0

for every x ∈ X (273J6 ). If x ∈ X, then u(x) ≤

R

gx dν ≤ lim inf n→∞

1 Pn gx (yi ) n+1 i=0

≤ lim inf n→∞

1 Pn hy (x). n+1 i=0 i

So

Z Z

Z u dµ ≤

f (x, y)ν(dy)µ(dx) =

1 lim inf n→∞ n+1

n Z X

hyi dµ

i=0

(by Fatou’s Lemma) 1 n→∞ n+1

≤ lim inf

n X

Z v(yi ) =

v dν

i=0

Z Z f (x, y)µ(dx)ν(dy),

= as required.

(b) Now suppose that µ is totally finite and that X has a non-empty subset A of full outer measure with #(A) < cov N (ν N ). Let µA be the subspace measure on A. Then for any q : X → [0, ∞] we have

R

q dµ ≤

R

(qA)dµA ≤

R

(qA)dµA ≤

R

q dµ

(214N). So, writing fA for the restriction of f to A × Y , Z Z

Z Z f (x, y)ν(dy)µ(dx) ≤

fA (x, y)ν(dy)µA (dx) Z Z

≤

fA (x, y)µA (dx)ν(dy)

(by (a)) Z Z ≤

f (x, y)µ(dx)ν(dy).

R (c) For the general case, let u : X → [0, ∞] be a measurable function such that u(x) ≤ f (x, y)ν(dy) for every R RR R x ∈ X and u dµ = f (x, y)ν(dy)µ(dx). Take any γ < u dµ. Because µ is semi-finite, there is a non-empty set R F ∈ Σ of finite measure such that F u dµ > γ. Now let E be the family of measurable sets E ⊆ F of finite measure for which there is a non-empty set A ⊆ E, of cardinal less than cov N (ν N ), such that µ∗ A = µE, that is, A has full outer measure for the subspace measure µE , that is, E is a measurable envelope of A. Then E is closed under finite unions and every non-empty member sequence hEk ik∈N in E such R is a non-decreasing R S S of Σ includes a member of E. So there that k∈N Ek ⊆ F and F \ k∈N Ek is negligible. In this case, γ < F u dµ = limk→∞ Ek u dµ, so there is a k ∈ N such R that γ ≤ Ek u dµ. Set E = Ek . Consider the restriction fE of f to E × Y and the subspace measure µE on E. We have Z γ≤

Z u dµ =

Z Z (uE)dµE ≤

E

Z Z ≤

fE (x, y)µE (dx)ν(dy)

(because E ∈ E, so we can use (b)) Z Z ≤ 6 Later

editions only.

f (x, y)µ(dx)ν(dy)

fE (x, y)ν(dy)µE (dx)

537Q

Sierpi´ nski sets, shrinking numbers and strong Fubini theorems

251

R R because fE (x, y)µE (dx) ≤ f (x, y)µ(dx) for every y, by 214Na or otherwise. Since γ is arbitrary,

RR

f (x, y)ν(dy)µ(dx) ≤

RR

f (x, y)µ(dx)ν(dy)

in this case also. 537O Corollary Let (X, Σ, µ) and (Y, T, ν) be probability spaces, and ν N the product measure on Y N . shr+N (µ) ≤ cov N (ν N ) then

RR

f (x, y)ν(dy)µ(dx) ≤

RR

If

f (x, y)µ(dx)ν(dy)

for every function f : X × Y → [0, ∞[. RR proof Take any γ < f (x, y)ν(dy)µ(dx). By 537Ma, there are a measurable function u : X → [0, ∞[ and a set A of R R full outer measure in X such that u dµ ≥ γ and u(x) ≤ f (x, y)ν(dy)µ(dx) for every x ∈ A. Let µA be the subspace measure on A, and fA the restriction of f to A × Y . If B ⊆ A is any non-negligible relatively measurable set, there is a non-negligible D ⊆ B such that #(D) < shr+N (µ), so non(B, N (µA )) = non(B, N (µ)) ≤ #(D) < cov N (ν N ). So Z γ≤

Z

Z (uA)dµA ≤ int

u dµ =

fA (x, y)ν(dy)µA (dx)

R

(because uA is measurable and (uA)(x) ≤ fA (x, y)ν(dy) for every x ∈ A) Z Z ≤ fA (x, y)µA (dx)ν(dy) (by 537N) Z Z ≤

f (x, y)µ(dx)ν(dy)

R R because fA (x, y)µA (dx) ≤ f (x, y)µ(dx) for every y, by 214N again. As γ is arbitrary, we have the result. 537P Corollary Let (X, Σ, µ) and (Y, T, ν) be probability spaces, and ν N the product measure on Y N ; suppose that shr+N (µ) ≤ cov N (ν N ), and that f : X × Y → R is bounded. (a)

RR

f (x, y)ν(dy)µ(dx) ≤

RR

RR

f (x, y)µ(dx)ν(dy),

RR

f (x, y)µ(dx)ν(dy) ≤ f (x, y)ν(dy)µ(dx). RR R (b) IfRR f (x, y)µ(dx)ν(dy) is defined, and f (x, RR y)ν(dy) is defined for almost every x, then the other repeated integral f (x, y)ν(dy)µ(dx) is defined and equal to f (x, y)µ(dx)ν(dy). proof (a) Apply 537O to the functions (x, y) 7→ M + f (x, y), (x, y) 7→ M − f (x, y) for suitable M . (b) By (a), ZZ

Z Z f (x, y)µ(dx)ν(dy) ≤

f (x, y)ν(dy)µ(dx) Z Z

≤

ZZ f (x, y)ν(dy)µ(dx) ≤

f (x, y)µ(dx)ν(dy).

537Q We can extend the second part of 537Pa, as well as the first, to unbounded functions, if we strengthen the set-theoretic hypothesis. Proposition (Humke & Laczkovich 05) Let (X, Σ, ν) and (Y, T, µ) be probability spaces, and µN , ν N the product RR RR measures on X N , Y N respectively. If shr+N (µN ) ≤ cov N (ν N ) then f (x, y)µ(dx)ν(dy) ≤ f (x, y)ν(dy)µ(dx) for every function f : X × Y → [0, ∞[.

252

Topologies and measures III

537Q

proof ?? Suppose, if possible, otherwise. R RR (a) There is a measurable function u : Y → [0, ∞[ such that u(y) ≤ f (x, y)µ(dx) for every y and f (x, y)ν(dy)µ(dx) R R < u dν. Since u dν is the supremum of the integrals ofPthe non-negative simple functions dominated by u, we may m suppose that u itself is a simple function; express it as j=0 αj χFj where αj ≥ 0 for each i and (F0 , . . . , Fm ) is a partition of Y into measurable sets. Now m Z Z X

f (x, y)χFj (y)ν(dy)µ(dx) ≤

Z X m Z

j=0

f (x, y)χFj (y)ν(dy)µ(dx)

j=0

(133J(b-v)) Z Z ≤

f (x, y)ν(dy)µ(dx)

(because if x ∈ X and q : Y → [0, ∞] is measurable and f (x, y) ≤ q(y) for every y, then R Pm R at most j=0 q × χFj dν = q dν) Z m X < u dν = αj νFj .

Pm R j=0

f (x, y)χFj (y)ν(dy) is

j=0

There are therefore a j ≤ m and a γ < 1 such that

RR

f (x, y)χFj (y)ν(dy)µ(dx) < γαj νFj . R Now there is a measurable function v : X → [0, ∞[ such that v dµ ≤ γαj νFj and D = {x : x ∈ X,

R

f (x, y)χFj (y)ν(dy) ≤ v(x)}

has full outer measure in X, by 537Mb. x, y) = lim inf n→∞ (b) For y ∈ Y and x = hxi ii∈N ∈ X N , set h(x

1 n+1

Pn

i=0

f (xi , y). If y ∈ Y , then

R

f (x, y)µ(dx) ≤

x, y) for µN -almost every x . P h(x P We have a measurable function q : X → [0, ∞[ such that q(x) ≤ f (x, y) for every x and Z

Z

1 n→∞ n+1

q dµ ≤ lim inf

f (x, y)µ(dx) =

1 n→∞ n+1

≤ lim inf

n X

n X

q(xi )

i=0

x, y) f (xi , y) = h(x

i=0

for almost every x = hxi ii∈N . Q Q At the same time, V = {hxi ii∈N : lim inf n→∞ is conegligible in X N , because

R

1 Pn v(xi ) n+1 i=0

≤ γαj νFj }

v dµ ≤ γαj νFj .

(c) Set x, y) : x ∈ X N , y ∈ Fj , h(x x, y) ≥ αj } W = {(x R and consider the function χW : X N × Y → {0, 1}. If y ∈ Fj then f (x, y)µ(dx) ≥ αj so W −1 [{y}] is conegligible in X N . On the other hand, if x = hxi ii∈N belongs to V ∩ DN , Z

Z x, y)ν(dy) ≤ αj χW (x

x, y)χFj (y)ν(dy) h(x Z

=

1 n→∞ n+1

lim inf Z

≤ lim inf n→∞

(133Kb)

1 n+1

n X i=0 n X i=0

f (xi , y)χFj (y)ν(dy)

f (xi , y)χFj (y)ν(dy)

537S

Sierpi´ nski sets, shrinking numbers and strong Fubini theorems

≤

1 lim inf n→∞ n+1

n Z X

253

f (xi , y)χFj (y)ν(dy)

i=0

(133Jb) ≤

1 lim inf n→∞ n+1

n X

v(xi ) ≤ γαj νFj .

i=0

(d) As V is conegligible and DN has full outer measure (254Lb), Z Z

ZZ x, y)ν(dy)µN (dx x) ≤ γνFj < νFj = x, y)µN (dx x)ν(dy) χW (x χW (x Z Z x, y)µN (dx x)ν(dy). χW (x =

But we are supposing that shr+N (µN ) ≤ cov N (ν N ), so this contradicts 537P. X X So we have the result. 537R Lemma Let (X, Σ, µ) be a complete probability space and (Y, T, ν) a probability space such that shr+N (µ) ≤ N cov N (ν N ), where ν N is the product measure R on Y . Let f : X × Y → R be a bounded function which is measurable in each variable separately, and set u(x) = f (x, y)ν(dy) for x ∈ X. Then u : X → R is measurable. proof ?? Otherwise, there are a non-negligible measurable set E ⊆ X and α, β ∈ R such that α < β and µ∗ {x : x ∈ E, u(x) ≤ α} = µ∗ {x : x ∈ E, u(x) ≥ β} = µE (413G). Let A ⊆ {x : x ∈ E, u(x) ≤ α} and B ⊆ {x : x ∈ E, u(x) ≥ β} be sets of cardinal less than shr+N (µ) and outer measure greater than 21 µE (521Ca). Let hyi ii∈N be a sequence in Y such that u(x) = limn→∞

1 Pn f (x, yi ) n+1 i=0

for every x ∈ A ∪ B. Because x 7→ f (x, yi ) is measurable for each i, uA ∪ B is measurable; but this means that A and B can be separated by measurable sets, which is impossible, because µ∗ A + µ∗ B > µE. X X 537S Proposition Let (X, Σ, µ) and (Y, T, ν) be probability spaces such that shr+N (µ) ≤ cov N (ν N ), where ν N is the product measure on Y N , and cf([τ (ν)]≤ω ) < cov(E, N (µ)) for every E ∈ Σ \ N (µ), where τ (ν) is the Maharam RR type of ν. Let f : X ×RRY → [0, ∞[ be a bounded function which is measurable in each variable separately. Then f (x, y)µ(dx)ν(dy) and f (x, y)ν(dy)µ(dx) exist and are equal. ˜ ⊇ Σ⊗T b be the σ-algebra of sets W ⊆ X × Y such that all the vertical and horizontal sections of proof (a) Let Λ ˜ then x 7→ νW [{x}] : X → [0, 1] is measurable, by 537R. If W ∈ Λ ˜ and almost every W are measurable. If W ∈ Λ, horizontal section of W is negligible, then Z

Z Z χW (x, y)ν(dy)µ(dx)

νW [{x}]µ(dx) = Z Z ≤

χW (x, y)µ(dx)ν(dy) = 0

by 537Pa, so almost every vertical section of W is negligible. ˜ and there is a metrically separable subalgebra C of B (b) Let (B, ν¯) be the measure algebra of (Y, T, ν). If W ∈ Λ • 0 b such that W [{x}]4W 0 [{x}] is negligible for almost containing W [{x}] for every x ∈ X, then there is a W ∈ Σ⊗T every x. P P Note first that for every F ∈ T the map x 7→ ν(W [{x}]4F ) = ν((W 4(X × F ))[{x}] b such is measurable, by (a). So x 7→ W [{x}]• : X → C is measurable, by 418Bc. By 418T(b-ii), there is a W 0 ∈ Σ⊗T that W [{x}]• = W 0 [{x}]• for almost every x. Q Q

254

Topologies and measures III

537S

˜ there is a W 0 ∈ Σ⊗T b such that W [{x}]4W 0 [{x}] is negligible for almost (c) In fact we find that for any W ∈ Λ every x. P P Set κ = τ (ν) = τ (B), and let heξ iξ<κ generate B. Let K ⊆ [κ]≤ω be a cofinal set of size cf[κ]≤ω . For K ∈ K, let BK be the closed subalgebra of B generated by {eξ : ξ ∈ K} and AK the set {x : x ∈ X, W [{x}]• ∈ BK }. Note that K 7→ AK is non-decreasing and that the union of any sequence in K belongs to K. So there is a K0 ∈ K such that µ∗ AK0 = supK∈K µ∗ AK . If E is a measurable envelope of AK0 , then {AK \ E : K ∈ K} is a cover of X \ E by negligible sets. So cov(X \ E, N (µ)) ≤ cf[κ]≤ω and X \ E must be negligible, that is, AK0 has full outer measure. Taking a sequence hFn in∈N in T such that {Fn• : n ∈ N} is dense in BK0 , we see from (a) that the function x 7→ inf n∈N ν(W [{x}]4Fn ) is measurable, while it is zero on AK0 . So W [{x}]• ∈ BK0 for almost every x ∈ X, that is, AK0 is actually conegligible. Taking a measurable conegligible set E 0 ⊆ AK0 and applying (b) to W ∩ (E 0 × Y ), we see b such that W [{x}]4W 0 [{x}] is negligible for almost every x ∈ X. Q that there is a W 0 ∈ Σ⊗T Q (d) Now turn to the function f under consideration. Suppose for the moment that it is bounded. For q ∈ Q set ˜ By (c), we have Vq ∈ Σ⊗T b such that Vq [{x}]4Wq [{x}] is ν-negligible for µ-almost Wq = {(x, y) : f (x, y) ≥ q} ∈ Λ. every x, and therefore Wq−1 [{y}]4Vq−1 [{y}] is µ-negligible for ν-almost every y, by (a). If q ≤ q 0 then Wq0 \ Wq is empty, so Vq0 [{x}] \ Vq [{x}] is ν-negligible for µ-almost every x, and Vq0 \ Vq is (µ × ν)-negligible, where µ × ν is the T b product measure on X × Y . Similarly, q0
ZZ

Z

g d(µ × ν) ZZ ZZ = g(x, y)ν(dy)µ(dx) = f (x, y)ν(dy)µ(dx).

f (x, y)µ(dx)ν(dy) =

g(x, y)µ(dx)ν(dy) =

(e) Finally, if f is unbounded, set fk (x, y) = min(f (x, y), k) for each k ∈ N. Then ZZ ZZ f (x, y)µ(dx)ν(dy) = lim fk (x, y)µ(dx)ν(dy) k→∞ ZZ ZZ = lim fk (x, y)ν(dy)µ(dx) = f (x, y)ν(dy)µ(dx). k→∞

537X Basic exercises (a)(i) Let (X, Σ, µ) be a measure space such that singletons are negligible and cf N (µ) = ω1 . Show that there is a Sierpi´ nski subset of X. (ii) Show that if µ is Lebesgue measure on R and cf N (µ) = ω1 , then there is a strongly Sierpi´ nski subset of R. (b) Show that for any uncountable cardinal κ there is a purely atomic probability space with a strongly Sierpi´ nski set of size κ. (c) Let (X, Σ, µ) be a measure space. Show that the union of any sequence of Sierpi´ nski sets in X is again a Sierpi´ nski set in X. (d) Let (X, Σ, µ) be a measure space and Y any subspace of X. Show that a subset of Y is a Sierpi´ nski set for the subspace measure on Y iff it is a Sierpi´ nski set for µ. (e) Suppose that λ is an infinite cardinal and the usual measure νλ on {0, 1}λ has a Sierpi´ nski set of size κ. Show that νλ has a Sierpi´ nski set A such that #(A ∩ E) = κ whenever νλ E > 0. (f ) Let (X, ρ) be a non-separable metric space with r-dimensional Hausdorff measure, where r > 0. Show that X has a Sierpi´ nski subset of size equal to the topological density of X. > (g) Suppose that non N < cov N , where N is the null ideal of Lebesgue measure on R. Let (X, T, Σ, µ) and (Y, S, spaces of countable Maharam type, and f : X × Y → [0, ∞[ a function such that RRT, ν) be Radon probability RR I= f (x, y)µ(dx)ν(dy) and I 0 = f (x, y)ν(dy)µ(dx) are both defined. Show that I = I 0 . > (h) Let (X, Σ, µ) be a probability space in which there is a well-ordered family in N (µ) with union X; e.g., because non N (µ) = #(X) or add N (µ)R = cov N (µ). Show that there is a function f : X × X → [0, 1] such that R f (x, y)µ(dx) = 0 for every y ∈ X and f (x, y)µ(dy) = 1 for every x ∈ X.

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> (i) (In this exercise, all integrals are to be taken with respect to one-dimensional Lebesgue measure µ.) (i) Find RR RR a function f : [0, 1]2 → {0, 1} such that f (x, y)dxdy = 1 but f (x, y)dydx = 0. (Hint: there RR is a disjoint family hAy iy∈[0,1] of sets of full outer measure.) (ii) Find a function f : [0, 1]2 → {0, 1} such that f (x, y)dxdy = 1 but RR RR RR f (x, y)dxdy = 1 but f (x, y)dydx = 0. f (x, y)dydx = 0. (iii) Find a function f : [0, 1]2 → {0, 1} such that (Hint: enumerate [0, 1] as hxξ iξ
538 Filters and limits A great many special types of filter have been studied. In this section I look at some which are particularly interesting from the point of view of measure theory: Ramsey ultrafilters, measure-converging filters and filters with the Fatou property. About half the section is directed towards Benedikt’s theorem (538M) on extensions of perfect probability measures; on the way we need to look at measure-centering ultrafilters (538G-538K) nd iterated products of filters (538E, 538L). The second major topic here is a study of ‘medial limits’ (538P-538S); these are Banach limits of a very special type. In between, the measure-converging property (538N) and the Fatou property (538O) offer some intriguing patterns. 538A Filters For ease of reference, I begin the section with a list of the special types of filter on N which we shall be looking at later. Definitions Let F be a filter on N.

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(a) F is free if it contains every cofinite subset of N. (b) F is a p-point filter if it is free and for every sequence hAn in∈N in F there is an A ∈ F such that A \ An is finite for every n ∈ N. (Compare 5A1Va.) (c) F is Ramsey or selective if it is free and for every f : [N]2 → {0, 1} there is an A ∈ F such that f is constant on [A]2 . (d) F isPrapid if it is free and for every sequence htn in∈N of real numbers which converges to 0, there is an A ∈ F such that n∈A |tn | is finite. Note that a free filter F on N is rapid iff for every f ∈ N N there is an A ∈ F such that #(A ∩ f (k)) ≤ k for every k ∈ N. P P (i) If F is rapid and f ∈ N N , let g ∈ N N be a strictly increasing sequence such P 1 if g(k) ≤ i < g(k + 1); then there is an A ∈ F such that i∈A ti is finite; as F that f ≤ g. Set ti = 2 if i < g(0), k+1 P is free, there is an A ∈ F such that i∈A ti ≤ 1, in which case #(A ∩ f (k)) ≤ #(A ∩ g(k)) ≤ k for every k ∈ N. (ii) If F satisfies the condition and hti ii∈N → 0, take a strictly increasing f P ∈ N N such that ≤ 2−k whenever k ∈ N and P∞ |ti | −k i ≥ f (k); let A ∈ F be such that #(A ∩ f (k)) ≤ k for every k; then i∈A |ti | ≤ k=0 2 #(A ∩ f (k + 1) \ f (k)) is finite. Q Q (e) F is nowhere dense if for every sequence htn in∈N in R there is an A ∈ F such that {tn : n ∈ A} is nowhere dense. (f ) F is measure-centering or has property M if whenever (A, µ ¯) is a totally finite measure algebra and han in∈N is a sequence in A such that inf n∈N µ ¯an > 0, there is an A ∈ F such that {an : n ∈ A} is centered. (g) F is measure-converging S if it T is free and whenever (X, Σ, µ) is a probability space, hEn in∈N is a sequence in Σ, and limn→∞ µEn = 1, then A∈F n∈A En is conegligible. (h)S F has T the Fatou property if whenever (X, Σ, µ) is a probability space, hEn in∈N is a sequence in Σ, and X = A∈F n∈A En , then limn→F µEn is defined and equal to 1. (i) For any countably infinite set I, I will say that a filter F on I is free, or a p-point filter, or Ramsey, etc., if it is isomorphic to such a filter on N. Of course this usage is possible only because every property here is invariant under permutations of N. For ‘rapid’ and ‘measure-converging’ filters, we need an appropriate translation of ‘sequence converging to 0’; the corresponding notion on an arbitrary index set I is a function u ∈ c0 (I), that is, a real-valued function u on I such that {i : i ∈ I, |u(i)| ≥ } is finite for every  > 0; if we give I its discrete topology, c0 (I) is C0 (I) as defined in 436I. 538B We need a number of basic ideas which can profitably be examined in a rather more general context. I start with a fundamental pre-order on the class of all filters. The Rudin-Keisler ordering If F, G are filters on sets I, J respectively, we say that F ≤RK G if there is a function f : J → I such that F = f [[G]] = {A : A ⊆ I, f −1 [A] ∈ G}, the filter on I generated by {f [B] : B ∈ G}. Of course ≤RK is reflexive and transitive. If F ≤RK G and G is an ultrafilter, then F is an ultrafilter (2A1N). If F is a principal ultrafilter then F ≤RK G for every filter G. 538C Lemma (a) If I is a set, F is an ultrafilter on I and f : I → I is a function such that f [[F]] = F, then {i : f (i) = i} ∈ F. (b) If I is a set, F and G are ultrafilters on I, F ≤RK G and G ≤RK F, then there is a bijection h : I → I such that h[[F]] = G; that is, F and G are isomorphic. proof (a) It is enough to consider the case in which I = κ is a cardinal. (i) {ξ : ξ < κ, ξ ≤ f (ξ)} ∈ F. P P Define hDn in∈N , hEn in∈N by saying that Dn+1 = {ξ : ξ ∈ Dn , f (ξ) ∈ Dn , f (ξ) < ξ}, En = Dn \ Dn+1 T for n ∈ N. If ξ ∈ Dn then ξ > f (ξ) > . . . > f n (ξ), so n∈N Dn = ∅ and hEn in∈N is a partition of κ. If ξ ∈ En+1 then S S f n+1 (ξ) < f n (ξ) < . . . < ξ, f n+1 (ξ) ≤ f n+2 (ξ), so f (ξ) ∈ En . Set E = n≥1 E2n , E 0 = n∈N E2n+1 ; then f [E] ⊆ E 0 is disjoint from E, so E ∈ / F. Also f [E 0 ] ⊆ E ∪ E0 is disjoint from E 0 , so E 0 ∈ / F. Because F is an ultrafilter, E0 ∈ F, as claimed. Q Q D0 = κ,

