Topological Riesz Spaces and Measure Theory

Topological Riesz Spaces and Measure Theory Topological Riesz Spaces and Measure Theory D.H.FREMLIN Lecturer in Mathe...

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Topological Riesz Spaces and Measure Theory

Topological Riesz Spaces and Measure Theory D.H.FREMLIN Lecturer in Mathematics, University of Essex

WE

CAMBRIDGE AT THE UNIVERSITY PRESS 1974

CAMBRIDGE UNIVERSITY PRESS Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, Sao Paulo, Delhi Cambridge University Press The Edinburgh Building, Cambridge CB2 8RU, UK Published in the United States of America by Cambridge University Press, New York www. Cambridge. org Information on this title: www.cambridge.org/9780521201704 © Cambridge University Press 1974 This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 1974 This digitally printed version 2008 A catalogue recordfor this publication is available from the British Library Library of Congress Catalogue Card Number: 72—95410 ISBN 978-0-521-20170-4 hardback ISBN 978-0-521-09031-5 paperback

Contents Acknowledgements Preface Prerequisites 1 Riesz spaces 11 Partially ordered sets 12 Partially ordered linear spaces 13 Lattices 14 Riesz spaces 15 Archimedean Riesz spaces 16 Linear maps between Riesz spaces 17 Order-dense Riesz subspaces 18 The countable sup property IX Examples for Chapter 1 2 Topological Riesz spaces 21 Compatible topologies 22 Locally solid topologies 23 Fatou topologies 24 Lebesgue topologies 25 Complete metrizable topologies 26 L-spaces and M-spaces 2X Examples for Chapter 2 3 Dual spaces 31 The space E~ 32 The space E x 33 Perfect Riesz spaces 4 Riesz spaces on Boolean rings 41 Boolean rings 42 The space 5(91) 43 The space L°°(9l) 44 The space L # 45 Ring homomorphisms 4X Examples for Chapter 4

page vii ix

xiii 1 4 6 9 17 20 26 30 33

36 37 43 53 61 68 75 82 84 87 91 97 107 111 114 119

CONTENTS 5 Measure algebras 51 Measure rings 52 The space L\%) 53 Maharam algebras 54 Measure-preserving ring homomorphisms 5X Examples for Chapter 5

page 126 130 136 139 142

6 Measure spaces 61 Definitions and basic properties 62 Measurable functions; the space L° 63 Integration 64 Maharam measure spaces 65 Banach function spaces 6X Examples for Chapter 6

145 151 160 169 176 185

7 Representation of linear functionals 71 Sequentially smooth functionals 72 Smooth functionals: quasi-Radon measure spaces 73 Radon measures and Riesz' theorem 7X Examples for Chapter 7

193 202 210 216

8 Weak compactness 81 Weak compactness in E~ 82 Weak compactness in £ x 83 Weak compactness in L-spaces 8X Examples for Chapter 8

219 226 234 247

Appendix Al Linear topological spaces A2 Spaces of continuous functions

250 255

References

258

Index of special symbols

261

Index

263

VI

Acknowledgements The gestation of this book has extended over three appointments: as a lecturer at United College, The Chinese University of Hong Kong; as a Central Electricity Generating Board Junior Research Fellow at Churchill College, Cambridge; and as a lecturer at the University of Essex. I am grateful for the material support of all these institutions, and for the moral support of my colleagues there. Almost every mathematician I have spoken to in the last six years has influenced the book in some way; but I should like to give special thanks to Professor W. A. J. Luxemburg, who introduced me to its subject matter; to Dr N. N. Chan, who suggested that it should be written; and to Dr F. Smithies, who gently moulded it into publishable form. Finally I must mention Miss M. Mitchell, who typed the bulk of the MS, and the editors of the Cambridge University Press, for their skill and patience in transforming an idiosyncratic manuscript into print.

VII

Preface This book is addressed to functional analysts who would like to understand better the application of their subject to the older discipline of measure theory. The relationship of the two subjects has not always been easy. Measure theory has been the source of many examples for functional analysis; and these examples have been leading cases for some of the most important developments of the general theory. Such a stimulation is, of course, entirely welcome. But there have in addition been several cases in which special results in measure theory have been applied to prove general theorems in analysis. The ordinary functional analyst feels inadequately prepared for these applications, and is exasperated by the intrusion of a large body of knowledge in an unfamiliar style into his own concerns. My aim therefore is to identify those concepts in measure theory which have most affected functional analysis, and to integrate them into the latter subject in a way consistent with its own structure and habits of thought. The most powerful idea is undoubtedly that of Riesz space, or vector lattice. The principal Banach spaces which measure theory has contributed to functional analysis all have natural partial orderings which render them Riesz spaces. Many of their special properties can be related to the ways in which their orderings, their linear structures and their topologies are connected. For a clear understanding of the difference, for instance, between an L1 space and an L00 space, there is no substitute for an abstract analysis of their properties as ordered linear topological spaces. The other point at which measure theory has had an impact on functional analysis is in the representation of linear functionals as integrals, and the consequent deduction of surprising properties. I think that the analyst's instinctive rejection of such methods is a perfectly sound reaction. The difficulty is to provide techniques powerful enough to act as substitutes. Here again I believe that a thorough understanding of some quite simple topological Riesz spaces can see us through most of the difficulties. Now as soon as we begin to look at measure theory with a sceptical and abstract eye, a number of peculiarities strike us. The first is the ix

PREFACE relative unimportance of measure spaces themselves. They are used to construct spaces of equivalence classes of functions which then take on a life of their own; and quite different measure spaces can give rise to isomorphic function spaces. So it is natural to ask just what it is about a measure space that determines the associated function spaces. The answer is readily to hand: it is a construction called the measure algebra. Two measure spaces with isomorphic measure algebras will have isomorphic L 1 spaces, isomorphic L00 spaces, and so on; and to a large extent the converse is true. I think that the most convincing way of demonstrating these facts is to exhibit methods of constructing the function spaces (or, rather, isomorphic copies of them) directly from a measure algebra. These constructions are not particularly simple; in this book they take up Chapter 4 and the first half of Chapter 5. However, given some intuitive grasp of the nature of Riesz spaces, they are fairly straightforward, and provide invaluable insights. For instance, another curiosity of traditional measure theory is the unimportance of any notion of homomorphism between measure spaces. This distinguishes it from all comparably abstract branches of mathematics. I believe that this deficiency occurs because the natural and important homomorphisms of the theory are between measure algebras and not between measure spaces at all. In §§45 and 54 I discuss such homomorphisms and their effects on the constructions I have set up. The reader may be forgiven for wondering at this stage just how much he needs to know to cope with this book. A basic knowledge of functional analysis is essential. It would be possible to go a fair way with normed spaces alone; but this would shut off many of the most interesting ideas, and I have written on the assumption that I may call on the fundamental concepts of the theory of linear topological spaces. As for measure theory, in a formal sense I require none; I give every definition from that of measure space onwards. In an informal sense, I cannot pretend that this book is a genuine alternative to the traditional presentation of the elementary theory. §§61-3, while formally complete, are far too sketchy to be satisfactory as a first introduction. So if you are uncertain of your grounding in the subject, I refer you to the detailed advice in the Prerequisites section. I have preferred, in ordering the material of this book, to arrange results by the contexts in which they apply; thus propositions which refer to arbitrary Riesz spaces go into § 14 and those which refer only to Archimedean Riesz spaces go into § 15. Now this is not, of course,

PREFACE

a natural order for learning the subject. The beginner will certainly wish in the first instance to restrict his attention to Archimedean Riesz spaces, and leave any generalizations to one side for the moment. I think my decision is justified; it means that the hypotheses of successive theorems do not usually vary in arbitrary details, and are consequently much easier to recall precisely. But it makes a page-by-page progression inappropriate. To assist the reader to avoid spending too much time on material which will not be immediately essential, I have used asterisks to mark sections and propositions which can at first be passed over. They will be given references in the text when they come to be needed. Most of this book is taken up with fulfilling the objects I described at the beginning of this preface. But I think that topological Riesz spaces are inherently fascinating, and I have laid Chapter 2 out as a survey of their elementary properties; and in Chapter 8 I try to take a special subject up to the area of present research work. University of Essex January 1973

D. H. FREMLIN

XI

Prerequisites ZOBN'S LEMMA, TRANSFINITE INDUCTION, AND THE AXIOM OF CHOICE

The best introduction I know to this subject is in the appendix of KELLEY; HALMOS N.S.T. is also perfectly satisfactory. I shall apply the axiom of choice, and the principle of dependent choice for sequences, without comment, though when syntax allows I signal with such a word as * choose' or ' choice'. Zorn's lemma is used most often in the form If X is any collection of sets such that for any subset <W of X which is totally ordered by ordinary inclusion, we have \}®f = {*:3 YeW,teY}eX, then X has a maximal element. (This is nearly what KELLEY calls the 'maximal principle'.) When employing this form, I shall generally say merely 'by Zorn's lemma, X has a maximal element', and leave it to the reader to verify that the hypotheses on X are satisfied - supposing, of course, that the verification is straightforward. But if the lemma is to be applied in the more general form If X is a partially ordered set such that every totally ordered subset has an upper bound in X9 then X has a maximal element, then I shall give more details. FUNCTIONAL ANALYSIS I can recommend BOURBAKI V, ROBERTSON ROBERTSON, KELLEY & NAMIOKA, or SCHAEFER T.V.S. ; any of these

& is

adequate for the principal results. For details see §A1. AS I explained in the Preface, I make no formal requirements in this field. But a knowledge of the elementary theory up to and including the Radon-Nikodym theorem would certainly be helpful. The first point is that although abstract measure theory is not particularly difficult in its early stages, the only easy examples of measure spaces are essentially trivial. There is no substitute for Lebesgue measure on the real line as a leading example. So at some

MEASURE THEORY

xiii

PREREQUISITES stage the reader should make himself familiar with a proof that it exists. Secondly, several of the most important ideas in this book appeared first in measure theory. Pre-eminent among these are the arguments leading up to the Radon-Nikodym theorem. I give this theorem in § 63 with a proof which refers to almost every preceding chapter. In fact my proof is very closely related to one of the traditional ones, but its component ideas have been scattered. I think that it is illuminating to see how they can be joined more closely together in a direct proof. I shall give a concordance in the proper place. Meanwhile, I do not want the reader to feel that he must take a course in measure theory before proceeding further. Perhaps the books suggested below should be regarded as a starred section; optional on first reading, but likely to be useful later. R. G. Bartle, The Elements of Integration (Wiley, 1966) H. L. Royden, Real Analysis (Macmillan, 1963) H. Widom, Lectures on Measure and Integration (van Nostrand Reinhold, 1969) J. H. Williamson, Lebesgue Integration (Holt, Rinehart & Winston, 1962).

XIV

1. Riesz spaces The prerequisites for this chapter are few; an acquaintance with linear spaces is the principal requirement. A Riesz space is simply a linear space over the field R of real numbers which has a special kind of partial ordering, and all we need to know about partial orderings will be covered in §§11 and 13. But the theory of Riesz spaces is already rich, and some of the work in §§16 and 17 is far from trivial. It does, however, have to be taken seriously. These are the basic results which will enable us to handle Riesz spaces with assurance and facility.

11 Partially ordered sets This section is little more than a list of definitions. As such I suggest that it should be read carefully once, together with the associated examples; but that there is no need to consciously memorize anything. You will find the index perfectly reliable. Actually the concepts here have applications to every branch of mathematics, and they will mostly be familiar in everything but name. I think it is amusing and instructive to seek such applications out and consciously appreciate them. Now to work: 11A Definition A partially ordered set is a pair (A, ^ ) where A is a set and ^ is a binary relation on A such that: (i) a ^ a for every aeA; (ii) if a and b belong to A and a a for a ^ b and a #= 6. For examples see 1XA and IXC. 11B Suprema and infima Let (A, <) be a partially ordered set, and B a subset of A. There may, or may not, be an ae A with the property for every ceA, FTR

I

a^c

iff b^c

for every

beB, I

11]

RIESZ SPACES

that is, a ^ c iff c is an upper bound for J3. Clearly there can be at most one such a [HA(ii)]. If a has this property, we say that 6B has a least upper bound, which is a\ or 'supi? exists, and supl? = a\ or simply 'supi? = a\ Similarly, inf B, if it exists, is that one element of A such that for every ceA, c ^ inf B

iff c ^ b for every beB.

We observe that infJ5 = sup {a: a O V beB}, supJ5 = inf{a:6 < a V beB}, ix&A = sup 0 , sup A = inf 0 , all these inequalities being true in the sense that if one side is defined, so is the other, and they are then equal. For a warning of the dangers in this notation, see 1XB. 11C Directed sets Let (A, ^ ) be a partially ordered set. A set 5 9 i i s directed upwards, B f, if for every pair a, b of elements of B there is a c eB such that a < c and b ^ c. I shall write B f a to mean that B f and that sup B = a. Observe, for instance, that {a} f a for every aeA. Associated with a directed set is a filter. If B is a non-empty set which is directed upwards, I shall write 3F{B\) for the filter on A with base ^ ,> , ^ rr D {{a:aei?, a ^ b}:beB}. !F(B \) is sometimes called the filter of sections of the directed setS. B ^('B is directed downwards'), B ]> a and ^(2? •[) are defined in the same way, but upside down. 11D Functions Let (A, ^ ) and (B, B is increasing if a ^ 6 =>/a =^/6. When / is a sequence (flOweN say> I shall write (# n ) we nt to mean that it is an increasing sequence, that is, that an =^ an+\ for every %GN. Similarly, n<=N i s a decreasing sequence, neH \, if a ^ < an for every neN. By analogy with 11C, I shall write (fln}nev^a to mean that (an)neTx\ a n ( l that sup neN a n = a. (an}n€]ii^a is defined similarly. An increasing function/: JL -> J5 is order-continuous if sup/[O] = fa

whenever G\a and C is not empty

inf/[O] = fa

whenever C\a and C is not empty.

PARTIALLY ORDERED SETS

[11

(Of course, if/is increasing, then/[C] is directed whenever G is.) Note that it is only directed sets which must have their sups and infs preserved in this way. *There are two more concepts which will be useful to us in special circumstances. An increasing function/: A -> B is order-continuous on the left if it satisfies the first half of the condition for ordercontinuity, that is, if /[C]f/a whenever G\a and ( 7 + 0 . A n d / is

sequentially order-continuous if

*11E Order-closed sets j ? c i 5 l shall write

whenever

neM ta

whenever

<6 n > neH |6.

If (A, ^ ) is a partially ordered set and

B, C * 0 , C\a in A), B, 0 + 0 , C\a in A). Then B c JB and B c ®B. B is order-closed if JB = B = ^ 5 . *11F Order-bounded sets Let(^4, ^ ) be a partially ordered set. A set B c J. is order-bounded if it has both upper and lower bounds in A, i.e. if B c: [&? c] = {a: 6 ^ a ^ c} for some b and c in A. *11G Products Let((^4t, < t )X e I be an indexed family of partially ordered sets. Let A be the cartesian product ]JLeIAL, and define a relation ^ on A by a ^ 6iff a(t) ^ &(*) for every tel. Then (^4, < ) is a partially ordered set, the product of the family {(AL, ^ t))LeI. If, for each LEI, nL: A -> AL is the canonical map, then nL is increasing and order-continuous. *11H Exercises (a) Let (A, < ) be a partially ordered set. Then there is a topology on A for which the closed sets are precisely the order-closed sets of A, [See IXC] (b) Let (A, <) and (B,^,) be partially ordered sets. Then an increasing function / : A -> B is order-continuous iff/~1[C] is orderclosed in A for every order-closed C c B.

11]

RIESZ SPACES

Notes and comments We shall see a greatest lower bound or a least upper bound on every other page of this book; so it will be as well to grasp firmly the formal definitions in 11B. I generally take these formalities seriously; I want to know exactly what I mean by such expressions as inf 0. I cannot emphasize too strongly the fact that the least upper bound of a set depends on the partially ordered set in which it lies. There would be formal advantages in writing

to mean 'the least upper bound of B taken in the partially ordered set (A, < ) ' . Of course this would be absurdly cumbersome; but the abbreviation' sup B' must always be recognized as such. In 1XBI give a simple example of the danger. There's a more elaborate one in 4XF, and the distinctions become very important in Chapter 7. The same warning, of course, applies to the notations J and 3) in HE. There is a perfect symmetry in the notion of partially ordered set; if < is a partial ordering, so is its inverse ^ . Consequently all the associated definitions are doubled, as in 11B and 11C. I want to call attention to the concept of the filter of sections ^(B f), defined in 11C. I think that this filter is almost the most important thing associated with a directed set. You may have seen it already in the correspondence between net-convergence and filter-convergence in topological spaces [KELLEY 2L]. Note the technical point that ^{B\) is a filter on the whole space A, not on 2?. For another look at the material of this section, see BOURBAKI I, chapter in, §1.

12 Partially ordered linear spaces In this short section I discuss the simplest way in which a linear space structure and a partial ordering can be related. Although we shall very rarely have any reason to consider partially ordered linear spaces which are not Riesz spaces [§ 14], I think that the results here are clarified by being placed in their natural context. Here, and everywhere in this book, all linear spaces have the real numbers for their underlying field. There do exist applications in which it is more convenient to have the complex numbers; but I think that these are best approached by way of the real case. Otherfieldsare so far merely curiosities from the standpoint of this theory.

PARTIALLY ORDERED LINEAR SPACES

[12

12A Definition A partially ordered linear space is a quadruple ( £ , + , . , < ) where ( £ , + , - ) is a linear space over the field R of real numbers and ^ is a partial ordering on E such that (i) if x ^ y, thenx + z ^ y + z for every zeE; (ii) If x ^ 0 in E, then ax ^ 0 whenever a ^ 0 in R. 12B Positive cones From 12A(i), we see that x ^ yoO ^y-x. So ^ is determined entirely by £+ = {x: xe E, x ^ 0}, the positive cone of E. Given a linear space E over R and a set P ^ E, there is a partial ordering < on E such that (E, ^ ) is a partially ordered linear space and P = E+iS P n ( - P ) = {0} [for 11A (i) and (ii)], P + P^P aPc?

[for llA(iii)], V a ^ 0 [for 12A(ii)].

In particular, P = {0} will do [cf. 1XA]. 12C Lemma Let £ be a partially ordered linear space, xe E, A, B^ E. Then (a) sup (x + A) = x + sup A if either side exists. (b) sup ( — A) = — inf^4 if either side exists. (c) sup (A+B) = sup A + sup B if the right-hand side exists. (d) If a ^ 0, sup (ocA) = asup^l if the right-hand side exists. Proof of (c) Apply (a) twice, as follows: sup A + sup B = sup (A -}- sup B)

= sup {x + sup B: x eA} , yeB} = sup(,4+J3). Notes and comments The correspondence between an ordering and a positive cone [12B] is extremely important; it is one of the easiest ways of defining partial orderings on linear spaces. In 12C I prove only (c), because the other parts are direct consequences of the definitions. In a partially ordered linear space E, the map 2/h->o; + 2 / : E - > E i s a n order-automorphism for every x e £, and

12]

RIESZ SPACES

therefore must preserve suprema and infima. The same applies to the map y\-*ay: E->E for every oc > 0. On the other hand, the map yi-^ — y:E->E is order-reversing, that is, x ^ yo — y^—x, so it exchanges suprema for infima. Although I shall have no space to discuss them, many of the ideas of this book, set out for Riesz spaces, have extensions to partially ordered linear spaces. Compatible topologies [§21] are an obvious example; Lebesgue topologies [§24] are another. Some of these developments may be found in PERESSINI. 13 Lattices Once again we must have a section consisting mostly of definitions. I maintain a careful separation between general remarks on lattices and those concerning linear spaces because I wish to apply the former to Boolean lattices in § 41. Although it is possible to regard a lattice as a set with two binary algebraic operations, governed by certain identities, I prefer to take it as a special kind of partially ordered set. This makes it easier to think of the suprema and infima of infinite sets, which is something we must do continually. The important sections below are A-E, though G should be examined because it attaches a rather unusual meaning to the word 'disjoint'. 13A Definitions A lattice is a partially ordered set (A, <) such that sup {a, 6} and inf {a, b} exist for all elements a and b of A. It follows at once (by induction on the number of elements in 2?) that sup B and inf B exist for every non-empty finite set B c A. In a lattice A, we write ayb and aAb for sup {a, b} and inf {a, b} respectively. Thus v and A may be thought of as binary operations on A, the lattice operations. Since a < 6 o avb = b o aAb = a, the lattice can be defined in terms of these operations. But apart from a few special instances [e.g. 22F below], I think it is easier to regard the partial ordering as fundamental. From this point of view, for instance, the identity. (a v b) v c = a v (6 V c) is trivial, as both sides are equal to sup {a, 6, c}. 6

LATTICES

[13

13B Dedekind completeness (a) A lattice A is Dedekind complete, or order-complete, if every non-empty B £ A which has an upper bound in A has a least upper bound. (b) A lattice A is Dedekind a-complete, or sequentially Dedekind complete, if every non-empty countable B c A which has an upper bound or a lower bound has a least upper bound or greatest lower bound respectively. 13C Notes These definitions are extremely important. Observe: (a) If A is a lattice, then A is Dedekind complete iff every non-empty subset of A which has a lower bound has a greatest lower bound. P Suppose that A is Dedekind complete and that B ^ A has a lower bound and is not empty. Let C = {a:aeA, a is a lower bound for B}. Then C is a non-empty subset of A with an upper bound, so sup C exists. But sup C = miB. The reverse implication is proved in exactly the same way. Q. Consequently a Dedekind complete lattice is indeed Dedekind cr-complete. The definition of Dedekind cr-completeness must be given in terms of both suprema and infima because there is no way to get a countable C from a countable B. (b) Let A be any lattice, and B c A. Let Bx = {sup C: C <= B, C is finite and not empty}. Then B± is closed under the lattice operation v and has the same upper bounds as B. So 13Ba could be rewritten as: 'a lattice A is Dedekind complete if every non-empty set B <= A, which is directed upwards and has an upper bound, has a least upper bound \ This will often be easier to prove [e.g. 16Db]. 13D Distributive lattices

A lattice A is distributive if

(a v 6) A c = (a A c) v (6 A c),

(a A 6) V c = (a v c) A (6 V c)

for all a, 6, ceA; that is, if each of V and A is distributive over the other. All the lattices considered in this book are distributive, and I shall make no attempt to discuss any other kind, BIRKHOFF is still an excellent introduction to general lattice theory.

13]

RIESZ SPACES

Exercise Show that each of the two halves of the condition above implies the other, so that either would be sufficient for the definition. 13E Lattice homomorphisms

If A and B are lattices, a function

/ : A -> B is a lattice homomorphism if f(a vb)=fa v/6, f(a A b) = fa A/6 for all a, beA. A lattice homomorphism must be increasing. Observe that an order-continuous lattice homomorphism [11D] is precisely a function which preserves all suprema and infima of non-empty sets; to see this we have to use the trick of 13Cb to transform an arbitrary supremum or infimum into the supremum or infimum of a directed set, so as to apply the hypothesis of order-continuity. 13F Sublattices Let A be a lattice. A sublattice of A is a subset of A closed under v and A. *A a-sublattice is a sublattice B of A such that if C ^ B is countable and not empty, and inf C or sup C exists in A, then it belongs to B. 13G Disjoint sets and functions Let A be a lattice with a least member a0.1 shall call a set B £ A disjoint (in A) if b A c = a0 whenever b and c are distinct members of B. (Observe that possibly aQ GB.) If / is any set, a function / : I-+A is disjoint if ft A//C = a0 whenever i and K are distinct members of / . *I shall often use the following phrase: 'C is disjoint in £+', where £ is a Riesz space and E+ is its positive cone. Now E+ is a sublattice of the lattice E, and has a least member, viz. 0; so what I mean is that xAy = 0 whenever x and y are distinct members of G. *13H Products Let<(^lt, ^ t)>tej be a family of partially ordered sets, and suppose that none of the AL is empty. Let < be the product partial ordering defined on A = YlieIAt [11G]. Then (a) (A, <) is a lattice iff every (At, < t) is. (6) (A, <) is distributive, or Dedekind complete, or Dedekind cr-complete, iff every (AL, < t) is. (c) The canonical maps from A to each AL are always lattice homomorphisms. Notes and comments As I have rigorously expunged from this section everything which will not be used later, it is not an adequate introduction to lattice theory. The only two kinds of lattice which we 8

LATTICES

[13

shall be thinking about, Riesz spaces and Boolean lattices, share manyspecial properties; to begin with, they are distributive, which makes them rather uninteresting as lattices. The distributive property is so natural that one is liable to use it unconsciously; which is dangerous if there are any nondistributive lattices about. But I think that for now we can accept these simplifications cheerfully. Since most of the objects of discussion of this book are lattices, I shall not trouble with many examples at this point. The obvious ones to glance at, before continuing to the next section, are 1XB, IXC and 4XA.

14 Riesz spaces As soon as we join §§12 and 13 together, and consider partially ordered linear spaces which are lattices, the character of our study changes completely. Riesz spaces have a remarkable wealth of properties. Some of the most elementary, identities relating the lattice and linear space operations, are in 14B. These are so numerous that one naturally seeks an alternative to memorization; and such an alternative exists in the 'R-test', which runs 'if an identity is true in R, it is true in all Riesz spaces'. Actually, there is a perfectly valid metatheorem along these lines. But the metatheorem is a good deal more complex then the work it eliminates; besides, direct proofs are more illuminating of associated structures such as lattice-ordered groups. So I give the identities most commonly required in an order which makes each an easy consequence of the preceding ones. The really surprising thing about Riesz spaces is the strong distributive law which they obey [14D]. After this, the rest of the section is fairly straightforward, with the expected relationships between products, subspaces, quotients and homomorphisms. 14A Definitions A Riesz space, or vector lattice, is a partially ordered linear space ( £ , + , . , < ) such that (£, ^ ) is a lattice. If £ is a Riesz space, we write ar- = ( - # ) v O ,

\x\= xv

{-x)

for any xeE. Exercise Let £ be a partially ordered linear space such that sup {x, 0} exists for every xe £. Then £ is a Riesz space. [Use 12Ca and 12Cb.]

14]

RIESZ SPACES

14B Elementary identities Let E be a Riesz space, x, y and z members of E, and a and /? real numbers. Then (a) (x v y) + z — (x + z)v (y + z) [special case of 12Ca]. (b) (x A y) + z = (x + z) A (y + z) [similarly]. (c) — (x A y) = (— x) v (— y) [special case of 12Cb]. (d) If a ^ 0, ax v ay — a(x v y) [special case of 12Cd]. (e) If a ^ 0, ax A ay = a(x A y) [similarly]. (f) 2(xvy) = 2xy2y = x + y+\x-y\ [definition of | | and (a) above]. (g) 2(XA y) = x + y- \x — y\ [similarly, using (b) and (c)]. (h) x + y = xvy + xAy [adding (f) and (g)]. (i) x = x+ — x~ [apply (a) to x + x~]. (j) \x\ = x+ + x~ [apply (d) and (a) to x + 2x~], (k) x+v x~ = |x| [from the definitions, since |#| ^ 0 by (j)]. (1) * + A Z - = 0 [by(j), (k)and(h)]. (m) \ax\ = |a| \x\ [consider a ^ 0 and a ^ 0 separately]. (n) |# + 2/| ^ |a;| + |y| [since x + y < x+ \y\ ^ |^| + |2/|; and simi-

larly -x-y ^ \x\ + \y\]. 14C Definitions Let £ be a Riesz space. A set A c E is solid if ye A whenever there is an x e A such that \y\ ^ \x\. The intersection of any set of solid sets is solid. If A is any subset of E, the set {y: 3xeA, \y\ ^ \x\} is solid; it is the smallest solid set including A, and is called the solid hull of A. 14D The distributive law Let £ be a Riesz space and A a subset of E such that sup A exists. Then, for any x e E, sup {x A y: y e A} exists and is equal to x A sup A. Proof Plainly x A sup A is an upper bound for {x A y :yeA}. Conversely, suppose that z ^ x Ay for every ye A. Then

z ^ (x-y)A0 + y ^ (x — sup A) A 0 + 2/, for every So Similarly 10

z ^ (#-sup^4) AO + sup J. = x v inf ^1 = inf {x yy\yeA}iH

the left-hand side exists.

RIESZ SPACES

[14

Corollary £ is a distributive lattice. 14E Linear maps We shall be concerned with several classes of operators between Riesz spaces; the important ones are: (a) Increasing linear maps. A linear map T is increasing iff Tx ^ 0 whenever x ^ 0. (For this reason, increasing linear maps are often called positive.) (b) Riesz homomorphisms. A Riesz homomorphism is a linear map which is also a lattice homomorphism. A linear map T is a Riesz homomorphism iff

(Tx)+ = T(x+)

iff

\Tx\ = T\x\

iff

Tx A Ty = 0 whenever x A y = 0.

for every x; for every #;

(Use 14Bf, g and j . For the last condition, observe that in this case (Tx)+ = (T(x+) - T(xr)) v 0 = T(x+)V

T(xr)-T(x-)

= T{x+) + T(xr) - T(x+) A T(xr) - T(x~) = T(x+), using 14Bi, a, h and 1.) (c) Order-continuous increasing linear maps. A linear map T is increasing and order-continuous iff A\Q and A * 0 = iff

0 c A c £+ and ^4 f «

[Use 12Ca and 12Cb. Note that in the second condition we consider only sets A which are bounded below by 0. These are enough, for any set which is directed upwards can be truncated by one of its own members; I mean that if A f and A 4= 0 , we can consider A' = {XIXEA,

x ^ y]

for some y e A; and now A' is bounded below and has the same upper bounds as A, so the action of T on A' will tell us all we need to know. And by a simple translation we can move y to 0.] *(d) Sequentially order-continuous increasing linear maps. 11

14]

RIESZ SPACES

(e) Order-continuous (or sequentially order-continuous) Riesz homomorphisms. (£) Maps expressible as the difference of two increasing linear maps [see 16JS]. (g) Maps expressible as the difference of two order-continuous increasing linear maps [see 16G]. Much of the richness of the more advanced theory of Riesz spaces derives from this multiplicity; each kind of operator is associated with a different class of results. The following chart shows the relationships between these seven classes:

(f)

where the arrow (e) -» (b) signifies that every operator of type (e) is also of type (6). 14F Subspaces Let £ be a Riesz space. A Riesz subspace of £ is a linear subspace which is also a sublattice [13F]. A solid linear subspace is always a Riesz subspace. A Riesz subspace F of £ is order-dense if, for every xe E + , x = sup{2/:0 ^ y ^ x, yeF}. F is super-order-dense if for every xeE+ there is a sequence (yn}nevt in F+ such that (yn)nejx \x in E. F is locally order-dense if it is order-dense in its solid hull (which must be a solid linear subspace of E); i.e. if 0 ^ x < XOGF => x = sxip{y:yeF, 0 ^ y < x}. A solid linear subspace is always locally order-dense. A band or normal subspace of £ is a solid linear subspace F such that whenever i g f and sup A exists in E, then supAeF; i.e. an order-closed solid linear subspace [HE]. The intersection of any set of Riesz subspaces is a Riesz subspace; the intersection of any set of solid linear subspaces is a solid linear subspace; the intersection of any set of bands is a band. The range of a Riesz homomorphism is a Riesz subspace; the kernel 12

RIESZ SPACES

[14

of a Riesz homomorphism is a solid linear subspace; the kernel of an order-continuous Riesz homomorphism is a band. 14G Quotient spaces: proposition Let £ be a Riesz space and F a solid linear subspace of £. Then the linear space quotient E/F can be given a partial ordering such that it is a Riesz space and the canonical map 0: £ -> E/F is a Riesz homomorphism. Proof Let P = {<j>x: x e £+} s £/F. (a) Given X G £ , ^ 6 P O x r e F . P If areF, then Conversely, if <j>x e P, there is a y e £ + such that 02/ = 0#, i.e. y — xeF. Now — a < y ~~ x> s o and #- e F as .F is solid. Q (b) Now P satisfies the conditions of 12B. P If <j>x e P n ( - P), then x~eF by (a) above. But also 0( — #)eP, so #+ = ( — #)~eF, and a; = a ; + - r G F . Thus P n ( - P ) = {0}. But the other conditions are trivially satisfied. Q (c) So E/F can be given a partial ordering under which it is a partially ordered linear space and P = (£/F)+. Now if a; e £, sup {0#, 0} exists in E/F and is equal to x,

and 0#+ ^ 0. On the other hand, if x and 0y ^ ^> then y~eF and (y-ai)-eF, by (a) above. So (y-»+)-

as a solid linear subspace is always a sublattice. So 0y > 0# + . As y is arbitrary, 0(#+) = sup {<j>x, 0}. Q (d) By the exercise in 14A, E/F is a Riesz space, and 0 is a Riesz homomorphism by 14Eb. Remarks Thus not only is the kernel of every Riesz homorphism a solid linear subspace, but every solid linear subspace is the kernel of some Riesz homomorphism. For this reason some authors call solid linear subspaces ' ideals'. Although it will not be used in this book, the result of 14Lb is very important in understanding the nature and significance of bands. 13

14]

RIESZ SPACES

*14H Products Let {Et)ieI be an indexed family of Riesz spaces. On YlieIEt there are natural linear space and lattice structures [13H] which render it a Riesz space. The canonical projections to each EL are all order-continuous Riesz homomorphisms. *14I Order units Two concepts we shall want to use later are the following. Let £ be a Riesz space. An order unit of E (sometimes called a strong order unit) is an e e E+ such that for every xeE there is an ne N such that x ^ ne. A weak order unit of E is an ee E+ such that x = supneN# A ne V xe E. *14J At various points of the book we shall want to use the following results. Lemma Let £ be a Riesz space, (a) If x, y and z belong to £ + , (b) If (xi)i J<m are finite sequences in £+ such that %i t h e n t h e r e i s a double sequence j<m in £+ such that xi = Tik<mzik a n d Vj = 2fc
(a) Set w = z A {X + y) - z A X. Then by 14Bh ^ y.