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S (ii) If A ⊆ I and A ∈ / F then B = n∈N (f n )−1 [A] does not belong to F. P P For ξ ∈ B set m(ξ) = min{n : n ∈ N, f n (ξ) ∈ A}. If m(ξ) > 0 then m(f (ξ)) = m(ξ) − 1. So setting C = {ξ : m(ξ) is even and not 0}, C 0 = {ξ : m(ξ) is 0 odd} we have f [C] ∩ C = ∅, f [C 0 ] ∩ CS = ∅ and B ⊆ A ∪ C ∪ C 0 ; so / F. Q Q TB ∈ Turning this round, if A ∈ F then n∈N (f n )−1 [κ \ A] ∈ / F and n∈N (f n )−1 [A] ∈ F. (iii) For ξ < κ set g(ξ) = min{ζ : there is some n ∈ N such that f n (ζ) = ξ}. T Then g[[F]] = F. P P If A ∈ F then F contains n∈N (f n )−1 [A] ⊆ g −1 [A], so g −1 [A] ∈ F. Thus F ⊆ g[[F]]; as F is an ultrafilter, F = g[[F]]. Q Q Now g(ξ) ≤ ξ for every ξ < κ; applying (i) to g, we see that G = {ξ : g(ξ) = ξ} ∈ F. But consider H = {ξ : ξ < f (ξ)}. Then g(η) < η for every η ∈ f [H], so f [H] ∈ / F and H ∈ / F. Since we already know that {ξ : ξ ≤ f (ξ)} ∈ F, we see that {ξ : f (ξ) = ξ} belongs to F, as claimed. (b) Let f , g : I → I be such that f [[F]] = G and g[[G]] = F. Then (gf )[[F]] = g[[f [[F]] ]] = F, so J0 = {i : g(f (i)) = i} ∈ F, by (a). Similarly, J1 = {i : f (g(i)) = i} belongs to G. Set J = J0 ∩ f −1 [J1 ] ∈ F; then g(f (i)) = i for every i ∈ J and f (g(j)) = j for every j ∈ f [J], so f J and gf [J] are inverse bijections between J ∈ F and f [J] ∈ G. If J is finite, then certainly #(I \ J) = #(I \ f [J]) and there is an extension of f J to a bijection from I to itself. If J is infinite, let J 0 ⊆ J be a set such that #(J 0 ) = #(J \ J 0 ) = #(J) and J 0 ∈ F; then #(I \ J 0 ) = #(I \ f [J 0 ]) = #(I) so there is an extension of f J 0 to a bijection from I to itself. Thus in either case we have a bijection h : I → I and a K ∈ F such that K ⊆ J and hK = f K. But now h[[F]] = G and h is an isomorphism between (I, F) and (I, G). 538D Finite products of filters (a) Suppose that F, G are filters on sets I, J respectively. I will write F n G for {A : A ⊆ I × J, {i : A[{i}] ∈ G} ∈ F}. It is easy to check that F n G is a filter. (Compare the skew product I n J of ideals defined in 527Ba.) (b) If F and G are ultrafilters, so is F n G. P P If A ⊆ I × J and A ∈ / F n G, then {i : A[{i}] ∈ G} ∈ / F} and {i : ((I × J) \ A)[{i}] ∈ G} = {i : i ∈ I, J \ A[{i}] ∈ G} = I \ {i : A[{i}] ∈ G} ∈ F, so (I × J) \ A ∈ F n G. Q Q (c) If F, G and H are filters on I, J, K respectively, then the natural bijection between (I × J) × K and I × (J × K) is an isomorphism between (F n G) n H and F n (G n H). P P If A ⊆ I × (J × K) and B = {(i, j), k) : (i, (j, k)) ∈ A}, then A ∈ F n (G n H) ⇐⇒ {i : A[{i}] ∈ G × H} ∈ F ⇐⇒ {i : {j : (A[{i}])[{j}] ∈ H} ∈ G} ∈ F ⇐⇒ {(i, j) : (A[{i}])[{j}] ∈ H} ∈ F n G ⇐⇒ {(i, j) : B[{(i, j)}] ∈ H} ∈ F n G ⇐⇒ B ∈ (F n G) n H. Q Q (d) It follows that we can define a product F0 n . . . n Fn of any finite string F0 , . . . , Fn of filters, and under the natural identifications of the base sets we shall have (F0 n . . . n Fn ) n (Fn+1 n . . . n Fm ) identified with F0 n . . . n Fm whenever F0 , . . . , Fn , . . . , Fm are filters. (e) For any filters F and G, F ≤RK F n G and G ≤RK F n G. P P Taking the base sets to be I, J respectively and f (i, j) = i, g(i, j) = j for i ∈ I and j ∈ J, we have F = f [[F n G]] and G = g[[F n G]]. Q Q Inducing on n, we see that Fn ≤RK F0 n. . .nFn whenever F0 , . . . , Fn are filters; consequently Fm ≤RK F0 n. . .nFn whenever F0 , . . . , Fn are filters and m ≤ n. (f ) If F, F 0 , G and G 0 are filters, with F ≤RK F 0 and G ≤RK G 0 , then F n G ≤RK F 0 n G 0 . P P Let the base sets of the filters be I, I 0 , J and J 0 , and let f : I 0 → I and g : J 0 → J be such that F = f [[F 0 ]] and G = g[[G 0 ]]. Set h(i, j) = (f (i), g(j)) for i ∈ I and j ∈ J. If A ⊆ I × J, then h−1 [A] ∈ F 0 n G 0 ⇐⇒ {i : (h−1 [A])[{i}] ∈ G 0 } ∈ F 0 ⇐⇒ {i : g −1 [A[{f (i)}]] ∈ G 0 } ∈ F 0 ⇐⇒ {i : A[{f (i)}] ∈ G} ∈ F 0 ⇐⇒ {i : A[{i}] ∈ G} ∈ F ⇐⇒ A ∈ F n G.

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So F n G = h[[F 0 n G 0 ]] and F n G ≤RK F 0 n G 0 . Q Q Accordingly F0 n . . . n Fn ≤RK G0 n . . . n Gn whenever Fi ≤RK Gi for every i ≤ n. (g) It follows that if F0 , . . . , Fn are filters and k0 < . . . < km ≤ n, then Fk0 n . . . n Fkm ≤RK F0 n . . . n Fn . P P Induce on m to see that Fk0 n . . . n Fkm ≤RK F0 n . . . n Fkm . Q Q 538E There are many variations on the construction here. A fairly elaborate extension will be needed in 538L below. S Iterated products of filters (a) First, a scrap of notation. Set S ∗ = i∈N N i . If i, j ∈ N, σ ∈ N i and τ ∈ N j , define σ a τ ∈ N i+j by setting (σ a τ )(k) = σ(k) if k < i, = τ (k − i) if i ≤ k < i + j. For k ∈ N write for the member of N 1 with value k. Fix on a family hθ(ξ, k)i1≤ξ<ω1 ,k∈N such that each hθ(ξ, k)ik∈N is a non-decreasing sequence running over a cofinal subset of ξ. (You will probably prefer to suppose that when ξ = η + 1 is a successor ordinal, then θ(ξ, k) = η for every k ∈ N.) (b) Now suppose that ζ is a non-zero countable ordinal. Let hFξ i1≤ξ≤ζ be a family of free filters on N. For ξ ≤ ζ, define Gξ ⊆ PS ∗ as follows. Start by taking G0 to be the principal filter generated by {∅}. For 1 ≤ ξ ≤ ζ, set Gξ = {A : A ⊆ S ∗ , {k : k ∈ N, {τ : a τ ∈ A} ∈ Gθ(ξ,k) } ∈ Fξ }. It is elementary to check that every Gξ is a filter. Moreover, if every Fξ is an ultrafilter, so is every Gξ . (c) Continuing from (b), we find that Fξ ≤RK Gξ whenever 1 ≤ ξ ≤ ζ and Gη ≤RK Gξ whenever 0 ≤ η ≤ ξ ≤ ζ. P P Induce on ξ. Define f : S ∗ → N by setting f (τ ) = τ (0) if τ 6= ∅, f (∅) = 0. (i) If ξ ≥ 1, then for B ⊆ N, f −1 [B] ∈ Gξ ⇐⇒ {k : {τ : a τ ∈ f −1 [B]} ∈ Gθ(ξ,k) } ∈ Fξ ⇐⇒ B ∈ Fξ , so Fξ = f [[Gξ ]] ≤RK Gξ . (ii) If η = ξ ≤ ζ then of course Gη ≤RK Gξ . (iii) If 0 ≤ η < ξ then there is a k0 such that η ≤ θ(ξ, k) for k ≥ k0 . For k ≥ k0 , Gη ≤RK Gθ(ξ,k) by the inductive hypothesis; let gk : S ∗ → S ∗ be such that Gη = gk [[Gθ(ξ,k) ]]. Now define g : S ∗ → S ∗ by setting g(τ ) = gk (σ) if k ≥ k0 and τ = a σ, = ∅ otherwise. For B ⊆ S ∗ , g −1 [B] ∈ Gξ ⇐⇒ {k : {σ : a σ ∈ g −1 [B]} ∈ Gθ(ξ,k) } ∈ Fξ ⇐⇒ {k : k ≥ k0 , {σ : g(a σ) ∈ B} ∈ Gθ(ξ,k) } ∈ Fξ (because Fξ is free) ⇐⇒ {k : k ≥ k0 , {σ : gk (σ) ∈ B} ∈ Gθ(ξ,k) } ∈ Fξ ⇐⇒ {k : k ≥ k0 , B ∈ Gη } ∈ Fξ ⇐⇒ B ∈ Gη , so Gη = g[[Gξ ]] ≤RK Gξ . Q Q (d) It follows that if 1 ≤ ξ0 < . . . < ξn ≤ ζ then Fξn n . . . n Fξ0 ≤RK Gξn . P P Induce on the pair (ξn , n). If ξn = 1 then n = 0 and we just have F1 ≤RK G1 , as in part (i) of the proof of (c). For the inductive step to ξn = ξ > 1, if n = 0 then again we need only note that Fξ0 = Fξ ≤RK Gξ . If n > 0, let k0 ≥ 1 be such that ξn−1 ≤ θ(ξ, k) for every k ≥ k0 . For k ≥ k0 , Fξn−1 n . . . n Fξ0 ≤RK Gξn−1 ≤ Gθ(ξ,k) by the inductive hypothesis, so we have a function gk : S ∗ → N n such that Fξn−1 n . . . n Fξ0 = gk [[Gθ(ξ,k) ]]. Define g : S ∗ → N n+1 by setting g(τ ) = a gk (σ) if k ≥ k0 and τ = a σ, = the constant function with value 0 otherwise.

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Then, for B ⊆ N n+1 , g −1 [B] ∈ Gξ ⇐⇒ {k : {σ : g(a σ) ∈ B} ∈ Gθ(ξ,k) } ∈ Fξ ⇐⇒ {k : k ≥ k0 , {σ : a gk (σ) ∈ B} ∈ Gθ(ξ,k) } ∈ Fξ ⇐⇒ {k : k ≥ k0 , {σ : gk (σ) ∈ Bk } ∈ Gθ(ξ,k) } ∈ Fξ (writing Bk = {σ : a σ ∈ B} ⊆ N n for k ∈ N) ⇐⇒ {k : k ≥ k0 , Bk ∈ Fξn−1 n . . . n Fξ0 } ∈ Fξ ⇐⇒ {k : k ∈ N, Bk ∈ Fξn−1 n . . . n Fξ0 } ∈ Fξ ⇐⇒ B ∈ Fξn n . . . n Fξ0 . Thus g witnesses that Fξn n . . . n Fξ0 ≤RK Gξn , and the induction proceeds. Q Q Consequently Fξn n . . . n Fξ0 ≤RK Gζ whenever 1 ≤ ξ0 < . . . < ξn ≤ ζ. (e) The following special remark will be useful in Theorem 538L. Suppose that we are given Aξ ∈ Fξ for each ξ ∈ [1, ζ]. Define T ⊆ S ∗ and α : T → [0, ζ] as follows. Start by saying that ∅ ∈ T and α(∅) = ζ. Having determined T ∩ N n and αT ∩ N n , where n ∈ N, then for τ ∈ N n+1 say that τ ∈ T iff τ is of the form σ a where σ ∈ T ∩ Nn,

α(σ) > 0,

k ∈ Aα(σ) ,

σ(i) < k for every i < n,

and in this case set α(τ T) = θ(α(σ), k). Continue. S ∗ Suppose that D ∈ 1≤ξ≤ζ Fξ . Then TD = {τ : τ ∈ T ∩ n∈N Dn , α(τ ) = 0} belongs to Gζ . P P I aim to show by S ∗ induction on ξ that if τ ∈ T ∩ n∈N Dn and α(τ ) = ξ then {σ : τ a σ ∈ TD } belongs to Gξ . If ξ = 0 then of course ∗ } = {∅} ∈ G0 . For the inductive step to ξ > 0, {σ : τ a σ ∈ TD ∗ {k : {σ : τ a a σ ∈ TD } ∈ Gθ(ξ,k) }

⊇ {k : k ∈ D, τ a ∈ T, α(τ a ) = θ(ξ, k)} (by the inductive hypothesis) ⊇ {k : k ∈ Aξ ∩ D, τ (i) < k for every i < dom τ } ∈ Fξ , ∗ } ∈ Gξ . At the end of the induction, we can apply this to τ = ∅ and ξ = ζ. Q Q so {σ : τ a σ ∈ TD

538F Ramsey filters There is an extensive and fascinating theory of Ramsey filters; see, for instance, Comfort & Negrepontis 74. Here, however, I will give only those fragments which are directly relevant to the other work of this section. Proposition (a) A Ramsey filter on N is a rapid p-point ultrafilter. (b) If F is a Ramsey ultrafilter on N, G is a non-principal ultrafilter on N, and G ≤RK F, then F and G are isomorphic and G is a Ramsey ultrafilter. (c) Let F be a Ramsey filter on N. Suppose that hAn in∈N is any sequence in F. Then there is an A ∈ F such that n ∈ Am whenever m, n ∈ A and m < n. (d) Let F be a Ramsey filter on N. Let S ⊆ [N]<ω be such that ∅ ∈ S and {n : I ∪ {n} ∈ S} ∈ F for every I ∈ S. Then there is an A ∈ F such that [A]<ω ⊆ S. (e) If F is a countable family of distinct Ramsey filters on N, there is a disjoint family hAF iF ∈F of subsets of N such that AF ∈ F for every F ∈ F. <ω (f) Let F be a countable family of non-isomorphic Ramsey ultrafilters a function. Suppose T on N, and h : N → [F] that we are given an AF ∈ F for each F ∈ F. Then there is an A ∈ F such that whenever i, j ∈ A, F ∈ h(i) and i < j, there is a k ∈ AF such that i < k < j. (g) If cov M = c, where M is the ideal of meager subsets of R, there is a Ramsey ultrafilter on N. proof (a) Let F be a Ramsey filter on N. (i) F is an ultrafilter. P P Let A be any subset of N. Define f : [N]2 → {0, 1} by setting f (I) = 1 if #(I ∩ A) = 1, 0 otherwise. Then we have an I ∈ F such that f is constant on [I]2 . As F is free, #(I) ≥ 3 and the constant value of f cannot be 1. So either I ⊆ A and A ∈ F, or I ∩ A = ∅ and N \ A ∈ F. As A is arbitrary, F is an ultrafilter. Q Q

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T (ii) F is a p-point filter. P P Let hIn in∈N be a sequence in F. Set Kn = (N \ n) ∩ i 0 and whenever i ∈ N and F ∈ h(i), there is a k ∈ AF such that i < k < g(i). Set lm = g m (0) and Jm = lm+1 \ lm for each m, so that hJm im∈N is a partition of N. Let haξ iξ<ω1 be a family S of infinite subsets of N, all containing 0, such that aξ ∩ aη is finite for all distinct ξ, η < ω1 (5A1Fa), and set Mξ = m∈aξ Jm for each ξ; then Mξ ∩ Mη is finite for all distinct ξ, η < ω1 . It follows that each member of F can contain at most T one Mξ , and there is a ξ < ω1 such that Mξ does not belong to any member of F, that is, M = N \ Mξ belongs to F. (ii) Define f : N → N by setting f (n) = max{m : m ∈ aξ , lm ≤ n} for n ∈ N. For each F ∈ F, f [[F]] is isomorphic to F, by (b). It follows that if F, F 0 are distinct members of F, f [[F]] 6= f [[F 0 ]]. Because F is countable, there is a disjoint family hKF iF ∈F of sets such that KF ∈ f [[F]] for every F ∈ F ((e) above). Set LF = f −1 [KF ] ∈ F for each F ∈ F. (iii) Set S = {{i, j} : i, j ∈ N, i < j < g(i)}. For each F ∈ F, there is an L0F ∈ F such that L0F ⊆ LF and [L0F ]2 is either included in S or disjoint from S. The former is surely impossible, since L0F is infinite, so [L0F ]2 ∩ S = ∅, that is, g(i) ≤ j whenever i, j ∈ L0F and i < j. S T (iv) Consider A = F ∈F L0F ∩ M . Then A ∈ F. Suppose that i, j ∈ A and i < j; then g(i) ≤ j. P P Let F, F 0 ∈ F be such that i ∈ L0F and j ∈ L0F 0 . case 1 If F = F 0 , then both i and j belong to L0F , so g(i) ≤ j by (iii). case 2 If F 6= F 0 , then i ∈ LF and j ∈ LF 0 , so f (i) ∈ KF and f (j) ∈ KF 0 and f (i) 6= f (j). Let m, n ∈ N be such that i ∈ Jm and j ∈ Jn ; since j ∈ / Mξ , n ∈ / aξ and f (j) < n. As KF and KF 0 are disjoint, f (i) < f (j). It follows that m < f (j) < n, so

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g(i) ≤ g(lm+1 ) ≤ g(lf (j) ) ≤ ln ≤ j and g(i) ≤ j in this case also. Q Q By the choice of g, this means that if F ∈ h(i) there must be a k ∈ AF such that i < k < j, as required. (g)(i) Suppose that E ⊆ PN is a filter base, containing N \ n for every n ∈ N, and of size less than cov M. Let f : [N]2 → {0, 1} be a function. Then there is an F ⊆ N such that f is constant on [F ]2 and F meets every member of E. P P Set E + = {J : J ⊆ N, J ∩ E 6= ∅ for every E ∈ E}, Sn = {n} ∪ {i : i ∈ N \ {n}, f ({i, n}) = 1}, Sn0 = {n} ∪ {i : i ∈ N \ {n}, f ({i, n}) = 0} for n ∈ N. case 1 Suppose that {n : n ∈ J, J ∩ Sn ∈ E + } belongs to E + for every J ∈ E + . Set T I = {I : I ∈ [N]<ω , f (K) = 1 for every K ∈ [I]2 , N ∩ i∈I Si ∈ E + }. T If I ∈ I, J = N ∩ i∈I Si and E ∈ E, then J ∈ E + ; because E is a filter base, J ∩ E ∈ E + ; by hypothesis, {n : n ∈ J ∩ E, J ∩ E ∩ Sn ∈ E + } belongs to E + and is not empty. There is therefore some n ∈ J ∩ E such that J ∩ Sn ∈ E + , in which case I ∪ {n} ∈ I. In particular, there is some k ∈ N such that {k} ∈ I. Set C = {α : α ∈ N N , {α(i) : i < m} ∈ I for every m ∈ N}. Then C is compact, and it is non-empty because the constant function with value k belongs to C. Moreover, if α ∈ C and m ∈ N and E ∈ E, there is an n ∈ E such that {α(i) : i < m} ∪ {n} ∈ I, so there is a β ∈ C such that β(i) = α(i) for i < m and β(m) = n. Thus {β : β ∈ C, E ∩ β[N] 6= ∅} is a dense open subset of C. Writing M(C) for the ideal of meager subsets of C, cov M(C) is either ∞ (if C has an isolated point) or cov M, by 522Vb; in either case, it is greater than #(E). There is therefore some α ∈ C such that F = α[N] meets every member of E; in this case, f is equal to 1 everywhere in [F ]2 , so we have an appropriate F . case 2 Otherwise, there is a K ∈ E + such that {n : n ∈ K, K ∩ Sn ∈ E + } does not belong to E + . Let E0 ∈ E be disjoint from {n : n ∈ K, K ∩ Sn ∈ E + }. Set G = E ∪ {K ∩ E : E ∈ E}, so that G is a filter base and #(G) < cov M. If n ∈ E0 then there is an En0 ∈ E disjoint from K ∩ Sn . So if J ∈ G + , J ∩ Sn0 ⊇ (J ∩ K ∩ En0 ) \ {n} belongs to G + for every n ∈ E0 ; accordingly {n : n ∈ J, J ∩ Sn0 ∈ G + } ⊇ J ∩ E0 belongs to G + . We can therefore apply the argument of case 1 to G and the function 1 − f to see that there is an F ⊆ N, meeting every member of G ⊇ E, such that f = 0 on [F ]2 . Q Q (ii) Enumerate the set of functions from [N]2 to {0, 1} as hfξ iξ 0, there is an A ∈ F such that {an : n ∈ A} is centered; T (iii) whenever hFn in∈N is a sequence in Tω such that inf n∈N νω Fn > 0, there is an A ∈ F such that n∈A S T Fn 6= ∅; (iv) whenever (X, Σ, µ) is a perfect totally finite measure space and hFn in∈N is a sequence in Σ, µ∗ ( A∈F n∈A Fn ) ≥ lim inf n→F µFn ; (v) whenever µ is a Radon probability measure on PN, then µ∗ F ≥ lim inf n→F µEn , where En = {a : n ∈ a ⊆ N} for each n. proof (i)⇒(ii) is trivial. not-(iv)⇒not-(ii) Suppose there sequence hFn in∈N in S are Ta perfect totally finite measure space (X, Σ, µ) and S aT Σ such that lim inf n∈N µFn > µ∗ ( A∈F n∈A Fn ). Let F be a measurable envelope of A∈F n∈A Fn . Let T be the σ-subalgebra of Σ generated by {F } ∪ {Fn : n ∈ N}; then µ T is a compact measure (451F). Let ν be its

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1 µ T; µX

538G

then ν is a compact probability measure. We see that lim inf n→F νFn > νF ; take γ such that

νF < γ < lim inf n→F νFn , and set C = {n : νFn > γ}, so that C ∈ F. Let K be a compact class such that ν is inner regular with respect to K. For n ∈ C, let Kn ∈ K ∩ T be such that Kn ⊆ Fn \ F and νKn ≥ γ − νF ; for n ∈ N \ C set Kn = X. The measure algebra (B, ν¯) of ν is a probability algebra with countable Maharam type, so there is a measurepreserving Boolean homomorphism π : B → Bω (332P or 333D). Set an = πKn• for each n. Then ν¯ω an = νKn ≥ γ − νF > 0 T T for every n. On the other hand, if A ∈ F, then A ∩ C ∈ F so n∈A∩C Kn ⊆ n∈A∩C Fn \ T F is empty. As Kn belongs to the compactTclass K for every n ∈ A ∩ C, there must be a finite set I ⊆ A ∩ C such that n∈I Kn = ∅, in which case inf n∈I an = π( n∈I Kn )• = 0. This shows that {an : n ∈ A} is not centered. So han in∈N witnesses that (ii) is false. (iv)⇒(i) Suppose that (iv) is true. Take a totally finite measure algebra (A, µ ¯) and a sequence han in∈N in A such that inf n∈N µ ¯an > 0. Let (Z, µ) be the Stone space of (A, µ ¯), so that Z is compact and µ is a Radon measure on Z (411Pe), therefore perfect (416Wa). For n ∈ N let b an ⊆ Z be the open-and-closed set corresponding to an ∈ A. Then lim inf n→F µb an ≥ inf n∈N µb an = inf n∈N µ ¯an > 0, S T T ∗ so µT ( A∈F n∈A b an ) > 0 and there is an A ∈ F such that n∈A b an is non-empty. In particular, (inf n∈I an )b = Z ∩ n∈I b an is non-empty for every finite I ⊆ A, so {an : n ∈ A} is centered. As (A, µ ¯) and han in∈N are arbitrary, F is a measure-centering filter. (iv)⇒(v) The point is simply that µ is perfect (416Wa again) and that S T S A∈F n∈A En = A∈F {a : A ⊆ a ⊆ N} = F. (v)⇒(iii) Suppose that (v) is true, and that hFn in∈N is a sequence in Tω such that inf n∈N νω Fn > 0. Define φ : {0, 1}N → PN by setting φ(x) = {n : x ∈ Fn } for each n. Then φ is almost continuous (418Dd or otherwise), so the image measure µ = νω φ−1 is a Radon probability measure on PN (418I). Defining En as in (v), we have µEn = νω φ−1 [En ] = νω Fn for every n ∈ N, so 0 < inf n∈N ν¯ω an ≤ lim inf n→F µEn ≤ µ∗ F = ν ∗ φ−1 [F] (451Pc7 ). In particular, there must be an x ∈ φ−1 [F], so that A = {n : x ∈ Fn } belongs to F, and non-empty.

T

n∈A

Fn is

(iii)⇒(ii) Assume (iii). Let han in∈N be a sequence in Bω such that inf n∈N ν¯ω an > 0. Let θ : Bω → Tω be a lifting (341K), and set Fn = θan for each n. ThenTνω Fn = ν¯ω an for every n, so (iii) tells us that there is an A ∈ F such that T n∈A Fn 6= ∅. In this case, θ(inf n∈I an ) = n∈I Fn 6= ∅ for every non-empty finite I ⊆ A, so {an : n ∈ A} is centered. 538H Proposition (a) Any measure-centering filter on N is an ultrafilter. (b) If F is a measure-centering ultrafilter on N and G is a filter on N such that G ≤RK F, then G is measure-centering. (c) Every Ramsey ultrafilter on N is measure-centering. (d) (Shelah 98) Every measure-centering ultrafilter on N is a nowhere dense ultrafilter. (e) (Benedikt 99) If cov N = c, where N is the Lebesgue null ideal, then there is a measure-centering ultrafilter on N. proof (a) Let (A, µ ¯) be any totally finite measure algebra such that A has disjoint non-zero elements a, b. Given I ⊆ N, set an = a if n ∈ I, b if n ∈ N \ I. Then inf n∈N ν¯ω an > 0, so there is a J ∈ F such that {an : n ∈ J} is centered, in which case either J ⊆ I or J ∩ I = ∅; so that one of I, N \ I must belong to F. (b) Let f : N → N be such that f [[F]] = G. Let (A, µ ¯) be a totally finite measure algebra and han in∈N a sequence in A with inf n∈N µ ¯an > 0. Then haf (n) in∈N has the same property, so there is an A ∈ F such that {af (n) : n ∈ A} is centered. Now f [A] ∈ G and {am : m ∈ f [A]} is centered. (c) Let F be a Ramsey ultrafilter, (A, µ ¯) a totally finite measure space, and han in∈N a sequence in A such that γ = inf n∈N µ ¯an is greater than 0. Set b = inf A∈F supn∈A an ; then µ ¯b ≥ γ. Set S = {I : I ∈ [N]<ω , b ∩ inf n∈I an 6= 0}. Then ∅ ∈ S. If I ∈ S, set c = b ∩ inf n∈I an and C = {n : c ∩ an = 0}. Then supn∈C an does not meet c so does not include b, and C ∈ / F. Accordingly {n : I ∪ {n} ∈ S = N \ C ∈ F. 7 Formerly

451Oc.

538I

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263

By 538Fd, there is an A ∈ F such that [A]<ω ⊆ S, in which case {an : n ∈ A} is centered. As (A, µ ¯) and han in∈N are arbitrary, F is measure-centering. (d) Let F be a measure-centering ultrafilter, S and htn in∈N a sequence in R. Let F ⊆ [0, 1[ be a nowhere dense set with non-zero Lebesgue measure, and set H = k∈Z F + k, so that H is nowhere dense in R; let µ be Lebesgue measure on [0, 1]. For n ∈ N set S En = {x : x ∈ [0, 1], x + tn ∈ H} = [0, 1] ∩ k∈Z F − tn + k, T T so that µEn = µF > 0. By 538G(iv), there is an A ∈ F such that n∈A En is non-empty; take x ∈ n∈A En , so that tn ∈ H − x for every n ∈ A, and {tn : n ∈ A} is nowhere dense. As htn in∈N is arbitrary, F is a nowhere dense filter. (e)(i) Let han in∈N be a sequence in Bω such that inf n∈N ν¯ω an > 0, and C ⊆ PN a filter base such that #(C) < cov N . Then there is an A ⊆ N such that A meets every member of C and {an : n ∈ A} is centered. P P Set  = inf n∈N ν¯ω an . For C ∈ C set bC = supn∈C an ; because C 6= ∅, ν¯ω bC ≥ . Set b = inf C∈C bC ; because C is downwards-directed, ν¯ω b ≥  (321F) and b 6= 0. S Let θ : Bω → Tω be a lifting (341K). For C ∈ C, set FC = n∈C θan ; then FC• = bC ⊇ b, T so θb \ FC is negligible. Because b 6= 0, θb is not negligible; because #(C) < cov N , θb ∩ C∈C FC is non-empty (apply 522Va to the subspace measure on θb). Take any x in the intersection, and set A = {n : x ∈ θan }. For every T C ∈ C, there is an n ∈ C such that x ∈ θan , so A ∩ C 6= ∅. If I ⊆ A is finite and not empty, then θ(inf n∈I an ) = n∈I θan contains x, so inf n∈I an 6= 0; thus {an : n ∈ A} is centered. Q Q (ii) Since #(Bω ) = c (524Ma), we can enumerate as hhaξn in∈N iξ 0. Choose hCξ iξ
| lim µEn − lim µFn | = lim |µEn − µFn | ≤ lim µ(En 4Fn ) ≤ µ∗ ( lim En 4Fn ) n→F

n→F

n→F

n→F

n→F

(538G(iv)) = µ∗ ( lim En 4 lim Fn ) = µ∗ ∅ = 0. Q Q n→F

n→F

(c) We therefore have a functional φ : A → [0, 1] defined by setting φ(limn→F En ) = limn→F µEn for every sequence hEn in∈N in Σ. Clearly φ extends µ. Also φ is additive. P P If hEn in∈N , hFn in∈N are sequences in Σ such that limn→F En and limn→F Fn are disjoint, then

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φ( lim En ∪ lim Fn ) = φ( lim En ∪ Fn ) = lim µ(En ∪ Fn ) n→F

n→F

n→F

n→F

= lim µEn + µFn − µ(En ∩ Fn ) n→F

= lim µEn + lim µFn − lim µ(En ∩ Fn ) n→F

n→F

n→F

= φ( lim En ) + φ( lim Fn ) − φ( lim En ∩ Fn ) n→F

n→F

n→F

= φ( lim En ) + φ( lim Fn ). Q Q n→F

n→F

(d) T Next, if hAm im∈N is a non-increasing sequence in A, and 0 ≤ γ < inf m∈N φAm , there is an A ∈ A such that A ⊆ m∈N Am and φA ≥ γ. P P We can suppose that A0 = X. For each m ∈ N, let hE Tmn in∈N be a sequence in Σ such 0 that Am = limn→F Emn , starting with E0n = X for every n. For m ∈ N, set Emn = i≤m Ein for n ∈ N; then T 0 Am = i≤m Ai = limn→F Emn ; T 0 0 0 set Im = {n : n ∈ N, µEmn ≥ γ}. Since limn→F µEmn = φAm > γ, Im ∈ F. For n ∈ N, set Fn = {Emn : m ∈ N, 0 0 µEmn ≥ γ}; set A = limn→F Fn . Then µFn ≥ γ for every n, so φA ≥ γ. Also, for m ∈ N, Fn ⊆ Emn whenever n ∈ Im , 0 so A ⊆ limn→F Emn = Am . Q Q (e) In particular, inf m∈N φAm must be 0 whenever hAm im∈N is a non-increasing sequence in A with empty intersection. By 413J, there is a complete measure λ on X extending φ and inner regular with respect to Aδ , the set of intersections of sequences in A. But λC = sup{λA : A ∈ A, A ⊆ C} for every C ∈ Aδ . P P Suppose that 0 ≤ γ < λC. There is a sequence hAm im∈N in A with intersection C; because A is an algebra of sets, we can suppose that hAm im∈N is non-increasing. Now γ < λC = inf m∈N λAm = inf m∈N φAm , so (d) tells us that there is an A ∈ A such that A ⊆ C and γ ≤ φA = λA. Q Q It follows at once that λ is inner regular with respect to A. (f ) If E ∈ Σ and we set En = E for every n ∈ N, then E = limn→F En belongs to A and λE = φE = limn→F µEn = µE. So λ extends µ. Finally, we see from 412L, as usual, that λ is uniquely defined. 538J Proposition Let F be a measure-centering ultrafilter on N and (X, Σ, µ) a perfect probability space; let λ be the corresponding extension of µ as defined in 538I. ¯ the measure algebra of λ, and (C, ν¯) the probability algebra (a) Let (A, µ ¯) be the measure algebra of µ, (B, λ) N 8 reduced power (A, µ ¯) |F (328C ). Then we have a measure-preserving isomorphism π : B → C defined by saying that π((limn→F En )• ) = hEn• i•n∈N for every sequence hEn in∈N in Σ. (b) Let (X 0 , Σ0 , µ0 ) be another perfect probability space, and φ : X → X 0 an inverse-measure-preserving function. Let λ0 be the corresponding F-extension of µ0 . Then φ is inverse-measure-preserving for λ and λ0 . (c) Let F 0 be a filter on N such that F 0 ≤RK F, and λ0 the F 0 -extension of µ. Then λ extends λ0 . proof (a)(i) I had better check first that the formula for π defines a function. If hEn in∈N , hFn in∈N are sequences in Σ such that (limn→F En )• = (limn→F Fn )• in B, then 0 = λ( lim En 4 lim Fn ) = lim µ(En 4Fn ) n→F

n→F

n→F

= lim µ ¯(En• 4 Fn• ) = ν¯(hEn• i•n∈N 4 hFn• i•n∈N ), n→F

so hEn• i•n∈N = hFn• i•n∈N in C. (ii) Setting B0 = {E • : E ∈ A}, where A is as in 538I, it is now routine to check that π : B0 → C is a surjective measure-preserving Boolean homomorphism. (Recall that C is, by definition, the quotient of AN by the ideal {han in∈N : limn→F µ ¯an = 0}.) But of course this means that B0 is isomorphic to C, therefore Dedekind complete. Since λ is inner regular with respect to A (538I), B0 is order-dense in B, and must be the whole of B. 8 Later

editions only.

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265

(b) Setting A = {limn→F En : En ∈ Σ ∀ n ∈ N},

A0 = {limn→F Fn : Fn ∈ Σ0 ∀ n ∈ N}

as in 538I, φ−1 [C] ∈ A and λφ−1 [C] = λ0 C for every C ∈ A0 . P P Let hFn in∈N be a sequence in Σ0 such that C = limn→F Fn ; then λφ−1 [C] = λφ−1 [ lim Fn ] = λ( lim φ−1 [Fn ]) n→F

n→F

= lim µφ−1 [Fn ] = lim νFn = λ0 C. Q Q n→F

n→F

0

By 412K, φ is inverse-measure-preserving for λ and λ . (c) By 538Hb, F 0 is measure-centering. Let f : N → N be such that F 0 = f [[F]]. Setting A = {limn→F En : En ∈ Σ ∀ n ∈ N},

A0 = {limn→F 0 En : En ∈ Σ0 ∀ n ∈ N},

A0 ⊆ A and λA = λ0 A for every A ∈ A0 . P P Let hEn in∈N be a sequence in Σ such that A = limn→F 0 En ; then A = limn→F Ef (n) , so λA = limn→F µEf (n) = limn→F 0 µEn = λ0 A. Q Q By 412K again, the identity map from X to itself is inverse-measure-preserving for λ and λ0 , that is, λ extends λ0 . 538K Having identified the measure algebra of a measure-centering-ultrafilter extension λ as a probability algebra reduced product (538Ja), we are in a position to apply the results of §3778 . Proposition Let (X, Σ, µ) be a perfect probability space, F a measure-centering ultrafilter on N and λ the corresponding extension of µ as constructed in 538I. (a)(i) Let hfn in∈N be a sequence in L0 (µ) such that {fn• : n ∈ N} is bounded in the linear topological space L0 (µ). Then (α) f (x) = limn→F fn (x) is defined in R for λ-almost every x ∈ X; (β) f ∈ L0 (λ). (ii) For every f ∈ L0 (λ) there is a sequence hfn in∈N in L0 (µ), bounded in the sense of (i), such that f = limn→F fn λ-a.e. (b) Suppose that 1 ≤ p ≤ ∞, and hfn in∈N is a sequence in Lp (µ) such that supn∈N kfn kp is finite. Set f (x) = limn→F fn (x) whenever this is defined in R. (i)(α) f ∈ Lp (λ); (β) kf kp ≤ limn→F kfn kp . (ii) Let g be a conditional expectation of f on Σ. (α) If p = 1 and {fn : n ∈ N} is uniformly integrable, then then kf k1 = limn→F kfn k and g • = limn→F fn• for the weak topology of L1 (µ). (β) If 1 < p < ∞, then g • = limn→F fn• for the weak topology of Lp (µ). (c) Suppose that 1 ≤ p ≤ ∞ and f ∈ Lp (λ). (i) There is a sequence hfn in∈N in Lp (µ) such that f =a.e. limn→F fn and kf kp = supn∈N kfn kp . (ii) If p = 1, we can arrange that hfn in∈N should be be uniformly integrable. (d) Let (X 0 , Σ0 , µ0 ) be another perfect measure space, and λ0 the extension of µ0 corresponding to the measurecentering filter F. Let S : L1 (µ) → L1 (µ0 ) be a bounded linear operator. ˆ • = g • whenever hfn in∈N , hgn in∈N (i) There is a unique bounded linear operator Sˆ : L1 (λ) → L1 (λ0 ) such that Sf 1 1 are uniformly integrable sequences in L (µ), L (ν) respectively, f = limn→F fn λ-a.e., g = limn→F gn λ0 -a.e., and gn• = Sfn• for every n ∈ N. (ii) The map S 7→ Sˆ : L∼ (L1 (µ); L1 (µ0 )) → L∼ (L1 (λ); L1 (λ0 )) is a norm-preserving Riesz homomorphism. proof We shall find that most of the work for this result has been done in §377. The only new step is in (a)(i), but we shall have some checking to do. (a)(i) Let hf˜n in∈N be a sequence of Σ-measurable functions from X to R such that f˜n = fn µ-a.e. for every n ∈ N. R 1 α) Let  > 0. Then there is a γ > 0 such that min(1, |fn (x)|)µ(dx) ≤ , so that µEn ≤  for every n ∈ N, (α γ

where En = {x : |f˜n (x)| ≥ γ}. Set E = limn→F En , so that λE ≤ . For x ∈ X \ E, {n : |f˜n (x)| ≤ γ} ∈ F, so limn→F f˜n (x) is defined in R. As  is arbitrary, limn→F f˜n (x) is defined in R for λ-almost every x. Since {x : x ∈ dom fn and fn (x) = f˜n (x) for every n ∈ N} is µ-conegligible, therefore λ-conegligible, limn→F fn is defined in R λ-a.e.

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538K

β ) For any α ∈ R, (β S {x : limn→F f˜n (x) > α} = k∈N limn→F {x : fn (x) ≥ α + 2−k } ∈ dom λ. So f =a.e. limn→F f˜n belongs to L0 (λ). (ii) At this point I seek to import the machinery of §377. ¯ be the measure algebras of µ, λ respectively; recall that we can identify L0 (µ) α) Let (A, µ (α ¯) and (B, λ) 0 0 and L (λ) with L (A) and L (B) (364Jc). Write (C, ν¯) for the probability algebra reduced power (A, µ ¯)N |F; let N −1 N φ : A → C be the canonical surjection, and π : B → C the isomorphism of 538Ja; set ψ = π φ : A → B. Then ¯ λψ(ha ¯an for every sequence han in∈N in A, and ψ is surjective. n in∈N ) = limn→F µ 0

β ) Let W0 ⊆ L0 (A)N be the set of sequences bounded for the topology of convergence in measure, and W0 ⊆ (β L (µ) the set of sequences hfn in∈N such that hfn• in∈N ∈ W0 . Then we have a Riesz homomorphism T : W 0 → L0 (B) defined by saying that T (hfn• in∈N ) = (limn→F fn )• whenever hfn in∈N ∈ W0 . P P We know from (i) that (limn→F fn )• is 0 0 ∼ defined in L (λ) = L (B) whenever hfn in∈N ∈ W0 . (I am taking the domain of limn→F fn to be {x : limn→F fn (x) is defined in R}.) Since 0

N

limn→F fn =a.e. limn→F gn whenever fn =a.e. gn for every n, T is well-defined. Since limn→F fn + gn =a.e. limn→F fn + limn→F gn , limn→F αfn =a.e. α limn→F fn ,

limn→F |fn | =a.e. | limn→F fn |

whenever hfn in∈N , hgn in∈N ∈ W0 and α ∈ R, T is a Riesz homomorphism. Q Q (γγ ) If han in∈N is any sequence in A, T (hχan in∈N ) = χψ(han in∈N ). P P Express each an as En• , where En ∈ Σ, and set F = limn→F En . In the language of 538Ja, ψ(han in∈N ) = π −1 φ(han in∈N ) = π −1 (han i•n∈N ) = F • , so T (hχan in∈N ) = (limn→F χEn )• = (χF )• = χ(F • ) = χψ(han in∈N ). Q Q (δδ ) This means that we can identify T : W0 → L0 (B) with the Riesz homomorphism described in 377B9 . By 377D(d-i)9 , T [W0 ] = L0 (B), which is what we need to prove the immediate result here. ⊂ L0 (µ) is continuous to see that a k kp -bounded (b)(i) As in 377D, we need only recall that the embedding Lp (µ) → p sequence in L (µ) will belong to W0 . So we can use 377Db. (ii) Use 377Ec9 . (c) Use 377Dd. (d) Use 377F9 . 538L Theorem Suppose that ζ is a non-zero countable ordinal and hFξ i1≤ξ≤ζ is a family of Ramsey ultrafilters on N, no two isomorphic. Let hGξ iξ≤ζ be the corresponding iterated product system, as described in 538E. Then Gζ is a measure-centering ultrafilter. proof (a) Define h(Aξ , µ ¯ξ )iξ≤ζ inductively, as follows. (A0 , µ ¯0 ) = (Bω , ν¯ω ) is to be the measure algebra of the usual N measure on {0, 1} . Given h(A , µ ¯ )i , where 0 < ξ ≤ ζ, let (Aξ , µ ¯ξ ) be the probability algebra reduced product η η η<ξ Q ¯θ(ξ,k) )|Fξ , as described in 328A-328C9 . At the end of the induction, write (C, ν¯) for (Aζ , µ ¯ζ ). k∈N (Aθ(ξ,k) , µ (b) We have a family hφξη iη≤ξ≤ζ defined by induction on ξ, as follows. The inductive hypothesis will be that φη0 η is a measure-preserving Boolean homomorphism from Aη to Aη0 , and that φη00 η = φη00 η0 φη0 η whenever η ≤ η 0 ≤ η 00 < ξ. For the inductive step to ξ, take φξξ to be the identity map on Aξ . If ξ > 0, set φ˜kj = φθ(ξ,k),θ(ξ,j) for j ≤ k in N; then 328D9 tells us that we have measure-preserving Boolean homomorphisms φ˜k : Aθ(ξ,k) → Aξ such that φ˜j = φ˜k φ˜kj for j ≤ k. If j ≤ k and η ≤ θ(ξ, j), then φ˜k φθ(ξ,k),η = φ˜k φ˜kj φθ(ξ,j),η = φ˜j φθ(ξ,j),η whenever k ≥ j; so we can take this common value for φξη . If η ≤ η 0 < ξ, then take k such that η 0 ≤ θ(ξ, k), and see that 9 Later

editions only.