But also w ^ z, so w < t/ A 2, as required. (b) (c) We prove these by induction on m and n. Let us begin with (c), with m = 2, so that yo + y1 = Hin+l, when we have Vo + Vl = I set zw0 = # n A y0, znl = xn-zn0. Now by (a) above, so z nl < xn Ayx ^ yx. So if we set w0 = yo-znO and ^ = y1 — znl, we shall have WQ + W! = ^ii
RIESZ SPACES

[14

hypothesis we can choose the zi0 and ziv for i < n, such that zi0 + zix = xi for every i < n and 2i
T ii
Z

i0

=

V i
+

V>

set ^ = Zfa — Zft for each i < n. Then ^i
by 14Bm and n. Now by (b) above there is a sequence (jj^)i
(a) We know that x+ AX~ = x+ Ay~ = y+ Ax~ = y+ Ay~ = 0.

So by 14Ja uAv = 0, where u = x+ + y+ and v = a;~ + 2/~. Now = (u — v)+ = (i£ — v) v 0 = w v v - f

15

14]

RIESZ SPACES

Similarly (x + y)~ = v. So \x + y\ = u + v = \x\ + \y\. (b) axAfy^ (oc+j5)xA(a+j3)y= (a+j3)(xAy) = 0. (c) For n = 2, |a o ^ o | A [ o ^ l = |a o | |a?0| A \ax\ \x±\ = 0 by (b). So il = |ao^ol + l a i^i| by (a), and also

\aoxo\ + l ^ ^ l = \aoxo\ v j a ^ l + \ocoxQ\ A K ^ The inductive step to general n is now easy, using 14Ja to see that \<*nxn\ ^i
= °-

14L Exercises (a) Let E be a Riesz space and F a linear subspace. Then these are equivalent: (i) x+eF for every xeF (ii) |x\ eFfor every xeF (iii) Fis a Riesz subspace. [Use 14Bf, gand j.] *(b) Let £ be a Riesz space and F a solid linear subspace. Let <j>: E -> EjF be the canonical Riesz homomorphism [14G]. Then (j) is order-continuous iff F is a band, and
ARCHIMEDEAN RIESZ SPACES

[15

only one of much interest at this stage is the quotient space construction [14G]. Although such a construction is an obvious thing to look for, it is far from clear that it will appear so simply.

15 Archimedean Riesz spaces All the most familiar Riesz spaces, and all those with which we shall be concerned in this book, possess the 'Archimedean' property. A Riesz space is Archimedean if, as in the real numbers, {nxineN} is unbounded above for any strictly positive x. This property is so common that one's intuitive idea of Riesz spaces is liable inadvertently to include it. In our present inquiry such an unconscious assumption will do little harm, as all spaces considered are Archimedean. However, for the benefit of readers who may wish later to examine nonArchimedean spaces, I have segregated here the elementary special consequences of the Archimedean property; and elsewhere in the book I try to include the hypothesis 'Archimedean' only when necessary. The short paragraphs up to 15E are technical, but extremely useful. 15F gives a characterization of bands in Archimedean Riesz spaces, which is more important for a general understanding than for the special needs of this book. 15G sorts out one-dimensional Riesz spaces, and will be needed only in Chapter 8. 15A Definition belonging to E,

A Riesz space E is Archimedean if, for x and y x ^ ny V w e N = > 2 / < 0 .

Clearly, a Riesz subspace of an Archimedean Riesz space is Archimedean. 15B Proposition A Dedekind cr-complete Riesz space is Archimedean. Proof

If x ^ ny for every n e N, set z = sup {ny: n e N}. Then

y + z = &up{(n+l)y:neN} < z [using 12Ca], so y < 0. 15C Lemma Let E be an Archimedean Riesz space. Let A be a non-empty subset of E and suppose that the set B of upper bounds of A is also non-empty. Then inf (B — A) = 0. 17

15]

RIESZ SPACES

Proof z is a lower bound for B — AiRz^y — x for every ye B and xeA, i.e. iff y — zeB for every yeB. Consequently, if yQeB and xoeA, yo — nz ^ x0 for every neN. Hence z < 0. 15D Lemma Let £ be an Archimedean Riesz space. Then for any xeE+ the map at->ax: R-> E is order-continuous, so that if ^4 f a inR, {/fa:/? Proof

The point is that if y ^fixfor every /?eA, (a — n^x

^ y V w ^ 1,

i.e. n(ax — y) ^ a; for every w ^ 1; so a# < ?/. 15E Lemma Let £ be an Archimedean Riesz space and F a Riesz subspace of E such that for every non-zero xeE+ there is some yeF such that 0 < y ^ x. Then F is order-dense. Proof Let xeE+, and let ^4 = {y.yeF, 0 ^ i/ < x}- ? Suppose, if possible, that # is not the least upper bound of A, that is, that there is an upper bound xx of A such that xx % x. Then 0 < x — x A #1? so there is a i/0GF such that 0 < y0 ^ X — XAXV But now

for every ^/G^4. Inducing on n, we see that nyQeA for every which is impossible. X *15F Bands in Archimedean Riesz spaces space, and i g E , Let us write Ad = {X:XGE,

\X\ A \y\ = 0

Let £ be a Riesz

V ye A).

Then (a) Ad is a band in E (b) if £ is Archimedean, and F is a band in £, then F = (Fd)d. Proof (a) By 14Bn and 14Ja, Ad is closed under addition; by 14Bm and 14Kb it is a linear subspace; clearly it is solid; now by 14D it is a band. (b) Of course Fdd ^ F. On the other hand, suppose that xeE\F. As F is solid, \x\ ^ F ; so \x\ cannot be the least upper bound of B = {y:yeF,O^y^ 18

\x\}9

ARCHIMEDEAN RIESZ SPACES

[15

that is, there is an upper bound xx of B in E such that x1^ \x\. Let z = \x\ — \x\ AX± > 0. Now for any yeB, y + z < \x\ AX1-\-Z = \x\. But

for any y e F and n e N (rc+l)(|y| A2) 0, dd

so x<£F . This shows that F* 1 c F; so F** = F, as required. *15G Lemma Let £ be a Riesz space. Then the following are equivalent: (i) E = {0} or E ~ R; (ii) E is Archimedean and totally ordered; (iii) E is Archimedean and, whenever x, yeE+\{0}, x A y > 0; (iv) E has no non-trivial bands; (v) E has no non-trivial solid linear subspaces. Proof (a) (i) =>• (v) => (iv) are elementary. (b) (iv) => (iii) Suppose that E satisfies (iv). (a) Let x > 0 in E. Then it is easy to see that {y: \y\^ocx

V a > 0}

is a band in £, not containing x; so it is {0}. It follows that E is Archimedean. (p) If x,yeE and xAy = 0, then F = {Z:\Z\AX

= 0}

is a band in E [15Fa]. If F = {0}, then y = 0;ifF=E, then a; = 0. (c) (iii)=>(ii) Suppose that E satisfies (iii). If x, yeE, then (x — y)+ A (y — x)+ = 0 [14B1], so either (x — y)+ = 0 or (y — x)+ = Oi.e. either x ^ y or y ^ x. (d) (ii) => (i) Suppose that E satisfies (ii), and is not {0}. Let x0 G E+\{0}. Given y e £ + , let a 0 = sup {a: a e R, a# 0 < £/} < oo as £ is Archimedean. Thenocoxo ^ y [15D]. Also, a > a 0 => a# 0 4 2/ ^ a:r o > 2/? so aoa:o ^ y [15D again]. Thus £ = £+ - £+ [14Bi] = {ax0: a e R}, and the map a h-> aa;0 is the required isomorphism between R and £. 19

15]

RIESZ SPACES

*15H Exercises (a) A product of Archimedean Riesz spaces is Archimedean. (b) If £ is a finite-dimensional Archimedean Riesz space, E £ Kn [1XD] for some weN. Notes and comments As we shall not have any need to consider non-Archimedean spaces, I give only one example, the lexicographic ordering of R2 [1XF]. I think that what is important about 15C-15E is the fact that they fail for spaces of this type (indeed, with a little thought one can see that each of these results is a necessary and sufficient condition for the Archimedean property). This failure gives perhaps a salutary shock. Because R x is Archimedean [1XD], all its Riesz subspaces are Archimedean, that is, all true function spaces are Archimedean. The spaces of linear maps in the next section, which will loom large in this book, are plainly Archimedean because they are defined from Archimedean spaces. However, many of the Riesz spaces of measure theory are defined as quotient spaces, and quotient spaces do not in general have to be Archimedean (see 1XE, and recall the theorem of Amemiya quoted at the end of § 14, which can be used to show that every Riesz space can be represented as a quotient of an Archimedean Riesz space). I think that a reason why the important examples should be Archimedean is to be found in 22Eb; they are defined so as to be normed spaces, and only Archimedean spaces can carry norms of the right type. It is the case that every Archimedean Riesz space is isomorphic to a Riesz subspace of a Dedekind complete Riesz space; by 15B, this is another necessary and sufficient condition. From this stems Ogasawara's representation of an arbitrary Archimedean Riesz space as a space of continuous functions on a suitable topological space, each function being permitted to take the values ± oo on a nowhere dense closed set. I have no space to discuss this further; the results are to be found in VULIKH, theorems IV. 11.1 and V. 7.1, and LUXEMBURG & ZAANEN R.S., theorem 50.8.

16 Linear maps between Riesz spaces In 14E we glanced briefly at a number of classes of operators between Riesz spaces. The natural ones were those of (a)-(e), for they preserve aspects of the order and linear structures of the spaces. Thus Riesz homomorphisms [14Eb], which preserve both the linear and the lattice 20

LINEAR MAPS BETWEEN RIESZ SPACES

[16

structures, are associated with an effective notion of quotient space [14G]. However, the set of all Riesz homomorphisms between two Riesz spaces does not ordinarily have a simple structure of its own. (Consider, for example, the set of Riesz homomorphisms from R2 to R, which is not closed under ordinary addition.) Similarly, the set of order-continuous increasing linear maps, though closed under addition [see 16H below], is not closed under ordinary multiplication by a negative scalar. If therefore we seek a set of linear maps which will be even a linear space, we must go outside these classes. The set of increasing linear maps, for instance, is closed under addition and multiplication by nonnegative scalars; consequently, the set of functions expressible as the difference of two increasing linear maps will be a linear space. Now it happens that if the range space is Dedekind complete, this space of linear maps is a Riesz space. Since much the most important range space is of course R, this strong condition is acceptable. The first six paragraphs of this section are accordingly devoted to proving this result, and providing some tools for handling the Riesz space we have constructed; it turns out itself to be Dedekind complete. This space which I shall call L~(E; F), is not actually the most exciting space of linear maps from E to F. More interesting is LX(E; F), the space of differences of order-continuous increasing linear maps. This also is a Dedekind complete Riesz space, being in fact a band in L~(E; F) [16H]; and L~(E; F) will be important to us principally because of this relationship, which is a valuable source of results about L><(£; F). In the applications in this book, the range space F will usually, though not always, be the real numbers; but the proofs are made very little different by the generalization to arbitrary Dedekind complete Riesz spaces. 16A Lemma Let E and F be Riesz spaces, and T: E+ -> F a function. Then T extends to a linear map from E to F iff

T(x + y) = Tx + Ty Vx,yeE+, T(ax) = aTx V xeE+,ae R+. [For by 14Bi E = £ + - £ + . ] 16B Definition Let E and F be Riesz spaces. Let L~(E; F) be the set of all linear maps from E to F which can be expressed as the difference of two increasing linear maps. 21

16]

RIESZ SPACES

Then L~(E; F) has a natural linear structure, being a linear subspace of the set of all linear maps from E to F, and in addition a partial ordering given by

8 < T iff Sx < Tx

VxeE+,

which renders it a partially ordered linear space; its positive cone being just the set of increasing linear maps. I shall write E~ for L~(E; R). 16C Theorem Let E be any Riesz space and F a Dedekind complete Riesz space. Then a linear map T: E -> F is in L~(E; F) iff is bounded above in F for every xe E+. Proof (a) If T = Tx — T2, where T± and T2 are increasing, then {Ty : 0 < y ^ x) is bounded above by T7^ for every a;e E + . (b) Conversely, if the condition is satisfied, define S: E+ -> F by & = sup{2fy : 0 ^ y ^ #} V xeE+, which always exists because F is supposed Dekekind complete. Clearly 8(ax) = ocSx whenever x e E+ and a e R+. Now if xv x2 e E+, Sx1 + 8x2i = sup{T(y1 + y2): 0 ^ yx ^ xv 0 ^ y2 < o;2} : 0 ^ z ^ ^ + ^2}

< 8x1 + 8x2. (Because if 0 < z ^ xt + x2, 0 < z — z A x1 ^ x2; cf. 14Ja prf. Note also that in the first line I used 12Cc.) So 8 extends to a linear map from E to F [16A]. Now 8 and 8-T are increasing, so TeL~(E; F). 16D Theorem Let E be a Riesz space and F a Dedekind complete Riesz space. Then L~(E\ F) is a Dedekind complete Riesz space and (a) if TeL~{E; F) then T+ is given by T+x = sup{jP?/: 0 < y ^ #} V # e E + ; (b) if A is a non-empty set in L~(E; F) and A f, then sup A exists iff sup {Tx :TeA} exists for every x e E+, and in this case (supA)(x) = sxip{Tx:TeA) 22

V XEE+.

LINEAR MAPS BETWEEN RIESZ SPACES

[16

Proof (a) We know that L~(E; F) is a partially ordered linear space. But we have seen in the course of proving 16C that if TeL~(E; F) there is a linear map 8: E-> F such that Sx = sup{2ty: 0 ^ y ^ x) V xe JS+, and now it is obvious that S = sup{3P, 0} in L~(E; F). So L~(E; F) is a Riesz space and 8 = T+. (b) Suppose that Af in L~(E; F) and that 8x = &up{Tx:TeA} exists for every xe E+. Because A is directed upwards, S satisfies the conditions of 16A, and extends to a linear map So: £ -> F.A cannot be empty (unless F = {0}), and, if ToeA, So — To is increasing; so in any case $ o eL~(E; F), and plainly So = sup A. It follows at once, by the device of 13Cb, that L~(E; F) is Dedekind complete; for if A is bounded above by To, then {Tx: T e A} is bounded above by Tox, and therefore has a least upper bound, for every x e E+. *16E Corollary

Let E and F satisfy the conditions of 16D.

(a) If TeL~(E; F), then \T\ is given by \T\ (x) = mp{Ty: \y\ ^ x} V xeE+. (b) If Tv T2eL~(E; F), then T± v T2 and 2\ A T2 are given by (2i V T2) (») = sup l ? ; ^ + T2x2: »x, o;2e JS+, ^ + o;2 = x}9 {Tx A T2) (a;) = inf {T1x1 + T2x2: xv x2e E+, x± + x2 = x}

for every x eE+. (c) If A is a non-empty subset of L~ (£; F) and -41, then A\ 0 iff inf {To;: TeA} = 0 for every # E £+. Proofs (a) and (b) follow from the formula in 16Da by applying the identities (c) follows by applying 16Db to the set — A. [See also 161.] *16F Lemma Let £ be a Riesz space and F a Dedekind complete Riesz space. Let T: E -> F be a linear map. For each xe E+, set Ax = {L%
= a:}. 23

16]

RIESZ SPACES

Then (a) Ax \ for each x e £+; (b) T e L~(E; F) iSAx is bounded above in F for each xe £ + , and in this case V xeE+. Proof (a) If (x^}i
=

^i ^Or e v e r y * < n

anc

*

for every j < m. Now it follows that by 14Bn. So Si
< \Ty+\ + \Ty~\ + \T(x- \y\)\,

which belongs to Ax. Thus {Ty: 0 < y ^ x) is bounded above by the upper bounds of ^ and TeL~(E; F). Also, by 16Ea, \T\ (x) = sup{%: |2/| ^ x) ^ for any xe E+; so | y | (x) = sup J^. 16G Definition Let £ and F be Riesz spaces. Then L x (£; F) is the set of linear maps from £ to F which are expressible as the difference of two order-continuous increasing linear maps. I shall write £ x for L><(£; R).

16H Theorem Let £ be a Riesz space and F a Dedekind complete Riesz space. Then L x (£; F) is a band in L~(£; F). Proof (a) Suppose that Tx and T2 are increasing order-continuous linear maps from £ to F. Then so is T± + T2. P Of course Tx + T2 is linear and increasing. Suppose that A is a non-empty set in £ such that^410.

Then {Ti + TJIA]),. But

24

LINEAR MAPS BETWEEN RIESZ SPACES

[16

[because A is directed downwards and Tx and T2 are increasing] = inf TJA] + inf T2[A] [as in 12Cc] =0 because Tx and T2 are both order-continuous. Because Tx + T% is linear, this is enough to show that it is order-continuous [14Ec]. Q (b) So let P be the set of all order-continuous increasing linear maps from E to F, a subset of L~(E; F)+. Then P + P £ p . Also, it is clear that aP c P for every aeR+; so that P-P = LX(E; F) is a linear subspace of L~(E; F). Next, 0 ^ 8 ^ TeP=>8eP, again using the first condition in 14Ec. So L x (£; F) is solid. P If |#| ^ \TX-T2\, then 0 ^ S+ ^ \S\ < |21| + \T2\ EP, SO / S + G P . Similarly 8~eP so S = S+-S-eL*(E;F). Q x

In particular, L (£; F) is a Riesz subspace of L^(£; F). (c) Finally, suppose that A is a non-empty subset of LX(E;F) and that # = sup A exists in L~(E; F). Fix ^ e ^ . Following 13Cb, set B = {sup (C U {To}) : 0 g i , C finite}. Thus B c L X (E; F), B is bounded below by TOi and £ f >S. We know that S — To is increasing; I shall show that it is order-continuous with the second criterion of 14Ec. For if 0 c / ) c £ + and D f x, then (S-To) (x) = suvTeBTx-Tox

[16Db] x)

[12Ca]

- To) (y) [as T — To is order-continuous for every TeB] = su P2/eI) supTGjB (T - To) (y) = suVy

eD(S-T0)(y)

(repeating the arguments of the first two lines); and this is the required result. Consequently, 8 = (S-To) + ToeLx(E; F). As A is arbitrary, LX(E; F) is a band in L~(E; F). *16I Exercise Let £ be a Riesz space and F a Dedekind complete Riesz space. Let A be a non-empty subset of L~(E; F). Then A is bounded above in L~(E; F)ifF

25

16]

RIESZ SPACES

is bounded above in F for every xe £+, and in this case (sup A) (x) == +

for every x e E . Similarly, (inf A) (x) = inf Bx provided the right-hand side is always defined. Notes and comments The results of this section are the first justification we see for the serious study of abstract Riesz spaces. The spaces L~(E; F) and LX(E; F) are defined in such a way that their structure is not immediately accessible, and we must deduce as much as possible from the theorems 16D and 16H. It is in this section also that I begin to assume familiarity with the basic properties of Riesz spaces laid out in §§ 12 and 14. Consequently the proofs above contain a number of elisions in which elementary but non-trivial results have to be applied. I t is a valuable exercise to search out these applications. An outstanding problem in the subject is that of salvaging something from the above work if F is not Dedekind complete. I t seems that within the methods and concepts I have introduced, not much can be done; but in a suitably enlarged context of partially ordered linear spaces there ought to be interesting extensions.

17 Order-dense Riesz subspaces A number of important results are associated with the concepts of order-dense or locally order-dense Riesz subspace. [For definitions, see 14F.] The ones here are those which have applications to the work of this book. In particular, they can be thought of as the core of the Radon-Nikodym theorem; but this will become apparent much later. Since the results of this section are of interest primarily because of their applications, I have starred all the paragraphs except the first two. 17A gives a valuable sufficient condition for greatest lower bounds (and, of course, least upper bounds) to be the same whether taken in the whole space or in a subspace; you will recall that this cannot be taken for granted [1XB]. 17B is an important general theorem on the extension of order-continuous increasing linear maps. 17A Lemma If £ is a Riesz space and F is a locally order-dense Riesz subspace of E, then whenever A is a non-empty subset of F and x is the greatest lower bound of A in F, x is the greatest lower bound of A in E. 26

ORDER-DENSE RIESZ SUBSPACES

[17

Proof ? For otherwise A has a lower bound xx in E such that xx ^ x, i.e. x v xx > x. Let y be any member of A; then 0 < x\i xx — x ^ y — xeF. As F is locally order-dense, there is a z e F such that 0 < z ^ a : v ^ 1 - ^ But now seva^ ^ # + 2, soa: + 2 i s a lower bound for A in F strictly greater than x; which is impossible. X Remark The same result will be true of least upper bounds. This is sometimes expressed by saying that Fis regularly embedded in E. 17B Theorem Let E and G be Riesz spaces, and F an order-dense Riesz subspace of E. Suppose that T: F -> G is an order-continuous increasing linear map such that Ux = sup{Ty:yeF, 0 ^ y < x) exists in G for every x e E+. Then T has a unique extension to an ordercontinuous increasing linear map from E to G. Proof (a) I show first that if A is a non-empty set in F+, and A \ x in E, then sup T[A] exists and is equal to Ux. P Of course Ty < Ux for every ye A. On the other hand, suppose that w is any upper bound for T[A], and suppose that zeF and that 0 ^ z ^ x. Then B = {zAy:yeA}^zAx = z in E, by 14D. A fortiori, B\z in F. So, as T is order-continuous, T[B] f Tz. But we know that for any ye A, T(zAy) ^Ty^w. So w is an upper bound for T[B], and w ^ I7^. As z is arbitrary, w ^ Ux; which proves the result. Q (b) Now it follows that U(x± + x2) = Uxx + Ux2 V xl9 x2 E E+. PForlet

Ai = {y:yeF, 0 ^ y ^ xJ

for i = 1,2. Because F is order-dense, Ai\xi Al + A2yf xt + x2 [12Cc]. Now by (a) above,

in £ for both i. So

Ufa + xJ = sap T[At +AJ = sap (T[A^] + T[A^) = sup T[A^ + sup T[^l2] = Uxx + Z7^2, using 12Cc again. Q 27

17]

RIESZ SPACES

(c) Since it is easy to see that U(ax) = ocUx whenever a ^ 0 and xe E+, U has an extension to a linear map U: E-> G [16A], Clearly Ux = Tx for every x e E+, so U is an extension of T. Finally, U is ordercontinuous. P For suppose that 0 c= A £ £+ and that A\x. Then JB = { z : z e F , 3 y e A , 0 ^ z ^

y}\x,

since clearly A and i? have the same upper bounds. So Ux = sup I 7 ^ ] by (a). But any upper bound of U\A\ is an upper bound of T[B], so sup U[A] exists and is equal to Ux. As A is arbitrary, U is ordercontinuous. Q (d) Of course the extension of T is unique, because it must agree with U on E+ and therefore on E. *17C Proposition Let E and G be Riesz spaces, and F an orderdense Riesz subspace of E. Let U: E-> G be an order-continuous increasing linear map, and let T be the restriction of U to F. (For example, U might have been constructed from T as in 17B above.) Then: (a) if T is a Riesz homomorphism, so is U; (b) if I7 is a one-to-one Riesz homomorphism, so is U. Proof (a) Suppose that I7 is a Riesz homomorphism, and that x A y = 0 in E. Set A = {z: ZEF, 0 ^ z ^ x} and B = {w: weF, 0 ^ w ^ i/}. Then^4 f # a n d B \ y [because Fis order-dense], so, because Uis ordercontinuous, UxAUy = ( s u p ^ ?7z) A Uy A(

(Uz A using 14D twice. But if zeA and weB, 0 ^ ZAW ^ xAy = 0, so UZAUW = TZATW = T(ZA«>) = 0. Thus UxAUy = 0. As a; and y are arbitrary, £7 is a Riesz homomorphism, by 14Eb. (b) If now T is one-to-one, then for any non-zero xeE there is a zeF such that 0 < 2 ^ |a?|. So \Ux\ = U(\x\)^Tz> 0. Thus Ux =)= 0. As x is arbitrary, (7 is one-to-one. 28

ORDER-DENSE RIESZ SUBSPAGES

[17

*17D Lemma Let £ be a Dedekind complete Riesz space, F any Riesz space, and T: E -> F an order-continuous Riesz homomorphism such that T[E] is locally order-dense in F. Then T[E] is a solid linear subspace of F. Proof As T is a Riesz homomorphism, T[E] is a Riesz subspace of F. It is therefore enough to show that if w e T[E] and 0 < y ^ w, then ]. Let xe E be such that To; = w, and consider

A = {zA\x\:zeE,Tz^y}. A is bounded above by \x\ and 0 e 4 , so sup J. exists; also, it is clear that A f, so, as T is order-continuous, = sup£F(z A \x\):ze£,

= sup{Tz A \Tx\

TZ < y)

:zeE,Tz^y}

= 8Vip{Tz:zeE,Tz ^ y) because T[E] is locally order-dense. *17E Lemma If E and F are Riesz spaces and T: E -> F is a oneto-one Riesz homomorphism such that T[£] is locally order-dense in F, then T is order-continuous. Proof Since T is a one-to-one Riesz homomorphism, it is a Riesz space isomorphism between E and T[E], Now if 0 cz^ljO in £, inf T|>1] = 0 in T[E]. By 17A, inf T[^l] = 0 in F. So by 14Ec T is order-continuous. *17F Corollary If £ is a Dedekind complete Riesz space, F is any Riesz space, and T: E -> F is a one-to-one Riesz homomorphism such that T[E] is locally order-dense in F, then T[E] is solid in F. Proof

By 17E, T is order-continuous; now by 17D T[E] is solid.

17G Exercises (a) Suppose in 17B, that F is locally orderdense but not order-dense in E. Show that T has an extension, as in the theorem, but that the extension may not be unique. (b) Let £ be a Riesz space and F an order-dense Riesz subspace of £ 29

17]

RIESZ SPACES

such that every member of E is dominated by a member of F (that is, V XEE 3 yeF such that \x\ ^ y). Show that Ex and Fx may be identified as Riesz spaces. *(c) Let E and G be Riesz spaces, and F a, super-order-dense Riesz subspace of E. Suppose that T:F->G is a sequentially ordercontinuous increasing linear map such that swpneJiTyn exists in G whenever <^> neN is an increasing sequence in F+ with a least upper bound in E. Then T has a unique extension to a sequentially ordercontinuous increasing linear map from E to G. [Hint: show that if (2/w)neN i s a sequence in F+ such that (yn}nejsi t # in E, then SUPneN^Vr. = SUp {Tl/: 2/G F, 0 < i/ ^ x}. Now use the method of 17B.] (d) If E is any Riesz space and F is a locally order-dense Riesz subspace of E which is (in itself) Dedekind complete, then F is solid. [Use 17F, or otherwise.] *(e) If E is any Riesz space, F is a locally order-dense Riesz subspace of E, and G is a locally order-dense Riesz subspace of F, then G is locally order-dense in E. Notes and comments The extension theorem 17B is used in various places in Chapter 6. Like many results concerning ordercontinuous maps it has a 'sequential' form referring to sequentially order-continuous maps [17Gc]; for some applications in measure theory, this is really the natural one to use [see 62K/62Me]. Note that the sequential form of the concept' order-dense' is' super-order-dense'. *The hypotheses of 17D and 17E should be examined carefully; they are not quite natural, and applications often fulfil them with little to spare. On the other hand, it is not clear that they can be relaxed in any useful way. These two results, of course, tend to be employed together, as in 17F.

*18 The countable sup property A large proportion of the elementary examples of Riesz spaces share the special property that every set with a least upper bound has a countable subset with the same least upper bound. (It follows, of course, that the same will be true for greatest lower bounds.) While this is not vitally important to us, it is of considerable interest, especially in view of the elegant characterization of Archimedean Riesz spaces of this type in 18D below. 30

THE COUNTABLE SUP PROPERTY

[18

18A Definition A Riesz space E has the countable sup property if, whenever A is a subset of E with a least upper bound, there is a countable B c A such that supl? = sup A. 18B Lemma Let E be an Archimedean Riesz space with the countable sup property. Let A c; E be bounded above. Then there is a countable set B c; A with the same upper bounds as A. Proof If A = 0, set B = 0 . Otherwise, let C be the set of upper bounds of A. As E is Archimedean, inf (C — ^4) = 0 [15C]. So there is a countable set D $z C — A such that infD = 0. Now we can choose a countable B^A such that D g C - ^ s o inf ((7-5) = 0. Suppose that xeA and that w is an upper bound for B. Then for any ze<7 and */<=£, z-y>x-W, so x — w ^ inf (C — J5) i.e. a; < w. Thus every upper bound for B is an upper bound for A; but the converse is trivial, so the result is proved. 18C Corollary If E is an Archimedean Riesz space with the countable sup property, every Riesz subspace of E has the countable sup property. 18D Theorem Let E be an Archimedean Riesz space. Then E has the countable sup property iff every disjoint set in E+ which is bounded above is countable. Proof (a) Suppose that £ has the countable sup property, and that A is a disjoint set in £+ bounded above by x0. (Recall that by ' disjoint' I mean that x A y = 0 whenever x and y are distinct members of A; 13G.) Then there is a countable set B c A with the same upper bounds as -4[18B]. But A c B u {0}. P ? If y is a non-zero member of A\B, then y A Z = 0 for every zeB. Suppose that x is any upper bound for A; then for any zeB [using 14Ja], so x — y is also an upper bound for B and therefore for A. Inducing on n, x — ny ^ yforevery neN. X Q So A is countable. (b) Suppose that E satisfies the condition, and that A c: £+ is a non-empty set with a least upper bound x. Let 0 ^ oc < 1; then there is a countable Aa<^ A such that ax = sup{j/ Aotxiye Aa}. 31

18]

RIESZ SPACES

PSet £ = {z:3 ye A, 0 < z ^

(y-ax)+},

and let X be the family of all subsets of B which are disjoint in £+. By Zorn's lemma, X has a maximal element G say; as G is bounded above by (1 — a) x, it is countable, and we can choose a non-empty countable set Aa^ A such that V zeC 3 yeAa

such that

0 ^ z ^ (y —ax)*.

? Suppose that there is a w > 0 such that ax — w^yAax i.e.

V +

0 < w < (ax — y)

V

Then w;A2 = 0 for every ze(7, so (G being maximal) ^ A 2 = 0 for every zel? i.e. wA(y — a # ) + = 0 V But now w A (l-oc)x = w A (x-ax)+

= sup^^w A (2/ — ocx)+ = 0

[using 12Ca and 14D], and as 1 — a > 0, tu A ocx = 0 [14Kb]. But 0 < w ^ ax. X So a# = sup{y Aax:yeAa}. Q Now let a(w) = 1 - 2~n and D = \Jn^A^n).

Then

x = sup {oc(n) x: n e N} [15D] = supD. Finally, suppose that A is an arbitrary non-empty subset of E with a least upper bound x. Fix # o e,4 and set A' = {t/ v# 0 :2/e-4}. Then sup A' = x and A' has a lower bound. I t is clear that, applying the result above to A' — x0, A' has a countable subset B1 with least upper bound x. Now choose a countable JS c J. such that and see that x = sup ({#0} U 5), which is a countable subset of A. 18E Exercises *(a) Show that an Archimedean Riesz space E has the countable sup property iff whenever F is an Archimedean 32

THE COUNTABLE SUP PROPERTY

[18

Riesz space and T: E -> F is increasing, linear and sequentially ordercontinuous, then T is order-continuous. (b) Let £ be a Riesz space with an order unit [definition: 141], and F a Dedekind complete Riesz space with the countable sup property. Then L~(E; F) has the countable sup property. (c) Let E be a Riesz space with a weak order unit [definition: 141], and F a Dedekind complete Riesz space with the countable sup property. Then LX(E; F) has the countable sup property. Notes and comments There are simple examples of Riesz spaces with the countable sup property in 2XA-2XC; see also 25Lc. Theorem 18D is clearly a very powerful method of identifying such spaces, as well as being an important property of them. From this theorem we can see that the countable sup property is intimately related to the ' countable chain condition' on topological spaces and Boolean lattices. It is important, however, to appreciate that the countable sup property is not the same as the property: 'every disjoint set in the positive cone is countable', which is shared by many of the elementary examples. The latter is of course much stronger. IX Examples for Chapter 1 Because the structures dealt with in this chapter will recur throughout the book, I shall give very few examples at this stage, and most of them will be slightly off the main line of thought. Thus 1XA-1XC are designed to illustrate technical aspects of the concepts I have introduced, rather than their practical applications. In contrast, 1XD and 1XE discuss a fundamental class of Riesz spaces, the spaces R x , which also have important topological properties. 1XA A partially ordered set < be the relation on A given by

Let A be any set (even 0 ) ; let

a ^ b o a = 6. Then (A, <) is a partially ordered set [11A]. If B c A, then sup2? exists iff inf B exists iff either B has just one element or B is empty and A has just one element. If (G, ^ ) is any other partially ordered set, every function from A to C is increasing and order-continuous [11D]. Conversely, every increasing function from G to A is order-continuous. FTR

2

33

IX]

RIESZ SPACES

1XB A lattice Let X be any set, and let ^ be the restriction of c to 0>X, the set of subsets of X. Then (0>X, <) is a Dedekind complete distributive lattice; if A is a non-empty subset of £PX,

sup A = \JA

and inf A = C\A.

[See also 4XA.] Now let % be any topology on X. Then % is a sublattice of SPX [13F]. Let ^ be the restriction of ^ or g to %, so that (X, <J) is a lattice in its own right. The embedding of % in ^ X is a lattice homomorphism (because % is a sublattice of &>X); and it is order-continuous on the left [11D], because if A c %9 then U^.> which is sup A in ^ X , belongs to %, and therefore must be sup A in %. We see also from this that % is a Dedekind complete lattice [13B]. But suppose that A is a non-empty subset of %. Then inL4 in S£ is int (fl^)> while inf A in ^ X is f|^l itself. Of course these are not in general the same. Thus we have here an example in which the meaning of the expression inf A depends on which lattice we consider A as a subset of. *1XC Totally ordered sets Let (A9 <) be a totally ordered set, i.e. a partially ordered set such that for every a and b belonging to A, either a ^ b or b < a. Then A must be a lattice. On A the order topology is the topology generated by all sets of the forms {a: a > 6} and {a: a < b} as b runs through A. This topology is always Hausdorff and regular. A subset of A is closed for the order topology iff it is orderclosed in the sense of 1 IE. *The order topology is the coarsest topology which is ' compatible' in the sense of the note at the end of § 21; and it is the finest topology which is 'Lebesgue' in the sense of the introduction to § 24. A is Dedekind complete iff every closed interval [6? c] = {a: b ^ a < c} is compact in the order topology. (In the special case A = R, this is the Heine-Borel theorem; cf. also 24Lb.) A is compact in the order topology iff either it is empty or it is Dedekind complete and has greatest and least elements. 1XD The space Rx. Let X be any set. Then Rx, the set of all functions from X to R, has a natural Riesz space structure given by

(f+9)(t)=fi + 9t V teX, (otf)(t) = aft V teX, f>gofl>gt V teX; 34

EXAMPLES FOR CHAPTER 1

[IX

that is to say, Kx is regarded as a product of copies of the Riesz space R [14H]. Rx is Dedekind complete [13Hb, or directly]. Regarded as a product of copies of R, Kx has a natural topology, which is of course a complete Hausdorff locally convex linear space topology. This topology is Fatou [23A], Levi [231] and Lebesgue [24A]; so the completeness of Rx is an example of Nakano's theorem [23K]. The linear topological space dual (R^)' is equal to (R x ) x . 1XE Quotients of R x . Let X be a non-empty set, and ^ a filter on X. Let E be Kx with its usual Riesz space structure, and set F = {x : x e E, {t:x(t) = 0}e Then F is a solid linear subspace of E. Now the quotient Riesz space E/F [14G] is Archimedean iff 3F is closed under countable intersections, and in this case the canonical map from E to E/F is sequentially ordercontinuous [1 ID, 14Lb].F is a band iff !F is generated by a single set Y, and in this case E/F £ R F . For example, let X = N and let 3* be any filter on X including the Fr6ehet filter, i.e. containing the complement of every finite set. Then E/F cannot be Archimedean. *1XF A non-Archimedean Riesz space Give R2 its usual linear space structure, and define the lexicographic ordering < on R2 by (a, /?) ^ (y, S) iff either

a
2

As this is a total ordering, R is a lattice; clearly it is a partially ordered linear space; therefore it is a Riesz space. If x = (1,0) and y = (0,1), then ny ^ x for every WGN, but y > 0; so R2 is not Archimedean [15A]. The usual linear space topology on R2 is Hausdorff, Lebesgue and Levi, but not of course compatible [2lBa]. Further reading for Chapter 1 For an elegant elementary discussion of partially ordered sets, see BOURBAKI I , chapter in, § 1. For an alternative introduction to the theory of Riesz spaces, see BOUKBAKI VI, chapter n; there is a deeper approach in BIRKHOFF, chapter xv. A large part of the work done for Riesz spaces in this chapter can be applied to lattice-ordered groups; for hints see BIRKHOFF, chapter XIII. For the more advanced theory of Riesz spaces, see LUXEMBURG & ZAANEN R.S.

35

2.

Topological Riesz spaces

This chapter and the next are an exposition of the elementary theory of Riesz spaces with linear space topologies on them; they form a natural confluence of the Riesz space theory of Chapter 1 with ordinary abstract functional analysis. The present chapter proceeds by studying a series of properties that Riesz space topologies can have (most of them being, of course, relations between the topology and the order structure). The properties are all chosen to be ones possessed by important special cases; the theory never gets far from the applications that will be made of it, except perhaps in the work leading up to Nakano's theorem in § 23. In § 26 the conditions imposed are so strong that they become significantly less abstract; this section almost forms a layer intermediate between the first five sections of this chapter and the concrete examples from measure theory that will follow later. Throughout this chapter I avoid the hypothesis of local convexity as far as possible. This is for both positive and negative reasons. The negative justification is that it can be done without an excessive amount of extra work. On the positive side, the most important examples of non-locally-convex spaces in analysis are Riesz spaces and are particularly accessible by the methods of this chapter [see, for example, 63K]. But of course it is quite possible to read this chapter with only locally convex spaces in mind. Although we shall not study dual spaces thoroughly until the next chapter, I shall occasionally use the symbol E' for the space of continuous linear functionals on a linear topological space E, as well as E~ and Ex, defined in 16B and 16G respectively.

21 Compatible topologies The simplest requirement we can make of a linear space topology on a Riesz space is that the positive cone be closed. Every Hausdorff topology we study will be of this type. It is perhaps slightly surprising that this already has non-trivial consequences. 21A Definition If £ is a Riesz space, a linear space topology on E is said to be compatible with the order structure of E if £+ is closed. 36

COMPATIBLE TOPOLOGIES

[21

21B Proposition Let E be a Riesz space and % a compatible linear space topology on E. Then: (a) E is Archimedean; (b) {x: x ^ x0} and {x: x ^ x0} are closed for every x0 e E; (c) % is Hausdorff; *(d) Let 0 <= A f (i.e. let ^4 be a non-empty set which is directed upwards) in E, and suppose that ^(A \) [definition: 11C] has a cluster point x0. Then x0 = sup A. *(e) In particular, if A is a non-empty relatively compact subset of E which is directed upwards, then sup.4 exists and is \vav3F{A \). Proofs (a) If x ^ 2ny for every neN i.e. 2~nx-y ^ 0 for every weN, then — y = lim^^^ 2~nx — y ^ 0 i.e. 2 / ^ 0 . (b) and (c) are easy. *(d) (i) If xeA, then {y: y ^ x} is a closed set belonging to ^"(^4 f), so contains x0; thus #0 is an upper bound for A. (ii) Now if x is any upper bound for A, {y: y ^ x} is a closed set belonging to ^"(^1 f), so contains x0; thus #0 is the least upper bound of A. *(e) is now easy. *21C Exercise Let £ be a Riesz space and % a locally convex compatible linear space topology on E. Then the weak topology %S(E, E') is also compatible, and if xeE\E+ there is an increasing continuous linear functional/on E such that/a; < 0. *Notes and comments It is clear that (with a suitable definition of 'Archimedean' space) everything in this section can be applied to general partially ordered linear spaces. Indeed, 2lBb provides a definition of ' compatible' topology on any partially ordered set, for which 21Bd and 2lBe will still be true [cf. IXC]. 22 Locally solid topologies A linear space topology on a Riesz space is 'locally solid' if the solid neighbourhoods of 0 form a basis. This is true of most of the 'strong' topologies which interest us, though not of 'weak' topologies. The results of this section are mostly straightforward. Locally solid topologies are really very natural and easy to work with, largely because the technical result 22C enables us to use pseudo-norms effectively. 37

22]

TOPOLOGIGAL RIESZ SPACES

22A Definitions If £ is a Riesz space, a linear space topology on £ is locally solid if 0 has a neighbourhood basis consisting of solid sets. If £ is a Riesz space, a Riesz pseudo-norm on £ is a function p: E -» R such that (i) px&O, (ii) p(x + y) t^px+py, (iii) \x\ ^ \y\=>px ^ py, (iv) Kma^0 p(ax) = 0 for all x,yeE. Observe that in this case |a| < 1 => \<xx\ < \x\ => p(ocx) ^ p(x)

V xeE.

A Riesz norm or Riesz seminorm on £ is a Riesz pseudo-norm which is a norm or seminorm respectively. 22B Proposition Let £ be a Riesz space with a locally solid linear space topology. Then: (a) the function x »-> \x\: £ -> £ is uniformly continuous; (b) the lattice operations v and A on £ are uniformly continuous; (c) the solid hull of a bounded set is bounded; (d) order-bounded sets [definition: 11F] are bounded; *(e) the closure of a solid set is solid. Proofs

(a) This is because

IN-Ml ^ \x~y\

v x

>ytE>

since by 14Bn \x\ — \y\ ^ \x — y\ and \y\ — \x\ ^ \y — x\ = \x — y\. So if

U is a solid neighbourhood of 0 and x — yeU, \x\ — \y\eU. (b) now follows from 14Bf and 14Bg. (c) Supposed ^ £ is bounded, i.e. is absorbed by every neighbourhood of 0. Let B be the solid hull of A. If U is any neighbourhood of 0, there is a solid neighbourhood V of 0 such that V £= U; as A is bounded, there is an a ^ 0 such that A ^ aV. But now aV is solid, so

As U is arbitrary, B is bounded, (d) follows, since [x, y] ^ x + (solid hull of {y — 38

LOCALLY SOLID TOPOLOGIES

[22

*(e) Letw4 c E be a solid set. Suppose that x eA and that \y\ < \x\. Then the map (j>: E-> E given by 0Z = ( * / A | Z | ) V ( - | Z | )

is continuous [(a) and (b) above]. Now <j>x = y, so

But [A] c ^4 (because |$£z| < |«| for every z e E), so 1/ eA, as required. 22C Proposition Let £ be a Riesz space and X a locally solid linear space topology on E. Then X is defined by the continuous Riesz pseudo-norms on E, that is, for every neighbourhood U of 0 there is a continuous Riesz pseudo-norm p on E such that {x\px < 1} c J7. Proof Choose a sequence weN of solid neighbourhoods of 0 such that V0^U and Vn+l + Vn+1 + Vn+1 ^ Vn for every weN. Define 2ifxeVn\Vn+i> i£xenneiVn. d(x) = 0 Then |o;| ^ \y\ => d(x) ^rf(i/),because every Vn is solid. Define/?: £ - > R b y p(x) = inf {S i < w d(^) :a; = 2 i < n a J . Now (i) plainly /o(a;) ^ 0 for every a;e E; (ii) and /)(# + y) ^ />(o;) +yo(^/) for every x,yeE; (iii) and |#| ^ |y| =>yo(a;) < p(y). P For if 2/ = S ^ n ^ , then M < |y| <Si(i/). Q (iv) Also, p(x) < d(o;) for every xeE, so that p(x) ^ 2~w whenever o:eT^. Hence [using (ii) above] p is continuous. As p(0) = 0, lima_Qp(ax) = 0 V xeE, and p is a continuous Riesz pseudo-norm. 39

22]

TOPOLOGIGAL RIESZ SPACES

Finally, observe that for every finite sequence (x^)i
* -1

n-1

with the convention that 2 and 2

are both zero. Now by the

inductive hypothesis r-l

ti-l

and of course xreVm+1 as d(# r ) < 2~ m . So W

Fm. Q

In particular, 2i lji xeU.

— ^* ^ u ^ ^ follows at

22D Proposition Let E be a Riesz space with a locally solid linear space topology. Then its dual E' is a solid linear subspace of E~, and the solid hull of an equicontinuous set in E' is equicontinuous. Proof From the condition in 16C it is obvious that a linear functional on E is in E~ iff it is bounded on order-bounded sets; so 22Bd above is enough to show that £' c: £~. As for the equicontinuous sets in E', consider the sets

U° = {fife Er, \fx\^l

V xe U}

for solid neighbourhoods U of 0 in E. I f / e U° and \g\ ^ | / | in E~, then N

< \g\ ( N ) ^ | / | (N) = Bup{/i/: |y| ^ |^|} < 1 V xeU

[using 16Ea]. So U° is a solid subset of E~. Now (i) Ef is a union of solid sets, so is solid (ii) every equicontinuous set is included in a solid equicontinuous set. [See also 26Hc] 40

LOCALLY SOLID TOPOLOGIES

[22

22E Proposition Let £ be a Riesz space and X a locally solid Hausdorff linear space topology on E. Then (a) X is compatible with the order structure of E [because £+ = {a;:a;A0 = 0}]; (b) E is Archimedean [21Ba]; (c) Bands in Bare closed [15Fb]; *(d) If A c E and xe A, then xe{xAy:yeA}, x = m$ueAx Ay = vc&v^Ax v y. 22F Proposition Let £ be a Riesz space with a Hausdorff locally solid linear space topology %. Then the uniform space completion £A of E has a natural Riesz space structure, and its topology %A is a locally solid Hausdorff linear space topology; and the embedding E -> EA is a Riesz homomorphism. Proof We use 22Bb. (a) Because the operation v: Ex £-> E is uniformly continuous, it has a (unique) continuous extension to an operator v: EA x EA -> EA. The following identities are true for all x, y, ZEE: (i) xvy = yvx, (ii) xv(yvz) = (xvy)vz, (iii) xvx = x, (iv) s+(yvz) = (a; + y)v(a: + z), (v) a(x vy) = ocxv oty V a ^ 0, and therefore, by continuity, for all #, y, zeEA. (b) These are enough to make v define a Riesz space structure on EA, as follows. Say that x ^ y in £Aiff # \i y = y. Now # ^ y and «/^z => xvz = xv(yvz) => x ^ z; x ^y

= (xvy)vz

= yvz = z

and i/ < x => a? = y;

A

xeE => xvx = x => x ^ x; thus (ii) and (iii) show that ^ is a partial ordering. Next, x ^y => xvy = y => (a; + z) v (y+z) = => x + z ^ y + z V zeEA;

41

22]

TOPOLOGICAL RIESZ SPACES

thus (iv) and (v) show that (EA, ^ ) is a partially ordered linear space. Finally, y ^ x and z ^ x => (y vz) vx = y v {zvx) = y vx = x and of course yv{yvz) = (yvy)vz zv (yvz) = zv (zvy) = zvy = y\/z, from which yy z = sup {2/, 2} in EA; as y and 2 are arbitrary, EA is a Riesz space [14A, exercise]. (c) Any of the conditions in 14Eb can be used to show that the identity map E -> EA is a Riesz homomorphism. (d) Finally, let W be any closed neighbourhood of 0 in E \ Then W n E is a neighbourhood of 0 in E, so by 22C there is a continuous Riesz pseudo-normp on E such that {#: x e E, px < 1} c TF n E. Asp is uniformly continuous, it has a (unique) continuous extension p on E \ Now p is still a Riesz pseudo-norm. P Consider the conditions of 22A. We can rewrite (iii) as p{x) =p(xv(-x)) ^p{xv(-x)vy)

V x,yeE;

which, like (i) and (ii), will be satisfied by p if satisfied by p. And condition (iv) is a consequence of the continuity of p. Q But it is easy to see that {x:xeE\p(x)< 1} c W^E C Tf. So XA is defined by the continuous Riesz pseudo-norms on E and is locally solid. Remarks Observe that the conditions (i)-(v) of the proof above are necessary and sufficient conditions for a binary operation v to define a Riesz space structure on a given linear space. An alternative proof of the proposition in the locally convex case can be got from 31C in the next chapter. Note also 22Gb; the positive cone of EA is precisely the closure in EA of the positive cone of E. 22G Exercises (a) Let E be a Riesz space and || || a Riesz norm on E. Then the dual norm on E' is also a Riesz norm. (b) Let E be a Riesz space with a locally solid linear space topology, and F a Riesz subspace of E. Then F is also a Riesz subspace of E, and 42

LOCALLY SOLID TOPOLOGIES

[22

the positive cone of F is precisely the closure in £ of the positive cone ofF. *(c) Let £ be a Riesz space with a locally convex locally solid linear space topology %. Then % is defined by the continuous Riesz seminorms on £. [Hint: 14Jd.] *(d) Let £ be a Riesz space. Then there is a finest linear space topology on £ for which all order-bounded sets are bounded, and this topology is locally solid. [Hint: 25F, proof.] (e) Let £ be a Riesz space with a metrizable linear space topology % such that the solid hull of a bounded set is always bounded. Then % is locally solid. Notes and comments It is so natural to require a topology on a Riesz space to be locally solid that some authors use the phrase 'topological Riesz space' to mean a Riesz space endowed with a locally solid linear space topology. But we shall also be interested in weak topologies, which are hardly ever locally solid; consider, for instance, the examples 2XA-2XC. It is well known that any linear space topology can be defined by pseudo-norms; a simple adaptation of this result gives 22C. Actually we really need this for the proof of 23B below rather than for itself. For the time being, though, we know just that locally solid linear space topologies are precisely those which can be defined by Riesz pseudonorms. Note that, as hinted in 22Gc, the locally convex case is rather simpler, though it also relies on 14Jb (via 14Jd). Clearly it is 22Ba/b that give rise to 22F, as well as 22Be and 22E; these results suggest that it might be interesting to study topologies for which the lattice operations are continuous or uniformly continuous. Another generalization, which can be applied to any partially ordered linear space, has been extensively studied; this is the notion of a 'normal' positive cone, treated in PERESSINI and SCHAEFEET.V.S.

The essential idea of 22D and 22Ga is that the polar in £~ of a solid set in £ is solid; this recurs in other contexts.

23 Fatou topologies All the locally solid topologies in which we are interested actually satisfy a further condition: 0 has a base of order-closed neighbourhoods. This leads to some much deeper results, the outstanding one being 43

23]

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Nakano's theorem [23K], which gives a powerful sufficient condition for a topological Riesz space to be complete. Most of the section is devoted to proving this result through a series of lemmas, many of which have other applications; but as the theorem itself will not be applied in its full strength except in 64E, this work is all starred. The unstarred parts are Theorem 23B, which is an essential tool for coping with Fatou topologies, and the definition of 'Levi' topology [231], which will be used later in other contexts. 23A Definitions Let £ be a Riesz space. A Fatou topology on £ is a locally solid linear space topology such that 0 has a neighbourhood basis consisting of sets U such that JTJ c JJ [definition: HE]. A Fatou pseudo-norm on £ is a Riesz pseudo-norm p such that 0 cz A t x in E+ => px = sup{pt/: ye A}; that is, p is order-continuous on the left on E+ [11D]. Following 22A, a Fatou seminorm or Fatou norm is a Fatou pseudo-norm which is a seminorm or norm respectively. 23B Theorem Let £ be a Riesz space and % a Fatou topology on £. Then % is defined by the continuous Fatou pseudo-norms on £. Proof (a) Let U be any neighbourhood of 0 for S£, let V be a solid neighbourhood of 0 such that V £ U, and let W be a neighbourhood of 0 such that JW s W c V. By 22C, there is a continuous Riesz pseudo-norm 0x on £ such that dx{x)< 1 =•

xeW.

+

Let O be the set of all functions/?: £ -> R + such that (a) 0 ^ px ^ dxx for every xeE+ and (ft) if 0 c: A f x in £ + then px ^ supye^/)2/. Then the function p0 given by pox = O V xeE+nW,

pox=l

V xeE+\W

belongs to O, so O 4= 0 . Define d0: £+ -> R+ by 60x = aixp{px:peQ}} V

xeE+.

Then it is plain that doe$) and that 60x < 1 =>pox < 1 => xe W. (b) #0 is increasing. P Given xeE+, set py = 60(xAy) for every y e £ + ; then a simple computation shows that p e <J>, so that for every Next, 0O(# + 2/) < 00x + 0O2/ for every x, y e £+. 44

FATOU TOPOLOGIES

[23

P (i) Let z e E+. Define />:£+-> R+ by pu = max (0,60(z + u)-d1z)

V u e E+.

Then p e O so py < #02/, i.e. 0O(2 + ^) ^ ^iz + ^oVl a n ( i this is true for every z G 25+. (ii) So now define p: E+ -> R+ by

the remarks so far are just what we need to see that 0 ^ pz ^ dxz for every zeE+, so p e O and px ^ dox, i.e. 60(x + y) ^ 60x + d0y, as required. Q (c) Now define 0O: E\E+ -> R+ by (90x = 5 0 (N)- T h e n

for all a;, y e E. As moreover 60x < ^xa; for every xeE,60 is continuous; hence, it is a Riesz pseudo-norm; being order-continuous on the left on E+, it is a Fatou pseudo-norm. Finally dox < 1 => 0o(|a|) < 1 => \x\ e W => O;G F c ?7. [See also 23Na.] *23C Lemma Let £ be a Riesz space and p a Fatou pseudo-norm on E. Suppose that 0 c: A f x in 25; then p# Proof Consider S = {y+ + x~:yeA}.

Then J5f because -41, and

sup-B = So

px =p(|#|) = supg6jB/9z = sup {p{y+ + x~):ye A}.

But

yeA => y ^x => y- ^ xr => 2/+ + a:- ^ |y|,

so

/a*<sup{/>(|y|)

:I/G^1}

= sup{/>2/:i/G-4}.

*23D Lemma Let £ be a Dedekind tr-complete Riesz space and p a Fatou pseudo-norm on E. Let (xn}nex be an order-bounded sequence in E such that p{%n-%n+i) < 2 ~ n f° r every %eN. Then f0m^) < 2 " ^

V neN. 45

23] Proof

TOPOLOGICAL RIESZ SPACES p(xn-suvmej,infi>mzt)

= p(sup m6H inf i>m a; i -ag

[using 23C for the first time]

[using 23C for the second time]. Now however, if k > m > n, m-X

k-1

m-1

so

|^n-i fc1

and

P t ^ - i n ^ ^ i ) < S P(^+i-^) ^ 2~n+1>

whence the result. *23E Notation In the rest of this section we shall have occasion to use some manipulations of filters which, while fairly straightforward, seem to have no generally accepted description. The principle I use is the following. If X is a (non-empty) set, &> a filter on X, Y another set, a n d / : I - > 7 a function, t h e n / ^ is the filter on Y i.e. the filter on Y with base

If X and Y are topological spaces a n d / i s continuous, t h e n / ^ ->ft whenever J*" -> t. It X and Y are uniform spaces and / is uniformly continuous, t h e n / J ^ is Cauchy whenever iF is. 46

FATOU TOPOLOGIES

[23

Two special cases we shall be concerned with are: (i) Let £ be a linear space, p: £ -> R a function, u G £, and &> a filter on £. Then />(J^ - u) is the filter on R {A:{x:p(x-u)eA}e&r},

Ve>0 3 , | p ( ) | (ii) Let £ be a Riesz space, we £, and #* a filter on £. Then is the filter on £ uis defined similarly. *23F Lemma Let £ be a Riesz space and p a Fatou pseudo-norm on £. Then (a) F = {u :pu = 0} is a band in £. (b) If xeE, {u:p(x — u) = 0} is an order-closed [definition: H E ] sublattice of £. (c) If !F is a filter on £, ^4 = {^ :/>(#* — w) -> 0} is an order-closed sublattice of £ (perhaps 0 ) . Proof (a) is a simple computation; F is solid and closed under addition, so is a solid linear subspace; now 23C shows that it is a band. Any translate of a band is an order-closed sublattice, which proves (b). As for (c), it is easy to see that if x e A, then A = x + F, reducing it to (b). *23G Lemma Let £ be a Dedekind cr-complete Riesz space and p a Fatou pseudo-norm on £. Let SF be a filter on £ which is '/)-0auchy', that is, ^piu-v) = 0. Suppose that A is a member of 3F which is order-bounded and closed under the lattice operation A. Then there is a ueJQiA such that Proof

Choose a sequence (An}ne1x in !F such that p{x-y) ^ 2-n V x,yeAn

V neN.

For each weN, choose an element

47

23]

TOPOLOGICAL RIESZ SPACES

Then {un:neN} n e N . So if

is order-bounded and p(un-un+1)

< 2rn for every

then p(un — u)^ 2~n+1 for every n e N by 23D, and p(x-u)

< 3.2~" V #e^l w V

neN,

But as J. is closed under A, infk>i>mUi eQiA

VmeN,

*23H Lemma Let £ be a Dedekind complete Riesz space with a Fatou topology. Let B <== E be order-bounded and closed under the lattice operation A. Let S? be a filter on E such that for every continuous Fatou pseudo-norm p on E

Then there is an xQeJ>S>B such that ^ -> x0. Proof If p is a continuous Fatou pseudo-norm on E, C(p) is closed under A [23Fc]. So if w(/>) = inf G(p), C(p)^w(p) and w{p)e@B. But also, again by 23Fc, p(0r-w{p))-+O. Now if /?! and p 2 are both continuous Fatou pseudo-norms on E, is also a continuous Fatou pseudo-norm on £, and so w(pl) < MPi+/>2). SO D = {t(;(/o): p is a continuous Fatou pseudo-norm on £} f; a s D c ^JS a n d B i s bounded above, D\x0 say, andxQe*f@B. Moreover, if p0 is any continuous Fatou pseudo-norm on £, D o = {t<;(yo +p 0 ): p is a continuous Fatou pseudo-norm on £} is cofinal with D, i.e. for every weD there is a veD0 such that t; ^ w, so DQ\XQ\

and

because (p +/>0) («^" — ^(/> +p0)) -> 0 for every continuous Fatou pseudonorm p. So by 23Fc again, P o ^ ~~ xo) -> 0; as p 0 is arbitrary, !F -> a:0 [using 23B]. 48

FATOU TOPOLOGIES

[23

231 Definition Let £ be a Riesz space. A linear space topology on E is Levi if every set A c= E which is bounded (in the linear topological space sense) and directed upwards, has an upper bound (in the ordering of £). *23J Lemma Let E be a Dedekind complete Riesz space with a Levi Hausdorff Fatou topology. Suppose that 3F is a Cauchy filter on E such that 3* t\z [notation: 23E(ii)] converges for each zeE. Then IF converges. Proof For each z e £, let x(z) be the limit of !F A z. (a) If zx ^ z2, x{z1) ^ x(z2). P & A zx = ^ A (z2 A z±) = (J^ A Z2) A zx and the function u i—• u A zx is continuous; so a ^ ) = lim (J^ A z2 A zx) = zx A lim (J^ A z2) = zx A a;(z2). Q (b) Set ^4 = {x(z):zeE+}. Then u4f, by (a) above. But also A is bounded. P Let U be any solid neighbourhood of 0 in £. Let V be a balanced neighbourhood of 0 such that V+V £ ?7, and let Be^ be such that B — B^ V. Let w e J5; there is an a e R + such that u e a V and then B^u+V ^aV+V £ ( a + i ) ( 7 + F ) c (a + l)?7. But for any «(2f) e {yAziyeB}

as C7 is solid. So ^L is absorbed by U. As C7 is arbitrary, A is bounded. Q So A is bounded above and x0 = sup J. exists. (c) Now !F -> #0. P Let p be a continuous Fatou pseudo-norm on E, and let e > 0. Let Be3F be such that p(u — v) ^ e V%, andfixw0 e JS. Now we know that A f x0, so [using (a) above], and by 23C there is a zx ^ wj}" such that But we know that

x(zx)E{UAZ1:UEB},

so there is a % e JS such that

49

23]

TOPOLOGICAL RIESZ SPACES

Now

l ^ i A ^ — uo\ = \u± AZX — u0Azx\ ^ \ux — uo\,

so/?(% Azx-u0)

^ p(u±-u0)

^ e, and

p(xo-uo)

So

P(#o~^) O e

V

;

as e and p are arbitrary, ^" -> x0 [using 23B again]. Q *23K Nakano's theorem Let E be a Dedekind complete Riesz space with a Levi Hausdorff Fatou topology. Then E is complete (as a uniform space). Proof Let & be a Cauchy filter on £. Then for any y,zeE,(&rvy)Az is Cauchy and contains the order-bounded set [y Az,z], so it converges [23G, 23H, setting A = B = [y AZ,Z]]. So #" vy converges [23J]. So J^ converges [23J, upside down], *23L Proposition Let E be a Riesz space with a Hausdorff Fatou topology. Let A c E be a solid set such that JA c A. Then ^4 is closed. Proof (a) Let O be the set of continuous Fatou pseudo-norms on E. ForpeO, set F(p) = {z:zeE,pz = 0}. Then F(p) is a band in E [23Fa]. Set where for each p e O F(p)d = {U:UEE,

\U\ A \Z\ = 0

V ^

d

as in 15F. We recall that each F(p) is a band [15Fa]. Also, if px andp 2 belong to O, so does Pi+p2, and F(p1+p2)d = (F(Pl) n F(pt))* 2 F(^)^ u Fta)«, from which it follows that F is a solid linear subspace of E. (b) In fact, F is order-dense. P Suppose that w > 0 in E. As the topology on E is Hausdorff, there is a neighbourhood U of 0 such that w $ U; by 23B, there is a p e O such that /w > 0. So w $ F(p) = F(p)dd [15Fb; recalling that by 22Eb E is Archimedean]. Let z e F(/>)d be such 50

FATOU TOPOLOGIES

[23

that u = \z\ A w > 0. Then ueF and 0 < u ^ w. As w is arbitrary, Fis order-dense by the criterion of 15E. Q (c) Let x be any member of A. Suppose that peO, yeF(p)d, and that 0 < y <: \x\. We know that A is solid [22Be], so ye A. Choose a sequence (xn}ne$ in A such that p{y — xn) ^ 2~w for every neN.Setyn = y A |# n |,sothat0 ^ ?/„ ^ y,yneA, and p(y-yn) < 2-» for every weN. Let £ = {z:ze£+,

3 neN, z ^ yt V i ^ n}9

so that J5 f, B c J. and 2/ is an upper bound for B. 1 Suppose, if possible, that y 4= sup B. Then there is a w < y which is also an upper bound for B.Asy — w>0 and y — we F(p)d, y — w<£ F(p) i.e. p(y — w) > 0. Let neN be such t h a t 2~n < \p{y — w).