538L

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267

φξη0 φη0 η = φ˜k φθ(ξ,k),η0 φη0 η = φ˜k φθ(ξ,k),η = φξη , so the induction proceeds. For each ξ ≤ ζ, write πξ for φζξ : Aξ → C, and Cξ for the subalgebra πξ [Aξ ]. Of course πξ φξη = πη , so that Cη ⊆ Cξ , whenever η ≤ ξ ≤ ζ. Q (c) For each ξ > 0, we have a canonical map hak ik∈N → hak i•k∈N : Q k∈N Aθ(ξ,k) → Aξ . Since every πξ : Aξ → Cξ is a measure-preserving isomorphism, we have a corresponding map ψξ : k∈N Cθ(ξ,k) → Cξ . Reading off the basic facts of 328B and 328D, we see that Q —– ν¯ψξ (hck ik∈N ) = limk→Fξ ν¯ck for every sequence Q hck ik∈N ∈ k∈N Cθ(ξ,k) , —– ψξ (hck ik∈N ) ⊆ supk∈A ck whenever hck ik∈N ∈ k∈N Cθ(ξ,k) and A ∈ Fξ (we can take the suprema in C because Cξ is regularly embedded in C, as noted in 314Ga). (d) Let haτ iτ ∈S ∗ be a family in A0 = Bω such that γ = inf τ ∈S ∗ µ ¯aτ > 0. By 538Fe, we can find a disjoint family hAξ i1≤ξ≤ζ of subsets of N such that Aξ ∈ Fξ for every ξ. Use these to define T ⊆ S ∗ and α : T → [0, ζ] as in 538Ee. Set cτ = 0 for τ ∈ S ∗ \ T . For τ ∈ T define cτ ∈ Cα(τ ) by induction on α(τ ), as follows. If α(τ ) = 0, set cτ = π0 aτ . For the inductive step to α(τ ) = ξ > 0, we know that τ a ∈ T and α(τ a ) = θ(ξ, k) whenever k ∈ Aξ and τ (i) < k for every i < dom τ . So cτ a ∈ Cθ(ξ,k) for every k, and ψ(hcτ a ik∈N ) ∈ Cξ ; take this for cτ . Note that ν¯cτ = limk→Fξ ν¯cτ a ≥ inf{¯ ν cτ a : k ∈ N, τ a ∈ T }. Inducing on α(τ ), we see that ν¯cτ ≥ γ for every τ ∈ T . In particular, ν¯c∅ ≥ γ. S (e) For I ⊆ N, set TI = T ∩ n∈N I n and eI = inf τ ∈TI cτ ; let S be the family of those finite subsets I of N such that eI 6= ∅. Then ∅ ∈ S. Moreover, if I ∈ S and 1 ≤ ξ ≤ ζ, then {k : I ∪ {k} ∈ S} ∈ Fξ . P P Set k0 = sup I + 1. If k ∈ Aξ and k ≥ k0 , set dk = inf{cτ a : τ ∈ TI , α(τ ) = ξ}. Set B = {k : k ∈ Aξ , k ≥ k0 , dk ∩ eI 6= 0}. If k ∈ B, then TI∪{k} = TI ∪ {τ a : τ ∈ TI , α(τ ) = ξ}, because every member of T is strictly increasing and τ a can belong to T only when k ∈ Aα(τ ) , that is, α(τ ) = ξ. So eI∪{k} = dk ∩ eI 6= 0 and I ∪ {k} ∈ S. ?? If B ∈ / Fξ , then B 0 = {k : k ∈ Aξ , k ≥ k0 , dk ∩ eI = 0} belongs to Fξ . So eI ⊆ inf{cτ : τ ∈ TI , α(τ ) = ξ} = inf ψξ (hcτ a ik∈N ) = ψξ (h inf cτ a ik∈N ) τ ∈TI α(τ )=ξ

τ ∈TI α(τ )=ξ

(because ψξ is a Boolean homomorphism and TI is finite) ⊆

sup inf cτ a

k∈B 0

τ ∈TI α(τ )=ξ

(by (c)) = sup dk . k∈B 0

But eI ∩ dk = 0 for every k ∈ B 0 and eI 6= 0. X X Thus {k : I ∪ {k} ∈ S} ⊇ B ∈ Fξ . Q Q (f ) For i ∈ N set Ci = {k : I ∪ {k} ∈ S whenever I ∈ S and I ⊆ i}, so that Ci ∈ Fξ for every ξ ∈ [1, ζ]. At this point, recall that every Fξ is supposed to be a Ramsey ultrafilter. So for each ξ ∈ [1, ζ] we have an A0ξ ∈ Fξ such that A0ξ ⊆ Aξ ∩ C0 and j ∈ Ci whenever i, j ∈ A0ξ and i < j (538Fc). Next, T for i ∈ N set Mi = {α(τ ) : τ ∈ T , τ (j) ≤ i whenever j < dom τ }; then Mi is finite, so there is a D ∈ 1≤ξ≤ζ Fξ such that whenever i, j ∈ D, i < j and ξ ∈ Mi , there is a k ∈ A0ξ such that i < k < j (538Ff). Of course we can suppose S that D ⊆ 1≤ξ≤ζ A0ξ . (g) I ∈ S for every finite subset I of D. P P Induce on #(I). We know that ∅ ∈ S. If i ∈ D, then {i} ∈ S because D ⊆ C0 . For the inductive step to #(I) ≥ 2, set j = max I, J = I \ {j} and i = max J. Then J ∈ S, by the inductive hypothesis; so if TI = TJ , we certainly have I ∈ S. Otherwise, there is a member of TI \ TJ , and this must be of the form τ a <j> where τ ∈ TJ and j ∈ A0α(τ ) . But this means that α(τ ) ∈ Mi and there is a k ∈ A0α(τ ) such that i < k < j. In this case, j ∈ Ck and J ⊆ k, so I = J ∪ {k} belongs to S, and the induction proceeds. Q Q

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Topologies and Measures III

538L

∗ ∗ ∗ (h) Thus {cτ : τ ∈ TD } is centered; setting TD = {τ : τ ∈ TD , α(τ ) = 0}, {cτ : τ ∈ TD } and therefore {aτ : τ ∈ TD } ∗ are centered. But TD belongs to Gζ , by 538Ee. Since haτ iτ ∈S ∗ was chosen arbitrarily in (d) above, Gζ satisfies the condition of 538G(ii), translated to the countably infinite set S ∗ , and is measure-centering.

538M Benedikt’s theorem (Benedikt 98) Let (X, Σ, µ) be a perfect probability space. Then there is a measure λ on X, extending µ, such that λ(limn→F En ) is defined and equal to limn→F µEn for every sequence hEn in∈N in Σ and every Ramsey filter F on N. proof (a) If there are no Ramsey filters, we can take λ = µ and stop; so let us suppose that there is at least one Ramsey filter. Let F be a family of Ramsey filters consisting of just one member of each isomorphism class, so that every Ramsey filter is isomorphic to some member of F, and no two members of F are isomorphic. Fix a well-ordering 4 of F and a family hθ(ξ, k)i1≤ξ<ω1 ,k∈N such that hθ(ξ, k)ik∈N is always a non-decreasing sequence of ordinals less than ξ such that {θ(ξ, k) : k ∈ N} is cofinal with ξ. (b)(i) For any non-empty countable set D ⊆ F, enumerate it in 4-increasing order as hFξ i1≤ξ≤ζ , and let GD be the measure-centering ultrafilter constructed from hFξ i1≤ξ≤ζ and hθ(ξ, k)i1≤ξ≤ζ,k∈N by the method of 538E-538L; let λD be the corresponding extension of µ as defined in 538I. (ii) For any non-empty finite set I ⊆ F, list it in 4-increasing order as F0 , . . . , Fn , and set HI = Fn n . . . n F0 as defined in 538D. By 538Ed, or otherwise, HI ≤RK GI , so HI is measure-centering (538Hb); let λ0I be the corresponding extension of µ. (c) If ∅ = 6 I ⊆ J ∈ [F]<ω , then HI ≤RK HJ , by 538Dg, and λ0J extends λ0I , by 538Jc. Thus hλ0I i∅6=I∈[F]<ω is an upwards-directed family of probability measures on X. S If I ⊆ [F]<ω \ {∅} is countable, we have a non-empty countable set D ⊆ F including I. Now 538Ed tells us that HI ≤RK GD for every I ∈ I, so that λD extends λ0I for every I ∈ I. Thus for every countable subset of {λ0I : I ∈ [F]<ω \ {∅}} there is a measure on X extending them all. By 457G, there is a measure λ on X extending every λ0I . (d) If F is a Ramsey ultrafilter and hEn in∈N is a sequence ˜ F 0 extends λ ˜ F , where isomorphic. In particular, F ≤RK F 0 , so λ 0 0 ˜ ˜F . and F . But λ extends λ{F 0 } = λF 0 and therefore extends λ ˜ F (limn→F En ) = limn→F µEn , as required. λ

in Σ, there is an F 0 ∈ F such that F and F 0 are ˜F , λ ˜ F 0 are the extensions of µ corresponding to F λ Accordingly λ(limn→F En ) is defined and equal to

538N Measure-converging filters: Proposition (a) Let F be a free filter on N. Let νω be the usual measure on {0, 1}N , and Tω its domain. Then the following are equiveridical: (i) F is measure-converging; S T (ii) whenever hFn in∈N is a sequence in Tω and limn→∞ νω Fn = 1, then A∈F n∈A Fn is conegligible; (iii) whenever (X, Σ, µ) is a measure space with locally determined negligible sets (definition: 213I), and hfn in∈N is a sequence in L0 (µ) which converges in measure to 0, then limn→F fn = 0 a.e.; (iv) whenever µ is a Radon measure on PN such that limn→∞ µEn = 1, where En = {a : n ∈ a ⊆ N} for each n, then µF = 1. (b) (M.Foreman) Every rapid filter is measure-converging. (c) (M.Talagrand) If there is a rapid filter, there is a measure-converging filter which is not rapid. (d) Let F be a measure-converging filter on N and G any filter on N. Then G n F is measure-converging. (e) If cov M = d, where M is the ideal of meager subsets of R, there is a rapid filter. proof (a)(i)⇒(iii) Suppose that F is measure-converging, and that (X, Σ, µ) and hfn in∈N are as in (iii). Let H ∈ Σ be a conegligible set such that H ⊆ dom fn and fn H is measurable for every n ∈ N. Let k ∈ N; set Hk = {x : x ∈ H, lim supn→F |fn (x)| > 2−k }. Then Hk ∩ F is negligible whenever F ∈ Σ and µF < ∞. P P If µF = 0 this is trivial. Otherwise, let ν =

1 µF µF

be the normalized subspace measure on F . For each n ∈ N, set Fn = {x : x ∈ F ∩ H,

|fn (x)| ≤ 2−k }. Then limn→∞ ν(F \ Fn ) ≤

2k µF

R

min(|fn |, χF )dµ = 0 S T because hfn in∈N → 0 in measure. So limn→∞ νFn = 1 and H 0 = A∈F n∈A Fn is ν-conegligible. But H 0 ∩ Hk = ∅, so µ∗ (Hk ∩ F ) = ν ∗ (Hk ∩ F ) = 0. Q Q S Since µ has locally determined negligible sets, Hk S is negligible. This is true for every k ∈ N, so H \ k∈N Hk is conegligible; and limn→F fn (x) = 0 for every x ∈ H \ k∈N Hk , so limn→F fn = 0 a.e., as required. limn→∞

538N

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269

(iii)⇒(iv) Assuming (iii), let µ and hEn in∈N be as in (iv). Set fn = χ(PN \ En ); then limn→∞ hfn in∈N → 0 in measure, and H = {a : limn→F fn (a) = 0} is conegligible. But for any a ∈ H,

R

fn dµ = 0, so

1 2

a = {n : a ∈ En } = {n : fn (x) ≤ } belongs to F, so H ⊆ F and µF = 1. (iv)⇒(ii) Assume (iv), and let hFn in∈N be as in (ii). Define φ : {0, 1}N → PN by setting φ(x) = {n : x ∈ Fn } for x ∈ {0, 1}N . Then φ is almost continuous (418Dd or otherwise), so the image measure µ = νω φ−1 on PN is a Radon measure (418I). Since Fn = φ−1 [En ] for each n, limn→∞ µEn = 1 and 1 = µF = νω φ−1 [F]. But now S T S −1 [F] a∈F n∈a Fn = a∈F {x : a ⊆ φ(x)} = φ is νω -conegligible, as required. (ii)⇒(i) S Assume T (ii), and take a probability space (X, Σ, µ) and a sequence hHn in∈N in Σ such that limn→∞ µHn = 1; set G = A∈F n∈A Hn . Let λ be the c.l.d. product measure on X × {0, 1}N , and set Wn = Hn × {0, 1}N ,

Vn = {(x, y) : x ∈ X, y ∈ {0, 1}N , y(n) = 1}

for n ∈ N. Let Λ1 be the σ-algebra of subsets of X × {0, 1}N generated by {Wn : n ∈ N} ∪ {Vn : n ∈ N}, and λ1 the completion of the restriction λΛ1 . Note that as the identity map from X × {0, 1}N is inverse-measure-preserving for λ and λΛ1 , it is inverse-measure-preserving for their completions (234Ba10 ); but λ is complete, so this just means that λ extends λ1 . Then λ1 is a complete atomless probability measure with countable Maharam type. Its measure algebra C is therefore isomorphic, as measure algebra, to the measure algebra Bω of νω ; let π : Bω → C be a measure-preserving isomorphism. By 343B, or otherwise, there is a realization φ : X × {0, 1}N → {0, 1}N , inverse-measure-preserving for λ1 and νω , such that φ−1 [F ]• = πF • in C for every F ∈ Tω . Because π is surjective, there is for each n ∈ N an Fn ∈ Tω such that φ−1 [Fn ]4Wn is λ1 -negligible. Since limn→∞ νω Fn = limn→∞ λ1 Wn = limn→∞ λWn = limn→∞ µHn = 1, S T S T −1 F = T [F ] = A∈F n∈A φ−1 [Fn ] is λ1 -conegligible. We have G × {0, 1}N = A∈F n∈A Fn is νω -conegligible, and φ S A∈F n∈A Wn , so S (G × {0, 1}N )4φ−1 [F ] ⊆ n∈N Wn 4φ−1 [Fn ] is λ1 -negligible. Thus G × {0, 1}N is λ1 -conegligible, therefore λ-conegligible. But this means that G is µ-conegligible, by 252D applied to G × {0, 1}N ; and this is what we needed to know. S (b) TLet F be a rapid filter on N, and hHn in∈N a sequence in Tω such thatPlimn→∞ νω Hn = 1. Set G = n∈A 1 − νω Hn < ∞; set H = SA∈F Tn∈A Hn . Since limn→∞ (1 − νω Hn ) = 0, there is an A ∈ F such that H ⊆ G. Then m∈N n∈A\m n P νω H ≥ supm∈N 1 − n∈A\m (1 − νω Hn ) = 1, so G is conegligible. Thus F satisfies (a-ii) and is measure-converging. (c) Let F be a rapid filter. limn→∞ #(In ) = ∞. Let G be

Let hIn in∈N be a disjoint sequence of non-empty finite subsets of N such that {A : A ⊆ N, limn→F

1 #(A ∩ In ) #(In )

= 1}.

Then G is a filter on N. S

(i) P Let hHi ii∈N be a sequence in Tω such that limi→∞ νω Hi = 1, and set G = T G is measure-converging. P H . Set i A∈G i∈A gn =

1 P #(In ) i∈In

χHi

for each n; then

R

limn→∞ 1 − gn ≤ limn→∞ supi∈In 1 − νω Hi = 0, R P P so there is a B ∈ F such that n∈B 1 − gn < ∞. Let H be the set of those x ∈ {0, 1}N such that n∈B 1 − gn (x) is finite; then H is conegligible. If x ∈ H, set Ax = {i : x ∈ Hi }. Then 10 Formerly

235Hc.

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Topologies and Measures III 1

P

n∈B #(I ) #(In n

\ Ax ) =

P

n∈B

538N

1 − gn (x) < ∞,

so lim supn→F

1 #(In #(In )

\ Ax ) ≤ inf m∈N supn∈B\m

1 #(In #(In )

\ Ax ) = 0,

and limn→F

1 #(In #(In )

∩ Ax ) = 1.

Thus Ax ∈ G. But this means that H ⊆ G, so G is conegligible. Once again, G satisfies (a-ii), so is measure-converging. Q Q (ii) G is not rapid. P P Define hti ii∈N by saying that 1 #(In )

if n ∈ N and i ∈ In , [ = 0 if i ∈ N \ In .

ti =

n∈N

If A ∈ G, then B = {n : #(A ∩ In ) ≥ 21 #(In )} belongs to F, and must be infinite. So P

i∈A ti

≥

P

1

n∈B 2 #(In )

·

1 #(In )

= ∞,

and G cannot be rapid. Q Q (d) Let hEij ii,j∈N h(in , jn )in∈N of S be aTfamily in Tω such that hνω Ein jn in∈N → 1 for some, orSany, enumeration T N × N. Set G = C∈GnF (i,j)∈C Eij . For each i ∈ N, limj→∞ νω Eij = 1, so Gi = A∈F j∈A Eij is conegligible; set T H = i∈N Gi . For x ∈ H, set Ax = {(i, j) : x ∈ Eij }. As x ∈ Gi , Ax [{i}] ∈ F for every i ∈ N; thus Ax ∈ G n F and x ∈ G. So G includes the conegligible set H, and is itself conegligible. As hEij ii,j∈N is arbitrary, G is measureconvergina. (e)(i) Suppose that E ⊆ [N]ω is a family with #(E) < cov M, and that f ∈ N N is non-decreasing. Then Q there is an A ⊆ N, meeting every member of E, such that #(A ∩ f (n)) ≤ n for every n ∈ N. P P Consider X = n∈N N \ f (n). Then X is a closed subset of N N , homeomorphic to N N , so (X, M(X)) ∼ = (R, M), where M(X) is the ideal of meager subsets of X (522Vb again). For E ∈ E, set GE = {x : x ∈ X, E ∩ x[N] 6= ∅}; then GE is a dense open subset of X. Since #(E) < cov M = cov M(X), there is an x ∈ X ∩ Q Q

T

E∈E

GE ; set A = x[N].

(ii) Let hfξ iξ
538O

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271

(b) If F and G are filters on N, G ≤RK F and F has the Fatou property, then G has the Fatou property. (c) If F and G are filters on N with the Fatou property, then F n G has the Fatou property. proof not-(iii)⇒not-(i) Suppose that (X, Σ, µ) is a measure space and hfn in∈N a sequence of non-negative funcR R tions in L0 such that lim inf n→F fn dµ > lim inf n→F fn dµ. Changing the fn on negligible sets does not change R lim inf n→F fn dµ, so we may assume that every fn is defined everywhere in X and is Σ-measurable. Take α such that R R R lim inf n→F fn dµ > α > lim inf n→F fn dµ; set A = {n : fn dµ ≤ α}; then A meets every member of F. Since fn is integrable for every n ∈ A, the set G = {x : supn∈A fn (x) > 0} is a countable union of sets of finite measure. Let λ be the c.l.d.Sproduct T measure on G × R, and consider the ordinate sets Wn = {(x, α) : x ∈ G, 0 ≤ α < fn (x)} for n ∈ A. Set W = C∈F n∈C∩A Wn ; setting g = lim inf n→F fn , {(x, α) : x ∈ G, 0 ≤ α < g(x)} ⊆ W . ˜ say. Now λ∗ W > α. Since λ is a product of two σ-finite measures it is σ-finite, and W has a measurable envelope W ˜ ≤ α. Now, writing µL for Lebesgue measure on R, P P?? Otherwise, λW

˜ = α ≥ λW

Z

˜ [{x}]µ(dx) µL W

G

(252D) Z ≥

g dµ > α. X XQ Q G

˜ such that α < λV < ∞, and now λ∗ (V ∩ W ) > α. Let ν be the subspace measure There is therefore a set V ⊆ W on V ∩ W . Set Vn = V ∩ W ∩ Wn if n ∈ A, = V ∩ W if n ∈ N \ A. Then lim inf νVn = sup inf νVn ≤ sup νVn n→F C∈F n∈C n∈A Z ≤ sup λWn = sup fn dµ ≤ α. n∈A

n∈A

On the other hand, S

C∈F

T

n∈C

Vn =

S

C∈F

T

n∈C∩A

V ∩ W ∩ Wn = V ∩ W

∗

and ν(V ∩ W ) = λ (V ∩ W ) > α. Moving to a normalization of ν, we see that (i) is false. (iii)⇒(iv) If F satisfies (iii) and µ is a Radon probability measure on PN, set g = lim inf n→F χEn . If a ∈ F, then {n : χEn (a) = 1} = a ∈ F, so g(a) = 1; thus Z

∗

µ F=

χ(F)dµ

(133Je11 ) Z ≤

Z g dµ ≤ lim inf n→F

χEn = lim inf µEn , n→F

as required. S T (iv)⇒(ii) Given (iv), suppose that hFn in∈N is a sequence in Tω and νω∗ ( A∈F n∈A Fn ) = 1. As in the corresponding part of the argument for 538Na, define φ : {0, 1}N → PN by setting φ(x) = {n : x ∈ Fn }, and let µ be the Radon measure νω φ−1 . Then S T µ∗ F = νω∗ ( A∈F n∈A Fn ) = 1 (451Pc), so limn→F νFn = limn→F µEn = 1. 11 Later

editions only.

272

S

Topologies and Measures III

538O

(ii)⇒(i) Assume (ii), and take a probability space (X, Σ, µ) and a sequence hHn in∈N in Σ such that X = T H A∈F n∈A n . As in the corresponding part of the argument for 538Na, let λ be the c.l.d. product measure on X × {0, 1}N , and set Wn = Hn × {0, 1}N ,

Vn = {(x, y) : x ∈ X, y ∈ {0, 1}N , y(n) = 1}

for n ∈ N. Let Λ1 be the σ-algebra of subsets of X × {0, 1}N generated by {Wn : n ∈ N} ∪ {Vn : n ∈ N}, and λ1 the completion of the restriction λΛ1 . As before, there is a function φ : X × {0, 1}N → {0, 1}N , inverse-measure−1 preserving for S T λ1 and νω , such that there is for each n ∈ N an Fn ∈ Tω such that φ [Fn ]4Wn is λ1 -negligible. Set G = A∈F n∈A Fn . S T S T Since X = A∈F n∈A Hn , X × {0, 1}N = A∈F n∈A Wn and S (X × {0, 1}N ) \ φ−1 [G] ⊆ n∈N φ−1 [Fn ]4Wn is λ1 -negligible. By 413Eh, νω∗ G ≥ λ1 φ−1 [G] = 1. By (ii), limn→F νω Fn = 1. But νω Fn = λ1 φ−1 [Fn ] = λ1 Wn = λWn = µHn for each n, so limn→F µHn = 1. As (X, Σ, µ) and hHn in∈N are arbitrary, F has the Fatou property. (b) Let h : N → N be such that G = h[[F]]. Let (X, Σ, µ) be a probability space and hHn in∈N a sequence in Σ such that [ \ X= Hn A∈G n∈A

[

=

\

[ \

Hn =

A⊆N,h−1 [A]∈F n∈A

Hh(n) .

A∈F n∈A

Then 1 = lim inf µHh(n) = sup inf µHh(n) n→F

A∈F n∈A

= sup inf µHn = lim inf µHn . A∈G n∈A

n→G

As (X, Σ, µ) and hHn in∈N are arbitrary, G has the Fatou property. S T (c) Let (X, Σ, µ) be a probability space and hEij ii,j∈N a family in Σ such that X = C∈F nG (i,j)∈C Eij . For each S T S T i ∈ N, set Fi = B∈G j∈B Eij , and let Gi ∈ Σ be a measurable envelope of Fi . Then A∈F i∈A Gi = X. P P If x ∈ X, there is a C ∈ F n G such that x ∈ Eij whenever (i, j) ∈ C. Set A = {i : C[{i}] ∈ G} ∈ F. If i ∈ A, then T x ∈ j∈C[{i}] Eij ⊆ Fi ⊆ Gi , T so x ∈ i∈A Gi . Q Q Accordingly limi→F µGi = 1. Take  > 0; then A = {i : µGi ≥ 1 − } belongs to F. For each i ∈ A,

1 −  ≤ µGi = µ∗ Fi =

Z

Z χFi =

Z lim inf χEij ≤ lim inf j→G

j→G

χEij

(by (a) above) = lim inf µEij , j→G

so {j : µEij ≥ 1−2} ∈ G. But this means that {(i, j) : µEij ≥ 1−2} ∈ F nG. As  is arbitrary, lim(i,j)→F nG µEij = 1. As (X, Σ, µ) and hEij ii,j∈N are arbitrary, F n G has the Fatou property. R 538P Theorem Let ν : PN → R be a bounded finitely additive functional. Write . . . dν for the associated linear functional on `∞ (see 363L), and set En = {a : n ∈ a ⊆ N}R for each n ∈ N. Then the followingRare equiveridical: (i) whenever µ is a Radon probability measure on PN, ν(a)µ(da) is defined and equal to µE RR RRn ν(dn); (ii) whenever µ is a Radon probability measure on [0, 1]N , x dν µ(dx) is defined and equal to x(n)µ(dx)ν(dn); (iii) whenever (X, Σ, µ) is a probability space and hf i is a uniformly bounded sequence of measurable real-valued n n∈N RR RR functions on X, then fn (x)ν(dn)µ(dx) is defined and equal to fn dµ ν(dn);

538P

Filters and limits

273

of {0, 1}N , R (iv) whenever hFn in∈N is a sequence of Borel subsets N νω Fn ν(dn), where νω is the usual measure on {0, 1} .