Since every lower bound of {yi :i ^ n} is less than or equal to w, ^ inf {ym — z:m^n, by 15C. So mfm>n{ym-w)+ So

z ^ yi V i ^ n} = 0

= 0, i.e.

y - w = sup m ^(i/ -ymvw)

= sup m>n Bupm>i>w(y -

and Vi) < So indeed JBf 2/ and yeJA c ^4. (d) Thus D = {y:yeF,0^y^

P(y-w). X

\x\) ^ A.

But as F is order-dense, Z)f |#| so |a;| G^4. As .4 is solid, ^ G 4 ; which proves the proposition. *23M Lemma Let E be any Riesz space. Then %S{E~, E) and %S(EX, E) are Levi topologies on E~ and £ x respectively. Proof Suppose that A is a non-empty subset of E~, directed upwards and bounded for %S(E~, E). Then sup {fx :feA} < 00 for every x e £ + . By 16Db, sup A exists in Er, and 4. is bounded above in E~. If 4 c Ex, then sup 4 G £ X because Ex is a band in E~; so 4. is bounded above infix. 51

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*23N Exercises (a) Let £ be a Riesz space with a locally convex Fatou topology %. Then % is defined by the continuous Fatou seminorms on E. (b) Let £ be a Riesz space with a Fatou topology. Let A be a closed totally bounded set in £. Then A is order-closed. (c) Let £ be a Riesz space with a Fatou topology. Then 0 has a neighbourhood basis consisting of solid order-closed sets. (d) A locally solid Levi linear space topology on an Archimedean Riesz space is Hausdorff. (e) Let £ be a Riesz space with a Levi linear space topology such that order-bounded sets are bounded. Then £ has the 'boundedness property5 i.e. If B c E+ is such that for every sequence (xn}nelx in B and every sequence <#w)weN decreasing to 0 in R, the set {<xnxn: n e N} is bounded above, then B is bounded above. (f) Conversely, if £ has the boundedness property, then any sequentially complete compatible linear space topology on £ must be Levi. (g) Let £ be a Dedekind complete Riesz space and F a band in £. Let % be a Levi linear space topology on £. Then the topology on F induced by % is Levi. (h) Let £ be a Riesz space with a locally solid topology. Then %b(E', E) is Fatou. In particular, if £ has a Riesz norm, the dualnorm on £' is Fatou. Notes and comments Fatou topologies are conspicuous for two reasons; first, they are very common (see the examples in 24C, §26, 2XA-2XC, 2XF-G, 23Nh above, 42Re, 63K); second, one can prove deep results about them. We are now beginning to see some really interesting interactions between topological, algebraic and order properties. The results of §22, while sometimes requiring a moment's thought, are all more or less to be expected. But in this section even the simplest arguments bear watching. *The name 'Levi topology' is suggested by B.Levi's theorem [26Hj/63Ma]. The term 'Fatou norm' is taken from LUXEMBURG & ZAANEN B.F.S. [note XIII, § 41], and the phrase 'Fatou topology' seems to follow naturally, PERESSINI [chapter 4, § 1, pp. 138 et seq.] uses both concepts, but only in Dedekind complete spaces; a space with a Fatou topology is called locally order complete, whereas one with a Levi topology is called boundedly order complete [PERESSINI, p. 139], 52

FATOU TOPOLOGIES

[23

The best tool we have for handling Fatou topologies is 23B. (Actually the proof seems odder than the result.) The same method, starting from 22Gc instead of 22C, produces 23Na. But one of the difficulties with Fatou topologies is that there is no effective dual characterization known in the locally convex case. Consequently we are denied the techniques which normally make locally convex topologies easier to handle. The lemmas leading up to Nakano's theorem, particularly 23G and 23H, are slightly more precise than is necessary for the arguments here; but I have left them in these forms because there exist applications in which the precision is useful. Nakano's theorem itself has a rather curious position. I t is a very striking result and is pleasing for the way in which it relates ordercompleteness to Cauchy completeness. Its conditions are satisfied by a great variety of important cases, including most of the examples cited above. But in essentially all of these much easier proofs are available. Suppose, for instance, that the topology on E is metrizable; then a simple adaptation of 23D proves the result; and in fact we can do rather better [25Nb]. And 81C is another example of a special case in which a quite different method can be applied. Indeed, general theorems about Fatou topologies are often of this type. Even 23B is usually bypassed by particular cases, because most Fatou topologies are specified directly by Fatou pseudo-norms. Or consider 23L. This is interesting in various ways, not least because of the construction of the order-dense solid linear subspace F in parts (a)-(b) of the proof, which is often useful in analysing Fatou topologies. But it will actually be applied only to a Dedekind complete normed space [65D], in which case 23D again gives the result at once.

24 Lebesgue topologies The last condition of this kind which we study is in some ways the most interesting. A topology on a partially ordered space is 'Lebesgue' if every closed set is order-closed in the sense of 11D. The consequences of this property are so important that in the rest of this book I shall tend to classify topological Riesz spaces according to whether or not they possess it. I begin by running through the elementary general properties of Lebesgue linear space topologies on Riesz spaces [up to 24G]. These are subtle but not difficult. The rest of the section hinges on an effective necessary and sufficient condition for a topology on an Archimedean 53

24]

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Riesz space to be Lebesgue [241]; 'effective' in the sense that its corollaries 24J and 24K are useful. 24A Definition Let £ be a Riesz space. A linear space topology on £ is Lebesgue if 0 c: A j 0 => OeZ. 24B Theorem Let £ be a Riesz space with a Lebesgue linear space topology. (a) I f 4 c £ , / i c l and 2A c A. (b) 0 has a base of neighbourhoods U such that JTJ £ [7. (c) If 0 ci J.10 in £, and C7 is a neighbourhood of 0, then there is an xeA such that [ — x,x] c JJ. (d) If 0 c 4 | 0 in £, then J ^ l ) -> 0. Proofs

(a) and (b) are easy.

(c) Let F be a closed neighbourhood of 0 such that V—V^ U. Then there is an xe A such that [0,x] c F. P ? Otherwise, for each XEA, set Ax = Aft [0, x] and Bx = {z:3yeAx,

y ^ z ^ x}.

Now ^ J , 0, so for each w e [0, a], {w;Vj/ :yeAx} | w, and we3)Bx c JS^. Thus [0, a:] c JJ^. But we are supposing that [0, #] $ V, and F is closed, so Bx $ F. Set C = {Z:ZEE\V,

Now (i) V ^e-4 (ii) V zeC

3 a,

3 zeC, z ^ x, 3 yEA,y ^ z,

(iii) ^4^0. It follows that Cj 0. But 0$C. X Q Now [ - x, x] = [0, x] - [0, #] c Y -Y <^ U, as required. (d) follows at once. 24C Corollary A Lebesgue locally solid linear space topology is Fatou. 54

LEBESGUE TOPOLOGIES

[24

24D Theorem Let E be an Archimedean Riesz space with a Lebesgue linear space topology. (a) If A is a non-empty subset of E which is directed upwards and bounded above, 3F{A f) is Cauchy. (b) Order-bounded sets in E are bounded. Proof (a) Let B be the set of upper bounds of A in £. Because A\, {B — A)y, as E is Archimedean, inf (B — A) = 0 [15C]. So, given any neighbourhood U of 0 in £, there is an x0e A and a,yoeB such that [24Bc]. Now suppose that xv x2eA and that xx ^ x0, x2 ^ #0. Then |#i — #2| ^ Vo — %o, &o x± — x2e U. As U is arbitrary, ^(A f) is Cauchy. (b) Suppose that 5 c £ is order-bounded; that is, there exist x < y such that B c [x,«/]. We know that {oc(y — x): a > 0}! 0 in £ because E is Archimedean [15D]. So for any neighbourhood U of 0 there is an a > 0 such that [0,oc(y-x)] c [7, i.e. [ 0 , y - a ] c a" 1 ?/ [24Bc]. Thus [0, y — x] is bounded. So £ c: a; + [0, y — x] must be bounded. 24E Corollary If £ is a Riesz space with a compatible Lebesgue linear space topology under which it is complete as a uniform space, it is Dedekind complete. Proof

21Ba, 24Da, 21Bd.

24F Corollary If £ is an Archimedean Riesz space with a Lebesgue linear space topology, then £ ' c fix. Proof L e t / e £ ' . Then/is bounded on order-bounded sets [24Db], s o / e E~[16C]. Consider | / | in £~. If 0 c A\ 0 in £, and e > 0, then U = {y:yeE, \fy\ < e}is a neighbourhood of 0, so there is an xeA such that [ - x, x] c C7 [24Bc]. Now |/1 (») < e [16Ea]. As e is arbitrary, infX€A | / | (x) = 0; as J. is arbitrary, | / | is order-continuous [14Ec], i.e. | / | efix.S o / e £ x , as required [16H]. 24G Proposition Let £ be a Riesz space and % a locally convex locally solid linear space topology on £ such that £ ' c £x. Then % is Lebesgue. Proof Suppose that 0 ^ A^OinE. Then 0 belongs to the closure of A for the weak topology %S(E, £'). P Suppose (fi}i
24]

TOPOLOGICAL RIESZ SPACES

sequence in £', and e > 0. Then/ = 2i<» \fi\ is defined in £ x ; so there is an xeA such that fx ^ e, and now \fix\ ^ e for every i ^ n. Q Consequently 0 must belong to the closed convex hull of A for %. Now let U be any solid neighbourhood of 0 for %; U meets the convex hull of A, so there exist finite sequences (x^)^neH -> 0; (ii) whenever 0 cz A\ in £, and A is bounded above, ^(A\) is Cauchy. Proof (a) Suppose that condition (i) holds. Let us say that a sequence (x^)ie^ has property Pr if it is bounded above and mfieJxi = 0 whenever « / c N has r +1 members. Then the sequences with property Pr are convergent to 0. P Induce on r. For r = 0, only the constant zero sequence has property Pr; so the result is trivial. ? So suppose, if possible, that (Xi}iex is a sequence in £+, bounded above by x say, which has property Pr+1 but is not convergent to zero. Then there is a neighbourhood U of 0 and a subsequence <2/iXeN o f <^XeN such that y^U for every i e N . Note that <^XeN still has property Pr+1. Choose inductively a sequence <1^X€N of neighbourhoods of 0 such that Vo + Vo £= U and Vi+1 + Vi+1 £ Vt for every ieN. I shall construct inductively sequences (w^)iE^ and <2#)^en in such a way that: (a) (z#)jeN has property i^.+1 and is bounded above by a; for every ieN; 56

LEBESGUE TOPOLOGIES (/?) (7) (5) (e)

[24

zit $ Vo + Yi for every i, j e N; 0 < wi ^ zi0 for every i e N ; wi^V0 for every i e N; wk A z,y = 0 whenever j e N and & < i e N.

Construction Set zoj = ^. for every j e N . Then (zoi}ielK has properties (a) and (/?) because TQ + T J C C7. NOW suppose J€N and (wk)k 0 such that y?[0, #] c j ^ . Set wi = (^0 — ^ ) + . Then zi0 — wi = ziQ so wi^V0 because zio^"Po + TJ. Now consider the sequence Because <%)^eN has property Pr+l9 this sequence has property Pr, and by the inductive hypothesis is convergent to 0. So there is an meN such that Zy hfi^^-ZioGVi+ifor every j ^ m. Set ^"^o)+

V j G N.

Then clearly {zi+1j)^e^ has property i^.+1. Next, and «ifi + m §t Fo + T^+1 + F£+1, so If & < i a n d j e N , while finally

so by 14Kb t^ A zi+lj = 0. Thus the process continues. But now, if k < i, 0 ^ wk A wi < ^ A zi0 = 0; so (w^tex is a disjoint sequence in £+, bounded above by x; but it is not convergent to 0. X Thus every sequence with property Pr+1 is convergent to 0; and the induction continues. Q (b) (i) => (ii) ? Suppose, if possible, that (i) is true but that there is a non-empty set A, directed upwards and bounded above, such that ^{A \) is not Cauchy. Then there is a neighbourhood U of 0 such that

57

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for any zeA. Choose a sequence (zn}neV in A such that, for each neN, zn+1 ^ zn and z n+1 -zn$U. Let w be an upper bound for A. Let V be a neighbourhood of 0 such that V+V c (7. Since [0, u - z0] is bounded, there is an integer reN such that Now define Then <^>ieN has property Pr. P Suppose that J g N has r + 1 members. Set y = infie j ^ . Then

so [using 14Ja] = tf A({U"Z0)"

So by (a) above <^) ieH -> 0 and there is an i e N such that ^ e F. But now

contradicting the choice of (z^)ie^. X (c) (ii) => (i) If (xnynex is a disjoint sequence in £+, bounded above by x, set

A=

{^ii
Then 0 c i f and JL is also bounded above by x [14Kc], So ^(A f) is Cauchy, i.e. (Si< n ^) n e N i s a Cauchy sequence. So (xn}ne]n -> 0. *24I Corollary Let £ be an Archimedean Riesz space, and % a linear space topology on E. Then % is Lebesgue iff (i) order-bounded sets are bounded (ii) % has a base of neighbourhoods U of 0 such that JTJ ^ U (iii) whenever (xn}neV is a disjoint sequence in E+ which is bounded above, (xn}nelx -> 0 for %. Proof (a) If % is Lebesgue, it satisfies conditions (i) and (ii) by 24Bb and 24Db, and condition (iii) by 24Da and 24H(ii) => (i). (b) Conversely, suppose that % satisfies the conditions. Suppose that 0 <= A\ 0 in E, and let U be any neighbourhood of 0 for %. Let 58

LEBESGUE TOPOLOGIES

[24

V be a neighbourhood of 0 such that JY c U. Now by 24H, inverted, ^{A\) is Cauchy; let x e A be such that # - y e F whenever ye A and y ^ x. Now as .

and x e / F c U. As C7 is arbitrary, OeA; as J. is arbitrary, £ is Lebesgue. *24J Corollary Let E be an Archimedean Riesz space and % a Fatou topology on E. Then % is Lebesgue iff whenever (xn}nels( is a disjoint sequence in £ + which is bounded above, (xn}neV -> 0 for %. Proof

Immediate from 241.

*24K Corollary Let E be an Archimedean Riesz space and % a locally convex linear space topology on E. Then % is Lebesgue iff (i) E' c £x? (ji) whenever (xn}neV is a disjoint sequence in JE+, bounded above, <#w>neH -> 0 for 2 . Proof (a) If % is Lebesgue, it satisfies the conditions (i) and (ii) by 24F and 241. (b) The point is that condition (i) above implies both (i) and (ii) of 241. If E' c £x, then order-bounded sets are %S{E, £>l>ounded, therefore ^-bounded. At the same time, if A c Ex has polar A0 c £, then JA° c A0. P Suppose that 0 c JS c ^o a n ( j ^ a t £ f ^ in £. Then 0, so for any/e-4, as-y) = 0, and |/a;| ^ sup^^l/^l ^ 1. A s / i s arbitrary, XG^4°. Q SO if ?7 is any closed absolutely convex neighbourhood of 0 for %,JTJ<^ U00 == U. So X satisfies the conditions of 241, and is Lebesgue. 24L Exercises (a) Let £ be a Riesz space with a Lebesgue linear space topology. Then any continuous Riesz pseudo-norm on E is a Fatou pseudo-norm. (b) Let £ be a Riesz space with a compatible linear space topology X for which order-bounded sets are relatively compact. Then % is Lebesgue and E is Dedekind complete. [Use 2lBe.] *(c) Let E be an Archimedean Riesz space with a metrizable Lebesgue linear space topology. Then E has the countable sup 59

24]

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property. [Show that if A f x, there is a sequence in A converging to x\ or, use 241 and 18D.] *(d) Let E be an Archimedean Riesz space with a Lebesgue linear space topology %. Then there is a locally solid linear space topology on E which is Lebesgue and finer than %. [Hint: 25F, proof.] *(e) Let £ be a Riesz space with a complete compatible linear space topology % satisfying the conditions of 24H. Then % is Lebesgue. [Use 21Bd.] *(£) Let £ be a Riesz space with a locally solid linear space topology % satisfying the conditions of 24H. Then the topology %A on its completion £A [22F] is Lebesgue. [Hint: show that %A also satisfies the condition 24H(ii).] (g) Let £ be a Dedekind cr-complete Riesz space with a Lebesgue linear space topology %. Let (xn}nBjx be an order-bounded sequence in (xn)ne^->x £ such that x = inf n6H sup <>n = s u p ^ i n f ^ a v Then for %. Notes and comments Lebesgue topologies are a perpetual source of surprises. The early results of this section show how quite simple arguments can have unexpected consequences; moreover, subject to suitable modifications, many of them can be generalized to much wider contexts. The subtlest appears to be 24Bc; this is the foundation of the most interesting results. 24F and 24G seem to sort out locally solid locally convex Lebesgue topologies, which are of course the most important ones. Note that the condition 'locally solid' in 24G cannot be dispensed with [8XB]. However, the general condition 241 seems to add something even in this case. In § 82 this condition is tied in with duality theory in some general remarks on locally convex Lebesgue topologies. 24H is a very powerful result. Apart from its application to 241 above, it has a striking consequence in 81H. Perhaps it would be profitable to study spaces satisfying the conditions of 24H in their own right. The locally solid ones are of course dealt with by 24Lf. Of course there are many important examples of Lebesgue topologies. Some are given in 1XD, 1XF, 26B, 2XB, 2XC, 2XF, 2XG, 63K, 6XF, 6X1 and 82H. *The name 'Lebesgue topology' is suggested by 24Lg, which is an abstract version of Lebesgue's Dominated Convergence Theorem [63Md]. It corresponds to 'condition A(ii)' of LUXEMBURG & ZAANEN B.F.S. [note x, § 33]. NAKANO L.T. [§ 6] gives a definition of' continuous 60

LEBESGUE TOPOLOGIES

[24

topology' which is equivalent to my' sequentially Lebesgue topology' [83 Ka], but in a context which ensures that it will be actually Lebesgue [NAKANO L.T., theorem 6.2]. 25 Complete metrizable topologies As in other branches of functional analysis, Banach spaces have a preeminent position in the theory of topological Riesz spaces. Banach lattices, that is, Riesz spaces endowed with Riesz norms under which they are complete, were the first topological Riesz spaces to be studied, and remain the most important examples. One of the purposes of this chapter is to exhibit their properties in such a way that it is clear which aspects of their structure are involved at each point. In the present section I give those results which are consequences of the closed graph theorem and therefore apply to all complete metrizable linear space topologies. We find that a Riesz space structure on a Banach space, if related to the topology at all, is very strongly related; the same is true, of course, of other kinds of algebraic structure. If we restrict our attention to spaces in which the topology is compatible, they seem to divide naturally into three types: (i) not locally solid; (ii) locally solid, not Lebesgue; (iii) locally solid and Lebesgue. The first class will not concern us, though I gave an example in 2XD. The other two are both important. Since we know from 25A that the topology can be regarded as defined by the Riesz space structure, it is natural to ask what Riesz space properties correspond to the classification above. The division between (i) and the others can be expressed directly in terms of a concept, 'uniform completeness', which will be useful elsewhere; so I include this result [25K]. In the locally convex case, the three classes are distinguished very simply by the nature of the dual space [25G,25M]. 25A Proposition Let £ be a Riesz space. Then there is at most one compatible complete metrizable linear space topology on E, and for any such topology (U n E+) — (U ft E+) is a neighbourhood of 0 whenever U is. Proof Let % and %' be two compatible complete metrizable linear space topologies on E. Choose sequences (U^)ne^ and (Vn}nex of balanced sets in E such that {Un: n e N} is a neighbourhood basis at 0 for X, 61

25]

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{Vn: n e N} is a neighbourhood basis at 0 for %', Un+1 + Un+1^Un Set

Wn =

and Vn+1 + Vn+lc:Vn V neN.

{Un(\VnnE*-)-(Un(\VnnE+).

Then for every n e N, T^ is balanced, absorbent [because E + — £+ = £], Wn+1 + Wn+X s TTn, and Tfn+1 <=ffwn Fn. So {TFn: n e N } is a neighbourhood basis at 0 for a metrizable linear space topology © finer than either % or %'. But © is complete. P Let (xn)neIS be a sequence in £ such that x n+i~~xnEWn for every n e N . Then we can choose sequences (yn}nem and neN such that

y n et7 w nJ;nlS+

zneUn(]Vn(]E+, yn-zn

= xn+1-xn.

V neN.

Now 2ieif2/t exists for both X and %'; let the limits be y and t/' respectively. Since <2i>n2/iXieNt and S is compatible, V = sup n e N S i < w % [2lBd]; but the same applies to 2/', soy = yf. Also2/ — Si which must be included in Un + Un + Un; so

y-^KnVi

= y'-I*
n (vn+vn+vn) n £ + s Tfn_2

for every n ^ 2, and 2/ = S ieN j/ i for ©. Similarly, z = S^ eN ^ exists for <3, and <^n>neN -> y - 2 + ir0 for ©. As © is metrizable, this is enough to show that it is complete. Q But now, by the closed graph theorem, © = %' = %; which proves both assertions of the proposition. 25B Corollary Let £ be a Riesz space with a compatible complete metrizable linear space topology, and suppose that (xn}nelx is a sequence in £ converging to 0. Then there exist non-negative sequences <2/n>neN a n d (zn)neK> a l s o converging to 0, such that xn = yn-zn for every n e N . Proof Let (Uk}ke1Si be a decreasing sequence of sets forming a neighbourhood basis at 0 in £, and let (nkykE^ be a strictly increasing sequence in N such that xie(Uk n E+) — (Uk 0 E+) whenever i ^ nk. Now, for nk ^ i < nk+1, choose yi and zi in Uk n £ + such that xi = yi — zi. (For i < n0, choose ^ and ^ arbitrarily such that xi = yi — z^) Then (y^)ie^ and <^)^eN are the required sequences. Note I t is not the case that <#+>neN and <^>neN necessarily converge to 0. 62

COMPLETE METRIZABLE TOPOLOGIES

[25

25C Corollary Let £ be a Riesz space with a compatible complete metrizable linear space topology, and let (xn}ne1!l be a sequence in £ converging to 0. Then <# n ) neN has a subsequence (j/n}n€^ such that {2n \yn\: neN} is bounded above. Proof As £ is metrizable, (xn)nelx has a subsequence neN such that <4*yw>well -> 0. Now let {wn)n^ and nGN b e non-negative sequences, convergent to 0, such that wn — zn — ^nyn for every neN. As {wn + zn: n e N} is bounded, and £ is complete, y = SneN %~n(wn + 2n) exists in £; as X is compatible, i/ ^ %~n{wn + zn) ^ ^n|2/n| f ° r ^very weN, as required. 25D Proposition Let £ be a Riesz space with a compatible complete metrizable linear space topology, and F a Riesz space with a linear space topology such that order-bounded sets are bounded. Then every increasing linear map T: E -> F is continuous. Proof Let (xn}n€ix be a sequence converging to 0 in £. Let <2/n>neN be a subsequence of <#n}neN s ^ h ^ a t {2 n |i/ n | :?ieN} has an upper bound y say [25C]. Then so i.e.

-Ty^2"Tyn^T n

Tyne2~ [-Ty,Ty]

V

Since, by hypothesis, [ — Ti/, Ti/] is bounded in F, (Tyn}neV -> 0. But as £ is metrizable, this is enough to show that T is continuous. 25E Corollary Let £ be a Riesz space with a compatible complete metrizable linear space topology. Then £~ c £'. 25F Proposition Let £ be a Riesz space with a compatible complete metrizable linear space topology % for which order-bounded sets are bounded. Then % is locally solid. Proof

If U is any neighbourhood of 0 for %, set

Wu = {x:[-\z\,\x\]<=:U}. Then Wv is absorbent [as order-bounded sets are bounded],

63

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[using 14Bn and 14Jb to show that [-|& + y|, |» + y | ] s [ - H , N ] + [-|y|, \y\] forevery

x,yeE]9

Wv is solid, and Wv c U. So {Wn: U is a neighbourhood of 0 for %} is a neighbourhood basis at 0 for a locally solid linear space topology © 2 %. But now order-bounded sets are still bounded for © [22Bd] so by 25D the identity map (E, %) -> (E, ©) is continuous, and © = £. So 3; must have been locally solid. 25G Corollary Let E be a Riesz space with a locally convex compatible complete metrizable linear space topology %. Then % is locally solid iff E' c E~, and in this case Ef = E~. Proof (a) If % is locally solid, then E' c E~ by 22D; but we also know that E~ c E' by 25E above, so E' = E~. (b) Conversely, if E' £ E~, then order-bounded sets, which are certainly bounded for the weak topology %S(E, E~), must be bounded for %S(E, E'), and therefore for %; so by 25F % is locally solid. 25H Let us now investigate a general condition for the compatible complete metrizable linear space topology (if any) on a given Riesz space to be locally solid. (a) Definition Let E be a Riesz space with an order unit e [definition: 141]. Define || ||e: E -» R by ||#||e = inf {a: a ^ 0, \x\ < ae}, this being defined because e is an order unit. Then || ||e is a Fatou seminorm, and if E is Archimedean it is a norm. The topology it defines is Levi; indeed, all bounded sets are order-bounded. (b) Definition Let E be an Archimedean Riesz space. For each e e E + , let Ee be the solid linear subspace of E generated by e, i.e. Ee = {x: 3 neN, \x\ < ne}. Then Ee is in itself an Archimedean Riesz space with an order unit, e. Now we say that E is uniformly complete if Ee is complete under the norm || ||e for every e e E + . 251 I give a couple of sufficient conditions for a space to be uniformly complete. 64

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[25

Lemma Let £ be a Riesz space with a compatible complete linear space topology for which order-bounded sets are bounded. Then E is uniformly complete. Proof Let eeE+, and let (xn}nex be a sequence in Ee which is Cauchy for || || e. As the unit ball of Ee is [ — e, e], which is bounded in E, the identity map Ee -> E is continuous; so (xn}nejx is Cauchy in E, and has a limit x say in £. Next, the unit ball of Ee is closed in E, because the topology on E is compatible; it follows at once that (xn}nelx converges to x for || ||e. [For given e > 0, there is an neN such that xie{z:\\z — Xj\\e ^ e} for every i, j ^ n; but this is closed in E, so xe{z: Iz-x^ ^ e} for every j ^ n.] As (xnyne1x is arbitrary, Ee is complete; as e is arbitrary, E is uniformly complete. 25J Lemma complete.

A Dedekind cr-complete Riesz space E is uniformly

Proof Let e e E+. Then Ee is also Dedekind or-complete. Let (xn)ne1x be a sequence in Ee such that \\xn+1 — xn\\e ^ 2~n for every n e N. Then {#w: w e N} is bounded, therefore order-bounded. So <#n}wen converges to s u p ^ i n f ^ ^ , by 23D. 25K Proposition Let E be a Riesz space with a compatible complete metrizable linear space topology X. Then % is locally solid iff E is uniformly complete. Proof (a) If X is locally solid, then E must be uniformly complete by 251. (b) Suppose that E is uniformly complete. Then each set [ — e,e], where eeE+, is bounded. P For consider the identity map Ee-> E. Suppose that (xn}neVS is a sequence in Ee which converges both for || ||e and for %. Let x be its limit for || ||e and x' its limit for %. Set

so that 0. Now we know that xiG[x — ane,x + ane] for every i ^ n, and this set is closed for % because % is compatible; so

x'e[x-ane, FTR

3

x + ane], 65

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i.e. |x — x' | ^ <xne for every n e N. Because £ is certainly Archimedean [21Ba], x = x'. This shows that the identity map (£e, || ||e) -> (£,£) has closed graph. By the closed graph theorem (this is where we use the hypothesis that E is both complete and uniformly complete), the identity map £ e -> E is continuous. So the unit ball of £e, [ —e, e], must be bounded in £. Q But this shows that all order-bounded sets in E are bounded for %; and % is locally solid by 25F. 25L I now proceed to examine Lebesgue compatible complete metrizable linear space topologies. Proposition Let £ be a Riesz space with a compatible complete metrizable Lebesgue linear space topology %. Then (a) SisFatou; (b) E is Dedekind complete; *(c) E has the countable sup property; (d) if F is any Archimedean Riesz space, L x (£; F) = L~(E; F) [definitions: 16B, 16G]. Proof (a) By 21Ba, E is Archimedean. So by 24Db, orderbounded sets are bounded. By 25F, % is locally solid; by 24C, it is Fatou. (b) follows from 24E. *(c) is given as an exercise in 24Lc. (d) Suppose that T: E -> F is an increasing linear map and that 0 c: A\ 0 in £. Then there is a sequence (xn}ne$ in A converging to 0 in E, simply because 0 eA. This has a subsequence <2/n)neN such that {2nyn: n e N} is bounded above by y say [25C]. So

mixeATx ^ infn^Tyn ^ inf n e N 2-% = 0 as F is Archimedean. As A is arbitrary, this shows that T is ordercontinuous and belongs to L x (£; F); as T is arbitrary, L~{E; F) c L><(£; F). 25M Corollary Let £ be a Riesz space with a locally convex compatible complete metrizable linear space topology %. Then % is Lebesgue iff £' c Ex, and in this case £' = £ x = £~. 66

COMPLETE METRIZABLE TOPOLOGIES

[25

Proof If % is Lebesgue, then it is locally solid [25La], so E' = E~ [25G]; also, E~ = Ex by 25Ld. Conversely, if E' c £x? then % is locally solid by 25G again, and therefore Lebesgue by 24G. 25N Exercises (a) If £ is a Eiesz space with a metrizable locally solid linear space topology for which increasing Cauchy sequences are convergent, it is complete. *(b) If £ is a uniformly complete Riesz space with a Levi metrizable locally solid linear space topology, it is complete. (c) If £ is a Riesz space with a Levi metrizable locally solid linear space topology, and F is a Riesz space with a linear space topology such that order-bounded sets are bounded, then every increasing linear map from £ to F is continuous. (d) Let £ be a Riesz space with a barrelled locally solid linear space topology. Then £ ' is a band in £~. (e) Let £ and F be Riesz spaces with compatible complete metrizable linear space topologies. Then every increasing linear map from £ to F is continuous. *(f) Let £ be a Riesz space with a compatible complete metrizable linear space topology. Then the lattice operations on £ have closed graphs. *(g) Let £ be a Riesz space with a locally solid complete metrizable linear space topology. Then a set A c £ is closed iff it is relatively uniformly closed in £, that is, if A n Ee is || |[e-closed for every ee £ + [notations as 25H]. *(h) Let £ be an Archimedean Riesz space and F a solid linear subspace of £. Then the quotient Riesz space E/F [construction: 14G] is Archimedean iff F is relatively uniformly closed in £. (i) A solid linear subspace of a uniformly complete Archimedean Riesz space is uniformly complete. *(j) A relatively uniformly closed Riesz subspace of a uniformly complete Archimedean Riesz space is uniformly complete. *(k) Let £ be a uniformly complete Archimedean Riesz space, and F a solid linear subspace of £ such that the quotient space E/F is Archimedean. Then EjF is uniformly complete. Notes and comments The closed graph theorem is employed in three places above; 25A, 25K (in part (b) of the proof), and 25Ne. 25A is really two separate propositions, run together merely because 3-2

67

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TOPOLOGICAL RIESZ SPACES

they can be proved simultaneously. The uniqueness of the topology is a special case of 25Ne; 25Ne is got from the closed graph theorem and 25C by the method used in 25K; 25C is a corollary of 25B; and 25B is the other half of 25A. Of course the use of the closed graph theorem in 25K is just another special case of 25Ne. At a number of points [25D, 251,25Nc] I refer to topologies for which order-bounded sets are bounded. The examples we have seen are locally solid topologies [22Bd], %8(E, Er) [25G, proof], andLebesgue topologies on Archimedean spaces [24Db]. The first two both have the property that the positive cone is 'normal' [see the end-note to § 22], which is sufficient to make order-bounded sets bounded. The property of uniform completeness is a half-way house on the way to Dedekind er-completeness which will turn up again in §81, where it has an application very similar to that of 25K. Apart from spaces satisfying the criteria of 251 to 25J, observe that every C(X) [A2A] is uniformly complete; more examples follow from 25Ni and 25Nk. In fact it is clear that most important Riesz spaces are uniformly complete; the major exceptions we shall see are the spaces [§42]. 26 L-spaces and M-spaces In this section we shall study two special classes of Banach lattices. They are both abstractly defined versions of spaces which are important in measure theory; Z-spaces are essentially the same as i>1-spaces [§§52, 63], while M-spaces with units are essentially the same as C6-spaces [A2B], and include i°°-spaces [§§43, 52, 62] as a subclass. In some ways they are opposite ends of the spectrum of spaces which interest use; at the same time they are dual to each other [26C, 26D]. Thus they are leading cases in the development of the general theory. They also have important special qualities; those of M-spaces are fairly straightforward, but .L-spaces have many remarkable properties, among which 26E-26G are outstanding. 26A Definitions (a) A Banach lattice is a pair (£, || ||) where E is a Riesz space and || || is a Riesz norm on E under which E is complete. (b) An L-space is a Banach lattice E such that

II*+ 2/11= INI + \\y\\ V*,s/eE+. (c) An M-space is a Banach lattice E such that

\\xvy\\=ma,x(\\x\\,\\y\\) V x,yeE+. 68

L-SPACES AND M-SPACES

[26

(d) An M-space with unit is an if-space E in which the unit ball has a least upper bound, e say, for which ||e|| < 1. In this case, the unit ball of E must be [ — e, e], and || || = || ||e in the notation of 25Ha. e is called the unit of E. Conversely, of course, if £ is a uniformly complete Riesz space with an order unit e, then under || \\e E is an M-space with unit. In an M-space with unit, the norm topology is Fatou and Levi [25Ha]. 26B Theorem Then

Let E be an L-space and % the norm topology on E.

(a) !£ is Levi andLebesgue; (b) E is Dedekind complete; *(c) E has the countable sup property; *(d) any order-dense Riesz subspace of E is super-order-dense. Proof (a) Let A be a non-empty set in E+ such that A f and A is bounded. Then ^{A\) is Cauchy. P Let a = supa.G^||#|| < oo. For every e > 0, there is an xe A such that \x\ ^ a — e. Now if ye A and y>x, ||y —o;|| = \y\ -1|#|| ^ a - ( a - e ) = e,by the additive property of || ||. So the diameter of y^ x} is less than or equal to 2e, and this set belongs to &"(A f). Q As E is complete, 3F[A \) has a limit xQ say; by 2lBd, x0 = sup A. (Of course % is compatible by 22Ea.) Thus % is Levi. Now suppose that B is a non-empty set in JS+ such that B\ 0. Let xQeB and set A — {x0 — x: x eB, x < xQ}. Then A\x0 and A is certainly bounded, so by the argument above ^(A^-^XQ and xoeA. Subtracting both sides from xQ we see that OeB. As B is arbitrary, % is Lebesgue. (b) and (c) now follow from 25L; (d) follows immediately from (c), or otherwise. 26C Theorem Let E be an £-space. Then E' = E~ = Ex is an M-space with unit; its unit is the functional J given by

69

26]

TOPOLOGICAL RIESZ SPACES

Proof. The three spaces are equal by 25M and 26Ba above. The additive property of the norm of E means that there is a linear functional j : E -> R as defined above, which is plainly in the positive cone of E~ = £'. Now suppose that/e £'. Then I/I < a => \fx\ < a||»|| = ajx V xeE+ => -ocj | / | < a j in E~

=> |M < |/| (N) < a/N= «ll N II =«NI V*eE - li/ll < «• So Il/H = inf {a:a ^ 0, | / | < aj}, and J must be an order unit of E~ which defines the norm of Er = £'. As £' is certainly complete under its norm, it is an M-space with unit. 26D Proposition Let G be any Riesz space with a Riesz norm || || such that ||a:Vy|| = max(|]a;||, \\y\\) for all x, yeG+. Then G' is an L-space. Proof Recall that G' is a solid linear subspace of G~ [22D], and the argument of 22D shows also that the norm of G' is a Riesz norm [22Ga], so Gr is a Banach lattice. Now suppose that/ and g are nonnegative elements of G'. Of course ||/+<7|| ^ ||/|| + 11^11- On the other hand, given e > 0, we canfindx and y in the unit ball of G such that \fx\>\f\-e, \gy\>\\g\\-e. Setz = \x\ v \y\; then ||z| = max(|| 1*11, || \y\\\) = max(||x||, \\y\\) < 1. So

\\f+9\\Xf+9)(z)=fz + 9z

>f(\*\)+g(\v\)>\ffl As e is arbitrary, \\f+g\\ = ||/|| +1|^||. Thus G' is an i-space. *26E We come now to a very special property of .L-spaces. Theorem Let E be an i-space, and F a Dedekind complete Riesz space with a locally convex locally solid Levi linear space topology. Then L~(£; F) is precisely the set of continuous linear maps from EtoF. 70

L-SPACES AND M-SPAGES

[26

Proof (a) We know already that every member of L~(E; F) is continuous, by 25D. (b) Now suppose that T: E -> Fis a continuous linear map. To show that it is in L~(E; F), I shall use the approach of 16F. Let xe E+ and write Ax = {Zi
= x}.

Then Ax\ by 16F. Moreover, it is bounded. P Let V be any convex neighbourhood of 0 in F, and let U be a solid neighbourhood of 0 included in V. Then there is an cc e R + such that Tyecc\\y\\U

V yeE,

sothat \Ty\ea\y\TJca\y\V V yeE. Now if z is any member of ^4^. there is a finite sequence (xi)i
in E+

af^ll F £ (£*
a

finite sequence in £+ such that Yii
Now ^ f |T| (x) [16F], and the norm on F is Fatou, so

\\\T\{x)\\=mV{\\z\\:zeAx}. On the other hand, if z e Ax, then z can be expressed as Si< »| Txi \, where Tn 0 for every i < w. So

IN < Si<»||^|| < S«»«|*il = «i*||. 71

26]

TOPOLOGICAL RIESZ SPACES

Thus

xeE+.

|| | T | (x) || ^ a || x\\ for every

Now suppose that \S\ ^ |T| in L~(E; F). Then

for every x e E. So || 8\\ < a = || T\\. As S and I7 are arbitrary, the norm of L~(E; F) is a Riesz norm. Q (b) Consequently it is a Fatou norm. P Suppose that 0 <= A f To in L~(E; F)+. Then, for any xeE, {T{\X\):TGA}\TQ{\X\)

by 16Db. So ||ro*|| = ||\T,X\\\ As x is arbitrary, || ro|| ^ sup T e ^ \\T\\. Of course, as the norm of L~( E; F) is a Riesz norm, ||ro|| = sup Te ^||T|j. Q (c) Finally, suppose that T belongs to the closure of L~(E; F) in L(E; F). Then there is a sequence (Tn)nel!l in L~(E; F) such that ||T n -T\ ^ 2~n for every

Now

neN.

IH2W - 1^1 II = III K+i\ - | r j 11| ^ || \Tn+1-Tn\ ||

so <|3Pj)nGN has a limit 8 say in L(E; F) [this is where we use the hypothesis that F is complete]. Now for any xe E+, and

Sx-Tx

= limf^fl0(|3Tw| (x)-Tnx)

> 0,

so 8 and 8 — T are both increasing linear maps, and T = 8—(S — T) belongs to L~(E; F). As T is arbitrary, L~(E; F) is closed. *26G Corollary Let £ be an i-space and F a Dedekind complete Riesz space with a Fatou norm which generates a Levi topology. Then L~(E; F) = LX(E; F) is precisely the space L(E; F) of continuous linear maps from E to F, and the norm on L(E; F) is Fatou; moreover, the topology it generates is Levi. Proof By 26E,L(£; F) = L~(E; F);by25Ld, L~{E; F) = L*{E; F). F is complete under its norm by Nakano's theorem [23K+] or 25Nb, +

72

See the remarks on Nakano's theorem in the end-note to §23.

L-SPACES AND M-SPACES

[26

so by 26F the norm on L(£; F) = L~(E; F) is Fatou. Finally, suppose that A is a non-empty bounded set in L(E; F) which is directed upwards. Then for every xeE+ the set {Tx:TeA} is bounded and directed upwards. As the topology on F is Levi, sixpTeATx always exists, and by the criterion of 16Db A is bounded above in L~( E; F). As A is arbitrary, this shows that the topology of L(E; F) is Levi. 26H Exercises (a) Let £ be a Eiesz space with a Fatou norm || || such that ||# + 2/|| = ||#||+||y|| for every x, yeE+. Then the norm topology on E is Lebesgue. (b) Let £ be a Eiesz space with a linear space topology %. Suppose that e eint £+ for %. Then e is an order unit of £. If moreover orderbounded sets of E are bounded for %, then % is precisely the topology generated by the seminorm || ||e. (c) Let G be a Dedekind complete If-space with unit and E any Riesz space with a locally solid linear space topology. Then every continuous linear map T: E-+ G belongs to L~(E; G). (d) Let G be a Dedekind complete M-space with unit and E a Banach lattice. Then L~(E; G) = L(E; G) and the norm on L(E; G)is Fatou and Levi. (e) Let G be an if-space. Let (xi)i 0. Then *(f) Let G be an if-space. Let i g G b e a non-empty compact set. Then sup A exists and {sup 2?: 0 c= B c A) is compact. [Hints: (i) show that {sup.5:0 c: B g A, Bfinite} is totally bounded; (ii) show that the map jBt-»supB is continuous for the Hausdorff uniformity on the closed subsets of A.] *(g) Let £ be a Riesz space with a Hausdorff linear space topology % for which £+ has a 'compact base' i.e. there is a n / e £* such that {x:xeE+,fx = 1} is non-empty and compact for X. Then, writing ||#|| =J(|o;[) for every xeE, (£, || ||) is an i-space. *(h) Let G be an M-space with unit in which the norm topology is Lebesgue. Then G is finite-dimensional. [Hint: use 15G to show that if G is an Archimedean Riesz space which is not finite-dimensional there is an infinite disjoint sequence in G + ; now use 241.] *(i) Use 25Na to shorten part (c) of the proof of 26F. 73

26] TOPOLOGICAL RIESZ SPACES (j) (B.Levi's theorem.) Let E be an i-space, and (xn}ne^ an increasing sequence in E such that a = sup neN jxn < oo. Then x = supneNo:n exists in E and jx = a. (k) (Fatou's lemma). Let E be an Z-space, and <#n>neN a bounded sequence in £ + . Then x = sup neN inf On a^ exists in E, and (1) (Lebesgue's Dominated Convergence Theorem.) Let E be an L-space, and (xn}neV an order-bounded sequence in £ such that ^

= x.

Then limn_^aD( xn exists and is equal to fx. [See 24Lg.] Notes and comments For simple examples of .L-spaces and M-spaces, see 2XB, 2XA and A2B. More sophisticated examples must await the chapters on measure theory, though the duals of these M-spaces are already non-trivial ^-spaces. Both i-spaces and M-spaces with units are Banach lattices with Fatou Levi topologies, //-spaces are always Dedekind complete; M-spaces with units need not be [A2B], though many are (e.g. the dual of any i-space, and the most important L°°-spaces [2XA, § 53]). i-spaces are always Lebesgue, M-spaces with units hardly ever are [26Hh]. For both L-spaces and M-spaces with units effective representation theorems are available. Every L-space can be represented as the L1-space of a suitable measure space (in fact a Radon measure space [§73]); see KAKTJTANI or SEMADENI [§26.3]. Thus the results 26Hj26H1, abstract versions of the famous convergence theorems, are in fact equivalent to them. While every M-space with unit is isomorphic to a space C(X), where X is a compact Hausdorff topological space; this result is rather easier, and is given by KELLEY & ISTAMIOKA [24.6] and by SCHAEFER T.V.S. [V.

8.5].

Many proofs of 26E have been given; see, for example, CHACON & KRENGEL. Clearly 26E and 26Hc are connected, for if T is a continuous linear map with domain an L-space, its transpose T' has range an M-space, and vice versa. The conditions of 26E cannot be relaxed at all easily; the counterexample 2XE shows some of the difficulties. In particular, the range space F does have to be locally convex, as is shown by an ingenious example due to PRYCE U.S., unfortunately too sophisticated for inclusion here. And it can also be shown that the result is characteristic of L-spaces, in that any Banach lattice E for 74

L-SPACES AND M-SPAGES

[26

which 26E is true has an equivalent norm under which it is an L-space. [Satz 7.1] has shown that, under hypotheses similar to those of 26E, the space of compact linear operators from E to F is also a Riesz space.

KRENGEL

2X Examples for Chapter 2 Most of this section will be taken up with descriptions of familiar sequence spaces, showing how their natural Riesz space structures interact with topologies that can be imposed on them. The spaces f00,11 and c0, with their norm topologies, are dealt with in 2XA-2XC. Many of their properties can be deduced from the general theorems in this book; however, I have tried here to set out more elementary proofs, as these show some of the same ideas in simplified forms, and a comparison is often edifying. In 2XF and 2XG I show how we may use the ideas of this chapter to throw light on other topologies which arise naturally in the theory of linear topological spaces. 2XD and 2XE are more in the nature of counterexamples. 2XD is just a simple example of a compatible complete metrizable linear space topology which is not locally solid. 2XE is more important; it marks one of the boundaries of the theorem 26E. 2XA The space /°° (X) Let X be any non-empty set. Let ^(X) be the set of all bounded functions from X to R, i.e. the solid linear subspace of Kx generated by e where e(t) = 1 for every teX. Then is a Dedekind complete Riesz space in which e is an order unit. Onl*(X), || IU or I| ||e, given by II^Hoo = Bvp{\x(t)\:teX} = inf{ot:\x\ ^ ae} is a Fatou norm [25Ha]. /°°(Z) is complete under || ||^[25J, or directly], so is an M-space with unit [26Ab]. The norm topology is Levi [25Ha]. (/G0(X))x can be identified with l\X) [2XB below], so Z°°(X) is a perfect Riesz space [33A]. */°°(X) has the countable sup property iff X is countable; the norm topology is Lebesgue iff X is finite [cf. 26Hh]. [See also 2XE.] 2XB The space P(JC) Let X be any non-empty set. Let l\X) be the set of all functions x: X -> R such that

H i = 2*6x1^(01 <°°75

2X]

TOPOLOGICAL RIESZ SPACES

Then l\X) is a solid linear subspace of R x , so is in its own right a Dedekind complete Riesz space. || || x is a Riesz norm on P-(X) such that

ll^ + ylli== Nli + iMli f o r a11 X>V ^ °- ^ W i s complete under || \r p. 100; or use 25Na], so it is an L-space. There is a natural duality between lx(X) and /°°(X) given by

[TAYLOR,

the sum being absolutely convergent for every x e V-(X) and y e l°°(X). This identifies /°°(X), as normed Riesz space, with [26C; the identification l°°(X) = I 1 ^ ) ' is proved by KOTHE, 14.7.8, and TAYLOR, theorem 4.32-A; or we may use 65B/6XD to prove that At the same time it identifies P(Z) with /°°(X)x. P This is a consequence of 33F. Alternatively, it can be proved directly as follows. (a) Suppose that x > 0 in P-(X) and that 0 c B\ 0 in /°°(Z). Then 0 = inf B in Rx, i.e. in%(0:2/e5} = 0 V teX. Fix y 0 G ^

an(

i ^ et ^ > 0. Then there is a finite I ^ X such that

Since, for each tel, there is a, yeB such that i/(£) < e, and since there is a yx e B such that ^x ^ y0 and j/1(^) ^ e for every £ e / . Now

As e is arbitrary, m£yeB(x, y} = 0. Thus x acts on Z°°(X) as a member of /CO(X)X. Since Z°°(X) certainly separates its points, /1(X) may be regarded as embedded in /°°(X)X. (b) Now suppose t h a t / ^ 0 in t°(X)\ For each teX, let et be the member of V°(X) given by et(t) = 1, et(
x{t)=f(et) 76

V teX.

EXAMPLES FOR CHAPTER 2 e

[2X

Then if / ^ X is finite, £ t e/ t is defined in /°°(X) and 0 ^ Yitei t ^ e> where e is the unit of /°°(X). So

As / is arbitrary

Stex^W

e

^fe
and x e P(X). Now consider gr G /°°(X) X defined by gr^/ = (#> 2/) f° r ©very yelQO(X). We know that gr(et) = f(et) for every i e l . But suppose that y is any member of /°°(X)+. Consider * = { S t « i y ( * K : J s Z f /finite}. Then/z = grz for every zeB, and 5 f i/. So, a s / a n d g are both ordercontinuous, , « /«/ = sup 0ei? /« = sup^^flfz = gy. As ?/ is arbitrary, / = g. Thus every positive element of /°°(iL)x can be represented by an element of lx{X)\ it follows that the duality between /°°(X) and induces a linear space isomorphism between /°°(X)X and V-{X). (c) Finally, we see that for xeP-(X)9

x^Oo

x(t) ^ 0 V teX o (x,et) ^ 0 V teX

so that the ordering on P-(X) induced by its embedding in /°°(X)X is correct, and the isomorphism is a Riesz space isomorphism. Q (The argument above is of course a particularly easy special case of thatin65A.) Hence l\X) is perfect [33A, 33F]. 2XG The space co(X) Let X be any non-empty set. Let cQ(X) be the set of all functions x: X -> R such that

{t:teX,\x(t)\ >e} is finite for every e > 0. Then cQ(X) is a solid linear subspace of /°°(X) [2XA above]. It is easy to see that co(X) is closed under || J^, so (co(X), I || oo) is a Dedekind complete Riesz space with a Fatou norm under which it is complete. The norm topology is Lebesgue; but it will be Levi only if X is finite. (So co(X) does not satisfy the hypotheses of Nakano's theorem, 23K.) Because /°°(X) is an M-space, so is co(X) [26Ac]. 77

2X]

TOPOLOGICAL RIESZ SPACES

*Note that a subset of cQ(X) is relatively compact for || H^ iff it is order-bounded; cf. 26Hf. [See also 2XF, 4XCb.] 2XD

Let E be the set of sequences x e RN such that ||a|| =m&x(\x(O)\,su-pie}x\x(i+l)-x(i)\)

< oo.

Then E is an order-dense Riesz subspace of RN (for if x e E then \x\ e E and || \x\ || ^ ||aj||). || || is a norm on E under which £ is a Banach space isomorphic (as Banach space) with f°(N). The topology % derived from || || is compatible (since for each neN \x{n)\ ^ (n+l)\\x\\

V XEE,

so the coordinate map x\->x(n) is continuous). £ has an order unit e givenby

e(n) = n + l V » 6 N .

But the interval [ — e, e] is not bounded, so % is an example of a compatible complete metrizable linear space topology which is not locally solid. Since [ — e,e] includes the unit ball, % must be Levi; but of course it is not Lebesgue [24Db]. Exercise

Show that the lattice operations on E are not continuous.

*2XE An example for 26E Recall that the space I2 = Z2(N), defined by 19 c J /2 = {x: is a Hilbert space under the inner product Clearly it is also a solid linear subspace of RN, and its norm is a Riesz norm. Moreover, it is easy to see (adapting the arguments of 2XB, or using 65A) that the inner product identifies I2 with (/2)x, and therefore that the norm topology is Lebesgue [25M]; alternatively, it is equally easy to show directly that the norm topology is Lebesgue, and therefore that /2 - (/2)' = (/2)x [25M]. [See also 6XG.] Now let us examine H = L~(/2; / 2 ). We know that H c L(/2; I2), the space of continuous linear endomorphisms of /2 [25D]. The norm of L(/2; /2) induces a metrizable linear space topology on H. The positive cone of H i s 78

{T:TeL(P;

I2), Tx > 0 V ^ O } ,

EXAMPLES FOR CHAPTER 2 [2X which must be closed in L(P; I2), and therefore in H, as the positive cone of l% itself is closed. Next, if

because the norm on I2 is a Riesz norm. So the interval [0, T] is bounded in H; it follows that all order-bounded sets are bounded. However, given neN, there is a 2n x 2n matrix An = (ocnij} such that \ani:}\ = 1 for all i, j < 2n and Yi]Coinikanjk = 2n i£i=j,

0 otherwise.

P Set Ao = (1) and

K for each neN.Q

Consider An as a linear transformation of /2, given by ( A * * ) (t) = i:j<2nocnijx(j) = 0

V i < 2 ^ w

Vi^2 .

2)

Then 2-^ A w is unitary, so V On the other hand, it is easy to see that | An\ is represented by the 2n x 2n matrix with all entries 1. So Thus is a bounded set in H which has a solid hull which is not bounded. So H cannot be complete in its norm [25F, 22Bc]. In particular, H + L(l2; I2). Thus L(l2; I2) $ L~(l2; I2). So we cannot substitute I2 for E in 26E. Exercise The norm topology on H is not Lebesgue or Fatou; but the unit ball is order-closed. *2XF I shall conclude this section with a description of a couple of less familiar topological Riesz spaces exemplifying the concepts of this chapter. 79

2X]

TOPOLOGIGAL RIESZ SPACES

The topology ^(Z00,11) For a non-empty set X, consider the Mackey topology on/00 = /°°(X) derived from its duality with/ 1 = ^(X) [2XB above]. There is a well-known characterization of the S£8(P, Incompact sets when X = N [seeKOTHE 22.4.3],which extends naturally to the general case in the form: A c I1 is relatively ^(Z 1 , Z°°)-compact iff (i) it is || [^-bounded, (ii) for every e > 0 there is a finite I ^ X such [This is 83F below.] This makes it obvious that the solid convex hull of A is still relatively weakly compact, from which it follows that ^(Z00,11) is locally solid. [See also 82H.] So it must be Lebesgue [24G, 82F]. A ^(Z 00 , Z1)-bounded set must be £S(Z°°, /^-bounded; by the uniform boundedness theorem (Z00 being (I1)'), it is || ^-bounded, i.e. orderbounded. Thus £&(Z°°, I1) must be Levi. [See also 33B.] I t is also well known that Z00 is complete under %k(lco, I1); this could be proved as a corollary of Nakano's theorem [23K], but the usual proof is from the fact that Z00 is the dual of the Banach space Z1 [see A IE]. Thus Sfc(Z°°, I1) is a complete HausdorfF Lebesgue Levi Fatou topology. *2XG The topology %k(P-, c0) Now let us consider the Mackey topology on Z1 = lx(X) derived from the duality between Z1 and c0 = co(X) [2XC]. I t is coarser than the Lebesgue topology derived from || || x [2XB], so it must be Lebesgue. It has the same bounded sets as %S{11, c0), which (because I1 is the dual of c0) has the same bounded sets as || || x; so, like || l^, it must be Levi. Finally, it is locally solid. P First method Let B be the unit ball of Z00 = Z°°(Z). Then B can be identified with [— 1, l ] x . Now it is easy to see that the product topology of B is the topology on B induced by %s{lco, Z1), under which B is compact. Consider the map (x9y)\-+ xy: B x B -> B, where we define (xy)(t) = x(t)y(t) for every teX. Then this is continuous for the product topology. So if A c B is compact for %s(lcc, Z1), {xy:xeA,yeB} is also compact. But this is precisely the solid hull of A in Z00. Now suppose that U c: I1 is a neighbourhood of 0 for S^Z1, c0). Then its polar U° c c0 is compact for %s{c0,11) or %s(lcoy I1). So certainly U° is bounded for ||fl^,and there is an a > 0 such that <xU° c B. Now the 80

EXAMPLES FOR CHAPTER 2

[2X

solid hull C of ocU° is compact, andD = arxG, which is the solid hull of U° itself, is compact, But D is still included in c0. We must now observe that by Krein's theorem [AlG], the convex hull of D is relatively compact for %s(cQ, I1). So D°, the polar of D in I1, is a neighbourhood of 0 for %k(P-9 c0); but of course it is solid, so it is the required solid neighbourhood of 0 included in U. Second method I t is in fact not difficult to show that a set A c c0 is relatively %s(c0, /^-compact iff: (i) A is bounded for || H^; (ii) whenever (^>^eN is a sequence of distinct points in X, = 0.

From this it follows at once that the solid hull of a relatively compact set in c0 is relatively compact. The rest of the work proceeds as above. Q Now we can say that I1 is complete under S^/ 1 , c0), either by Nakano's theorem [23K], or by AlE. (The results above have been generalized by MOORE & REBER.) Further reading for Chapter 2 The principal omission from this chapter has been the concept of 'normal cone'; at the same time, generalizations to the theory of partially ordered topological linear spaces have hardly been considered. Both of these subjects are tackled by SCHAEFER T.V.S. and PERESSINI. For further ideas on these lines, and for results concerning complete metrizable partially ordered linear spaces, see JAMESON. There is an alternative introduction to the elementary theory of L- and if-spaces in KELLY & NAMIOKA, with a few additional results. For more about normed Lebesgue and Fatou topologies, see LUXEMBURG & ZAANEN B.F.S. [notes x-xvi].

81

3 . Dual spaces In the theory of linear topological spaces, the concept of a dual space is of paramount importance. In this short chapter I supplement this theory with a brief discussion of the dual spaces E~ and £ x which we have already seen, and which are fundamental to the ideas of the rest of the book. The most important new results concern the canonical evaluation map from a Riesz space £ to the space of linear functionals on £x [32B below]. In the third section I give an equally abbreviated note on 'perfect' Riesz spaces, which include the majority of the spaces which interest us. Further general results concerning spaces in duality will be derived in the course of Chapter 8, while studying weak compactness.

31 The space £~ This is the basic dual space associated with a given Riesz space £. Recall that so far we know that E~ is the set of linear functionals on E which are bounded on order-bounded sets, and that it is a Dedekind complete Riesz space [16C, 16D]. The importance of the results below is that they apply to the evaluation map from E into G* for any solid linear subspace G of E~; their principal application is of course in the case G = £ x . 31A Let us begin with a list of definitions, repeating some given in Chapter 1. Definitions (a) Let E be any linear space over R. Then £* will denote the linear space of all linear maps from £ to R. (b) Let £ be a Riesz space. Then E~ is the Dedekind complete Riesz space L~(£; R) [16B, 16D]. (c) Let £ be a Riesz space. Then £ x is the band L x (£; R) in £~ [16G, 16H]. (d) Let £ be a linear topological space over R. Then £' is the linear space of continuous linear maps from £ to R. (e) Suppose that £ is a linear space over R and that F is a linear 82

THE SPACE JET

subspace of £*. For any xeE,I given by

[31

shall write £ for the member of F*

£(/)=/(*)

V feF.

31B Lemma Let £ be a Riesz space; suppose t h a t / e E~+ and that a; e £ + . Then there is a g e E~ such that 0 ^ g ^ f, gx = fx, and gy = 0 whenever a; A 2/ = 0 in £. Proof

Define gr: E+ -> R by jf2 = sup {f(z Aax):oce R+} ^ fz,

as/is increasing. Then without difficulty (using nothing stronger than 14Ja), we can see that g is linear on £+ (I mean that it satisfies the conditions of 16A), so extends to become a member of JE*. The rest of its properties are immediate [using 14Kb for the last]. 31C Proposition Let £ be a Riesz space and G a solid linear subspace of E~. Then the canonical map x\->$: £->G* is a Riesz homomorphism from £ into G x . Proof TfxeE+, £G G x by 16Ec [forif 0 c A],0in G, then A),0 in £~, so m£feA%{f) = mff€Afx = 0]. As G x is a linear subspace of G*, £ G G x for every # e £. The map a:h-> ^ is plainly linear and increasing. Finally, suppose that x A y = 0 in £ and that/G G+. Then there is a gr G £~ such that 0 ^ g ^ f, gx = fx, and gy = 0 [31B]. But G is solid in E~, so ^ G G . Now

(* A 0) (/)<£(/-?)+0(0) = 0 by 16Eb. As/is arbitrary, ^ A 0 = 0; by 14Eb, this is enough to show that the map x \-> $ is a Riesz homomorphism. 31D Corollary

Let E be a Riesz space and/G £~+. Then

r, \g\
x,yeE.

Proof These follow from applying 16Ea, 16Eb to the elements £, fi of E~~, and then going back to £ with 31C; or, of course, by applying 3 IB directly. 83

31]

DUAL SPACES

*31E Exercises (a) Let £ be a Riesz space and G a locally order-dense [definition: 14F] Riesz subspace of E~. Then the canonical map x i—• £: E -> G* is a Riesz homomorphism from E to G x . (b) Let E and F be Riesz spaces and T: E-+F a Riesz homomorphism which is onto. Define T: F* -> £* by (T'gr) 0*0 =fl^s)for every xe E and #eF*. Then T'[F~] c £~ and T':F~ -> E~ is a oneto-one Riesz homomorphism. Notes and comments Of course 3IB is the central idea of this section. In effect, this asserts that E~ contains enough functionals to make 31C and 31D true. 31D and 16E together indicate a curious symmetry in the relationship between E and £~.