RR

χFn (x)ν(dn)νω (dx) is defined and equal to

α) For t ∈ [0, 1] define ht : [0, 1]N → PN by setting ht (x) = {n : x(n) ≥ t} for x ∈ [0, 1]N , and let proof (i)⇒(ii)(α −1 µt = µht be the image measure on PN. Then µt is a Radon measure for each t. P P Because ht is Borel measurable and PN is metrizable, ht is almost continuous (418J), so µt is a Radon measure (418I). Q Q β ) For m ∈ N define vm ∈ [0, 1]N by setting (β vm (n) = 2−m

P2m

k=1

µ{x : x(n) ≥ 2−m k}.

Then kvm+1 − vm k∞ ≤ 2−m−1 . P P For any n ∈ N, m

−m

vm (n) − vm+1 (n) = 2

2 X

−m

µ{x : x(n) ≥ 2

−m−1

k} − 2

k=1

= 2−m−1

m+1 2X

µ{x : x(n) ≥ 2−m−1 k}

k=1

2m X

(2µ{x : x(n) ≥ 2−m k} − µ{x : x(n) ≥ 2−m k}

k=1

− µ{x : x(n) ≥ 2−m−1 (2k + 1)}) m

−m−1

=2

2 X

µ{x : 2−m k ≤ x(n) < 2−m−1 (2k + 1)} ≤ 2−m−1 . Q Q

k=1 ∞

So RR v = limm→∞ vm isR defined in ` x(n)µ(dx)ν(dn) = v dν.

R

and

v dν = limm→∞

R

vm dν. Also v(n) =

R

x(n)µ(dx) for every n ∈ N, so

(γγ ) Set

R

f (t) =

µt En ν(dn) =

R

ν(a)µt (da)

for each t ∈ [0, 1] (using (i)). Then, for any m ∈ N, m

Z

vm dν = 2−m

2 Z X

µ{x : x(n) ≥ 2−m k}ν(dn)

k=1 m

−m

=2

2 Z X

µ{x : h2−m k (x) ∈ En }ν(dn)

k=1 m

−m

=2

m

2 Z X

−m

µ2−m k En ν(dn) = 2

k=1

2 X

f (2−m k).

k=1

P2m (δδ ) Next, for m ∈ N and x ∈ [0, 1]N , set qm (x) = 2−m k=1 χh2−m k (x), so that hqm (x)im∈N is non-decreasing and kx − qm (x)k∞ ≤ 2−m for each m, while qm : [0, 1]N → [0, 1]N is Borel measurable. Now m

ZZ qm (x)dν µ(dx) = 2

−m

2 Z X

ν(h2−m k (x))µ(dx)

k=1 m

−m

=2

2 Z X

m

−m

ν(a)µ2−m k (da) = 2

k=1

R

Also h qm (x)dνim∈N →

R

2 X

f (2−m k).

k=1

RR

x dν uniformly for x ∈ [0, 1] , so N

x dν µ(dx) is defined and equal to m

ZZ lim

m→∞

−m

qm (x)dν µ(dx) = lim 2 m→∞

Z =

2 X k=1

f (2

−m

Z k) = lim

m→∞

vm dν

ZZ v dν =

x(n)µ(dx)ν(dn).

As µ is arbitrary, (ii) is true. (ii)⇒(iii) Assume (ii), and let (X, Σ, µ) be a probability space and hfn in∈N a Runiformly bounded sequence of measurable real-valued functions on X. As completing µ does not affect the integral . . . dµ, we may suppose that µ

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Topologies and Measures III

538P

is complete. Let γ > 0 be such that |fn (x)| ≤ γ for every n ∈ N and x ∈ X, and set q(x)(n) =

1 (γ 2γ

+ fn (x)) for all n

and x. Then q : X → [0, 1] is measurable, so there is a Radon probability measure λ on [0, 1] such that q is inversemeasure-preserving for µ and λ. P P Taking λ0 E = µq −1 [E] for Borel sets E ⊆ [0, 1]N , q is inverse-measure-preserving for µ and λ0 ; taking λ to be the completion of λ0 , q is inverse-measure-preserving for µ and λ, by 234Ba; and λ is a Radon measure by 433Cb. Q Q Now N

N

ZZ

ZZ q(x)(n)µ(dx)ν(dn) − γ

fn dµ ν(dn) = 2γ ZZ

ZZ z(n)λ(dz)ν(dn) − γ = 2γ

= 2γ

z(n)ν(dn)λ(dz) − γ

(by (ii)) ZZ

ZZ q(x)(n)ν(dn)µ(dx) − γ =

= 2γ

fn (x)ν(dn)µ(dx).

As µ and hfn in∈N are arbitrary, (iii) is true. (iii)⇒(iv) is elementary, taking fn = χFn and µ = νω . (iv)⇒(i) If (iv) is true and µ is a Radon probability measure on PN, there is an inverse-measure-preserving function φ from ({0, 1}N , νω ) to (PN, µ) (343Cd12 ). Set Fn = φ−1 [En ] for each n. Then Z ZZ ZZ ν(a) µ(da) = χa(n)ν(dn)µ(da) = χEn (a)ν(dn)µ(da) ZZ ZZ = χEn (φ(x))ν(dn)νω (dx) = χFn (x)ν(dn)νω (dx) is defined and equal to

R

νω Fn ν(dn) =

R

µEn ν(dn).

538Q Definition I will say that a bounded additive functional ν satisfying (i)-(iv) of 538P is a medial functional; if, in addition, ν is non-negative, νa = 0 for every finite set a ⊆ N and νN = 1, I will call νR a medial limit. I should remark that the term ‘medial limit’ is normally used for the associated linear functional . . . dν on `∞ , rather than the additive functionalR ν on PN; thus h ∈ (`∞ )∗ is a medial limit if h R≥ 0, h(w) = limn→∞ w(n) for every convergent sequence w ∈ `∞ and h(hfn (x)in∈N )µ(dx) is defined and equal to h(h fn dµin∈N ) whenever (X, Σ, µ) is a probability space and hfn in∈N is a uniformly bounded sequence of measurable real-valued functions on X. 538R Proposition Let M be the L-space of bounded additive functionals on PN, and Mmed ⊆ M the set of medial functionals. (a) Mmed is a band in M , and if T ∈ L× (`∞ ; `∞ ) (definition: 355G) and T 0 : M → M is its adjoint, then T 0 ν ∈ Mmed for every ν ∈ Mmed . (b) Taking Mτ to be the band of completely additive functionals on PN and Mm the band of measurable functionals, as described in §464, Mτ ⊆ Mmed ⊆ Mm . R (c) Suppose that hνk ik∈N is a norm-bounded sequence in Mmed , and that ν ∈ Mmed . Set ν˜(a) = νk (a)ν(dk) for a ⊆ N. Then ν˜ ∈ Mmed . (d) Suppose that ν ∈ M is a medial limit, and set F = {a : a ⊆ N, ν(a) = 1}. Then F is a measure-converging filter with the Fatou property. (e) Let (X, Σ, µ) and (Y, T, λ) be probability spaces, and T ∈ L× (L∞ (µ); L∞ (λ)). Let hfn in∈N , hgn in∈N be sequences • • • ∞ in L∞ (µ), L∞ (ν) respectively such that R T fn = gn for every n Rand hfn in∈N is norm-bounded in L (µ). Let ν ∈ M be a medial functional. Then f (x) = fn (x)ν(dn) and g(y) = gn (y)ν(dn) are defined for almost every x ∈ X and y ∈ Y ; moreover, f ∈ L∞ (µ), g ∈ L∞ (λ) and T f • = g • . proof (a)(i) Any of the four conditions of 538P makes it clear that Mmed is a linear subspace of M . We see also that Mmed is norm-closed in M . P P Let hνn in∈N R be a sequence R in Mmed which is norm-convergent to ν ∈ M . If µ is a Radon probability measure on [0, 1]N , then h x dνn in∈N → x dν uniformly for x ∈ [0, 1]N , so 12 Later

editions only.

538R

Filters and limits

275

ZZ

ZZ

x dνn µ(dx)

x dν µ(dx) = lim

n→∞

ZZ

ZZ x(i)µ(dx)νn (di) =

= lim

n→∞

x(i) µ(dx)ν(di).

As µ is arbitrary, ν ∈ Mmed . Q Q (ii) Before completing the proof that Mmed is a band, I deal with the second clause of (a). Recall from §355 that L× (`∞ ; `∞ ) is the set of differences of order-continuous positive linear operators from `∞ to itself. Since RM can be identified with (`∞ )∗ , any T ∈ L× (`∞ ; `∞ ) has an adjoint T 0 : M → M defined by saying that (T 0 ν)(a) = T (χa)dν for every a ⊆ N. If T : `∞ → `∞ is an order-continuous positive linear operator, it is a norm-bounded linear operator (355C), and all the functionals x 7→ (T x)(n) are order-continuous, therefore represented by members of `1 ; that is, we have a family hαni in,i∈N in [0, ∞[ such that P∞ (T x)(n) = i=0 αni x(i) whenever x ∈ `∞ and n ∈ N, P∞ supn∈N i=0 αni = kT k is finite. In this case, if ν ∈ M and ν 0 = T 0 ν in M ,

R

x dν 0 =

R

(T x)(n)ν(dn) =

R P∞

i=0

αni x(i)ν(dn)

∞

for every x ∈ ` . (iii) Continuing from (ii), suppose that T is positive, with associated matrix hαni in,i∈N , and that kT k ≤ 1, so P ∞ that i=0 αni ≤ 1 for every n. Consider the function φ = T [0, 1]N . This is a function from [0, 1]N to itself, and it is continuous for the product topology on N. Take any ν ∈ M and Radon probability measure µ onR [0, 1]N ; then the image measure µ1 = µφ−1 on [0, 1]N is a R Radon probability R measure (418I), and f (φ(x))µ(dx) = f (x)µ1 (dx) for any µ1 -integrable function f . In particular, setting f (x) = x dν,

RR

φ(x) dν µ(dx) =

RR

x dν µ1 (dx) =

RR

x(n)µ1 (dx)ν(dn)

because ν ∈ Mmed . Set ν 0 = T 0 ν. Then we can calculate

ZZ

x(n)µ(dx)ν 0 (dn) =

Z X ∞

Z αni

x(i)µ(dx)ν(dn) =

i=0

ZZ X ∞

αni x(i)µ(dx)ν(dn)

i=0

(the inner integral is with respect to a genuine σ-additive measure, so we have B.Levi’s theorem) ZZ ZZ = φ(x)(n)µ(dx)ν(dn) = x(n)µ1 (dx)ν(dn) ZZ ZZ ZZ = φ(x) dν µ(dx) = T x dν µ(dx) = x dν 0 µ(dx). As µ is arbitrary, ν 0 satisfies 538P(i), and is a medial functional. (iv) Thus T 0 ν ∈ Mmed whenever ν ∈ Mmed and T : `∞ → `∞ is positive, order-continuous and of norm at most 1. As Mmed is a linear subspace of M , the same is true for every positive order-continuous T and for differences of such operators, that is, for every T ∈ L× (`∞ ; `∞ ), as claimed. (v) I now return to the question of showing that Mmed is a band. The point is that if ν is a medial functional and b ⊆ N, then νb is a medial functional, where νb (a) = ν(a ∩ b) for every a ⊆ N. P P Define T : `∞ → `∞ by setting ∞ 0 T x = x × χb for x ∈ ` . Then T is a positive order-continuous operator, and T ν ∈ Mmed , by (iii) above. But (T 0 ν)(a) =

R

T (χa)dν =

R

χ(a ∩ b)dν = ν(a ∩ b) = νb (a)

for every a ⊆ N, so νb = T 0 ν is a medial functional. Q Q By 436M13 , this is enough to ensure that Mmed is a band in M . (b)(i) Recall that an additive functional on PN is completely additive iff it corresponds to an element of `1 , that is, belongs to the band generated by the elementary functionals δk where δk (a) = χa(k) for k ∈ N and a ⊆ N. To see 13 Formerly

436Xr.

276

Topologies and Measures III

538R

that δk belongs to Mmed , all we have to do is to note that δk = χEk where Ek is defined as in 538P; so if µ is a Radon probability measure on PN, we shall have

R

δk dµ = µEk =

R

µEn δk (dn).

Since Mmed is a band, it must include Mτ . (ii) On the other side, 538P(i) tells us that every member of Mmed is universally measurable, and therefore belongs to Mm , which is just the set of bounded additive functionals which are Σ-measurable, where Σ is the domain of the usual measure on PN. (c)(i) Because hνk ik∈N is norm-bounded, ν˜ is well-defined and additive; also it is bounded. P P If γ is such that kνk k ≤ γ for every k and kνk ≤ γ, then |˜ ν (a)| ≤ γ supk∈N |νk (a)| ≤ γ 2 for every a ⊆ N. Q Q Note that

R

χa d˜ ν = ν˜(a) =

R

νk (a)dν(k) =

RR

χa dνk ν(dk)

for every a ⊆ N, so that

R

x d˜ ν=

RR

x dνk ν(dk)

whenever x ∈ `∞ is a linear combination of characteristic functions, and therefore for every x ∈ `∞ . (ii) Now suppose that (X, Σ, µ) is a probability space and that hfn in∈N is a uniformly bounded sequence of ˆ ˆ) be the completion of (X, Σ, µ). For k ∈ N, x ∈ X set gk (x) = Rmeasurable real-valued functions on X. Let (X, Σ, µ R RR fn (x)νk (dn); because νk is a medial functional, we know that gk dµ = fn (x)µ(dx)νk (dn) is defined, so gk is RR RR ˆ Σ-measurable. Consequently gk (x)ν(dk)ˆ µ(dx) is defined and equal to gk (x)ˆ µ(dx)ν(dk). It follows that ZZ

ZZZ fn (x)µ(dx)˜ ν (dn) =

fn (x)µ(dx)νk (dn)ν(dk) ZZ

ZZ

=

fn (x)νk (dn)µ(dx)ν(dk) = gk (x)ˆ µ(dx)ν(dk) ZZZ = gk (x)ν(dk)ˆ µ(dx) = fn (x)νk (dn)ν(dk)ˆ µ(dx) ZZ ZZ = fn (x)˜ ν (dn)ˆ µ(dx) = fn (x)˜ ν (dn)µ(dx). ZZ

(Recall that µ and µ ˆ give rise to the same integrals, by 212Fb.) As (X, Σ, µ) and hfn in∈N are arbitrary, ν˜ ∈ Mmed . (d) Of course F = {N \ a : ν(a) = 0} is a filter. (i) If (X, Σ, µ) is a probability space, hEn in∈N is a sequence in Σ, and limn→∞ µEn = 1, then

RR

χEn (x)ν(dn)µ(dx) =

RR

χEn dµ ν(dn) =

R

µEn ν(dn) = 1.

R So E = {x : χEn (x)ν(dn) = 1} is µ-conegligible. But if x ∈ E and a = {n : x ∈ En }, then νa = χEn (x)ν(dn) = 1 T S T and a ∈ F and x ∈ n∈a En . Thus a∈F n∈a En ⊇ E is conegligible. As (X, Σ, µ) and hEn in∈N are arbitrary, F is measure-converging. S T (ii) If (X, Σ, µ) is a probability space, hEn in∈N is a sequence in Σ, and X = A∈F n∈A En , then {n : x ∈ En } ∈ F for every x ∈ X, and R

R

µEn ν(dn) =

RR

χEn (x)ν(dn)µ(dx) =

R

ν{n : x ∈ En }µ(dx) = 1.

So for any  > 0, ν{n : µEn ≤ 1 − } = 0 and {n : µEn ≥ 1 − } ∈ F; accordingly limn→F µEn = 1. As (X, Σ, µ) and hEn in∈N are arbitrary, F has the Fatou property. 0 (e)(i) For each n ∈ N, we can find a Σ-measurable function fn0 : X → R, equal almost R 0 everywhere to fn , and such 0 0 0 that supx∈X |fn (x)| = ess sup |fn |. Now hfn in∈N is uniformly bounded, so f (x) = fn (x)ν(dn) is defined for every x ∈ X; and f (x) is defined and equal to f 0 (x) for µ-almost every x. Since f 0 is integrable, f 0 and f are µ-virtually measurable and essentially bounded, and f ∈ L∞ (µ). Similarly, g ∈ L∞ (λ). R RR (ii) If h ∈ L1 (µ), then f ×h dµ = fn ×h P (α) If h is defined everywhere, measurable and bounded, R dµ ν(dn). P then, taking fn0 and f 0 as in (i), (f 0 × h)(x) = fn0 (x)h(x)ν(dn) for every x ∈ X, so

538S

Filters and limits

Z

Z

277

ZZ

f × h dµ = (fn0 × h)(x)ν(dn)µ(dx) ZZ ZZ = fn0 × h dµ ν(dn) = fn × h dµ ν(dn).

f × h dµ =

0

(β) In general, set γ = supn∈N ess sup fn . Given  > 0, there is a simple function h0 such that kh − h0 k ≤ , and now Z ZZ | f × h dµ − fn × h dµ ν(dn)| Z Z Z ZZ 0 0 ≤ | f × h dµ − f × h dµ| + | f × h dµ − fn × h0 dµ ν(dn)| ZZ ZZ +| fn × h0 dµ ν(dn) − fn × h dµ ν(dn)| Z Z ≤ kf k∞ kh − h0 k1 + sup | fn × h0 dµ − fn × h dµ| ≤ 2γ. n∈N

R

RR

As  is arbitrary, f × h dµ fn × h dµ ν(dn). Q Q R RR = Similarly, g × h dλ = gn × h dλ ν(dn) for every λ-integrable h. R • R ˜ ∈ L1 (µ) such that h ˜ × v = h• × T v for every v ∈ L∞ (µ). P (iii) If h ∈ L1 (λ) there is an h P Recall 1 1 ∞ × ∞ × that L (µ), L (λ) can be identified with L (µ) and L (ν) (365Mb); perhaps I should remark that the R • R R formulae ˜ × v, h• × T v represent abstract integrals taken in L1 (µ), L1 (λ) respectively (242B). Setting φ(w) = h• × w for h ˜ ∈ L1 (µ) such that w ∈ L∞ (λ), φ ∈ L∞ (λ)× , so φT ∈ L∞ (µ)× (355G) and there is an h

R

˜ • × v = φ(T v) = h

R

h• × T v

for every v ∈ L∞ (µ). Q Q ˜ as in (iii), and consider (iv) Take h and h Z

h• × g • =

Z

ZZ h × g dλ =

h × gn dλ ν(dn)

(by (ii)) Z Z Z Z ( h• × gn• )ν(dn) = ( h• × T fn• )ν(dn) Z Z ZZ Z • • ˜ ˜ ˜ × f dµ = ( h × fn )ν(dn) = h × fn dµ ν(dn) = h =

(by (ii) again) Z =

˜• × f • = h

Z

h• × T f • .

As h is arbitrary, and the duality between L∞ (µ) and L1 (λ) is separating, T f • = g • , as required. 538S Theorem If cov M = c, where M is the ideal of meager subsets of R, there is a medial limit. proof (a) Let M be the L-space of bounded additive functionals on PN. Let us say that a subset C of M is rationally convex if αν + (1 − α)ν 0 ∈ C whenever ν, ν 0 ∈ C and α ∈ [0, 1] ∩ Q; for A ⊆ M , write ΓQ (A) for the smallest rationally convex set including A. Set Q = ΓQ ({δn : n ∈ N}) R where Rδn (a) = χa(n) for a ⊆ N and n ∈ N. In the language of 538Rb, Q ⊆ Mτ ⊆ Mmed , so 538P(i) tells us that ν dµ = µEn ν(dn) for every ν ∈ Q, where En = {a : n ∈ a ⊆ N} as usual. (b) Suppose that F is a filter base on Q, consisting of rationally convex sets, with cardinal less than cov M. Let µ be a Radon probability measure on PN. Then there is a sequence hνk ik∈N in Q such that P∞ R k=0 |νk+1 (a) − νk (a)|µ(da) < ∞, {k : k ∈ N, νk ∈ F } is infinite for every F ∈ F. P P Each ν ∈ Q is a Borel measurable real-valued function on PN; let u ∈ L2 = L2 (µ) be a Ts (L2 , L2 )-cluster point of hν • iν∈Q along the filter generated by F. For any F ∈ F, the k k2 -closure of the rationally convex set {ν • : ν ∈ F } ⊆ L2 is convex, so includes the weak closure of {ν • : ν ∈ F } and therefore contains u. So {ν • : ν ∈ F } meets {v : v ∈ L2 , kv − uk2 ≤ } for every  > 0.

278

Topologies and Measures III

538S

Set Hk = {ν : ν ∈ Q, kν • − uk2 ≤ 2−k } for each k ∈ Q N; then every Hk meets every member of F. If we give each Hk its discrete topology, and take H to be the product k∈N Hk , then H is homeomorphic to N N . Taking M(H) to be the ideal of meager subsets of H, cov M(H) = cov M > #(F), while S k≥n {α : α ∈ H, α(k) ∈ F } is a dense open subset of H for every F ∈ F and n ∈ N. There is therefore an α ∈ H such that {k : α(k) ∈ F } is infinite for every F ∈ F; take νk = α(k) for each k. Since µ is a probability measure, Z

• |νk+1 − νk |dµ ≤ kνk+1 − νk• k2

(see 244Xd) ≤ 2−k−1 + 2−k for every k, and

P∞ R k=0

|νk+1 − νk |dµ is finite. Q Q

(c) Because a Radon probability measure on PN is defined by its values on the countable algebra B of openand-closed sets, the number of such measures is at most #(R B ) = c. Enumerate them as hµξ iξ
R

θ(a)µ(da) = limk→∞

R

νξk (a)µ(da) = limk→∞

R

µEn νξk (dn);

R

and because the latter limit is defined it is equal to µEn θ(dn). As µ is arbitrary, θ satisfies condition (i) of 538P, and is a medial functional; because Q ∈ F, θN = 1; and because F0 ⊆ F, θ(a) = 0 for every finite a ⊆ N, so θ is a medial limit. Q Q Remark It is possible to have medial limits when cov M  c; see 553N. 538X Basic exercises (a) Let F be a filter on N, and I an infinite subset of N such that N \ I ∈ / F; write FdI for the filter {A ∩ I : A ∈ F}. Let A be a Boolean algebra. Show that if F is free, or a p-point filter, or Ramsey, or rapid, or nowhere dense, or measure-centering, or measure-converging, or with the Fatou property, then so is FdI. (b) For A ∈ [N]ω let fA : N → A be the increasing enumeration of A. Let F be a free filter on N. Show that F is rapid iff {fA : A ∈ F} is cofinal with N N . (c) Let FFr be the Fr´echet filter, the filter of cofinite subsets of N, and Fd the asymptotic density filter, the filter of subsets of N with asymptotic density 1. (i) Show that FFr and Fd are p-point filters. (ii) Show that FFr ≤RK Fd but that FFr n FFr 6≤RK Fd . (d)(i) Let hFn in∈N be a sequence of filters on N, and F a filter on N. Write limn→F Fn for the filter {A : A ⊆ N, {n : n ∈ N, A ∈ Fn } ∈ F }. Show that if every Fn is rapid, then limn→F Fn is rapid. (ii) Let F be a rapid filter, and G any filter on N. Show that G n F is rapid. (iii) In 538E, suppose that F1 is rapid. Show that Gξ is rapid for every ξ ≥ 1. (e) Let F be a nowhere dense filter, and G a filter on N such that G ≤RK F. Show that G is nowhere dense. >(f ) Let F be a free filter on N. Show that the following are equiveridical: (i) F is a Ramsey filter; (ii) whenever K is finite, k ∈ N and f : [N]k → K is a function, there is an F ∈ F such that f is constant on [F ]k ; (iii) F is a p-point filter and whenever hEn in∈N is a disjoint sequence in [N]<ω , there is an F ∈ F such that #(F ∩ En ) ≤ 1 for every n; (iv) whenever hEn in∈N is a disjoint sequence in PN \ F, there is an F ∈ F such that #(F ∩ En ) ≤ 1 for every n.

538Yc

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(g) Let F be a countable family of distinct p-point ultrafilters on N. Show that there is a disjoint family hAF iF ∈F of subsets of N such that AF ∈ F for every F ∈ F. (h) Let F be a measure-centering ultrafilter on N, A a Boolean algebra, ν : A → [0, 1] an additive functional, and han in∈N a sequence in A such that inf n∈N νan > 0. Show that there is an A ∈ F such that {an : n ∈ A} is centered. (i) Let h(Ai , µ ¯i )iQ ¯) Q the probability algebra i∈I be a family of probability algebras, F an ultrafilter on I, and (A, µ reduced product of i∈I (Ai , µ ¯i )|F. For each i ∈ I, let ⊆ i be the order relation on Ai ; set P = i∈I Ai and let P |F be the partial order reduced product of h(Ai , ⊆ i )ii∈I modulo F as defined in 5A2A. Describe a canonical order-preserving map from P |F to A. (j) Let (A, µ ¯) be a homogeneous probability algebra with Maharam type κ, I a non-empty set, F an ultrafilter on I and (C, ν¯) the probability algebra reduced power (A, µ ¯)I |F. Show that C is homogeneous, with Maharam type the transversal number TrI (I; κ) (definition: 5A1L), where I = {I \ A : A ∈ F}. (k) Let (X, Σ, µ) be a perfect probability space and µ0 an indefinite-integral measure over µ which is also a probability measure. Let F be a measure-centering ultrafilter on N and λ, λ0 the F-extensions of µ and µ0 , as defined in 538I. Show that λ0 is an indefinite-integral measure over λ. > (l) (Benedikt 98) (i) Let F be any free filter on N. Show that F n F is not measure-centering. (Hint: let hen in∈N be the standard generating family in Bω , and consider amn = em \ en if m < n, 1 otherwise.) (ii) Let F be a measure-centering ultrafilter on N. Show that if f , g ∈ N N and {n : f (n) 6= g(n)} ∈ F, then f [[F]] 6= g[[F]]. (Hint: consider an = ef (n) \ eg(n) if f (n) 6= g(n).) (m)(i) Let hFn in∈N be a sequence of measure-converging filters, and F any filter on N. Show that limn→F Fn (538Xd) is measure-converging. (ii) In 538E, suppose that F1 is measure-converging. Show that Gξ is measure-converging for every ξ ∈ [1, ζ]. (n) Show that a non-principal ultrafilter on N cannot have the Fatou property. (Hint: 464Ca.) (o) Show that the asymptotic density filter (538Xc) has the Fatou property. (p)(i) Let hFn in∈N be a sequence of filters with the Fatou property, and F a filter with the Fatou property. Show that limn→F Fn (538Xd) has the Fatou property. (ii) In 538E, suppose that Fξ has the Fatou property for every ξ ∈ [1, ζ]. Show that Gξ has the Fatou property for every ξ ≤ ζ. R (q) Let ν : PN → R Rbe a bounded additive functional. Show that ν is a medial functional iff ν{n : x ∈ En }µ(dx) is defined and equal to µEn ν(dn) whenever (X, Σ, µ) is a probability space and hEn in∈N is a sequence in Σ. (r) Let ν be a medial limit, and h : N → N a function such that limn→∞ h(n) = ∞. Set ν 0 a = ν(h−1 [a]) for a ⊆ N. Show that ν 0 is a medial limit. > (s) Let (X, Σ, µ) be a probability space, and T a σ-subalgebra of Σ. Let hfn in∈N be a sequence in L∞ (µ) such that supn∈N ess sup |fn | is finite, and for each expectation of fn on T. Suppose that ν R n ∈ N let gn be a conditional R is a medial functional. Show that f (x) = fn (x)ν(dn) and g(x) = gn (x)ν(dn) are defined for almost every x, that f ∈ L∞ (µ), and that g is a conditional expectation of f on T. (t)(i) Show that if F and G are p-point filters on N, then F n G is a p-point filter. (ii) In 538E, show that if every Fξ is a p-point filter than Gζ is a p-point filter. (u) (V.Bergelson) Show that there are a probability algebra (A, µ ¯) and a sequence han in∈N in A such that inf n∈N µ ¯ an > 0 but am ∩ an ∩ am+n = 0 whenever m, n > 0. (Hint: for n ≥ 1, set En = {x : x ∈ [0, 1], b3nxc ≡ 1 mod 3}.) 538Y Further exercises (a) Let F be a free ultrafilter on N, and suppose that whenever G is a free filter on N and G ≤RK F, then F ≤RK G. Show that F is a Ramsey ultrafilter. (b) Show that if p = c then there are 2c Ramsey ultrafilters on N, and therefore 2c isomorphism classes of Ramsey ultrafilters. (c)(i) Let F, G be free filters on N such that F n G is measure-centering. Show that there is no free filter H such that H ≤RK F and H ≤RK G. (ii) Show that if there are two non-isomorphic Ramsey ultrafilters on N, then there are two non-isomorphic measure-centering ultrafilters F, G on N such that F n G is not measure-centering.

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(d) For an uncountable set I, let us say that a filter F on I is uniform and measure-centering if #(A) = #(I) for every A ∈ F and whenever (A, µ ¯) is a totally finite measure algebra and hai ii∈I is a family in A with inf i∈I µ ¯ai > 0, there is an A ∈ F such that {ai : i ∈ A} is centered. (i) State and prove a result corresponding to 538G for such filters. (Hint: in the part corresponding to 538G(iv), use ‘compact’ measures rather than ‘perfect’ measures.) (ii) State and prove a result corresponding to 538H. (Hint: set κ = #(I). In the part corresponding to 538Hc, suppose that you have a κ-complete ultrafilter on I, rather than a Ramsey ultrafilter; see 4A1L. In the part corresponding to 538He, suppose that κ is regular and that cov Nκ = 2κ , where Nκ is the null ideal of the usual measure on {0, 1}κ .) (iii) State and prove results corresponding to 538I-538K. (iv) State and prove results corresponding to 538L-538M, but with ‘normal ultrafilters’ in place of ‘Ramsey ultrafilters’. (e) Give an example of filters F, G on N such that F has the Fatou property, G ⊆ F and G does not have the Fatou property. (f )(i) Let F be a nowhere dense filter on N, and I the ideal {N \ A : A ∈ F}. Show that PN/I is finite. (ii) Show that a free filter with the Fatou property cannot be nowhere dense. R (g) (Meyer 73) Let ν be a medial limit.R Write U for theR set of sequences u ∈ R N such that sup{ v dν : v ∈ `∞ , v ≤ |u|} is finite; for u ∈ U , write u dν for limm→∞ med(−m, u(n), m)ν(dn) (see 364Xw14 ). Suppose that R (X, Σ, µ) is a probability space and hfn in∈N a sequence of µ-integrable real-valued functions on X such that h |fn |dµin∈N ∈ (x)in∈N ∈ U for µ-almost every x ∈ X. (ii) Show that if every fn is nonRR U . (i) Show that hfnRR negative then f (x)ν(dn)µ(dx) ≤ fn dµ ν(dn). (iii) Show that if {fn : n ∈ N} is uniformly integrable then n RR RR fn (x)ν(dn)µ(dx) = fn dµ ν(dn). (h) Let F be an ultrafilter on N. Show that F is measure-centering iff whenever A is a Boolean algebra, D ⊆ A \ {0} has intersection number greater than 0 (definition: 391H) and han in∈N is a sequence in D, then there is an A ∈ F such that {an : n ∈ A} is centered. (i)(i) Show that if cov N = c, there is a measure-centering ultrafilter on N including the asymptotic density filter (538Xc). (ii) Show that an ultrafilter on N including the asymptotic density filter cannot be a p-point filter. (iii) Show that a filter on N including the asymptotic density filter cannot be a rapid filter. ˙ on the set βN of ultrafilters on N defined by saying that F +G ˙ = (j)(i) Show that there is a semigroup operation + ˇ +[[F n G]] for all F, G ∈ βN, where + : N × N → N is addition. (ii) Show that if we identify βN with the Stone-Cech ˙ is continuous in the first variable. (iii) Show that there is a non-principal compactification of N (4A2I(b-i)14 ), then + ˙ = F. (Hint: consider a minimal closed sub-semigroup of the ultrafilter F on N which is idempotent, that is, F +F P set of non-principal ultrafilters.) (iv) For any function f ∈ N N , write FS(f ) for { n∈K f (n) : K ∈ [N]<ω }; say a finite sum set is a set of the form FS(f ) for some strictly increasing function f ∈ N N . Show that if F is a nonprincipal idempotent ultrafilter on N and I ∈ F, then I includes a finite sum set. (This is a version of Hindman’s theorem.) (v) Show that if I ⊆ N is a finite sum set there is an idempotent ultrafilter containing I. (vi) Show that if F is an idempotent ultrafilter on N, (A, µ ¯) is a probability algebra, and π : A → A is a measure-preserving Boolean homomorphism, then (α) limn→F µ(a ∩ π n a) ≥ (µa)2 for every a ∈ A (β) there is a finite sum set I ⊆ N such that {π n a : n ∈ I} is centered. (vii) Show that no idempotent ultrafilter is measure-centering. (Hint: 538Xu.) (viii) Show ˙ is not measure-centering. (ix) Repeat, as far as possible, for semigroups that if F is a Ramsey ultrafilter then F +F other than (N, +). (k) (V.Bergelson-M.Talagrand) Show that there are a probability algebra (A, µ ¯) and a sequence han in∈N in A such ¯(am ∩ an ) = 0 whenever I ⊆ N does not have asymptotic density 0. that µ ¯an = 21 for every n ∈ N but inf m,n∈I µ 538Z Problem Show that it is relatively consistent with ZFC to suppose that there are no measure-converging filters on N. (Note that in this case there are no medial limits, by 538Rd.) 538 Notes and comments This is a long section, and rather a lot of ideas are crowded into it, starting with the list in 538A. If you have looked at ultrafilters on N at all, you are likely to have encountered ‘p-point’, ‘rapid’ and ‘Ramsey’ ultrafilters, and most of 538B-538D and 538F will probably be familiar. The ‘iterated products’ of 538E will also be a matter of adapting known concepts to my particular formulation. Some of the slightly contorted language of 538Fe and 538Ff (with references to ‘#(F)’) is there because we do not know how many isomorphism classes of Ramsey filters there are. If there are none (as in random real models, see 553H), or one (Shelah 82, §VI.5), then things are very simple. If there are infinitely many then we could rephrase 14 Later

editions only.

538 Notes

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281

538Ff in terms of sequences of non-isomorphic filters. But as far as I know it is possible that there should be two, or seventeen. In 538H-538M I try to set out, and expand, some of the principal ideas of Benedikt 98. The starting point is the observation that a Ramsey ultrafilter gives us an extension of Lebesgue measure on [0, 1], indeed of any perfect probability measure. Observing that this property is preserved by iterations, we are led to ‘measure-centering’ ultrafilters. Once we have the idea of measure-centering-ultrafilter extension of a perfect probability measure, we can set out to look at its properties in terms of the (by now very extensive) general theory of this treatise. The first step has to be the identification of its measure algebra (538Ja, 538Xj), followed, if possible, by the identification of the corresponding Banach function spaces. It turns out that these can be reached by an alternative route not involving special properties of the ultrafilter or the probability space, which I have expressed in general forms in the new §§328 and 377, to appear in future editions of Volume 3. This gives a long list of facts, which I have written out in 538Ja and 538K. Minor variations of the measure and the filter are straightforward (538Jb, 538Jc, 538Xk). For iterated products of filters we have more work to do (538L), especially if we are to express them in a form adequate for the objective, the universal-extension result of 538M. S T You will have noticed that in the statement of 538G I speak of ‘ A∈F n∈A Fn ’ and ‘lim inf n→F µFn ’. Something of the sort is necessary since in that theorem I do not insist from the outset that F S shouldTbe an ultrafilter. Of course only ultrafilters are of interest in this context, by 538Ha, and for these we have A∈F n∈A Fn = limn→F Fn and lim inf n→F µFn = limn→F µFn , as in 538I. For most of this section I have kept firmly to the study of filters on N. For measure-centering filters, at least, there are interesting extensions to filters on uncountable sets, which I mention in 538Yd. We can do a good deal with the ideas of 538G-538K on cardinals less than c in the presence of (for instance) Martin’s axiom; but for anything corresponding to 538L-538M it seems that we must use a two-valued-measurable cardinal (541M below). Measure-converging filters (538N) and filters with the Fatou property (538O) form an oddly complementary pair. I have tried to emphasize the correspondence in the characterizations 538Na and 538Oa (compare 538G(v), 538Na(iv) and 538Oa(iv)), but after this they seem to diverge. The phrase ‘Fatou property’ comes from 538O(a-iii); if you like, Fatou’s Lemma says that the Fr´echet filter has the Fatou property. Note that any filter larger than a measure-converging filter is again measure-converging, so that if there is a measure-converging filter there is a measure-converging ultrafilter; but that no non-principal ultrafilter can have the Fatou property (538Xn). On the other hand, there are many free filters with the Fatou property, but it is not known for sure whether there have to be measure-converging filters. It is possible for a measure-converging filter to have the Fatou property (538Rd). In the last part of the section I look at a different kind of limit. A ‘Banach limit’ is an extension to `∞ of the ordinary limit regarded as a linear functional on the closed subspace of convergent sequences; a ‘medial limit’ is a Banach limit which commutes with integration in appropriate settings. To study these I use the R formulae of repeated integration to do some surprising things. In 363L I tried to explain what I meant by the formula ‘ . . . dν’ for a finitelyRRadditive functional ν. This defines linear functionals which are positive ν. In ‘repeated integrals’ like fn (x)µ(dx)ν(dn)
R R for non-negative (538P(iii)), we must interpret the formula as fn (x)µ(dx) ν(dn); the ‘inner integral’ is an ordinary integral with respect to the countably additive measure µ, and the ‘outer integral’ is a name for a linear functional. In the integral R R . . . dν we have no problem with measurability,R though we must check that the integrand n → 7 f dµ is bounded; but n RR RR when we look at the other repeated integrals, ν(a)µ(da) or x dν µ(dx) or fn (x)ν(dn)µ(dx), the conditions of 538P must explicitly assert that the outer integrals are defined. Because we don’t need to consider measurability, the ‘finitely additive integrals’ here are in some ways easy to deal R with; ‘disintegrations’ like ν˜ = νk ν(dk) (538Rc) slide past all the usual questions. However we must always be vigilant against the temptations of limiting processes. As with the Riemann integral, of course, we can integrate the limit of a uniformly P∞ convergent sequence of functions. But see the manoeuvres of part (a-iii) of the proof of 538R, where the sums i=0 αni ... demand different treatments at different points. And Fubini’s theorem nearly disappears; the point of ‘medial functionals’ is that something extraordinary has to happen before we can expect to change the order of integration. I have used the language of Volume 3 to express 538Re in a general form. Of course by far the most important example is when the operator T is a conditional expectation operator (538Xs). For more examples of operators in L× (L∞ ; L∞ ), see §§373-374. For most of the classes of filter here, there is a question concerning their existence. Subject to the continuum hypothesis, there are many Ramsey ultrafilters, and refining the argument we find that the same is true if p = c (538Yb). There are many ways of forcing non-existence of Ramsey ultrafilters, of which one of the simplest is in 553H below. With more difficulty, we can eliminate p-point ultrafilters (Wimmers 82) or rapid filters (Miller 80) or nowhere dense filters and therefore measure-centering ultrafilters (538Hd, Shelah 98). It is not known for sure that we can eliminate medial limits or measure-converging filters (538Z).

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539 Maharam submeasures Continuing the work of §§392-394 and 49615 , I return to Maharam submeasures and the forms taken by the ideas of the present volume in this context. At least for countably generated algebras, and in some cases more generally, many of the methods of Chapter 52 can be applied (539B-539J). In 539K-539M I give the main result of Balcar ˇkovic ´ 05: it is consistent to suppose that every Dedekind complete ccc weakly (σ, ∞)Jech & Pazak 05 and Velic distributive Boolean algebra is a Maharam algebra. In 539Q-539T I introduce the idea of ‘exhaustivity rank’ of an exhaustive submeasure. 539A The story so far As submeasures have hardly appeared before in this volume, I begin by repeating some of the essential ideas. (a) If A is a Boolean algebra, a submeasure on A is a functional ν : A → [0, ∞] such that ν0 = 0, νa ≤ νb whenever a ⊆ b, and ν(a ∪ b) ≤ νa + νb for all a, b ∈ B (392A16 ); it is totally finite if µ1 < ∞. If µ is a submeasure defined on an algebra of subsets of a set X, I say that the null ideal of µ is the ideal N (µ) of subsets of X generated by {E : µE = 0} (496Bc17 ). As elsewhere, I will write simple N for the Lebesgue null ideal. A submeasure ν on a Boolean algebra A is exhaustive if limn→∞ νan = 0 for every disjoint sequence han in∈N in A; it is uniformly exhaustive if for every  > 0 there is an n ∈ N such that there is no disjoint family a0 , . . . , an with νai ≥  for every i ≤ n (392B). A Maharam submeasure is a totally finite sequentially order-continuous submeasure (393A18 ); a Maharam submeasure on a Dedekind σ-complete Boolean algebra is exhaustive (393Bc19 ). (b) A Maharam algebra is a Dedekind σ-complete Boolean algebra with a strictly positive Maharam submeasure. Any Maharam algebra is ccc and weakly (σ, ∞)-distributive (393Eb20 ). A Maharam algebra is measurable iff it carries a strictly positive uniformly exhaustive submeasure (393D21 ). If µ is any Maharam submeasure on a Dedekind σ-complete Boolean algebra A, its Maharam algebra is the quotient A/{a : µa = 0} (496Ba17 ). (c) If ν is any strictly positive totally finite submeasure on a Boolean algebra A, there is an associated metric b of A under this metric is a Boolean algebra (392H22 ). If ν is exhaustive, then A b (a, b) 7→ ν(a 4 b); the completion A 23 is a Maharam algebra (393H ). If µ and ν are both strictly positive Maharam submeasures on the same Maharam algebra A, µ is absolutely continuous with respect to µ (393F24 ). Consequently the associated metrics are uniformly equivalent, and A has a canonical topology and uniformity, its Maharam-algebra topology and Maharam-algebra uniformity (393G17 ). (d) In any Boolean algebra A, a sequence han in∈N order*-converges to a if there is a partition B of unity in A such that {n : b ∩ (an 4 a) 6= 0} is finite for every b ∈ B (393Ma25 ). The order-sequential topology on A is the topology for which the closed sets are just the sets closed under order*-convergence (393L26 ). If A is ccc and Dedekind σ-complete, a subalgebra of A is order-closed iff it is closed for the order-sequential topology (393O17 ). If A is ccc and weakly (σ, ∞)distributive, then the closure of a set A ⊆ A for the order-sequential topology is the set of order*-limits of sequences in A (393Pb27 ). If A is a Maharam algebra, then its Maharam-algebra topology is its order-sequential topology (393N17 ). If A is a Dedekind σ-complete ccc weakly (σ, ∞)-distributive Boolean algebra, and {0} is a Gδ set for the order-sequential topology, then A is a Maharam algebra (393Q28 ). (e) It was a long-outstanding problem (the ‘Control Measure Problem’) whether every Maharam algebra is in fact a measurable algebra; this was solved by a counterexample in Talagrand 08?, described in §39417 . (f ) If X is a Hausdorff space, a Radon submeasure on X is a totally finite submeasure µ defined on a σ-algebra Σ of subsets of X such that (i) if E ⊆ F ∈ Σ and µF = 0 then E ∈ Σ (ii) every open set belongs to Σ (iii) if E ∈ Σ and  > 0 there is a compact set K ⊆ E such that µ(E \ K) ≤  (496C17 ). Every Radon submeasure is a Maharam submeasure (496Da17 ). If X is a Hausdorff space and µ is a Radon submeasure on X, a set E ∈ dom µ is self-supporting if µ(E ∩ G) > 0 whenever G ⊆ X is an open set meeting E. If E ∈ dom µ and  > 0, there is a compact self-supporting K ⊆ E such that µ(E \ K) ≤  (496Dd17 ). Let µ be a strictly positive Maharam submeasure on a Dedekind σ-complete Boolean algebra A. Let Z be the Stone space of A, and write b a for the open-and-closed subset of Z corresponding to each a ∈ A. Then there is a unique Radon submeasure ν on Z of A such that νb a = µa for every a ∈ A; the null ideal of ν is the nowhere dense ideal of Z (496G17 ). 15 All

four of these sections have been rewritten since the last printed editions of Volumes 3 and 4, so there will be repeated footnotes ‘Later editions only’ and ‘Formerly. . . ’. 16 Early editions did not allow ∞ as a value of a submeasure; the change is nearly trivial. 17 Later editions only. 18 Formerly 392G. 19 Formerly 392Hc. 20 Formerly 392I. 21 Formerly 392J. 22 Formerly 393B. 23 Formerly 393B. 24 Formerly 393E. 25 Formerly 392L. 26 Formerly 392K. 27 Formerly 392Mb. 28 Formerly 392O.