32 The space £ x This seems to be the most interesting of the Riesz space duals of a given Riesz space E, especially in view of the remarkable topological properties discussed in § 82 below. At the same time it turns out that, in particular cases, the ' natural' dual of £ is £ x as often as it is E~ [see, for instance, 2XA, 52E, 6XD]. Of course they are frequently identical [as in 25M], which leads to particularly powerful results. In this section I give the fundamental embedding theorem 32B and an elementary, but useful, characterization of members of £ x [32D]. 32A I begin with a vital technical result. Lemma Let £ be an Archimedean Riesz space. Let / > 0 in £ x . Then there is an x > 0 in £ such that: (i) fx > 0; (ii) 0 < y < x=>fy > 0; (iii) if g A / = 0 in £ x , gx = 0. Proof

(a) Let

F = {z:zeE,f(\z\) = 0}. Then F is a band in £ [cf. 23Fa]. But F + £ because/ # 0. So Fd * {0} [definition: 15F], because [by 15F] F = Fdd. Now if x > 0 in Fd, then 0 < y ^ x => x Ay = y > 0 => y$F => fy > 0, and of course fx > 0. 84

THE

SPACE £x

[32

(b) Now suppose that g A / = 0 in Ex; of course g > 0. Let e > 0. By 16Eb we may choose, for each weN, an element zn of E such that 0 ^ zn ^ # and For each weN, let wn = infi<w2$. Then and

g{x-wn) ^ S^nflrfc-z*) <

2e

for every neN. At the same time, (wnyne1x^, and if 0 ^ w < wn for every neN, - ^£ ^n n % so /w = 0; as 0 < w ^ x, w = 0. This shows that 0 = inf n e N ^. So = 0 because gr is order-continuous. So n

< 2e.

As e is arbitrary, gx = 0, as required. 32B Theorem Let E be an Archimedean Riesz space and G a solid linear subspace of £ x . Then the canonical map x i-» x is an ordercontinuous Riesz homomorphism from E onto an order-dense Riesz subspace of Gx. Proof (a) By 31C, xt->$: E->GX is a Riesz homomorphism. It is order-continuous because, if 0 c A\ 0 in J3, = mf^/z

= 0 V/eG+

and infjg^^ = 0 by 16Ec. So we are left considering the Riesz subspace {$ : x e E} in Gx. I shall use 15E to show that it is order-dense. (b) Perhaps I should first remark that G, and similarly G x , are Archimedean, because they are Riesz subspaces of the Dedekind complete Riesz spaces E~ and G~. Suppose that <j> > 0 in G x . Then there is a n / e G+ such that (i) / > 0, (ii) if tjr A $5 = 0 in G x , ifrf = 0 [32A]. Now there is an x e £+ such that /# > 0. Since £(/) = fx > 0, $ A ^ > 0. Because G x is Archimedean, inf{a£:a> 0} = 0 [15D], and there is an a > 0 such that > 0. 85

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DUAL SPACES

Let geG+ be such that (i)frg> 0, (ii) 0iJrh>0, (iii) if 6 A f = 0 in G x , #0 = 0 [32A again]. Let y e E+ be such that (i) gy > 0, (ii) 0gz>0, (iii) if g A h = 0 in G, hy = 0 [32A once more]. Now so $ A ^ > 0, by the choice of ^; as i/r < &, $A& > 0; by 14Kb, P A a£ > 0. Set z = y A a#; then £ = 0 A a£ by 31C, so 2 > 0. (c) Infact, 2 ^ cj). P ? Otherwise, (2 - 0)+j > 0, so there is an hoeG+ such that (i) (2 - 0)+ (fe0) > 0 (ii) 0 < h ^ h0 => (2 - 0)+ (A) > 0 [32A, for the last time]. Now examine But ^ A (a£ - §S A ^) + = 0, so (a£ - ^ A ^) + (^) = 0, by the choice of g. So h0 A g = 0, by the choice of ^0. Now, however, ho(y) = 0, by the choice of y. So (S-0)+(fto) ^ HK) = ho(z) < ho(y) = 0, which contradicts the orginal requirements on hQ. X Q (d) Thus we have 0 < 2 < <j>. As (j> is arbitrary, {$: we£} is orderdense in Gx, by 15E. 32G Corollary Let £ be a Dedekind complete Riesz space and G a solid linear subspace of JEX. Then the map x\-*& takes £ onto an order-dense solid linear subspace of Gx. Proof

Immediate from 32B and 17D.

32D The following is useful for proving that functionals belong to £ x . I t works only for Archimedean spaces. Lemma Let E be an Archimedean Riesz space, and let/e E* be such that i n f ^ \fx\ = 0 whenever 0 <= A\ 0 in E. Then/e E x . Proof Let % be the linear space topology on E defined by the seminorm x\-+\fx\\E-+ R + . The condition o n / states precisely that % is Lebesgue. So Ef c E x [24P]. But of course fe E'. Notes and comments The argument of 32A-32B is particularly concentrated, not to say opaque. I t can perhaps be clarified by following it through in a special case. Let E = co(X) [2XC], and set G = E x , which is identified with V-(X); in this case G x is identified 86

THE

SPACE E*

[32

with /°°(Z) [2XB]. Now suppose that 0 > 0 in G x , as in part (b) of the proof of 32B. Let A be the set {t: t e X, $(t) > 0}. Then/ must be such that { and x such that C = {t:x(t) > 0 } c 5 , Since x > 0, (7 4= 0 , and obviously ^ A & > 0, as (0 A £) (0 > 0 for every teC. Now if D = {t:teX, f{t) > 0}, we see that {t:z(t) > 0} c {*:*/(£) > 0} c {*:flr(*) > 0}

Since z(£) ^ a#(£) for every t, it is obvious that % ^ ^ A ^. Of course this sudden conclusion of the argument relies on the successful identification of £ as a subspace of Gx. But in fact the essential idea here, which is to identify the relationship in the lemma 32A with properties of the sets A, B, etc., above, is widely applicable. 65A exhibits a more general situation in which it can be used. Readers familiar with the Radon-Nikodym theorem will recognize, in the construction oftfr,an old device. It will be used again for the Radon-Nikodym theorem itself [52D below]. The ideas of this section are taken from LUXEMBURG & ZAANEN B.F.S. [note VII, §§ 27-8]. These authors are interested in a space intermediate between Ex and JE~, the space of functions fe £* such that <*»>n6H+0 in E => n6l->0 & ZAANEN B.F.S., note vi, § 20]. This also is a band in E~. In its properties it appears to be a pale shadow of £ x ; for this reason I ignore it in most of this book, though it is never far away [see 42P, 83K]. See also LUXEMBURG. [LUXEMBURG

*33 Perfect Riesz spaces Here we have a class of Riesz spaces which includes a large proportion of those arising naturally in analysis. A Riesz space E is 'perfect' iff the evaluation map from E to £ x x is an isomorphism between E and £ x x . This section concentrates on methods of showing that Riesz spaces are perfect. The consequences of the property will be dealt with later, especially in § 82. 87

33]

DUAL SPACES

33A Definition A Riesz space E is perfect if the canonical map \ E -> (JEX)X [31Ae, 32B] is a Riesz space isomorphism. Remarks If E is perfect, it is Dedekind complete (for Exx, being a solid linear subspace of the Dedekind complete space E x ~, is always Dedekind complete), and Ex must separate the points of E (as the evaluation map is one-to-one). These conditions are not enough, as they are satisfied by co(N), which is not perfect [2XC]. But the concept of 'Levi topology', already used in §23, provides us with a necessary and sufficient condition. 33B Proposition Let £ be a Dedekind complete Riesz space and G a solid linear subspace of Ex. Then the canonical map x H-> &: E -> G x is one-to-one and onto iff the weak topology %S(E, G) is Levi [definition: 231]. Proof (a) If the map X H J : E -> Gx is one-to-one and onto, it is a Riesz space isomorphism, by 31C. And we know by 23M that %S(GX, G) is Levi; so %S(E, G) must be Levi. (b) Conversely, suppose that %S(E, G) is Levi. Then the map x — i >$ is one-to-one. P If x 4= 0 in E, then \x\ > 0, so A = {n\x\ :neN} is not bounded above [because E is Dedekind complete and therefore Archimedean], But A is directed upwards, so it cannot be bounded for %S(E, G); and there is a n / e G such that/|#| =f= 0. By 31C,

|»|(/)=/(M),

so & 4= 0. Q At the same time, the map x h-> & is onto. P Suppose that ^ ^ 0 in Gx, and let , f ' i = {a Then for any/e(? s u p ^ |/r| < 0(|/|) < oo, so that A is bounded for £s(22, G); by 31C again, -4 f, so -4 is bounded above in E; as E is Dedekind complete, x0 = sup J. exists in E. But now using the full strength of 32B. Thus {%: x E E} includes G x+ ; but it is a linear subspace, so must be x G. Q 33C Corollary iff£ s (E,E x ) 88

A Dedekind complete Riesz space E is perfect

PERFECT RIESZ SPACES

33D Corollary

[33

If £ is any Riesz space, E~ and Ex are perfect.

Proof For %S(E~, £~ x ) is finer than the Levi topology %8(E~, E) [31C, 23M] and is therefore Levi. The same argument applies to £ x . 33E Corollary If £ is a Dedekind complete Riesz space with a locally convex Lebesgue Levi linear space topology %, E is perfect. Proof Since % and %8(E, E') have the same bounded sets, %8(E9 E') is Levi. But Ef c £ x [24F], so %8(E, £ x ) is Levi. 33F

Corollary

Any .L-space is perfect.

Proof 26B. 33G Exercises (a) Let E be any Riesz space and F a perfect Riesz space. Then L~(E; F) and LX(E; F) are perfect. (b) If £ is a perfect Riesz space and F is a band in £, then F is perfect. *(c) Let £ be a Dedekind complete Riesz space such that £ x separates the points of £, and suppose that there is a sequence <Xi>7ieN i n

E+

V xeE 3 weN, \x\ ^ xn. Then £ is perfect. Notes and comments The original' vollkommene Raume', studied in a series of papers by G. Kothe, were sequence spaces (that is to say, subspaces of RN); they were defined in a way which we can now recognize [see 6XD below] as the form £ x , for an arbitrary solid linear subspace £ of RN; so Kothe's first theorem was that £ x x x = £ x . His most important results are repeated in KOTHE, § 30. They were taken up by DIEUDOKNE, who generalized them to include spaces of equivalence classes of measurable functions, as in § 65 below. The abstract approach above comes from LUXEMBURG & ZAANEN B.F.S. [note vn, § 28]; the two systems are linked by FREMLIN A.K.S. II, in which I gave a representation theorem showing that an arbitrary perfect Riesz space is isomorphic to one of Dieudonne's 'espaces de Kothe'. Examples for Chapter 3 I give no special examples for this chapter. Whenever we encounter a Riesz space £, one of our first tasks is to seek to identify £~ and £ x . The latter is generally easier, especially 89

33]

DUAL SPACES

when we can use 65A or 65B. Important examples from other chapters include the elementary sequence spaces [2XA-2XC], function spaces [§ 65], and, for some insight into less familiar cases, 8XA-8XC. Further reading for Chapter 3 Much the most important reference for this work is LTJXEMBUBG & ZAANEN B.F.S. [notes vi-x].

90

4 . Riesz spaces on Boolean rings The main purpose of this book is to show how the abstract theory of Riesz spaces may be applied to the study of function spaces arising in measure theory. In Chapters 1-3 I have set out the most important concepts needed for this approach. But in this chapter I propose to open up another line of attack. My eventual aim is to describe the relationship between the measure algebra of a measure space and its function spaces. In order to do this, I demonstrate methods of constructing Riesz spaces from Boolean algebras which, when applied to measure algebras, will produce isomorphic copies of the basic function spaces L 1 and L00. At the same time I shall apply the concepts of the first three chapters to describe the properties of these Riesz spaces. The technical problems encountered along the way are considerable. An intuitive understanding, however, of the basic constructions S and L00, is not hard to attain; this is because some relatively easy examples already offer most of the principal aspects of the theory [4XA-4XD]. The construction L # is essentially deeper, and requires faith in Chapters 1-3 to be meaningful at all; its real significance will not appear until § 62. One of the advantages of this method is that all the constructions are functors, and behave reasonably when the right kind of homomorphism is applied. This will enable us to see deeper into one of the more puzzling aspects of measure theory; details are in §§45, 54 and 61E. For readers who have not encountered measure algebras before, I recommend taking § 61 in parallel with this chapter.

41 Boolean rings The theory of Boolean rings is an abstract framework for the study of rings of sets. If X is any set, then its power set 0*X, with the operations of symmetric difference and intersection, forms a Boolean ring [4XA]. Conversely, any Boolean ring is isomorphic to a 'ring of sets', i.e. a subring of some SPX [41D]. A ring of sets has a natural ordering <=, under which it is a lattice. Since A c BiSA ()B = A, this lattice structure can be applied to any Boolean ring [41F]. 91

41]

RIESZ SPACES ON BOOLEAN RINGS

41A Definition A Boolean ring is a ring (91, + , . ) such that a2 = a for every a e 31. [Examples: 4XA, 4XCb, 4XF.] 41B Lemma If 91 is a Boolean ring: (a) a + a = 0 V ae9l; (b) ab = ba V a, Proof (a) a + a = (a + a)2 = a (b) a + b = (a + b)2 = so

ab = —ba = ba.

41C Lemma Let 91 be a Boolean ring, and a a non-zero element of 91. Let / be an ideal of 91 which is maximal subject to a$I. Then (a) if ce9l and ace/, t h e n c e / ; (b) the ideal generated by a and / is 91; (c) / is maximal. Proof

If ce 91, the set

is an ideal (because 91 is commutative), includes / , and contains c (because c + c2 = c + c = 0). So it is the ideal generated by / and c. (a) If c$I, then a belongs to the ideal generated by / and c, so a + acel; but we are supposing that a$I, so ac$I. Conversely, if ac does belong to / , so does c. (b) In fact, ifc is any member of 91, then a(c + ca) = (a + a2)c = 0 e / . So c + ca e /, and c belongs to the ideal of 91 generated by a and / . (c) Now any ideal of 91 properly including / must contain a, by hypothesis, so is 91 itself, 41D M. H. Stone's theorem Let 91 be a Boolean ring. Then there is a set X such that 91 is isomorphic to a subring of 0*X [definition: 4XA]. Proof Let X be the set of all maximal ideals t of 91. Define ^:9t->^Xby a = {t:teX,a$t} V ae9l. 92

BOOLEAN RINGS

[41

Now if a, b e 91 and t e X, then (i) aet=>abet; (ii) a$t,b$t=>ab$t=>a + bet [applying 41Ca first to t and a, and then to t and ab, as (a + b)ab = (iii) (iv) a, So 0(a&) =
[using (ii), (iii) and (iv)]. Thus <j> is a ring homomorphism. If a is any non-zero element of 91, let #*be the set of all ideals of 91 not containing a. By Zorn's lemma, ^ h a s a maximal element t say; now by 41Cc, t is a maximal ideal of 91, i.e. teX. And t e (pa, so (pa 4= 0. Thus
^ X the Stone representation. In the next section I shall use <j> and X in further constructions, so it is important to note now that they are canonical. 41F The ordering of a Boolean ring may define a relation s on 91 by

If 91 is a Boolean ring, we

a c b o ab = a. Now we see that if ^ X is the Stone representation of 91, then a c j o ^ c ^ , From this it follows at once that £ is a partial ordering under which 91 is a lattice. The lattice operations on 91 are given by inf {a, 6} = a&, sup{a,6} = a + b + ab. Of course, these facts can readily be verified without appeal to Stone's theorem. 91 always has a least element, which is 0; it has a greatest element iff it has a multiplicative identity, and they are then the same. 41G Notation If 91 is a Boolean ring, I shall use the symbols n, U and \ for the operations on 91 corresponding to n, U and \, that is, a n b = inf (a, b} = ab, a\)b = sup{a, b} = a + b + ab, a\b = a + ab, 93

41]

RIESZ SPACES ON BOOLEAN RINGS

as well as £ defined above and its inverse 2 . 1 n a Boolean ring, the words sup and inf will always refer to the ordering c. So, for instance, 21 is Dedekind complete or cr-complete iff it is complete in the sense of 13B with respect to this ordering. If 21 is a Boolean ring, a subring 93 of 21 (i.e. a subset containing 0 and closed under addition and multiplication) is a Boolean ring in its own right. The ordering on 93 induced by the ordering on 21 coincides with the ordering on 93 derived from its own Boolean ring structure. As 93 is a sublattice of 21, the lattice operations on 93 are also those derived from 21. It is arbitrary, but convenient, to use the phrase 'Boolean algebra' to mean a Boolean ring with a 1, a multiplicative identity. A subalgebra of a Boolean algebra is now a subring containing 1. *If 21 is a Boolean ring, a a-subring of 21 is a subring which is a c-sublattice in the sense of 13F, i.e. one which is closed under countable sups and infs in so far as these exist in 21. A o-subalgebra, or o-ideal, of 21 is a (r-subring which is also a subalgebra or ideal respectively. *41H A lemma on suprema and infima Let 21 be a Boolean ring, and A a non-empty subset of 21. (a) If the set B of upper bounds of A is not empty, then inf{& + a : a e ^ , beB} = inf{6\a:ae^, beB} = 0. (b) Given a0 e 21, A f a0 iff a0 is an upper bound for A and {a + ao:aeA} = {aQ\a:aeA}^0. (c) Given a0 e 21, A ! a0 iff a0 is a lower bound for A and ao:aeA} = {a\a o :aei}|O. Proof I shall appeal to the correspondence between u, n, \ and U, fl, \ to justify intuitively the following arguments. They can, of course, be validated formally from the definitions. We observe first that, for a, b e 21, a^b

o a + 6 = b\a.

(a) If c is a lower bound for {b\a:beB, aeA}, then fix boeB. As c^bo\a V aeA,cna = 0 V a e A, So bo\c is an upper bound for A. As A #= 0 , c g bo\c and c = 0. (b) If a0 is an upper bound for A, then the map b h-> ao\b is an orderreversing involution on {6: b £ a0}. So A f a0 iff {ao\a: a e A} 10. 94

BOOLEAN RINGS

[41

(c) If a0 is a lower bound for A, then A | iff B = {a\a0: a e A} 4,. Now c is a lower bound for 5 iff c u a0 is a lower bound for A and c n a0 = 0. So if a0 = in£A, then for any lower bound c of B,c ^ aQ and coao = 0, so c = 0. Conversely, if 0 = inf B, then for any lower bound d of A, d\a0 = 0 and d £ a0.

*41I Corollary Let 31 be a Boolean ring. Then 91 is Dedekind cr-complete iff every countable non-empty set in 9t has a greatest lower bound. Proof For now if A ^ % is a non-empty countable set with an upper .. r r , .. bound aa, A u sup J. = ao\inl {ao\a: a e A). 41J Ideals and quotients of Boolean rings Observe first that if 91 is a Boolean ring, then a set / c 91 is an ideal (written / < 91) iff (i) OeJ; (ii) a,bel=>a[)bel; (iii) a c bel=>ael. For (iii) is precisely the condition that /9I £ / , and now (ii) is equivalent to / + / c: / . Obviously, if / <] 91, then 91// is a Boolean ring. Examine the identities , 7X. .,. (a6) = a o (a + 6 + afc)" = a +b' +a'b' where a\-±a: 91 -> 91// is the canonical homomorphism. These show that the canonical homomorphism is a lattice homomorphism [13E], when 91 and 91// are given their lattice structures. [See also 45A below.] Observe that a ' c j ' o a'b' = a' o ab + ael o a\bel. *41K Products If (9l i > teI is any family of Boolean rings, the product ring IK2t t Xej *s Boolean. Clearly, the lattice structure of IK2tiXe / is precisely the product of the lattice structures of the family <3t4X6l in the sense of 11G/13H. 41L Exercises (a) Let 91 be a Boolean ring. Then its lattice structure is distributive in the strong sense of 14D; that is, if A <= 91 is such that sup A exists, and if b e 91, then sup{6 n a:aeA} exists = b n sup^4. 95

41]

RIESZ SPACES ON BOOLEAN RINGS

Similarly, if inf A exists, then inf {b \ja:aeA} exists = b u inf A. *(b) If 51 is a Boolean ring and A c 2t, set Ad = {b:be%

6na = 0 V a e i } .

Then Ad is an order-closed ideal of 31. If A is an order-closed ideal of 91, then A = Add [cf. 15F]. *(c) Suppose that 21 is a Boolean ring and that 93 is a subring of 21 such that for every non-zero a e 2t there is a non-zero b e 93 such that b s a. Then a = sup{6:6e93, 6 £ a} V ae2l [cf. 15E]. *(d) If 21 is a Boolean ring and / <\ 21, the canonical map from 21 to 21// is order-continuous iff / is order-closed, and sequentially ordercontinuous iff/ is a cr-ideal [cf. 14Lb]. *(e) The topology of the Stone space Let 21 be a Boolean ring and <j>: 21 -> 0*X its Stone representation. Let % be the topology on X generated by ^[21]. Then % is Hausdorff, locally compact and totally disconnected, and ^[21] is precisely the ring of compact open subsets ofX. Notes and comments The reason for the abstract study of Boolean rings is our need to cope with quotient rings of rings of sets; just as abstract Riesz spaces are important because of the quotients of function spaces which we wish to study. Stone's representation theorem means that up to a point Boolean rings can safely be thought of as rings of sets. The point at which this conception breaks down is when we begin to discuss the suprema and infima of infinite sets. The Stone representation is hardly ever order-continuous, or even sequentially order-continuous. So, in general, if we see a supremum or aninfimum, we are safer if we forget about the representation. 41H is an example of the kind of technique we must employ; 4XF is an example of a ring in which care is necessary. In 4lLa-d is a little group of results showing analogies between Boolean rings and Archimedean Riesz spaces. Another is 41 Ha, corresponding to 15C. It appears that no satisfying account has yet been given of these.

96

THE SPACE 5(51)

[42

42 The space 5(91) This is the fundamental Riesz space associated with a given Boolean ring 91. When 91 is a ring of sets, 5(91) can be regarded as the linear space of 'simple functions' generated by the characteristic functions of members of 91 [4XB]. Its most important property is the universal mapping theorem 42C, which establishes a one-to-one correspondence between' additive' functions on 91 and linear functions on 5(91). Simple universal mapping theorems of this kind are cheap; what makes the construction important is the fact that 5(91) has a canonical normed Riesz space structure [42E]. From this we deduce that 5(91) is universal for many other classes of function. Particularly important are 'completely additive' and ' countably additive' real-valued functionals [42K et seq.].

42A Definition Let 91 be a Boolean ring, and E a linear space over R. A function v\ 91 -> E is additive if v(a + b) = va + vb whenever ab = 0 in 91. In this case i>(0 + 0) = v(0) = 0. Remark This definition is familiar when 91 is a ring of sets; of course, if a n b = ab = 0, then a + b = avb. It remains a curious concept, because the + signs on the two sides of the equation refer to very different operations. 42B Construction Let 91 be any Boolean ring, and <j>: 91 its Stone representation [41D-41E], For each ae% let %a be the characteristic function of (j>a, that is,

l if te$a,

(xa)(t) = 0 if

Let 5(91) be the linear subspace of Rx generated by {xa: a e 91}. Then it is easy to see that ^: 91 -> 5(91) is additive in the sense of 42A. 42C Theorem Let 91 be a Boolean ring, and suppose that 5 = 5(91) is constructed as above. Let E be any linear space over R. Then there is a one-to-one correspondence between additive maps v: 91 -> E and linear maps T: 5 -> E given by v = TxProof Let ^: 91 -> SPX be the Stone representation of 91. (a) Let us observe first that it a, be 91, then (j)(a n b) = xaxXb X(a ob) = FTR

4

97

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RIESZ SPACES ON BOOLEAN RINGS

where x is ordinary pointwise multiplication on Kx, i.e.

(xxy)(t) = x(t)y(t) V teX V

x,yeRx.

(b) The core of the proof is the following observation. Let v\ 91 -> E be additive. If (pci}ia0 = 0 and we consider

Now these can be re-expressed as *o = Si
Adding these three equations, 0= and the induction proceeds. Q (c) But this is precisely the condition for the existence of a function T: £ - > E given by For if "Zi
1*i E is linear, then Tx: 91 -> £ is additive, because x is additive. 98

THE SPACE 5(31)

[42

42D Notation If 91 is a Boolean ring and E a linear space, then for each additive function v: 91 -> EI shall write i> for the corresponding linear map from 5(91) to £. Thus v = vx* 42E The Riesz space structure of 5(91) Let 91 be a Boolean ring, and let 5 = 5(91) be denned as in 42B. (a) 5 is a Riesz subspace of V*(X) [2XA], where X is the Stone space of 91. (b) Every non-zero member x of 5 can be expressed in the form where (a?)i 0 o o^ > 0 V i < n,

(c) Every non-zero member x of 5+ can be expressed in the form x

= where (bjyj 0 for each j < n. In this case, *(d) Suppose that a, /?eR, a, 6e9t, and 0 < a ^ ^ fyb. Then 0 < a < /? and 0 4= a s 6. Proof Let $S: 91 -> ^ X be the Stone representation of 91. Let 5 0 be the set of those functions x: X -> R such that (i) x[X] is finite, (ii) for each a #= 0, {£: #(£) = a} G ^[91]. Then it is clear that 5 0 is a linear subspace of Kx (because ^[91] is a subring of &X), that xae^o f° r each a e 91, and that each non-zero member x of 5 0 can be expressed in the form of (b) (the oct being j ust the non-zero values taken by the function x). Consequently 5 0 = 5. But So is clearly a Riesz subspace of /°°(X), which proves (a). The identities in (b) are now elementary. For (c), express x = S i < w a i / t a i a s in (b), but with the sum so arranged that 0 < a 0 < ... < ocn_1. Set JSO = a 0 and pi = o^ — a^_x for 1 < i < n. Set bj = sup J < i < r i a i for j < n, so that xbj = Tij^ia ^ ( 4-2

99

42]

RIESZ SPACES ON BOOLEAN RINGS

42F Increasing additive functions If 21 is a Boolean ring and E a partially ordered linear space [§ 12], then a function v: 21 -> E is increasing if va ^ vb whenever a £ b [HD]. Lemma (a) An additive function v: 21 -» E is increasing iff va > 0 for every a e 21. (b) In particular, x- 21 -> *S(9l) is increasing, (c) If £ is a partially ordered linear space and v: 21 -> JE is an additive function, then *> is increasing iff j>: 5(21) ~> E is increasing. Proof (a) If v is increasing, then 0 = v(0) ^ v(a) for every ae2I. Conversely, if va ^ 0 for every a e 21, then vb = vc + v(c + b) ^ vc

whenever c £ b in 21, so *> is increasing. (b) is trivial, and (c) follows at once from either 42Eb or 42Ec. 42G Locally bounded additive functions If 21 is a Boolean ring, let us call an additive functional v:%->R locally bounded if {vb: b s a} is bounded above for each a e 21. 42H Proposition Let 21 be a Boolean ring and v: 21 -> R an additive functional. Then v is locally bounded iff v e 5(2t)~, and in this case ) where y+: 21 -> R is given by y+(a) = sup{vb:b ^ a) V ae21, and is an increasing additive functional. Proof

(a) Suppose first that v eS~. In this case, for any a e 21,

is bounded above by v+(xa), since b £ a => #& ^ ^a. Thus y is locally bounded and v+: 21 -> R is defined, with v+(a) ^ v+(X<*>) V « G 2 1 . (b) Conversely, suppose that v is locally bounded. Suppose that x > 0 in S, and consider

100

THE SPACE 5(51)

[42

We know that there is an a o e2t such that x < Moo^o [42Ec], Now if 0 < y ^ x,y can be expressed as Hn 0 and %ieS~ [16C]. And this argument shows at the same time t h a t \>+(X
[using 16D].

(c) Putting (a) and (b) together, we see that if v is locally bounded, v+(a) = v+(xa) f° r ©very a e 91. So y+ is additive and increasing and *42I Lemma Let 91 be a Boolean ring, and ^: 21 -> R an additive functional which is not locally bounded. Then there is a decreasing sequence (an}neV in 21 such that \van\ ^ n and \v(an\an+l)\ ^ ^ + 1 for every ne N. Proof

We know that there is an aQ e 21 such that sup{vb:b ^ a0} = oo.

Now we can choose (fln}ne^ inductively such that, for each n9 a

n+i s an9

and

\van\ ^ n,

\v(an\an+1)\ >n + l,

sup{vb:b ^ an} — oo.

P We already have a 0 . Given aw, choose ac^

an such t h a t

Then for 60 c a n , sup { ^ : 6 g c } + sup {^6:6g an\c}. So one of these sups must be infinite. Let an+1 be either c or an\c, such that sup {vb:b ^ an+1} = oo. Now van+1 is either vc or mw — vc; in either case, | ^ a w + i | ^ ^ + l ; at the same time, \v(an\an+i)\ ^ n+1, as required. Q

42J Order-continuous increasing additive functions: proposition

Let 21 be a Boolean ring.

101

42]

RIESZ SPACES ON BOOLEAN RINGS

(a) For any Riesz space J3, an increasing additive function v\ 21 -> E is order-continuous iff miaeAva = 0 in £ whenever 0 5(91) is order-continuous. (c) If y: 21 -> R is an increasing additive function, then it is ordercontinuous iff v: 5(21) -> R is order-continuous. Proof (a) [cf. 14Ec]. This follows at once from 41Hb and 41Hc and the corresponding results 12Ca and 12Cb. (b) ? Otherwise, there is a non-empty set A \ 0 in 2t and an x > 0 in 5 = 5(31) such that x ^ xa f° r every aeA. Now by either 42Eb or 42Ec, there is a /? > 0 and a b 4= 0 such that jiyb < x. But now, by 42Ed, b s a for every aeA, which is impossible if inf A = 0. X (c) From (b) we see that if v: 5 - > R is order-continuous, then v = \>X'- 21 -> R is order-continuous. Conversely, suppose that *> is order-continuous. To show that v is order-continuous, we must examine an arbitrary non-empty set A ^ 0 in 5. Let e > 0. F i x # 0 e A . By 42Ec, there is an ocoeR+ and an a0e91 such that #0 < ao%ao. Let ^ > 0 be such that Sva0 ^ e. For each x e S+, define bx e 21 as follows. Let ^: 21 -> ^ Z be the Stone representation of 31. Then xeRx. Let

Ex = {t:x(t)>8}. Then Ex^ X and in fact ^ e ^[21] (see the characterization of S given in the proof of 42E). So there is a unique bxe 21 such that <j)bx = JS

whenever 0 < a; < a;0. Observe also that

so that B = {6X: xeA, x < ^ 0 }\. If 6 is a lower bound for B in 21, then Sxbx < x whenever

xeA, x ^ x0.

As .410, Sxb = 0; as 5 > 0, 6 = 0 [42Ed]. Thus 5 1 0 in 21. As v is ordercontinuous, there is an xeA such that x ^ x0 and a o ^ ^ e. Now vx < ^ K A + % 0 ) = ^vbx + Sva0 ^ 2e. As e is arbitrary, infxeJpx = 0; as A is arbitrary, v is order-continuous. 102

THE SPACE S(%) [42 42K Completely additive functionals Let 21 be a Boolean ring. An additive functional v\ 2i -> R is called completely additive if infae^ \va\ = 0 whenever

0 c: A \ 0 in51.

(The reason for the phrase 'completely additive' lies in 42Rj.) 42L Theorem Let 21 be a Boolean ring. Then an additive functional v: 21 -> R is completely additive iff v e 5(2l)x. Proof (a) If v: 21 -> R is completely additive, it is locally bounded. P Use 421. ? Otherwise, there is a decreasing sequence <#n>neN in 21 such that | van\ ^ n for each neN. Let B be the set of lower bounds of {an:neN}; then Bf. By the argument of 41Ha, C = {an\b:neN, beB}^0 Let

in2l.

D = {an\b: beB, n > \vb\ + 1} c C.

Then D is cofinal with C [i.e. V c e C 3rfeD, d s c] so D j 0. But | ^ | ^ 1 for every deD. X Q (b) If y:2l->R is completely additive, y+:2t->R is ordercontinuous. P The argument is similar to that of 24Bc. Suppose that 0 <= A j 0 in 21. 1 Suppose, if possible, that there is an e > 0 such that v+a > e for every a e A. Then for every a e A there is a b c a such that z>6 > e. But So there is a C E . 4 such t h a t c^a y(& \jc) = vb + v(c\b) ^ e.

and |^(c\6)| ^ vb — e, so t h a t

Now let B = {b : b G 21, vb ^ e, 3 a,ceA such that c c ^ g a } , Since for every ae^4 there is a beB such that 6 c a, 5 ^ 0 . But inf6eB |^61 ^ e > 0, which is impossible. X Thus inf ae ^p + a must be 0; as A is arbitrary, y+ is order-continuous [42Ja]. Q (c) Now if v\ 21 -> R is completely additive, 0eS(2l)~ by (a) and 42H, and D+ = (z^+)A. But v+ is order-continuous by (b), so (v+)A is ordercontinuous [42Jc], and 0+G5(21) X . Similarly, v~= (-*>)+ = ( - ^ and

VGSX.

(d) Conversely, if j> e 5 X , and 0 c A \ 0 in 21, as |0| and ^ are both order-continuous. So y is completely additive. 103

42]

RIESZ SPACES ON BOOLEAN RINGS

*42M Countably additive functionals A slightly weaker condition than 'completely additive' is 'countably additive'. If 21 is a Boolean ring, an additive functional v: 21 -> R is countably additive if (yan}nen -> ® whenever (ctn)ne^ 0 in 21. [See also 42Rj below.] Clearly, a completely additive functional is countably additive, and an increasing additive functional is countably additive iff it is sequentially order-continuous. *42N Proposition Let 21 be a Boolean ring and v: 21 -> R a locally bounded countably additive functional. Then v is the difference of two increasing countably additive functionals. Proof We know that v+: 21->R is an additive functional [42H]. Suppose that <#w>neN ^ 0 in 21. ? Suppose, if possible, that Since certainly (y+an}n€x | , there must be an e > 0 such that v+an > e for every neN. Now for each WGN there is a b £ an such that vb > e. Since (v(b n #m))meN -* 0> there must be an m e N such that and now v(b\am) ^ e. So v+(an\am) ^ e.

Consequently, we can find a strictly increasing sequence (n(i)}iels such that +

K#))

> e,

i.e.

for every i e N. But now he V

which is impossible. X As (an)neji is arbitrary, this shows that v+ is countably additive. But similarly v~ = (— v)+ is countably additive, and v = v+ — v~ [using 42H, or otherwise]. *42O Proposition Let 21 be a Boolean ring and v\ 21->R an increasing additive functional. Then v is countably additive iff i>: 5(21) -> R is sequentially order-continuous. Proof The argument is exactly the same as that of 42Jc; the only difference is that arbitrary directed sets are replaced throughout by monotonic sequences. 104

THE SPACE 5(31)

[42

*42P These arguments can be elaborated along the lines of 42L to set up a one-to-one correspondence between locally bounded countably additive functionals v\ 91 -> R and linear functional/: 5(91) -> R such that (fxn}nelSt -> 0 whenever (^n>7ieN ^ 0 in 5(91) [see the remark at the end of §32, and 83K]. The principal difference is that, unlike completely additive functionals, countably additive functionals need not be locally bounded [42L, proof, part (a), and 4XG]. However, the most important situations are covered by the following lemma. *42Q Lemma Let 91 be a Dedekind c-complete Boolean ring, and v\ 91 -> R a countably additive functional. Then v is locally bounded. Proof ? Otherwise, there is a decreasing sequence (an)nex in 91 such {an\a:neN}^0 that (\van\}n€}i^> oo [421]. Let a = infnelxan. Then [4lHc]. But now {v{an\a))n^ -> 0, i.e. n6l -> va, which is impossible. X 42R Exercises (a) Let 91 be a Boolean ring, £ a Riesz space, and v: 91-> £ an additive function. Then v: 5(91) -> E is a Riesz homomorphism iff v is a lattice homomorphism. (b) Let 91 be a Boolean ring, E an Archimedean Riesz space, and v: 91 -» E an increasing additive function. Then 0: 5(91) -> £ is ordercontinuous iff v is order-continuous. *(c) Let 91 be a Boolean ring, E a normed linear space over R, and v\ 91 -» £ an additive function. Then i>: 5(91) -> E is continuous for the norm || H^ on 5(91)iff v is bounded. (d) Let 91 be a Boolean ring, E a Dedekind complete Riesz space, and v\ 91 -> E an additive function, (i) Then v\ 5(91) -» £ belongs to L~(S; E) iff *>+a = sup {V6:6 ^ a) exists in E for every a e 91, and in this case j>+ = (i;+)A. *(ii) And now i>eL x (5; 15)iffirdaGA\va\ = 0 in £ whenever 0 <= A \ 0 in 91. (e) For any Boolean ring 91, || |oo is a Fatou norm on 5(91). (f) Let 91 be a Boolean ring. Then these are equivalent: (i) 5(91) is Dedekind complete; (ii) 5(91) is uniformly complete; (iii) for every ae 91, {b: b £ a] is finite; 105

42]

RIESZ SPACES ON BOOLEAN RINGS

(iv) the Stone representation of 91 expresses it as the ring of finite subsets of its Stone space; (v) Every real-valued additive functional on 21 is locally bounded. (g) Let 91 be a Boolean ring, and /i and v locally bounded realvalued functionals on 31. Then 6a = sup {/ib + v(a\b) :b<^a) exists in R for every ae%. 6: 91 -> R is locally bounded and additive, and 6 = ft V v in 5(91)^. (h) Let 91 be a Boolean algebra (i.e. a Boolean ring with a 1), and /JL and v locally bounded real-valued additive functionals on 91 of which [i is increasing. Then v is in the band of S(%)~ generated by pi iff V e > 0 3 # > 0 such that /ia < S=> \va\ ^ e. [Hint: show that the set of v satisfying the condition is a band in S~ containing ft. Now use 15F and a lemma analogous to (g) above, with A replacing v.] Under these conditions, we say that v is absolutely continuous with respect to fi *(i) Let 91 be a Dedekind cr-complete Boolean ring, and /i and v locally bounded real-valued additive functionals on 91 of which /A is countably additive and increasing. Then v is in the band of 5(9l)~ generated by ju, iff it is countably additive and fia = 0 => va = 0. *(j) Let 91 be a Boolean ring and v: 91 -> R an additive function. Then (i) v is countably additive if 2nGNmw, exists and is equal to ^(sup^eua^) whenever (anyneJSS is a disjoint sequence in 91 such that sup neN a n exists (ii) v is completely additive iff TiaeAva exists and is equal to j^(sup^l) whenever A is a disjoint set in 91 such that sup A exists. [Hint for (ii): show first that if v satisfies the condition, it is locally bounded, and v+ also satisfies the condition.] Notes and comments The basic property of 5(91) is the first universal mapping theorem, 42C. It is clear that this defines 5(91), and the function x> UP to linear space isomorphism. What is remarkable is that S is endowed with a natural partial order for which the next universal mapping theorem, 42Fc, is true; that this natural order is a lattice order [42Ea]; and that the remaining universal mapping theorems [42Jc/42Rb, 420, 42Ra, 42Rc] are all true. In fact we have six distinct theorems based on a single construction. 106

THE

SPACE S(%)

[42

In 16C we saw how functions which were the difference of increasing linear maps could be characterized; this result corresponds exactly to 42H. Similarly, functionals which are the difference of order-continuous increasing additive functionals are associated with the completely additive functionals [42L, corresponding to 32D]. In the analysis here of completely and countably additive functionals we are beginning the Radon-Nikodym theorem. The reason for studying both is that in the most important applications we are presented with countably additive functionals on rings of sets; but that (as I shall endeavour to show) their most striking properties are a result of the fact that they can be regarded as completely additive functionals on quotient rings. *Of the arguments above, two are familiar from elementary measure theory. 42C is just the theorem that an additive measure on a ring of sets defines an integral on the space of 'simple functions' [cf. 4XB]. 420 is the theorem that a countably additive measure defines a sequentially order-continuous integral; the same methods yield the more thoroughgoing 42Jc. *I ought to remark there that the method of constructing 5(91) given in 42B is far from the only one. Indeed, an essentially simpler algebraic construction, not using Stone's theorem, quickly sets up a space for which 42C is true. From this point, the positive cone of 5(91) can be defined as the cone generated by {#a: a e 91}, and the other results will follow. However, such an abstract approach introduces substantial additional complications into the proofs; and the temporary avoidance of the axiom of choice seemed an inadequate compensation. 43 The space L°°(9l) If we complete 5(91) with respect to the norm || fl^, we obtain the space L°°(9l). This is a Banach lattice, in fact an M-space. It is a model for the L00 spaces of ordinary measure theory, and shares many of their properties. The main question to which we shall address ourselves here is the Dedekind completeness of L°°(9l) [43D]. 43A Definition Let 91 be a Boolean ring and X its Stone space. Then 5(91) is a Riesz subspace of/°°(Z) [42Ea]. Let £^(91) be the closure of 5(91) in /°°(X) for ||fl*,.Then L°°(9l) is a closed Riesz subspace of /°°(-X"), because the lattice operations on /°°(JL) are continuous. Under || || a,, L°°(9l) is an M-space, because /°°(.X) is. For examples, see 4XB-4XF and 43Ec below. 107

43]

RIESZ SPACES ON BOOLEAN RINGS

43B Lemma Let 21 be a Boolean ring and X its Stone space, (a) If 1: X -> R is the function with constant value 1, then for every yeL™ and a ^ 0. (b) If XELCC+, there is a sequence (xn}nGjj( in 5(9t)+ such that <Xn>neV t * and <*»>neH ">

x f

° r || || ••

(c) 5(91) is super-order-dense in L°°(9t). *(d) If x eL°°+ and 5 > 0, then {a:aeSt, ##« ^ x} is bounded above in 91. Proof (a) Observe first that if y e 5 = 5(91) and a ^ 0, ?/ A a l e 5. P For if 2/ = 2* where {ai)i
Q 00

Consequently j/AaleL for every t/eL , because the map y H-> y A a l : /°° -> /°° is continuous. (b) Given x e L°°+, there must be a sequence <2/n>neN in 5+ such that | a; —yJloo < 2>~n for every weN. Set ^ = (^-3.2-^l)+ = ^ - 2 / n A 3 . 2 - M e 5 for each WGN. Since t; also, ||a?n-a;||oo < 4 - 2 ~ 7 1 for each n, so n€H->«. By

^ n > n e P N to:.

(c) By definition, 5 is super-order-dense in L00. *(d) We know that there is a y e 5+ such that ||x - y\\ „ < S. Express V a® YiKnPjXbj where (bj}:j is the Stone representation of 91. Now x

=> #(0 ^ ^ V tefia 0 V

So {a: 5^a ^ #} is bounded above by 60. 108

THE SPACE L°°(3l) 43C Corollary norm on L°°(2t).

[43

For any Boolean ring SI, the norm || ||«, is a Fatou

Proof We know that || || & is a Riesz norm. So suppose that 0 c A \ x in L°°+. Let a = sup^^H^/Ha,. Then x A aleL 0 0 [43Ba], and clearly x A a l is an upper bound for A. So x = # A a l and [[scl^ ^ a. As J. is arbitrary, this shows that || |oo is a Fatou norm. 43D Theorem Let 21 be a Boolean ring. Then (a) L°°(2l) is Dedekind complete iffSl is; (b) L°°(2l) is Dedekind cr-complete iff 91 is. Proof The proofs of the two halves of the theorem are formally independent, but almost identical; so I shall give a proof of (a), with occasional words in brackets which may be inserted to give a proof of (b). (i) Suppose first that L00 = L°°(2l) is Dedekind (cr-) complete and that A £ 21 is a non-empty (countable) set. Then { p : a e i } has an infimum x in L00. Let yeS = 5(21) be such that 0 ^ y ^ x and \x"~ y\oo < 1 [using 43Bb]. Then there is a /? > 0 and a 60 e 21 such that [using 42Ec; if y = 0, set 60 = 0]. Now, ifaeA, J$xb0 ^ y ^ x ^ xa> ^J 42Ed, b0 £ a; thus 60 is a lower bound for A. Conversely, if 6 is a lower bound for A, ^6 is a lower bound for {xa: as A), so yb < #; now

So 6\6O = 0, i.e. b £ &0. As & is arbitrary, 60 = inf^l; as A is arbitrary, 21 is Dedekind (cr-) complete [13Ca or 411]. (ii) Now suppose that 21 is Dedekind (cr-) complete, and that A £ L°°+ is a non-empty (countable) set with an upper bound weL™ say. For each x e A, we can choose a non-empty (countable) set Bx c 5 + such that x = supl?^ [43Bb]. Now B = \JxeABx is a non-empty (countable) set bounded above by w, and A and i? have the same upper bounds in L00. To save space, let us write an = 2~~n||w\\^ for eachneN, and let G be the set of upper bounds of B in L00. We can construct inductively a sequence (yn}nejn in 5+ with the following properties: 109

43]

RIESZ SPACES ON BOOLEAN RINGS

(<*) Vn+l > Vn\

(P) ll^ + i-^IU<^ +1 ; (7) Vn <

v

for ever P Set 2/0 = 0. Having found yn, consider for each z eB the set

Ez =

{t:(z-yn)(t)>an+1}.

Since z — yneS, i^G^ [31], where
II( z -y n - a

because we are supposing that || (z - yn)+\\ < an = 2aw+1. Now D = K:2G£}

is a non-empty (countable) set in 31, and D c {a: a e 31, aw which is bounded above in 31 by 43Bd. So d = sup D exists in 31. And Xd = s\xp{xa:aeD} in 5 by 42Jb; because S is order-dense in L00 [43Bc], #d = sup{^a: aeD} in L00 [17A]. Set

yn+i =

Then of course y n+1 ^ yw and ||yw+1-yn[|oo < a n+1 . Next,

But if vG C and

as 2 is arbitrary, yn+1 ^ v. Finally,

for every 2 G £ . Thus the induction continues. Q Now (yn}nejs is an increasing Cauchy sequence in S, so has a limit yeL™; by 21Bd, y = supneN2/n. So, for any zeB, ||(2-y) + |U = lim n ^ fl0 ||( 2 -y n )+|| w = 0, 110

THE SPACE L°°($0

[43

i.e. z ^ y; thus y is an upper bound for B. On the other hand, if v is any upper bound for B,yn ^ v for every n e N, so y ^ v. Thus y = supiS = sup A. As A is arbitrary, L00 is Dedekind (or-) complete. 43E Exercises (a) If 21 is a Boolean ring, the following are equivalent: (i) 21 has a 1; (ii) 5(21) has a weak order unit; (iii) L°°(2l) has an order unit. *(b) If 21 is a Boolean ring, there is a natural one-to-one correspondence between the elements of 21 and those bands in L°°(2l) which are generated by a single element. *(c) Let 21 be a Boolean ring, and <j>: 21 -» SPX its Stone representation. Let % be the topology on X generated by ^[21], as in 4lLe. Then LG0(2l) = C0(X) [definition: A2C]. [Hint: use the Stone-Weierstrass theorem.] Notes and comments The result 43Da will be basic to the discussion of Maharam algebras [§ 53] and measure spaces [§ 64], after the relationship between measure algebras and ordinary L00 spaces has been established [62H]. *An alternative proof of 43D can be got from the ideas of 41Le/43Ec. The Stone space X of 21 has a canonical topology % under which 21 is represented as the ring of compact open sets in X [41Le]. Now it can be shown that 21 is Dedekind complete iff % is extremally disconnected (i.e. the closure of every open set is open), and that this is so iff C0(X) is Dedekind complete; but C0(X) = L°°(2l) [43Ec]. A similar characterization of Dedekind (T-complete Boolean rings is not hard to find. When 21 has a 1, it is Dedekind complete iff ^ [21] is the algebra of regular open sets of X [see 4XF]. *43Eb shows that the Boolean ring 21 is uniquely determined by the Riesz space L°°(2l). 44 The space L # Continuing our search for abstract versions of the spaces arising in measure theory, we find that approximate equivalents of the L 1 spaces can be set up as duals of the L00 spaces. The equivalence is not in general perfect, so I shall use a new symbol, and write L#(2l) for L°°(2l)x. For the moment we shall examine Lfl spaces alone. They are 111

44] RIESZ SPACES ON BOOLEAN RINGS related to L1 spaces by the Radon-Nikodym theorem, as will appear in §§52 and 63. 44A Definition Let 31 be a Boolean ring. Then I shall write for L°°(8[)x. 44B Theorem Let 21 be a Boolean ring. (a) L#(2l) is a band in L°°(9t)' = L°°(2l)~; with the induced norm, it is an L-space. (b) There is a one-to-one correspondence between members / of L#(2I) and bounded completely additive functionals v: 21 -> R, given by v = fx> / ^ 0 iff ^ is increasing, and in this case

Proof (a) We know that L00' = L°°~ is an Z-space because L00 is an if-space [43A, 25G, 26D]. Now L# = L°°x is a band in L°°~ so is a closed Riesz subspace [22Ec], and therefore is in its own right an L-space. (b) (i) Because L # c L00', and S = 5(21) is dense in L00, members of # L are determined by their values on S; so the map/ \-+fx is one-to-one [42C]. (ii) Suppose t h a t / E L # . Then/# is an additive function on 21. For anyaG9l, so/^ is bounded. Now suppose that 0 cz A \ 0 in 21. Then [xa :a in 5[42Jb], so (because S is order-dense in L00, 43Bc), {^a: a e A] \ 0 in L00 [17A]. It follows that

totatA \f(X*)\< ^aeA |/| (^) = 0, because | / | : Z/0 -> R is order-continuous. So/^ is completely additive. (iii) Conversely, suppose that v: 21 -» R is bounded and completely additive. In this case, of course, v is locally bounded and A = v+ is completely additive [42L proof, part (b)]; since Xa = sup {vb :i»cfl}^ sup [vb: 6 e 21} < oo for every a e 21, A is also bounded; let a = sup {Aa: a e 21}. Consider X: S -> R, the linear functional derived from A. Then X is continuous for || H^ and || A|| = a. P Suppose that xs S+. Then x can be expressed as izi
112

THE SPACE Lft

[44

In general, for any xeS, \Xx\ = \X(x+)-X(x-)\ ^ max(A(s+), X(x~))

^ a m a x (1^+11^ U^"-^) = oc\\ \x\ {{„ = aH Thus X is continuous, and \\X\\ < a; now, of course, || A|| = a. Now it follows that, for any x e L°°+,

Q

{Xy.yeS, 0 ^ y ^ x} is bounded above by a ||#||oo- Since X: 5-> R is order-continuous [42Jc], and S is order-dense in L00 [43Bc], there is an order-continuous increasing linear functional/: L00 -> R extending X, by 17B. Thus v+ =fx, w h e r e / G L # . Similarly, there is a geL# such that g% = (_ v)+\ and now v = (f—g) X- This completes the proof. 44C Exercises (a) Let 21 be a Boolean ring. Then there is a one-to-one correspondence between bounded additive functionals v on 21 and members/of L°°(2l)', given by v — fx*(b) In (a) above, v is countably additive iff/has the property that 00 O A E N ~> 0 whenever (xn)nelx J, 0 in L . (c) Let 21 be a Boolean ring. Then L°°(2t)~ can be identified, as normed Riesz space, with 5(21)', which is a solid linear subspace of (d) Let 21 be a Boolean ring and/GL°°(2ir.Then/+^ = (/y)+, where / + is taken in L°°(2{)~ and (/#) + is defined as in 42H. (e) Given a Boolean ring 21, define j:L# (21) -> R by J / = ||/+|| - 1 | / - | | for every/eL # . Then in the sense that V e > 0 3 6e2l

JJ/

such that

|/C\;a)-j/| < e V a ^ b.

Notes and comments The fundamental idea here is part (iii) of the proof of 44Bb;if A: 21 -> R is a bounded order-continuous increasing additive functional, then there is a corresponding order-continuous increasing linear functional on LG0(2l). This is a fairly straightforward extension of 42Jc, in which arbitrary order-continuous increasing additive functionals were associated with order-continuous increasing linear functionals on 5(21). Now the bounded completely additive functionals are identified with members of LooX, just as general completely additive functionals are identified with members of Sx [42L]. 113

44]

RIESZ SPACES ON BOOLEAN RINGS

The value of this refinement lies in the fact that, because L00 is an M-space, Ifl = Lcox is an L-spaee. Thus we shall be able to apply some of our most powerful results about Biesz spaces to the space of bounded completely additive functionals. These expressions represent Ift as a subspace of 5 X c 5~. I t is important to note that L # is actually a solid linear subspace of 5~ (though not in general a band). For L# is a solid linear subspace of 00 JL°°~ = X, ', which can be identified with 5', which is a solid linear subspace of S~ by 22D. The point of this is that the lattice operations on L # can be calculated in the same way as those on 5~, as in 42Rg or 44Cd. When the ring 21 has a 1, then obviously every locally bounded additive functional on 21 is actually bounded; so S~ = 5" £ L°°~ and 5 X ~ Lft. There are further results concerning L # in § 83. It is difficult to give convincing examples of Ifi spaces, because simple Boolean rings give rise to trivial cases [see 4XC; also 4XFj]. In fact, as Kakutani's theorem shows, all Lft spaces can be based on measure algebras [see the end-note to § 26]. So further examples must await Chapters 5 and 6. 45 Ring homomorphisms An important aspect of the constructions of the last three sections is their behaviour relative to homomorphisms between the underlying rings. This is particularly significant because the rings which are most important to us are defined as quotient rings, that is, as homomorphic images. The fundamental results are that a homomorphism from 21 to 93 gives rise to maps from 5(21) to 5(93), from L°°(2l) to L°°(93), and from L#(93) to L#(2l). In the language of category theory, 5 and L00 are covariant functors, while L # is contravariant. If 93 is a quotient of 21, 5(93) and L°°(93) appear as quotients of 5(21) and L°°(2l) respectively; this is the result we need to identify the L00 space of an ordinary measure algebra with the usual function space. 45A Definition Let 21 and 93 be Boolean rings. A map n: 21 -> 93 is a ring homomorphism if n(ab) = na.nb and n(a-\-b) = na + nb for all a, 6 G 21. In this case, the formulae of 41G show that 7r(a[)b) = n(a + b + ab) = na + nb+ na.nb = na[)nb, n(a n b) = n(ab) = na.nb = nan nb, n(a\b) = n(a + ab) = na + na.nb = na\nb 114

RING HOMOMORPHISMS

[45

for all a,fee31, so that TT is a lattice homomorphism; also, of course, TT(O) = 0. 45B Proposition Let 21 and 93 be Boolean rings, and TT: 21 -> 93 a ring homomorphism. Then there is a unique Riesz homomorphism na r TT: 5(21) -> 5(93) defined by TT(X^) = X( ) f° every ae% and now ||^#|| oo < || #|| oo for every x e 5(21). So TT has a unique extension to a norm-decreasing Riesz homomorphism TT: L°°(2l) -> L^ffi). Proof The map a\-+x(na): 21 -> 5(93) is easily seen to be additive; so there is a unique linear map fr: 5(21) -> 5(93) such that TT(XCI) = x(na) for every a G 21 [42C]. Now if (ai)i
n(x+) = Si<7la^(77'^) = (nx)+

and II^HU = niaxda^ :i < n, n^ + 0} < ||»||oo. Thus n is a norm-decreasing Eiesz homomorphism from 5(21) to 5(93). Now n has a unique continuous extension to a map from L°°(2l) to I/°°(93), since these are the completions of 5(21) and 5(93) respectively. Since the lattice operations on both L°°(2l) and L°°(93) are continuous, TT: L°°(2l) -> L°°(93) is a Riesz homomorphism; and of course it is still norm-decreasing. 45C Proposition Let 21 and 93 be Boolean rings, and TT: 21 -> 93 a ring homomorphism. Let TT: L°°(2l) -> L°°(93) be the associated Riesz homomorphism. Then (a) if TT is one-to-one, then n is norm-preserving; (b) if n is onto, then TT[5(21)] = 5(93) and 7r[L°°(2l)] = L°°(93); moreover, for each 2/e5(93), U^oo = inf{||#|| 00 :#e5(2t), nx = y}, and for ), nx = y}. Proof (a) On examination of the formula for |TT#||OO given in the proof of 45B above, it is clear that if TT is one-to-one then || nx\\ ^ = || x\\ ^ for every x e 5(21). Now by continuity ||7ro;|| ^ must be equal to ||#|| oo for every XGL°°(21). (b) (i) If n: 21 -> 93 is onto, then for every disjoint sequence (bi)i
45]

RIESZ SPACES ON BOOLEAN RINGS

for each i < n. P Certainly there is a sequence (ci}i
Hkl

|||

SOTT: 5(91) -> 5(93) is onto, and \\y\\,, > inf{||a:||00:xG5(9l), m = y} for each y G 5(93). The reverse inequality is trivial, because n is normdecreasing. (iii) The corresponding result for n: L°°(9l) -> L°°(93) follows at once from the general theory of normed linear spaces. If yeL™(93) and e > 0, then there is a sequence {yn)n^ in 5(93) such that

for each neN. Now we can choose a20G5(91) such that nz0 = i/0 and INI oo = Hs/oll oo and, for each n ^ 1, a zn e 5(21) such that 7rzn = y n - yn_1 and I ^ J . = \\yn-yn^\U < 3.2-^e. But z = 2 w e H s n exists in L«(«) and II^Hoo ^ SWeH||«nlU < IMU + 4e, whileTO= SneN7T^ = y. As e is arbitrary, this shows that ||y|| ^ ^ inf {||x||«,: fix = y}; again, the reverse inequality is trivial. 45D Corollary (a) If 93 is a Boolean ring and 91 is a subring of 93, we may regard 5(91) and L°°(9l) as Riesz subspaces of 5(93) and L°°(93) respectively. (b) Now if 91 is an ideal of 93,5(91) and L°°(9l) are solid linear subspaces of 5(93) and L°°(93) respectively. (c) If 91 is a Boolean ring, / an ideal of 91, and 93 the quotient ring 91//, then 5(93) and L°°(93) can be identified, as normed Riesz spaces, with 5(9l)/5(/) and LQO(91)/LQO(/) respectively. Proof (a) Let n: 91 -> 93 be the identity map. Then is a norm-preserving Riesz homomorphism, by 45Ca, so embeds the normed Riesz space L°°(9l) as a closed Riesz subspace of L°°(93). And of course TT[5(9I)] <= 5(93). 116

RING HOMOMORPHISMS

[45

(b) Consider first 5(21) as a Riesz subspace of 5(58). Suppose that a;e5(21), ye5(93) and 0 < \y\ < \x\. Then there is an a o e2t such that M < INIIooAT^o [42Ec]. Express y as 2i w h e r e <^X<TI i s a disjoint sequence in 93 and faxbi 4= 0 for each i < n [42Eb]. Then 0

< |AI xh < |y| < NUaoo

v

*<^

so 6^ c a0 [42Ed] and 6^ e 21 for each i < n. Thus y e 5(21); as x and ?/ are arbitrary, 5(21) is a solid linear subspace of 5(93). So \x\ A \y\ e5(21) for every a;e5(21) and ye5(93). By continuity, \x\ A \y\ belongs to the closure of 5(21), which is L°°(2l), for every xeL™{%) and yeL™08); so L°°(2l) is solid in L°°(93). (c) Let n: 21 -> 93 be the canonical map. Since n is onto, n: 5(21) -> 5(93) and n: L°°(2l) -> L«(») are onto [45Cb]; as TT is a Riesz homomorphism, 5(93) s 5(2l)/£,

L°°(93) s I»°°(«)/F

as Riesz spaces, where nx = 0},

F = {x:xeL<*>(W), nx = 0}.

[See 14G for the construction of quotient Riesz spaces.] 45Cb shows also that the norm on 5(93) is the same as the quotient norm on 5(2l)/£, and similarly the norm on L°°(93) is the quotient norm of L°°(2l)/F. So we have only to determine E and F. Clearly 5(7) c= E. On the other hand, if a: e £, express x as Y^Kn^iX^i where (a^)i n6ir t |«| and <xn>neN -> |a;| for || ^ [43Bb]. So 0 < nxn < n\x\ = 0, and xneS(I) for each TIGN, by the argument just above. Hence \x\ belongs to the closure of 5(7) which is L°°(7); since L°°(7) is solid [(b) above], xeL^U). As x is arbitrary, F = L°°(7), as required. 117

45]

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45E Since n: L°°(2l) -» L°°(93) is norm-decreasing, it always has a norm-decreasing transpose n: L°°(93)' -> L°°(2t)'. But in order to ensure that7r'[L#(93)] c L#(2I), we must impose an order-continuity condition on?r. Proposition Let 21 and 33 be Boolean rings, and TT: 21 -> 93 an ordercontinuous ring homomorphism. Then there is an order-continuous increasing linear map TT': L#(93) ->* L#(2t) given by (rr'g) (x) = jr^s)

V a e L»(«),

or by V Moreover, TT' is norm-decreasing. Proof If g e L#(93), then gr^Tr: 91 -> R is completely additive, because n: 21 -> 93 is order-continuous and gr^: 93 -> R is completely additive; and gxn is bounded because ^ is bounded. So there is a unique / e L#(2t) such that =f(Xa)

V

Now gn = f on 5(21), because ^TT and / are both linear, and gn = / on L°°(2l) because gn a n d / a r e both continuous. T h u s / = n'g. Of course UTT'II ^ ||TT|| ^ 1. If g ^ 0, then /(^a) = gx^a ^ 0 for every ae2l, so / ^ 0; so TT' is increasing. Becauxe L#(93) is an L-space, it follows from 25Ld, or otherwise, that TT' is order-continuous. 45F Proposition Let 21, 93 and (£ be Boolean rings, and let TT: 21 -> 93 and 6: 93 -> © be ring homomorphisms. Then #TT: 21 -> © is a ring homomorphism, and

If 7T and 0 are order-continuous, so is On, and in this case

Proof I t is easy to see that Qn = (Onf on 5(21). The rest follows at once. 45G Exercises (a) Let 21 and 93 be Boolean rings, and n: 21 -> 93 a lattice homomorphism such that TT(O) = 0. Then TT is a ring homomorphism. 118

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[45

(b) Let 91 and 93 be Boolean rings and n: 91 -> 93 a ring homomorphism, with n: ^(91) -> L°°(93) the associated Riesz homomorphism. Let / = {a:ae% na = 0}< 91. Then TT[5(91)]

- 5(7r[9i]) - 5(91//) ^

7r[L°°(8l)] ~ /^fatSt]) - L°°(«//) and

TT[5(91)]

= 5(93) n 7T[L°°(9t)].