539C

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539B Proposition Let A be a Maharam algebra, τ (A) its Maharam type and dT (A) its topological density for its Maharam-algebra topology. Then τ (A) ≤ dT (A) ≤ max(ω, τ (A)). proof Recall that the Maharam-algebra topology is the order-sequential topology (539Ad). A is ccc and weakly (σ, ∞)-distributive (539Ab), so if D ⊆ A is topologically dense, then every element of A is expressible as the order*limit inf n∈N supm≥n am of some sequence han in∈N in D (539Ad). In this case D τ -generates A and τ (A) ≤ #(D); accordingly τ (A) ≤ dT (A). If D ⊆ A τ -generates A, let B be the subalgebra of A generated by D and B its topological closure. Then B is order-closed (because A is ccc), so is the whole of A (539Ad), and dT (A) ≤ #(B) ≤ max(ω, #(D)); accordingly dT (A) ≤ max(ω, τ (A)). 539C Theorem (a) Let A be a Maharam algebra. Then (A+ , ⊇ 0 , [A+ ]≤max(ω,τ (A)) ) 4GT (Pou(A), v∗ , Pou(A)), where (A+ , ⊇ 0 , [A+ ]≤κ ) is defined as in 512F, and (Pou(A), v∗ ) as in 512Ee. (b) Let A be a Maharam algebra. Then Pou(A) 4T Nτ (A) , where Nκ is the null ideal of the usual measure on {0, 1}κ . proof (a) If A = {0} this is trivial; suppose otherwise. Fix a strictly positive Maharam submeasure µ on A such that µ1 = 1. Let B be a subalgebra of A which is dense in A for the metric (a, b) 7→ µ(a 4 b) and has cardinal at most κ = max(ω, τ (A)) (539B). 2

(i) For a ∈ A+ choose φ(a) ∈ Pou(A) as follows. Start by taking dn ∈ B, for n ∈ N, such that µ(dn 4 (1 \ a)) ≤ µa for each n; set bn = dn \ supi
−n−2

µ(a0 \ a) ≤ inf n∈N µ((1 \ dn ) \ a) ≤ inf n∈N µ(dn 4 (1 \ a)) = 0, P∞ µ(a \ a0 ) ≤ n=0 µ(a ∩ dn ) < µa, so 0 6= a0 ⊆ a. Now set φ(a) = {a0 } ∪ {bn : n ∈ N}. (ii) For C ∈ Pou(A), set ψ(C) = {c ∩ b : c ∈ C, b ∈ B} \ {0} ∈ [A+ ]≤κ . (iii) Suppose that a ∈ A+ , C ∈ Pou(A) and φ(a) v∗ C. Then there is a b ∈ ψ(C) such that b ⊆ a. P P Let c ∈ C be such that c ∩ a0 6= 0, where a0 is defined as in (i) above. Then B = {b : b ∈ φ(a) \ {a0 }, c ∩ b 6= 0} is a finite subset of B, so sup B ∈ B and c \ sup B ∈ ψ(C). But c \ sup B = c ∩ a0 ⊆ a. Q Q Thus a ⊇ 0 ψ(C). + 0 As a is arbitrary, (φ, ψ) is a Galois-Tukey connection from (A , ⊇ , [A+ ]≤κ ) to (Pou(A), v∗ , Pou(A), and (A+ , ⊇ 0 , + ≤κ [A ] ) 4GT (Pou(A), v∗ , Pou(A). (b)(i) Set κ = τ (A). If κ is finite, then A is purely atomic and Pou(A) has an upper bound in itself, as does Nκ ; so the result is trivial. Accordingly we may suppose henceforth that κ is infinite. (ii) Fix a strictly positive Maharam submeasure µ on A and a topologically dense subalgebra B ⊆ A of size κ (539B). If C ∈ Pou(A), there is a sequence hbn in∈N in B such that µbn ≤ 4−n for every n ∈ N and {c : c ∈ C, c 6⊆ supi≥n bi } is finite for every n ∈ N. P P If C is finite this is trivial. Otherwise, set n = 4−n /(n + 2) for each n ∈ N, and enumerate C as hcn in∈N . Let hk(n)in∈N be a strictly increasing sequence such that µc0n ≤ n for every n, where c0n = supi≥k(n) ci ; choose hbn in∈N in B inductively so that µ(bn 4 supj≤n (c0j \ supj≤i
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539C

1 \ supi≥n bφ(C)(i) meets only finitely many members of C, for every n ∈ N. Next, define ψ : Sκ → Pou(A) as follows. Given S ∈ Sκ , set a0 (S) = 1, an+1 (S) = supm≥n sup{bξ : (m, ξ) ∈ S, µbξ ≤ 4−m } P∞ for each n; then µan+1 (S) ≤ m=n 2−m = 2−n+1 for every n, so ψ(S) = {an (S) \ an+1 (S) : n ∈ N} is a partition of unity in A. (iv) Suppose that C ∈ Pou(A) and S ∈ Sκ are such that φ(C) ⊆∗ S. In this case there is an m ∈ N such that (n, φ(C)(n)) ∈ S for every n ≥ m. Since µbφ(C)(n) ≤ 4−n for every n, supi≥n bφ(C)(i) ⊆ an+1 (S) and 1 \ an+1 (S) meets only finitely many members of C, for every n ≥ m. Thus every member of ψ(S) meets only finitely many members of C, and C v∗ ψ(S). This shows that (φ, ψ) is a Galois-Tukey connection from (Pou(A), v∗ , Pou(A)) to (κN , ⊆∗ , Sκ ), and (Pou(A), v∗ , Pou(A)) 4GT (κN , ⊆∗ , Sκ ). On the other side, we know already that (κN , ⊆∗ , Sκ ) 4GT (Nκ , ⊆, Nκ ) (524G); so (Pou(A), v∗ , Pou(A)) 4GT (Nκ , ⊆, Nκ ), that is, Pou(A) 4T Nκ . 539D Corollary Let A be a Maharam algebra. (a) π(A) ≤ max(cf[τ (A)]≤ω , cf N ). (b) If τ (A) ≤ ω, then wdistr(A) ≥ add N . proof (a) Set κ = τ (A). If π(A) is countable, or π(A) ≤ cf[κ]≤ω , we can stop. Otherwise, max(ω, κ) ≤ max(ω, cf[κ]≤ω ) < π(A) = cov(A+ , ⊇, A+ ) ≤ max(ω, κ, cov(A+ , ⊇ 0 , [A+ ]≤κ )) (512Gf), so π(A) ≤ cov(A+ , ⊇ 0 , [A+ ]≤κ ) ≤ cov(Pou(A), v∗ , Pou(A)) (539Ca, 512Da) = cf Pou(A) ≤ cf Nκ (539Cb) = max(cf[κ]≤ω , cf N ). (b) In this case, wdistr(A) = add Pou(A) ≥ add N by 539Cb. ˇkovic ´ 05, Balcar Jech & Paza ´ k 05) If A is an atomless Maharam algebra, there is 539E Proposition (Velic a sequence han in∈N in A such that supn∈I an = 1 and inf n∈I an = 0 for every infinite I ⊆ N. proof (a) Fix a strictly positive Maharam submeasure µ on A. If han in∈N is a sequence in A such that δ = inf n∈N µan is greater than 0, there are a non-zero d ∈ A and an infinite I ⊆ N such that d ⊆ supi∈J ai for every infinite J ⊆ I. P P?? Otherwise, set bJ = supi∈J ai for J ⊆ N. Choose hIξ iξ<ω1 , hcξ iξ<ω1 and hdξ iξ<ω1 inductively, as follows. I0 = N. The inductive hypothesis will be that Iξ is an infinite subset of N, Iξ \ Iη is finite whenever η ≤ ξ, and cξ ∩ bIξ+1 = 0 for every ξ < ω1 . Given hIη iη≤ξ where ξ < ω1 , set dξ = inf n∈N bIξ \n . Since µbJ ≥ δ for every non-empty J ⊆ N, µdξ ≥ δ and dξ 6= 0. By hypothesis, there is an infinite Iξ+1 ⊆ Iξ such that cξ = dξ \ bIξ+1 is non-zero. Given hIη iη<ξ where ξ < ω1 is a non-zero limit ordinal, let Iξ be an infinite set such that Iξ \ Iη is finite for every η < ξ, and continue. Now observe that if η < ξ < ω1 , Iξ \ Iη is finite, so that there is an n ∈ N such that Iξ \ n ⊆ Iη+1 , and cξ ⊆ dξ ⊆ bIξ \n ⊆ bIη+1 is disjoint from cη . But this means that hcξ iξ<ω1 is disjoint, which is impossible, because A is ccc. X XQ Q (b) Let us say that a Boolean algebra A splits reals if there is a sequence han in∈N in A such that supn∈I an = 1 and inf n∈I an = 0 for every infinite I ⊆ N. Now if A is a Maharam algebra, the set of those d ∈ A such that the principal ideal Ad generated by d splits reals is order-dense in A. P P Let a ∈ A+ . case 1 If µ Aa is uniformly exhaustive, then Aa is measurable (539Ab). Let ν¯ be a probability measure on Aa ; because Aa , like A, is atomless, there is a stochastically independent family han in∈N in Aa with ν¯an = 21 for every n, and now han in∈N witnesses that Aa splits reals.

539G

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285

case 2 If µ Aa is not uniformly exhaustive, let hbni ii≤n∈N be a family of elements ofQAa such that hbni ii≤n is disjoint for each n and  = inf i≤n∈N µbni is greater than 0. There is a family hfξ iξ<ω1 in n∈N {0, . . . , n} such that {n : fξ (n) = fη (n)} is finite whenever η < ξ < ω1 . (For each ξ < ω1 let θξ : ξ → N be injective. Now define hfξ iξ<ω1 inductively by saying that fξ (n) = min(N \ {fη (n) : η < ξ, θξ (η) < n}) for every ξ < ω1 and n ∈ N.) ?? If for every ξ < ω1 and I ∈ [N]ω there is a J ∈ [I]ω such that inf i∈J bi,fξ (i) 6= 0, choose hIξ iξ<ω1 inductively so that Iξ ∈ [N]ω , Iξ \ Iη is finite for every η < ξ, and cξ = inf i∈Iξ bi,fξ (i) is non-zero for every ξ < ω1 . Then whenever η < ξ the set Iξ ∩ Iη is infinite, so there is an i ∈ Iξ ∩ Iη such that fξ (i) 6= fη (i); now cξ ∩ cη ⊆ bi,fξ (i) ∩ bi,fη (i) = 0. But this means that we have an uncountable disjoint family in Aa , which is impossible, because A is ccc. X X Thus we have a ξ < ω1 and an infinite I ⊆ N such that inf i∈J di = 0 for every infinite J ⊆ I, where di = bi,fξ (i) for i ∈ I. Next, applying (a) to hdi ii∈I , we have an infinite K ⊆ I and a d 6= 0 such that d = supi∈J di for every infinite J ⊆ K. But this means that hd ∩ di ii∈K witnesses that Ad splits reals; while d ⊆ a. As a is arbitrary, we have the result. Q Q (c) By 313K, there is a partition D of unity in A such that Ad splits reals for every d ∈ D; choose a sequence hadn in∈N in Ad witnessing this for each d ∈ D. Set an = supd∈D adn for each n. If I ⊆ N is infinite, then supn∈I an = supd∈D supn∈I adn = sup D = 1, while d ∩ inf n∈I an = inf n∈I adn = 0 for every d ∈ D, so inf n∈I an = 0. Thus han in∈N witnesses that A splits reals, as claimed. 539F Definition For the next result I need a name for one more cardinal between ω1 and c. The splitting number s is the least cardinal of any family A ⊆ PN such that for every infinite I ⊆ N there is an A ∈ A such that I ∩ A and I \ A are both infinite. 539G Proposition Let X be a set, Σ a σ-algebra of subsets of X, and µ an atomless Maharam submeasure on Σ. Let M be the ideal of meager subsets of R. (a) non N (µ) ≥ max(s, cov M). (b) cov N (µ) ≤ non M. proof If µX = 0, these are both trivial; suppose otherwise. (a)(i) Suppose that D ⊆ X and #(D) < cov M. For any  > 0, there is an F ∈ B(X) such that D ⊆ F and µF ≤ . P P By 393I29 , there is for each n ∈ N a finite partition En of X into members of Σ such that µE ≤ 2−n−1  for each Q E ∈ En . Express each En as {Eni : i < k(n)}. For x ∈ D, let fx ∈ n∈N k(n) be such that x ∈ En,fx (n) for every n. Because #(D) < cov M, there S is an f ∈ N N such that f ∩ fx 6= ∅ for every x ∈ D (522Rb); we may suppose that f (n) < k(n) for every n. Set F = n∈N En,f (n) ; this works. Q Q Applying this repeatedly, we get a sequence hF i in B(X) such that D ⊆ Fn and µFn ≤ 2−n for every n; now n n∈N T F = n∈N Fn includes D and belongs to N (µ). As D is arbitrary, non N (µ) ≥ cov M. (ii) Set A = B(X)/B(X) ∩ N (µ), and define µ ¯ : A → [0, ∞[ by setting µ ¯E • = µE for every E ∈ B(X). Then µ ¯ is a strictly positive atomless Maharam submeasure on A. By 539E, there is a sequence han in∈N in A such that supn∈I an = 1 and inf n∈I an = 0 for every infinite I ⊆ N. For each n ∈ N, let En ∈ B(X) be such that En• = an . Suppose that D ⊆ X and #(D) < s. For x ∈ D, set Ax = {n : x ∈ En }. Because #(D) < s, there is an infinite I ⊆ N such that one of I ∩ Ax , I \ Ax is finite for every x ∈ D. Set
S S T F = m∈N (X \ n∈I\m En ) ∪ ( n∈I\m En ) ; then
F • = supm∈N (1 \ supn∈I\m an ) ∪ (inf n∈I\m an ) = 0, so F ∈ N (µ), while D ⊆ F . As D is arbitrary, non N (µ) ≥ s. Q (b) Let hk(n)in∈N , hEni ii
editions only.

286

Topologies and measures III

T

S

539G

: g ∈ Z, g(n) = f (n)} S is comeager in Z, so contains some gξ ; turning this round, Z = ξ<non M H(gξ ). Consider the sets Fξ = {x : x ∈ X, S fx ∈ H(gξ )}; then X = ξ<non M Fξ , while P∞ µFξ ≤ inf m∈N n=m µEn,gξ (n) = 0 H(f ) =

m∈N

n≥m {g

for every ξ. So cov N (µ) ≤ non M. 539H Corollary Let A be an atomless Maharam algebra, not {0}. Then d(A) ≥ max(s, cov M). proof Let Z be the Stone space of A and ν the Radon submeasure on Z corresponding to a strictly positive Maharam submeasure µ on A (539Af), so that N (ν) is the ideal of meager subsets of Z. Note that the meager sets of Z are all nowhere dense, because A is weakly (σ, ∞)-distributive (316I). Because A is atomless, so are µ and ν. As every meager subset of Z is nowhere dense (and Z 6= ∅), no dense set can be meager, and d(A) = d(Z) ≥ non N (ν) ≥ max(s, cov M) by 539Ga. 539I Theorem (a) Let µ be a Radon submeasure on a Hausdorff space X (539Af) and A its Maharam algebra. Let (Pou(A), v∗ ) the pre-ordered set of partitions of unity in A under refinement (512Ee). Then N (µ) 4T Pou(A). (b) Let µ be a Radon submeasure on a Hausdorff space X and A its Maharam algebra. (i) wdistr(A) ≤ add N (µ). (ii) τ (A) ≤ w(X). (iii) cf N (µ) ≤ max(cf[τ (A)]≤ω , cf N ). (iv) If τ (A) ≤ ω (e.g., because X is second-countable), then add N (µ) ≥ add N and cf N (µ) ≤ cf N . proof (a) For E ∈ N (µ), let KE be a maximal disjoint family of compact sets of non-zero submeasure disjoint from E, and set CE = {K • : K ∈ KE }. Because µ is inner regular with respect to the compact sets, CE ∈ Pou(A). Now E 7→ CE : N (µ) → Pou(A) is a Tukey function. P P Suppose that E ⊆ N (µ) and D ∈ Pou(A) are such that CE v∗ D for every E ∈ E; take any  > 0. Because D is countable, we have a countable partition H of X into measurable sets such that D = {H • : H ∈ H}. Because µ is inner regular with respect to the self-supporting compact sets (539Af), we can find a self-supporting compact set K ⊆ X such that µ(X \ K) ≤  and K is covered by finitely many members of H; consequently K • meets only finitely many members of D. 0 If E ∈ E, then K • meets S 0only finitely many members of CE , so there is a finite set KE ⊆ KE such that K \ KE is negligible, where KE = KE . But KE is compact and K is self-supporting, so K ⊆ KE and K ∩ E = ∅. S S This means that E ⊆ X \ K is included in an open set of submeasure at most . This is true for every  > 0, so E is included in a negligible Gδ set and belongs to N (µ); that is, E is bounded above in N (µ). As E is arbitrary, E 7→ CE is a Tukey function. Q Q (b)(i) Putting (a) and 513E(e-ii) together, wdistr(A) = add Pou(A) ≤ add N (µ). (ii) If U is a base for the topology of X with #(U) = w(X), consider D = {U • : U ∈ U } and the order-closed subalgebra B of A generated by D; note that B is closed for the order-sequential (or Maharam-algebra) topology of A (539Ad). Let E be the algebra of sets generated by U. If F ∈ dom µ and  > 0, there are compact sets K ⊆ F , L ⊆ X \ F such that µ(X \ (K ∪ L)) ≤ . There is an E ∈ E such that K ⊆ E ⊆ X \ L, so µ(E4F ) ≤ . Now E • ∈ B and µ ¯(F • 4 E • ) ≤ ; as  is arbitrary, F • ∈ B; as F is arbitrary, B = A and A is τ -generated by D. This means that τ (A) ≤ #(D) ≤ w(X), as required. (iii) Setting κ = τ (A), (a) and 539Cb tell us that N (µ) 4T Nκ , where Nκ is the null ideal of the usual measure on {0, 1}κ . So add N (µ) ≥ add Nκ and cf N (µ) ≤ cf Nκ ≤ max(cf[κ]≤ω , cf N ) (513Ee, 523N). (iv) If κ ≤ ω then Nκ 4T N so add N (µ) ≥ add N and cf N (µ) ≤ cf N . 539J We can approach precalibers by some of the same combinatorial methods as before.

539K

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Proposition Let A be a Boolean algebra and ν an exhaustive submeasure on A. (a) Let hai ii∈N be a sequence in A such that inf i∈N νai > 0. (i) There is an infinite I ⊆ N such that {ai : i ∈ I} is centered. (ii) For every k ∈ N there are an S ∈ [N]ω and a δ > 0 such that ν(inf i∈J ai ) ≥ δ for every J ∈ [S]k . (b) Suppose that haξ iξ<κ is a family in A such that inf ξ<κ νaξ > 0, where κ is a regular uncountable cardinal. Then for every k ∈ N there are a stationary set S ⊆ κ and a δ > 0 such that ν(inf i∈J ai ) ≥ δ for every J ∈ [S]k . (c) If ν is strictly positive, then (κ, κ, n) is a precaliber triple of A for every regular uncountable cardinal κ and every n ∈ N; in particular, A satisfies Knaster’s condition. proof (a)(i) This is 392J30 . (ii) Induce on k. The cases k = 0, k = 1 are trivial. For the inductive step to k + 1, let M ∈ [N]ω and δ > 0 be such that ν(inf i∈J ai ) ≥ δ for every J ∈ [M ]k . ?? Suppose, if possible, that for every S ∈ [M ]ω and γ > 0 there is a J ∈ [S]k+1 such that ν(inf i∈J ai ) < γ. Using Ramsey’s theorem (4A1G) repeatedly, we can find hIn in∈N such that I0 ∈ [M ]ω , In+1 ∈ [In ]ω , rn = min In ∈ / In+1 and ν(inf i∈J ai ) ≤ 2−n−2 δ for every n ∈ N and J ∈ [In ]k+1 . Set k S = {rn : n ∈ N}. If J ∈ [S] and min J = rn , then J ∪ {rm } ∈ [Im ]k+1 , so ν(inf i∈J ai ∩ arm ) ≤ 2−m−2 δ, for every m < n. It follows that ν(inf i∈J ai ∩ supm 0 and S ∈ [M ]ω such that ν(inf i∈J ai ) ≥ γ for every J ∈ [S]k+1 , and the induction continues. (b) Again induce on k. The cases k = 0, k = 1 are trivial. For the inductive step to k + 1 ≥ 2, write cJ = inf i∈J ai for J ∈ [κ]<ω . We know from the inductive hypothesis that there are a stationary set S ⊆ κ and a δ > 0 such that νcJ ≥ 3δ for every J ∈ [S]k . For each ξ ∈ S, choose m(ξ) ∈ N and hJξi ii<m(ξ) as follows. Given hJξi ii<j , where j ∈ N, choose, if possible, Jξj ∈ [S ∩ ξ]k such that ν(cJξj ∩ cJξi ) ≤ 2−i δ for every i < j and ν(aξ ∩ cJξj ) ≤ 2−j δ; if this is not possible, set m(ξ) = j and stop. Now the point is that we always do have to stop. P P?? Otherwise, set di = cJξi for each i ∈ N. Because Jξi ∈ [S]k , νdi ≥ 3δ for each i; also ν(di ∩ dj ) ≤ 2−i δ for i < j; so νd0j ≥ δ, where d0j = dj \ supi<j di for each j. But now hd0j ij∈N is disjoint and ν is not exhaustive. X XQ Q At the end of the process, we have m(ξ) and hJξi ii<m(ξ) for each ξ ∈ S. By the Pressing-Down Lemma (4A1Cc), 0 there are m ˜ and hJ˜i ii<m ˜ Jξi = J˜i for every i < m} ˜ is stationary in κ. ?? Suppose, ˜ such that S = {ξ : ξ ∈ S, m(ξ) = m, 0 k+1 −m ˜ if possible, that I ∈ [S ] and νcI ≤ 2 δ. Set ξ = max I, J = I \ {ξ}, η = min I ∈ J. Then J ∈ [S ∩ ξ]k . For each i<m ˜ = m(ξ), ν(cJ ∩ cJξi ) ≤ ν(aη ∩ cJξi ) = ν(aη ∩ cJηi ) ≤ 2−i δ, while ˜ ν(aξ ∩ cJ ) = νcI ≤ 2−m δ.

But this means that we could have extended the sequence hJξi ii<m X ˜ by setting Jξ m ˜ = J. X ˜ So S 0 and 2−m δ provide the next step in the induction. (c) This is now immediate from (b). ´ k 05, based on the characterizations of Maharam algebras 539K I come now to the work of Balcar Jech & Paza set out in §39331 . Lemma (Quickert 02) Let A be a Boolean algebra, and I the family of countable subsets I of A for which there is a partition C of unity such that {a : a ∈ I, a ∩ c 6= 0} is finite for every c ∈ C. (a) I is an ideal of PA including [A]<ω . (b) If A ⊆ A+ is such that A ∩ I is finite for every I ∈ I, and B = {b : b ⊇ a for some a ∈ A}, then B ∩ I is finite for every I ∈ I. (c) If A is ccc, then there is no uncountable B ⊆ A such that [B]≤ω ⊆ I. (d) If A is ccc and weakly (σ, ∞)-distributive, I is a p-ideal (definition: 5A1V). proof (a) Of course every finite subset of A belongs to I. If I0 , I1 ∈ I and J ⊆ I0 ∪ I1 , then J ∈ [A]≤ω . For each j, we have a partition Cj of unity in A such that {a : a ∈ Ij , a ∩ c 6= 0} is finite for every c ∈ Cj . Set C = {c0 ∩ c1 : c0 ∈ C0 , c1 ∈ C1 }; then C is a partition of unity in A and {a : a ∈ J, a ∩ c 6= 0} is finite for every c ∈ C. (b) Set J = B ∩ I. For each b ∈ J, let ab ∈ A be such that ab ⊆ b. Let C be a partition of unity such that {b : b ∈ I, b ∩ c 6= 0} is finite for every c ∈ C; then {ab : b ∈ J, ab ∩ c 6= 0} is finite for every c ∈ C, so {ab : b ∈ J} belongs to I and must be finite. ?? If J is infinite, there is an a ∈ A such that K = {b : b ∈ J, a = ab } is infinite; but in this case there is a c ∈ C such that a ∩ c 6= 0 and b ∩ c 6= 0 for every b ∈ K. X X So J is finite, as claimed. 30 Later

editions only. §392.

31 Formerly

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Topologies and measures III

539K

b be the Dedekind completion of A (314U). Let B ⊆ A be an uncountable set, and hbξ iξ<ω a family of (c) Let A 1 b Then (because A b is ccc, by 514Ee) d = sup distinct elements of B. Set d = inf ξ<ω1 supξ≤η<ω1 bη , taken in A. ξ≤η<ω1 bη for some ξ (316E); in particular, d 6= 0. Next, we can find a strictly increasing sequence hξn in∈N in ω1 such that d ⊆ supξn ≤η<ξn+1 bη for every n ∈ N. Set I = {bη : η < supn∈N ξn } ∈ [B]≤ω . If C is any partition of unity in A, there must be some c ∈ C such that c ∩ d 6= 0, and now {a : a ∈ I, a ∩ c 6= 0} is infinite. So I ∈ / I. (d) For each n ∈ N, let Cn be a partition of unity such that {a : a ∈ In , a ∩ c 6= 0} is finite for every c ∈ Cn . Let D be a partition of unity such that {c : c ∈ Cn , c ∩ d 6= 0} is finite for every d ∈ D and n ∈ N. Then S {a : a ∈ In , a ∩ d 6= 0} ⊆ c∈Cn ,c ∩ d6=0 {a : a ∈ In , a ∩ c 6= 0} S is finite for every d ∈ D and n ∈ N. Let hdn in∈N be a sequence running over D ∪ {∅} and set I = n∈N {a : a ∈ In , a ∩ di = 0 for every i ≤ n}. Then S In \ I ⊆ i≤n {a : a ∈ In , a ∩ di 6= ∅} is finite for each n. Also {a : a ∈ I, a ∩ dn 6= 0} ⊆

S

i
: a ∈ Ii , a ∩ dn 6= 0}

is finite for each n, so I ∈ I. Remark In this context, I is called Quickert’s ideal. 539L Lemma Let S A be a weakly (σ, ∞)-distributive ccc Dedekind σ-complete Boolean algebra, and suppose that A+ is expressible as k∈N Dk where no infinite subset of any Dk belongs to Quickert’s ideal I. Then A is a Maharam algebra. proof The point is that if han in∈N is a sequence in A which order*-converges to 0, then {an : n ∈ N} ∈ I. So no sequence in any Dk can order*-converge to 0. Because A is weakly (σ, ∞)-distributive and ccc, 0 does not belong to S the closure Dk of Dk for the order-sequential topology on A (539Ad). So A+ = k∈N Dk is Fσ and {0} is Gδ for the order-sequential topology. It follows that A is a Maharam algebra (539Ad). ´ k 05, Velic ˇkovic ´ 05) Suppose that Todorˇcevi´c’s p-ideal dichotomy 539M Theorem (Balcar Jech & Paza (5A1V) is true. Then every Dedekind σ-complete ccc weakly (σ, ∞)-distributive Boolean algebra is a Maharam algebra. proof Let A be a Dedekind σ-complete ccc weakly (σ, ∞)-distributive Boolean algebra. Let I be Quickert’s ideal on A; then I is a p-ideal (539Kd). By 539Kc, there is no B ∈ S [A]ω1 such that [B]≤ω ⊆ I. We are assuming that Todorˇcevi´c’s p-ideal dichotomy is true; so A must be expressible as n∈N Dn where no infinite subset of any Dn belongs to I. By 539L, A is a Maharam algebra. 539N Corollary Suppose that Todorˇcevi´c’s p-ideal dichotomy is true. Let A be a Dedekind complete Boolean algebra such that every countably generated order-closed subalgebra of A is a measurable algebra. Then A is a measurable algebra. proof (a) A is ccc. P P?? Otherwise, let haξ iξ<ω1 be a disjoint family of non-zero elements of A. Let f : ω1 → {0, 1}N be an injective function, and set bn = sup{aξ : ξ < ω1 , fξ (n) = 1} for each n; let B be the order-closed subalgebra of A generated by {bn : n ∈ N} ∪ {supξ<ω1 aξ }. Then aξ ∈ B for every ξ < ω1 , so B is not ccc; but B is supposed to be measurable. X XQ Q (b) A is weakly (σ, ∞)-distributive. P P Let hCn in∈N be a sequence of partitions of unity in A. As A is ccc, every S Cn is countable; let B be the order-closed subalgebra of A generated by n∈N Cn . Then B is measurable, therefore weakly (σ, ∞)-distributive, and there is a partition D of unity in B such that {c : c ∈ Cn , c ∩ d 6= 0} is finite for every n ∈ N and d ∈ D. As B is order-closed, D is still a partition of unity in A. As hCn in∈N is arbitrary, A is weakly (σ, ∞)distributive. Q Q (c) By 539M, A is a Maharam algebra; let µ be a strictly positive Maharam submeasure on A. Now µ is uniformly exhaustive. P P?? Otherwise, there are  > 0 and a family hani ii≤n∈N in A such that hani ii≤n is disjoint for every n ∈ N and µani ≥  whenever i ≤ n ∈ N. Let B be the order-closed subalgebra of A generated by {ani : i ≤ n ∈ N}. Then B is a measurable algebra; let ν¯ be a functional such that (B, ν¯) is a totally finite measure algebra. Since ν¯ and µB are both strictly positive Maharam submeasures on B, µ is absolutely continuous with respect to ν (539Ac). But µani ≥  for every n and i, while inf i≤n∈N ν¯ani must be zero. X XQ Q (d) So A is a Dedekind σ-complete Boolean algebra with a strictly positive uniformly exhaustive Maharam submeasure, and is a measurable algebra (539Ab).

539Pc

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539O I should say at once that 539M-539N really do need some special axiom. In fact the following example was found at the very beginning of the study of Maharam algebras. Souslin algebras: Proposition Suppose that T is a well-pruned Souslin tree (5A1Dd), and set A = RO↑ (T ). (a) A is Dedekind complete, ccc and weakly (σ, ∞)-distributive. (b) If B is an order-closed subalgebra of A and τ (B) ≤ ω, then B ∼ = PI for some countable set I; in particular, B is a measurable algebra. (c) (Maharam 47) The only Maharam submeasure on A is identically zero. proof (a)(i) A is Dedekind complete just because it is a regular open algebra. (ii) T is upwards-ccc, so A is ccc, by 514Nc. t : t ∈ T } is order-dense in A. Let r be the rank function of T (iii) For t ∈ T , set b t = int [t, ∞[ ∈ A; then {b (5A1Da). For each ξ < ω1 , Aξ {b t : t ∈ T , r(t) = ξ} is a partition of unity in A. P P If r(t) = r(t0 ) and t 6= t0 then 0 [t, ∞[ ∩ [t , ∞[ = ∅ so b t ∩ tb0 = 0 in A; thus Aξ is disjoint. If a ∈ A \ {0}, there is an s ∈ T such that sb ⊆ a; if r(s) ≥ ξ, there is a t ≤ s such that r(t) = ξ, and a ∩ b t 6= 0; if r(s) < ξ, there is a t ≥ s such that r(t) = ξ (because T is well-pruned), and b t ⊆ a. Q Q If A ⊆ A is a partition of unity, there is a ξ < ω1 such that Aξ refines A in the sense of 311Ge32 . P P B = {b t : t ∈ T, b t ⊆ a for some a ∈ A} is order-dense in A, so there is a partition C of unity included in B; C is countable; let D ⊆ T be a countable set such that C = {b t : t ∈ D}; set ξ = supt∈D r(t). Q Q Of course Aη refines Aξ whenever ξ ≤ η < ω1 . So if hCn in∈N is a sequence of partitions of unity in A, there is a ξ < ω1 such that Aξ refines Cn for every n ∈ N, and then {c : c ∈ Cn , a ∩ c 6= 0} is finite for every a ∈ Aξ and n ∈ N. As hCn in∈N is arbitrary, A is weakly (σ, ∞)-distributive. (b) If B ⊆ A is a countable set τ -generating B, there is a countable set D ⊆ T such that b = sup{b t : t ∈ D, b t ⊆ b} for every b ∈ B. Now ξ = sup{r(t) : t ∈ D} is countable, and b = sup{a : a ∈ Aξ , a ⊆ b} for every b ∈ B, so B is included in the order-closed subalgebra C of A generated by Aξ . Of course Aξ is order-dense in C. For a ∈ Aξ , set ba = inf{b : b ∈ B, b ⊇ a}; then every ba is an atom in B and {ba : a ∈ Aξ } is order-dense in B, so B is purely atomic. As B is ccc, the set I of its atoms is countable; being Dedekind complete, B is isomophic to PI. (c) Let µ be a Maharam submeasure on A. Then for every  > 0 there is a ξ < ω1 such that νa ≤  for every a ∈ Aξ . P P Set T 0 = {t : ν b t ≥ }. Then T 0 is a subtree of T and {t : t ∈ T 0 , r(t) = ξ} is finite for every ξ < ω1 , because ν is exhaustive. Also T 0 , like T , can have no uncountable branches. It follows that the height of T 0 is countable (5A1Db), that is, that there is a ξ < ω1 such that r(t) < ξ for every t ∈ T 0 and νa ≤  for every a ∈ Aξ . Q Q As this is true for every  > 0, there is actually a ξ < ω1 such that νa = 0 for every a ∈ Aξ . But as Aξ is a countable partition of unity and ν is a Maharam submeasure, ν1 = 0 and ν is identically zero. 539P Reflection principles In 539N, we have a theorem of the type ‘if every small subalgebra of A is . . . , then A is . . . ’. There was a similar result in 518I, and we shall have another in 545G. Here I collect some simple facts which are relevant to the present discussion. (a) If A is a Boolean algebra and every subset of A of cardinal ω1 is included in a ccc subalgebra of A, then A is ccc. (For there can be no disjoint set of cardinal ω1 .) (b) If A is ccc and every countable subset of A is included in a weakly (σ, ∞)-distributive subalgebra of A, then A is weakly (σ, ∞)-distributive. P P If Cn is a partition of unity in A for every n, set D = {d : {c : c ∈ Cn , c ∩ d 6= 0} is finite for every n ∈ N}. ?? If D is not order-dense in A, Stake a ∈ A+ such that d 6⊆ a for every d ∈ D. Let B be a weakly (σ, ∞)-distributive subalgebra of A including {a} ∪ n∈N Cn . Then every Cn is a partition of unity in B, so there is a partition B of unity in B such that B ⊆ D. But now a ∈ B+ so there is a b ∈ B such that a ∩ b 6= 0 and a ∩ b ∈ D. X X So D is order-dense in A and includes a partition of unity in A. As hCn in∈N is arbitrary, A is weakly (σ, ∞)distributive. Q Q (c) If A is ccc and every countable subset of A is included in a Dedekind σ-complete subalgebra of A, then A is Dedekind complete. P P Suppose that A ⊆ A is non-empty; let B be the set of upper bounds of A. By 316E, there are 32 Formerly

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a countable A0 ⊆ A with the same upper bounds as A, and a countable B 0 ⊆ B with the same lower bounds as B. Let B be a Dedekind σ-complete subalgebra of A including A0 ∪ B 0 ; as B is ccc, it is Dedekind complete; let c be the supremum of A0 in B. Since a ⊆ c ⊆ b for all a ∈ A0 and b ∈ B 0 , a ⊆ c ⊆ b for all a ∈ A and b ∈ B, that is, c is the supremum of A in A. Q Q (d) If A is a Maharam algebra and every countably generated closed subalgebra of A is a measurable algebra, then A is measurable. (This is part (c) of the proof of 539N.) (e) Suppose that Todorˇcevi´c’s p-ideal dichotomy is true. Let A be a Boolean algebra such that every subset of A of cardinal at most ω1 is included in a subalgebra of A which is a Maharam algebra. Then A is a Maharam algebra. P P By (a), A is ccc; by (c), A is Dedekind complete; by (b), A is weakly (σ, ∞)-distributive; by 539M, A is a Maharam algebra. Q Q (f ) Suppose that Todorˇcevi´c’s p-ideal dichotomy is true. Let A be a Boolean algebra such that every subset of A of cardinal at most c is included in a subalgebra of A which is a measurable algebra. Then A is measurable. P P By (a), A is ccc. So if B is a countably generated order-closed subalgebra, it has cardinal c, and is included in a measurable subalgebra C of A. Now B is order-closed in C, so is itself a measurable algebra. By 539N, A also is measurable. Q Q ˇkovic ´ 06 show that if κ is an infinite cardinal such that 2κ = κ+ , κ (g) On the other hand, Farah & Velic ω (5A1S) is true and the cardinal power κ is equal to κ, then there is a Dedekind complete Boolean algebra A, of cardinal κ+ , such that every order-closed subalgebra of A of cardinal at most κ is a measurable algebra, but A is not a measurable algebra (and therefore is not a Maharam algebra, by (d) above). In particular, this can easily be the case with κ = c. 539Q Exhaustivity rank While we now know that there are non-measurable Maharam algebras, we know practically nothing about their structure. I introduce the following idea as a possible tool for investigation. Definitions Suppose that A is a Boolean algebra and ν an exhaustive submeasure on A. For  > 0, say that a 4 b if either a = b or a ⊆ b and ν(b \ a) > . Then 4 is a well-founded partial order on A (use 5A1Cc; if han in∈N were strictly decreasing for 4 , then han \ an+1 in∈N would be disjoint, with ν(an \ an+1 ) ≥  for every n). Let r : A → On be the corresponding rank function, so that r (a) = sup{r (b) + 1 : b ⊆ a, ν(a \ b) > } for every a ∈ A (5A1Cb). Now the exhaustivity rank of ν is sup>0 r (1). 539R Elementary facts Let A be a Boolean algebra with an exhaustive submeasure ν and associated rank functions r for  > 0. (a) rδ (a) ≤ r (b) whenever ν(a \ b) ≤ δ − . P P Induce on r (b). If r (b) = 0, then νb ≤  so νa ≤ δ and rδ (a) = 0. For the inductive step to r (b) = ξ, if c ⊆ a and ν(a \ c) > δ then ν(b \ c) >  and r (b ∩ c) < ξ. Also ν(c \ b) ≤ δ −  so, by the inductive hypothesis, rδ (c) ≤ rδ (b ∩ c) < ξ; as c is arbitrary, rδ (a) ≤ ξ and the induction continues. Q Q In particular, r (a) ≤ r (b) if a ⊆ b,

rδ (a) ≤ r (a) if  ≤ δ.

(b) If a, b ∈ A are disjoint and  > 0, then r (a ∪ b) is at least the ordinal sum r (a) + r (b). P P Induce on r (b). If r (b) = 0, the result is immediate from (a) above. For the inductive step to r (b) = ξ, we have for any η < ξ a c ⊆ b such that ν(b \ c) >  and η ≤ r (c) < ξ. Now r (a ∪ c) ≥ r (a) + η, by the inductive hypothesis, and ν((a ∪ b) \ (a ∪ c)) > , so r (a ∪ b) > r (a) + η; as η is arbitrary, r (a ∪ b) ≥ r (a) + ξ and the induction continues. Q Q 539S The rank of a Maharam algebra Note that the rank function r associated with an exhaustive submeasure ν depends only on the set {a : νa > }. In particular, if µ and ν are exhaustive submeasures on a Boolean algebra (µ) (ν) A and µa ≤  whenever νa ≤ δ, then r (a) ≤ rδ (a) for every a ∈ A. If A is a Maharam algebra, then any two Maharam submeasures on A are mutually absolutely continuous (539Ac), so have the same exhaustivity rank; I will call this the Maharam submeasure rank of A, Mhsr(A). Note that if a ∈ A then Mhsr(Aa ) ≤ Mhsr(A). (µ) If A is a measurable algebra, Mhsr(A) ≤ ω, because if µ is an additive functional and  > 0, then µa > r (a) for (ν) every a ∈ A. More generally, for any uniformly exhaustive submeasure ν and  > 0, r (a) is finite, being the maximal size of any disjoint set consisting of elements, included in a, of submeasure greater than .

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539T Theorem Suppose that A is a non-measurable Maharam algebra. Then Mhsr(A) is at least the ordinal power ωω . proof Let ν be a strictly positive Maharam submeasure on A. (a) For the time being (down to the end of (d) below), assume that A is nowhere measurable (definition: 391Bc33 ). For a ∈ A, set νˇa = inf n∈N sup{mini≤n νai : a0 , . . . , an ⊆ a are disjoint}. Then νˇ is a Maharam submeasure. P P Of course νˇ0 = 0 and νˇa ≤ νˇb whenever a ⊆ b. If a, b ∈ A and  > 0, then there are n0 , n1 ∈ N such that whenever hci ii∈I is a disjoint family in A, then #({i : ν(ci ∩ a) ≥ νˇa + }) ≤ n0 and #({i : ν(ci ∩ b) ≥ νˇb + }) ≤ n1 . So #({i : ν(ci ∩ (a ∪ b)) ≥ νˇa + νˇb + 2}) ≤ n0 + n1 . It follows that νˇ(a ∪ b) ≤ νˇa + νˇb + 2; as , a and b are arbitrary, νˇ is a submeasure. Because νˇ ≤ ν, νˇ is a Maharam submeasure. Q Q (b) Because A is nowhere measurable, νˇ is strictly positive. P P If a ∈ A \ {0}, the principal ideal Aa is not measurable, so the Maharam submeasure ν Aa cannot be uniformly exhaustive; that is, there is an  > 0 such that there are arbitrarily long disjoint strings hai ii≤n in Aa with νai ≥  for every i ≤ n. But this means that νˇa ≥  > 0. Q Q (c) Let r , rˇ be the rank functions associated with ν and νˇ. Then r (a) is at least the ordinal product ω · rˇ (a) whenever a ∈ A and  > 0. P P Induce on rˇ (a). If rˇ (a) = 0, the result is trivial. For the inductive step to rˇ (a) = ξ + 1, take b ⊆ a such that νˇb >  and rˇ (a \ b) = ξ. Then for every n ∈ N there are disjoint b0 , . . . , bn ⊆ b such that νbi >  for every i, and r (b) ≥ ω; by the inductive hypothesis, r (a \ b) ≥ ω · ξ; by 539Rb, r (a) ≥ ω · ξ + ω = ω · (ξ + 1), and the induction proceeds. The inductive step to non-zero limit ξ is elementary. Q Q (d) Now

Mhsr(A) = sup r (1) ≥ sup ω · rˇ (1) = ω · sup rˇ (1) >0

>0

>0

(5A1Bb) = ω · Mhsr(A); as Mhsr(A) > 0, Mhsr(A) ≥ ω ω (5A1Bc). (e) For the general case, let a ∈ A+ be such that the principal ideal Aa is nowhere measurable. Then Mhsr(A) ≥ Mhsr(Aa ) ≥ ω ω . 539X Basic exercises (a) Let A be a Maharam algebra. Show that linkn (A) ≤ max(ω, τ (A)) for every n ≥ 2. (b) Show that, in the language of §522, p ≤ s ≤ min(non N , non M, d). (c) Let A be a Maharam algebra. (i) Show that if (α) cf[λ]≤ω ≤ λ+ for every cardinal λ ≤ τ (A), (β) λ is true for every uncountable cardinal λ ≤ τ (A) of countable cofinality, then FN(A) ≤ FN(PN), with equality unless A is finite. (Hint: 518D, 518I.) (ii) Show that if #(A) ≤ ω2 and FN(PN) = ω1 , then A is tightly ω1 -filtered. (Hint: 518M.) (d) Let A be a Boolean algebra, ν an exhaustive submeasure on A, and hai ii∈N a sequence in A such that inf i∈N νai > 0. Let F be a Ramsey ultrafilter on N. (i) Show that there is an I ∈ F such inf i,j∈I ν(ai ∩ aj ) > 0. (ii) Show that for every k ∈ N there is an I ∈ F such that inf{ν(inf i∈K ai ) : K ∈ [I]k } > 0. (iii) Show that there is an I ∈ F such that {ai : i ∈ I} is centered. (Hint: 538Hc.) (e) Let X be a set, Σ a σ-ideal of subsets of X, and I C PX a σ-ideal; suppose that Σ/Σ ∩ I is ccc. Let Y be a set, T a σ-algebra of subsets of Y , and ν : T → [0, ∞[ a Maharam submeasure; set J = N (ν), and let I n J be the b b ∩ (I n J ) is ccc. (Hint: 527L.) skew product as defined in §527. Show that (Σ⊗T)/(Σ ⊗T) 33 Later

editions only.

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539Y Further exercises (a) Let A be a Dedekind σ-complete Boolean algebra with a countable σ-generating set (331E), and µ a Maharam submeasure on A. Set I = {a : µa = 0}. Show that I 4T N . (Hint: apply 539Ia to an appropriate Radon submeasure on PN.) (b) Let X be a set, Σ a σ-algebra of subsets of X, and I a σ-ideal of subsets of X generated by Σ ∩ I; let ΣL be b L ) ∩ (I n N ) (527B), C = Σ⊗Σ b L /L. Show that C the algebra of Lebesgue measurable subsets of R. Write L = (Σ⊗Σ is ccc, or weakly (σ, ∞)-distributive, or Maharam, or measurable iff Σ/Σ ∩ I is. (c) Let A be a Boolean algebra with a strictly positive Maharam submeasure νˆ, and B a subalgebra of A which is dense for the associated metric (539Ac); set ν = νˆ B, so that ν is an exhaustive submeasure on B. For  > 0 let r : B → ω1 and rˆ : A → ω1 be the rank functions associated with ν and νˆ respectively. Show that rδ (a) ≤ rˆδ (a) ≤ r (a) whenever a ∈ B and 0 <  < δ. (d) Let A be an infinite Maharam algebra. Show that Mhsr(A) < τ (A)+ . (e) (J.Kupka) Let ν be a totally finite submeasure on a Boolean algebra A, and set νˇa = inf n∈N sup{mini≤n νai : a0 , . . . , an ⊆ a are disjoint}. for a ∈ A, as in the proof of 539T. Show that either νˇ ≥ 31 ν or there is a non-zero additive µ : A → [0, ∞[ such that µa ≤ νa for every a ∈ A. (Hint: 392D.) (f ) Show that the exhaustive submeasures constructed by Talagrand’s method, as described in §39434 , have exhaus2 tivity rank at most the ordinal power ω ω . 539Z Problems (a) Let µ be a non-zero Radon submeasure on a Hausdorff space X. Must there be a lifting for µ? that is, writing Σ for the domain of µ, must there be a Boolean homomorphism φ : Σ → Σ such that µ(E4φE) = 0 for every E ∈ Σ and φE = ∅ whenever µE = 0? 2

(b) Can a non-measurable Maharam algebra A have Maharam submeasure rank different from ω ω ? 539 Notes and comments During the growth of this treatise, the sections on Maharam submeasures have twice been transformed by new discoveries, and I hope that the work I have just presented will be similarly rapidly outdated. In the pages above I have tried in the first place to show how the cardinal functions of chapters 51 and 52 can be applied in this more general context. With minor refinements of technique, we can go a fair way. Because we know we have at least two non-trivial atomless Maharam algebras of countable type, we are led to a more detailed analysis, as in 539Ca and 539I. Equally instructive are the apparent limits to what the methods can achieve, which mostly point to remaining areas of obscurity. I say ‘remaining’; but what is most conspicuous about the present situation is our nearly total ignorance concerning the structure of non-measurable Maharam algebras. Talagrand’s construction, as described in §394, gives us a family of such algebras, but so far we can answer hardly any of the most elementary questions about them (539Yf, 394Z). ´ k 05 is that a Dedekind complete, ccc, weakly (σ, ∞)-distributive Boolean The message of Balcar Jech & Paza algebra is ‘nearly’ a Maharam algebra. Any further condition (e.g., the σ-finite chain condition, as in 393S34 ) is likely to render it a Maharam algebra; and with a little help from an extra axiom of set theory, it is already necessarily a Maharam algebra (539M). Similarly, much of the work of the last sixty years on submeasures suggests that exhaustive submeasures are ‘nearly’ uniformly exhaustive, and that an extra condition (e.g., sub- or super-modularity) is enough to tip the balance (413Yf34 ). At both boundaries, there are few examples to limit conjectures about further conditions on which such results might be based. Besides 539O and Talagrand’s examples, we have another important possibility of a not-quite-Maharam algebra in 555M below.

34 Later

editions only.