*(c) Let 91 and 93 be Boolean rings and n: 91 -> 93 a ring homomorphism, with TT: L°°(9l) -> L°°(93) the associated Riesz homomorphism. Then n is order-continuous iff n is order-continuous. [Hint: use 42Rb and 17B.] *(d) Let 91 be a Boolean ring and / an order-closed ideal of 91. Then the canonical map TT:91->91// is order-continuous [4lLd], and n': L#(9l//) -> L#(9l) is a norm-preserving Riesz homomorphism. (e) Let <9li>ieZ be a family of Boolean rings with product 91. Then 91 is Boolean [4IK], For each tel, let nL: 91 -> 9lt be the canonical ordercontinuous ring homomorphism. Then the associated Riesz homomorphisms nL: L°°(9l) -> L°°(9lJ induce a normed-Riesz-space isomorphism between L°°(9l) and the solid linear subspace {x:\\x\\ =sup t e Z 1 of the Riesz space product Iliei^CSlt)- *At the same time, the normpreserving Riesz homomorphisms n[: £ # (9l t ) -> L#(9l) induce an isomorphism between L#(9l) and the solid linear subspace

Notes and comments The results of this section are essentially algebraic. They rely on a careful analysis of functions on 5-spaces which is then extended to functions on L°°-spaces. So they all seem natural, despite the complexity of some of the proofs. 4X Examples for Chapter 4 Since Boolean rings usually appear as rings of sets or their quotients, these are the important cases to analyse. When 91 is a ring of sets, we have particularly accessible representations of 5(91) and L°°(9l) [4XB]. From these we can derive representations of 5(91//) and L°°(9l//), where / is an ideal of 91, as quotients of 5(91) and L°°(9l) respectively, as 119

4X]

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explained in 45Dc. These again are especially simple when 91 is a o*-algebra of sets [4XE]. An interesting special case is the algebra of regular open sets in the real line [4XF]. 4XG is an instructive counterexample. 4XA The algebra 0>X. Let X be any set, and 0>X its power set. On 0>X define A by AAB = (A\B)[)(B\A)

V i,BcI

Then {£PX, A, (1) is a Boolean ring [41 A]; its zero is the empty set. (The direct verification of the ring postulates is lengthy but elementary. Alternatively, we may identify SPX with {0, i}x, saying that a set i c j corresponds to the function with value 1 on A, 0 on X\A. In this case, {SPX, A, n) is isomorphic to (Z 2 ) x , where Z2 is the field with two elements 0 and 1. The extra Boolean postulate, that A f) A = A for every A ^ X, is trivial.) In fact SPX is a Boolean algebra [41G]; its multiplicative identity is X. The lattice operations on SPX are just (J and n. &X is Dedekind complete; if A c SPX, sup A = \JA. *For each teX, the set Jt — {A: A c X,t$A} is a maximal ideal of SPX. So the map t H->^ is an embedding of X in the Stone space of £PX. Note however that (unless X is finite) SPX has many more maximal ideals than these. 4XB Rings of sets If 7 is a set, a subring of ^ 7 is a family 21 of subsets of 7 containing 0 and closed under A and n; clearly, this is the same as being closed under U and \. Suppose that 91 is a subring of ^ 7 . Then S = 5(21) is isomorphic, as normed Riesz space, to the linear subspace E of /°°( 7) generated by the characteristic functions of members of 21. P S is defined as the linear subspace of l°°(X) generated by the characteristic functions of members of 0[2t], where l a : 2t-> /°°(7) is clearly additive; so, by 42C, there is a linear map T: 5-> Z°°(7) such that T{%a) = la for every ae2l. If we examine the description of members of S given in 42Eb, it is clear that T is a norm-preserving Riesz homomorphism, so S ^ ^[5] as normed Riesz space. Also, since S is spanned by {^a: ae2l}, T[S] must be the linear span of {la: ae 21}, which is E. Q 120

EXAMPLES FOR CHAPTER 4

[4X

This isomorphism means that we can identify S with E; since the set Y is already to hand, while the Stone space X of 31 must be constructed with the axiom of choice, this is a much more convenient representation of S. It follows also that L°°(9I) [43A] can be identified with the || H*,completion of E, which is [isomorphic to] the closure of E in ^(Y). Note that this is indeed a Riesz space isomorphism as well as a normed space isomorphism, because the lattice operations are continuous. *As a rule, Lf (91) is not of much interest in this case. If 91 contains all finite subsets of X, then L#(9I) can be identified with l\X). One proof follows the argument of 2XB; another observes that L°°(9l) is order-dense in /°°(X), and applies 17B and the result of 2XB; another uses 65B. 4XC 0>X and 0>fX (a) If 91 = 0>X, then 4XB gives us the identifications S{£PX) ~ {x: xe'Rx, x takes finitely many values},

[2XB]. Thus the Dedekind completeness of f°°(X) [2XA] is related to the Dedekind completeness of 0*X by 43Da. (b) For any set X, let 0*fX be the family of finite subsets of X. Then (PfX is an ideal of &X\ in its own right, it is a Dedekind complete Boolean ring. Now

S{0>fX) - so(X) = {x:xeRx, {t:x(t) * 0} is finite}, L»(^,Z) s co(X), L#(0>fX) - l\X) [as in 2XC]. 4XD o-algebras of sets Suppose, in 4XB, that 91 is a cr-subalgebra of 0>X [i.e. X e9t and 91 is closed under countable unions and intersections, as well as complementation]. In this case, the representation of L°°(9l) as a subspace of R x expresses L°°(9l) as V aeR}. P (i) Just as in the proof of 42E, 5(91), which we have identified as the linear subspace of R x generated by the characteristic functions of members of 91, is {x: x G E, x takes finitely many values}. 121

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(ii) If x GL°°(21), then there is a sequence <# w ) n6 n in S(%) such t h a t <||a? —a?n||00>weH -> 0, so t h a t x(t) = limn_^O0xn(t) for every teX. Now for any a G R,

{*:*(*) > a} = which must belong to 91. So #e JE. (iii) Conversely, if xe E, then for each neN define xn by xn{t) = 2-n[2nx(t)]

V teX

where [a] = s u p { m : m e Z , m < a } for each aeR. Then {*: a n 0) > a} = HmeN {*: *(*) > 2-*([2»a] + 1) - 2"-} G 21 for any a e R, so x^e 5. Since || a; — xn\^ < 2-nforeach^GN,a; 4XE Quotient rings For any Boolean ring 21 and ideal / of 21, L°°(2I//) can be identified with LOO(21)/LOO(/) [45Dc]. When 21 is a subring of ^ X , then L°°(2l) and L°°(/) are subspaces of Z°°(X). In the special case when 21 is a cr-subalgebra of SPX, as in 4XD, and / is a o*-ideal of 21 [i.e. / is an ideal closed under countable unions], then L°°(/) becomes {x:xeL™{%), {t:x(t) * 0}eI}. (For this is the solid linear subspace of Lc0(2l) generated by the characteristic functions of members of / , and it is closed for || H^. Remember that by 45Db L°°(I) is solid in L°°(2l).) Thus Lc0(2l//) becomes the space of equivalence classes in L°°(2l) under the relation x~y if {t:x(t) *4XF Algebras of regular open sets In any topological space, an open set G is regular if G = intG. The set ^ of all regular open sets is partially ordered by ^ . This ordering is induced by a Boolean ring structure on ^ under which ^ is a Dedekind complete Boolean algebra. When the underlying topological space is R, the algebra thus obtained has many facets; I shall present it as an example of an algebra on which there are no non-trivial countably additive functionals. Proofs of these remarks now follow. (a) Let X be a topological space. Suppose that E and F are closed subsets of X such that int E = int F = 0. Then int (E U F) = 0. [For X\E and X\F are dense open sets, so their intersection is dense.] 122

EXAMPLES FOR CHAPTER 4

[4X

(b) Let X be a topological space. Let 91 be the set {A:A^X,int(8A)

= 0} <= 0>X,

where dA = ^4\int^. is the boundary of A. Then 91 is a subalgebra of 0>X. P If A, B c X, then 8(A [)B)^8A[) 8B. So by (a) above, 91 is closed under u. As 8(X\A) = 8A for any A c X, 91 is closed under \. Finally, 0 e 91, so 91 is a subalgebra of 0>X. Q Observe that open and closed sets all belong to 91. (c) Continuing from (b) above, let / be {A:A <^X,mtA= 0} c 91. Using (a) again and 41J, / is an ideal of 91. So we may form the quotient 91//. For A e 91, let A' be the image of A in 91//; let £ be the canonical ordering in 91// [41F]. (d) Now A' ^B'ointA^intS. P If Ae% then 8AeI. So A' = A' = (int A)'. (Of course A and int ^4 belong to 91.) Now i ' s B ' ^ l ' c J ' ^ (intZ)* s B' => (int A)\BeI => (intZ)\5 = 0 = => (int J ) ' s (iatB)" => 2* s 5* => A' c J5*. Q (e) Thus 91// may be identified, as partially ordered set, with 9 = {int A :^4e9l}. Now an open set belongs to *& iff it is regular. P If G = int G, of course Ge@, since Ge 91. Conversely, if G = int A, where A is any set, G ^ G ^ A, so G ^ intG c int J. = 6?; thus G is regular. Q (f) Now ^ is a Dedekind complete lattice. P (i) IfG9He &, then G()He@, because GnHczmt(GnH)c: int (0 nfl)c intG n intH = G(]H. So G n -iff = inf {G, /f} in ^ . (ii) On the other hand, if S is any subset of ^, consider Go = intcl(|J g). For any EsS, E c |J
X\G = X + G = X\G = int (X\G) 123

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RIESZ SPACES ON BOOLEAN RINGS

[for G n (X\G) = 0, G u (X\0) = X]. (It is clear that there can be only one Boolean ring structure on ^ associated with the ordering £ , for this defines u and n and therefore \ and +.) Now ^ is a Dedekind complete Boolean algebra; its maximal element is X. (h) Let us now consider the case in which X is HausdorfF, separable and without isolated points (e.g. X = R). In this case, any increasing countably additive function v: &-> R is identically zero. P Let Q be a countable dense subset of X; then Q is infinite. Enumerate Q as each m, neN let Gmn be an open set in X such that Set Hmn = ink(\Ji 0. Clearly \jn^Hmn is dense for each m e N , that is, (Hmn}ne1x f X in ^ . So there is, for each m, an n(m) such that Set j ^ m = X\HmMm). Then r^ m ^ 2"we and qi^Em for every i < m. So is dense, and meN f X in ^ . So

As e is arbitrary, vJT = 0. Because v is increasing, v = 0. Q (i) I t follows at once that any countably additive functional on *3 is zero (for by 42Q and 42N, any countably additive functional on ^ is the difference of increasing countably additive functionals). At the same time we see that there can be no finite-valued measure on 3? [see 5IE]; indeed, it is easy to see that the only measure on ^ , according to the definition 51 A, is the one which takes the value oo on every non-zero element. (j) Consequently, L°°(^) is now a Dedekind complete Riesz spaceactually, an M -space with unit-such that L°°(^)x = {0} [44Bb]. Of course, L°°(#)~ = L°°(#)' [25G] is large. (k) Exercise Show that (i) there is a sequence (Gn)neV in the algebra ^(R) described above such that, for every non-empty Ge@, there is an n e N such that 0 a Gncz G (ii) any Dedekind complete Boolean algebra with this property is isomorphic to ^ . (1) Exercise Again supposing that X = R, show that L°°(9l) can be identified with the set of those bounded functions x: R->R such 124

EXAMPLES FOR CHAPTER 4 [4X that {t: x is continuous at t} is a dense G^ set. Consequently, L°°(^) is the Riesz space quotient of this by the solid linear subspace {x: x is zero on a dense Gs set}. Also, Cb(X) may be regarded as an order-dense Riesz subspace of the Dedekind complete Riesz space L°°(^). Hence, or otherwise, show that Cb(X)x = {0}. *4XG A counterexample for 420 and let 91 be

Let X be an uncountable set,

{A: A £ X, either A is finite or X\A is finite}. Then 91 is a subring of 0>X. Define v\ 91 -* R by vA — n

if A has n members, where neN;

= —n if X\A has n members. Now if (An)nelx | 0 in 91, one of the An must be finite; so in fact one of them is empty, and infweN \vAn\ = 0. Thus v is countably additive. But v is certainly not locally bounded. So we see that the condition 'Dedekind cr-complete' in 42Q is indeed necessary.

4XH Ring homomorphisms induced by functions Let X and Y be arbitrary sets, and/: X -> Y any function. Define n: SPY -> SPX by n(A) =f~x[A] for every A c 7. Then TT is an order-continuous ring homomorphism. We see that, for A ^ Y,

HxA) = X(nA) = Xif-W) = *4 o/, where here we identify %A with the characteristic function of A, as in 4XB. It follows that n: L«>(0>Y) -> L™(0>X) is given by 'fry = yof 00

for every ye/ s L*>(&Y) [see 4XCa]. Equally, identifying L#(0>X) with P(Z), TT': L#(^Z) -> L#(^7) is given by

for every ueY and a:elx{X).

125

5. Measure algebras In this chapter I shall give versions of those results in elementary measure theory which refer to measure algebras or to L1 and L00 spaces. The first two sections apply the concepts of §§ 42-4 to ' measure rings', that is, Boolean rings on which a strictly positive countably additive measure is defined. In this case, a true analogy of L 1 spaces can be found, and the correspondence between L 1 spaces and L # spaces is discussed. All measure rings of any significance are 'semi-finite', and consequently their L 1 and L # spaces can be identified; this is the basic idea of § 52. The next section deals briefly with Dedekind complete measure algebras, which seem to be central to ordinary measure theory. Finally, in § 54, the ideas of § 45 concerning homomorphisms are reviewed in the new context.

51 Measure rings In this section, we shall have only definitions and basic properties. A measure ring is a Boolean ring together with a strictly positive measure; this is an extended-real-valued functional which is additive and sequentially order-continuous on the left. The definition of measure ring which I have chosen [51 A] follows the ordinary definition of measure space [61 A] in allowing 'purely infinite5 elements, that is, non-zero elements of infinite measure such that every smaller element is either zero or also of infinite measure. It is an accident of the theory that these need not cause any inconvenience in the early stages; but fairly soon they must be outlawed, and accordingly I shall immediately introduce 'semi-finite' measure rings, that is, measure rings which have no purely infinite elements. In any measure ring, semi-finite or otherwise, the ideal of elements of finite measure is of great importance, and many of the results here are based on consideration of this ideal as a measure ring in its own right. All the most important examples of measure rings are derived from measure spaces, as in 6ID; so at this point it will be helpful to have in mind Lebesgue measure, as the simplest non-trivial example of a measure space. 126

MEASURE RINGS

[51

51A Definition A measure ring is a Boolean ring % together with a measure /i: % -» [0, oo] such that: (i) /i(a\jb) = fta+jbib if an& = 0in9l; (ii) ju,a = O o a = 0; (iii) if n6H f a in SI, then /ia = sup nels/ian. Note 'oo' here, and later, is regarded as an actual point adjoined to R+. The set [0, oo] has natural additive-semigroup and total-ordering structures, writing oo V a e [0, oo], a < oo V ae[0, oo]; and also a multiplicative-semigroup structure, writing O.oo = oo.O = 0, a.oo = oo.a = oo V a > 0, which we shall have occasion to use. Subtraction, however, is not fully defined, so we must be wary. 51B Definitions

(a) For any measure ring (91, ft), W will be {a:ae% [ia < oo}.

(b) A measure ring (%/i) is semi-finite if, whenever ae% and [ia = oo, there is a b s a such that 0 < jib < oo. 51C Proposition Let (21, fi) be a measure ring. (a) If a £ b in % jia < [ib. (b) Suppose that 0 c B c W and that B | 0 in SI. Then (i) there is a sequence (bn}ne1ii in J3 such that (bn}n€x 10; (ii) inf&ei?/*& = 0. Proof (a) For fib = fia+ju,(b\a) ^ /^a. (b) Let a = inf6eB/^6. Choose a sequence neN in B such that [icn
51]

MEASURE ALGEBRAS

Then for each neN there is a c e B such that c £ b n bn+1, so that a ^ fie ^ /£(& n 6W+1) for each WGN, and

for each r&eN. (Note that the subtraction here is legitimate because libn+1 < oo.) Now a\b £ 6W+1\6 for each UEN, SO fi(a\b) = 0 and a\6 = 0, i.e. a c ft. As 6 is arbitrary and inf £ = 0, a = 0; as a is arbitrary, Now however, <60\&n>neN t &o [41Hb]. So

Thus lim.n_^a0/ibn = 0 and a ^ infneVi/ibn = 0, as required. 51D Corollary Let (21,/i) be a measure ring. Let v\ W -> R be a countably additive functional. Then y is completely additive. Proof Suppose that 0 <= B10 in 9P. Of course any lower bound for B in 91 belongs to W, by 5lCa; so B10 in 31. By 5lCb there is a sequence (Pn)nen ^ B such that <6n)nGN ^ 0 in 91 and therefore (bn)neJsl | 0 in 81'. So mfa€B\va\ R is completely additive. Proof

If a, 6 e 91, then ju,(a u 6) = /ia+ju,(b\a) < /^a +/^6

[using 5lCa]. So 9l/ is closed under u; by 5lCa again, a<=beW => aeW; and certainly 0 e W, as /i(0) = 0. So 91^ is an ideal of 91. It follows that if <*n>n6it« in «/, then n6H ta i n » ; s o P* = s u P n e N ^ ? and 91^ is a measure ring. Of course it is semi-finite. Now /i is additive on W by 5lA(i), and it is completely additive by 51Cb. 128

MEASURE RINGS

[51

*51F Products If ((91,,/OXei is a family of measure rings, 9t = ULEI^C is a Boolean ring [41K]. Define /i: 9t-> [0, oo] by Then (9t,/0 is a measure ring; I will call it the product ILe/(9t*>/0 even though it is not a product in any obvious category. If for any KGI and aetyLK, we define a = (a,}ieI where aK = a and aL = 0 for t == | /c, we find that the map a n a represents 91* as an ideal of 91. Now /iK is just the measure on 91* induced by /i. From this it is clear that (91,/^) is semi-finite iff (9ti?/£t) is semi-finite for every teI. 51G Exercises (a) Let (91, p) be a measure ring. Then there is a natural metric p on W defined by p(a,b)=/i(a + b) V a,bsW. (b) For any measure ring (91, fi), fi is order-continuous on the left. [Hint: use the technique of 5lCb, upside down.] (c) Let (91,/0 be a measure ring. Then it is semi-finite iff a = sup {6: b e W, b s a] for every a e 91. [Cf. 41Lc] (d) Let (%ju,) be a measure ring. Then 5(910 and L°°(9P) can be thought of as solid linear subspaces of 5(91) and L°°(9l) respectively [45Db]. Now the following are equivalent: (i) (91,/^) is semi-finite; (ii) 5(910 is order-dense in 5(91); (iii) L»{W) is order-dense in L°°(3t). *(e) For any measure ring (91, /i), S(W) and L00(9l/) have the countable sup property. Notes and comments Although all the most important measure rings, being derived from measure spaces, are Dedekind cr-complete [6lDd], the work of the next section will appear in clearer perspective if we omit this condition for the time being. 5lCb echoes the corresponding result for i-spaces [26B]. The details of the proof have to be different for technical reasons, but the essential phenomenon is the same. In fact, the ring of elements of finite measure has a natural metric [5lGa], which behaves very like the norm of an L-space. In 5XA are some elementary examples of measure rings. These are essentially trivial. For a substantive example we do need a 'nonatomic' measure like Lebesgue's [see the references in 6XAb]. 5XAb is an extreme example of a measure ring which is not semi-finite; every non-zero element is purely infinite. FTR

5

129

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MEASURE ALGEBRAS

52 The space In this section we study the properties of the spaces S, L00 and Ifl from Chapter 4 when they are based on measure rings. The pivotal results are 52B and 52D, in which the measure is used to embed S(W) as a Riesz subspace of L#(2I) which, when 31 is semi-finite, is order-dense. This allows us to think of L#(2l) as a completion of S(W) with respect to the appropriate norm. We now have a construction L1(9l) which is very like the classical L 1 spaces. 52A The embedding of S{W) in L#(8t) Let (21,/*) be a measure ring. For the moment, fix ae W. Define va: 21 -> R by va{b) = /i(a(\b) V 6e2I. Then, because the map 6 -> a n b: 91 f-» 21* is an order-continuous ring homomorphism, and fi: 21/->R is completely additive [51E], va is completely additive. Moreover, sup {va{b): b e 21} = fia < oo,

so there is a unique/ a GL # (2l) such that and ||/a|| = /*a [44Bb]. Thus we have a map a\-+fa: W->Lft{%). But clearly this map is additive and increasing, since if a n c = 0 in W, fa+c(Xb) = /*((« + c ) & ) = Mab + cb) = Mab) +Mcb) = fa(Xb)+UXb) for every 6e2l, and/ a + c = / a + / c - So it induces a canonical increasing linear map (j>: S(W) -> I^f(2l), given by (j>{x<x>) =fa f° r every aeW. The basic properties of ^ are contained in the two propositions 52B and 52D. 52B Proposition For any measure ring (21, fi), the canonical map <j): 5(210 -*" £#(2l) defined in 52A is a one-to-one Riesz homomorphism. For any xeS(W), l%\\ =/*(|#|)> where/? is the linear functional on S(W) corresponding to the additive functional /i: W -> R. Proof (a) If a,beW and a n 6 = 0, faAfb = 0. CG 21, A? = A;(C n 6)+#(c\6) in L°°(2l). So

P For every

0 < (fa A/ 6 ) (^e) ^ /«^(c n 6) + / 6 x ( # ) = fi(a n c n b) +/i(bn c\b) = 0, 130

THE SPACE LH^l)

[52

remembering t h a t / a A/ 6 is defined in L#(«) = L«(«)x s L«(«)~ = L~(L°°(2i); R), and applying the formula of 16Eb. As c is arbitrary, / a A/& = 0. Q Thus ^(^a) A ^(#6) = 0 whenever a n 6 = 0 in 21. But now suppose that xAy = 0 in £(9P). Then we can find a, y# > 0 in R and a,beW such that a # a ^ x < \\x\nXa, Px*> ^ V ^ \\y\Uxh [42Ec], Then (a A /?)#(a n 6) ^ x A 2/ = 0, so a n 6 = 0. Now, because <j> is increasing,

0 < 0sA0y < |MU/a A ||y||oo/6 = 0 because/^ A/ 6 = 0 [14Kb]. Since x and t/ are arbitrary, this proves that ^ is a Riesz homomorphism [14Eb]. (b) Now l|0(#a)|| = ||/o|| = fia= p>(xa) f° r any aeSl^ recalling from 52A that ||/o|| = /^a. If next xeS(W)+, then a: can be expressed as Tti where each at ^ 0; since L#(2l) is an L-space,

Finally, for arbitrary a; e

(c) It follows that is one-to-one. P For if x + 0 in AS^SI') then \x\ > 0, so there is an a > 0 and an a + 0 such that a#a ^ \x\. Now ||0||

/(||)

/

0,

so <j)x + 0. Q 52G Lemma Let 21 be a Boolean ring, and v: 21 -> R a completely additive functional. Suppose that ae2l and that va > 0. Then there is a non-zero 6 g a such that vc ^ 0 for every ccj>, Proof Define A: 21 -> R by Ac = v(a n c) for every ce21. Then it is easy to see that A is completely additive, and Xa > 0. Now consider A G 5 ( 2 1 ) X [ 4 2 L ] . Since X+(x<*) > X{xa) = Aa > 0,

X+ > 0. So by 32A there is an x > 0 in 5(21) such that X+{x) > 0, 5-s

X~(x) = 0 131

52] MEASURE ALGEBRAS

[for certainly X+ A X~ = 0]. Now by 42Ec there is a b0 e 9t and a /? > 0 such that

o

y ^

^ ii I!

t

( A ) ( A ) Now set 6 = a n 60. If c £ 6, then 0 < ;\;c < #&0, so X~(xc) = 0 and PC = Ac = A(#c) = A+(#c) ^ 0. On the other hand, vb = A60 = X(Xb0) = X+(xbQ)-A-(xb0)

> 0,

so b =)= 0 as required. ^Exercise Prove this lemma without using the construction S( ). [Hint: adapt the argument of 32A so that it refers directly to completely additive functionals on Boolean rings.] 52D Proposition Let (91, /i) be a semi-finite measure ring. Then the canonical map : S(W) -> L#(9t) defined in 52A embeds 5(81') as an order-dense Riesz subspace of L#(8l). Proof We already know that ^ is a one-to-one Riesz homomorphism [52B], so I have only to prove that ^[5] is order-dense. Suppose t h a t / > 0 in L#(9l). Then, because (%/i) is semi-finite, there is an ae W such that/(^a) > 0. P Certainly there is an aoe91 such that/(^a 0 ) > 0. Let

B = {ao\a:aeW}. Then B | in 91. ? Suppose, if possible, that B has a non-zero lower bound 60 in 91. Then there is a non-zero c £ b0 such that c e 9tA So c s ao\c, which is impossible. X Accordingly, B10 in 91. Since fX: 91 -^ R is order-continuous, inibeBf(xb) = 0. In particular, there is an a e W such that fx(ao\a) < fxao- Now

fX<*> > fx(a n a0) = fXa0 -fx(Oo\a) > 0 as required. Q Now let a > 0 be such that/(^a) > aju,a. Set

in Lf (91) = L°°(9t)x, and let v = gx'. 91 -> R, so that v is a completely additive functional and "» = giXfl) =f(xa)-ot/ia 132

> 0.

THE SPACE L^W)

[52

By 52C, there is a non-zero b £ a such that vc ^ 0 for every c^b, i.e. /(# c ) = 9(Xc) + aMa for every c g j , Consequently, for every c e 91,

nc

) =

/ ( ^ ) > fx(c n 6) ^ a/t(c n 6) = Thus/ ^ (x,(j>(xb) in L # [44Bb]. Since a > 0 and 6 + 0 and <j> is one-to As/is arbitrary, this shows (using 15E) that ^[5] is order-dense in L # . 52E Proposition Let (91, /i) be a semi-finite measure ring. On S(W), a Riesz norm || || 1 can be defined by writing \\x\\ x = ju( \x\), where p, is the linear functional on 5(910 associated with the additive functional fi\ W -> R. Let L1(9l,/*) be the normed Riesz space completion of 5(910- Then the map 0: S(W) -> L#(9l) defined in 52A extends to a normed Riesz space isomorphism between L1(9lJ/^) andL # (9l). Consequently 5(910 i s super-order-dense in Proof (a) We know from 52B that S(W), with || ||1? can be identified with the Riesz subspace [S] is dense in L#(9l). P We know that L#(9I) is an i-space [44Ba], so its topology is Lebesgue [26Ba]. Now [S] is orderdense in L#(9l) [52D], that is, for a n y / ^ 0 in Lf, and feSftS] c 0 [5] [24Ba]. As L# = L#+ - L#+, ^[5] is dense. Q (c) So L#(9l), which is norm-preserving for the norms on 5 and L # , must extend to a normed space isomorphism between L 1 and L # , because L # is complete. Because the lattice operations on both sides are continuous, and ^ is a Riesz homomorphism on 5, (j>\ L1 -> h# is also a Riesz space isomorphism. Now 5 is order-dense in L1 because ^[5] is order-dense in Ifl. Consequently it is super-orderdense, because L 1 is an L-space [26Bd]. *52F L#(9t) and L#(W) If (91, /i) is any measure ring, then (W, fif) is a semi-finite measure ring, where jif is the restriction of /i to W, and of course (W)f = W. So we may apply the results above to show that 133

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MEASURE ALGEBRAS

), the completion of S(W) = S(Wf), is the same as L\W, fif), and is isomorphic to L#(W). Of course L1(2l, /i) can also be identified with the closure of (j)[S(W)] in L#(2I). Thus L#(W) is represented as a subspace of L#(2t). The exercise 52Hc goes into this further. 52G I conclude this section with a remark on the duality between L#(2l) and L°°(2t) when 21 carries a semi-finite measure. Proposition Let (21, /i) be a semi-finite measure ring. Then the norms on L°°(2l) andL#(2l) are dual in the sense that I/I = sup{|/x| :xeL» \\x\\a < 1} V feL#,

INU = sup{|M :feL#, \\f\\ < 1} V Proof The first part is just the definition of the norm on L # = LG0X c L00'. The second part is more interesting. Of course

for every a; e 2/°, and || ||', so defined, is a seminorm. Now if x is a nonzero member of 5(21), x can be expressed as Tn<nocix^i, where (a?)i 0. But

because a n ^ = 0 if i 4= J and ana^ = a. From this it is clear that

INI' ^ la;l = II^IU* Thus || ||' and || ^ agree on 5(21); since || ||' is

continuous with respect to || 1^, they agree on L°°(2l). *Remark We observe that the above result is a property of 21 alone; it is immaterial what the measure ji is, as long as it is semi-finite. It can be shown that this proposition characterizes those measure rings 21 on which a semi-finite measure can be defined. [See 4XFj.] *52H Exercises (a) Let 21 be a Dedekind R a countably additive functional. Then for every a e 21 there is a 6 ^ a such that vb = v+a, and v~b = 0. (If (bn)nelS( is a sequence such that bn^a and vbn ^ v+a — 2~n for each neN, set 134

THE SPACE L^W)

[52

x

(b) Let £ be a Riesz space and F a Riesz subspace of E such that the map x\->&: E -> F* is a Riesz homomorphism from E to F x . Use the method of 32B to show that the image of E is order-dense in F x . Now apply this result with E = S{W), F = L°°(2l) to prove 52D. (c) Let (%ju>) be any measure ring. Let F = {j:geL»(W)9 g(Xa) = 0 V aeW}. Then F is a band in L#(2l). Let /|AH=0

V

/

[cf. 15F]. Now the canonical map 0: 5(2l ) -> L#(9l) embeds S as an order-dense Riesz subspace of E [use the method of 52D]. Consequently <j> extends to an isomorphism between L1(2lJ/^) and E, which sets up a canonical isomorphism d: L#(W) -> E [see 52F]. At the same time, the order-continuous embedding of W in 21 induces a map n': L#(2l) -> L#(W) [45E]. Now TT'# is the identity on L#(W), and the kernel of TT' is F. Thus fr' can be thought of as the projection of L#(2l) onto E - L#(91O ^ LHSl^). with kernel F = Ed [ISP]. ^: L°°(W)X -> L°°(9l)x can also be given by (df)(x) = fsup{fy:yeL«>(W), 0 ^ y ^ x} for every ^GL°°(91)+ a n d / G L#{W)+, or by (^/) Ct«) = sup {/(^): 6 6 «/, 6 s a} for every ae% and/GL#(9l/)+. Notes and comments The reason for maintaining a careful distinction between L\%) and L#(9l), although they are canonically isomorphic in all significant cases, will appear in § 54. They are affected quite differently by homomorphisms between measure rings. The properties of L 1 can also be derived intrinsically, without considering Ift. It is easy to see that fi: S(W) -> R is strictly positive, because [i: W -> R is; so that || [^ = /«(| |) is a norm on S(W). Since of course ,. M n M H . > , ( ..

IIW + |y|||i = Hi+l|y|li for all x, y in 5(2I/), the same is true in the completion of 5, which is therefore an i-space. A less obvious point is the fact that S is orderdense in L1. This is clear from 52E, because <j)[S~\ is order-dense in IP. But it is also a consequence of the fact that, because /JL is completely additive on W [51E], fieS* [42L], so that the || Urtopology on S is 135

52]

MEASURE ALGEBRAS

Lebesgue. I t follows that S' c S\ Now S' is identified with L 1 ' as linear space, and since the positive cone of L1 is just the closure of the positive cone of 5, *S" and L 1 ' have the same ordering. Also, L 1 ' = Lx~, so L 1 can be identified with a Riesz subspace of (Lv)x = *S"X. The result now follows from 32B. However, the isomorphism between L1(9I) and L#(9I) is in itself one of the most important results of measure theory. In its more conventional forms, which will be touched on in § 63, it is the Radon-Mkodym theorem. The normal approach refers to countably additive functionals on cr-algebras of sets and uses 52Ha (the 'Hahn decomposition theorem') instead of 52C. Now an extra hypothesis must be added to ensure that, given a non-zero increasing countably additive functional v: 9t -» R, there is an a e W such that va > 0 (as in the first part of the proof of 52D). But such a hypothesis will always enable us to find a completely additive functional; see, for example, 63J.

53 Maharam algebras If the underlying Boolean ring of a semi-finite measure ring is a Dedekind complete Boolean algebra, then its L00 space turns out to be 'perfect' in the sense of § 33, that is, L00 is isomorphic to L°°xx or L # x . Thus the duality between L00 and L # or L 1 is symmetric in some respects. In many contexts, Maharam algebras appear to be the 'normal' measure rings, and all others 'unnatural'. 53A Definition A measure algebra is a measure ring (91, fi) in which 91 is an algebra, i.e. has a 1. A Maharam algebra is a semifinite measure algebra (91,/^) such that the Boolean algebra 91 is Dedekind complete. 53B Theorem Let (91, fi) be a semi-finite measure ring. Then the following are equivalent: (i) (91,/i) is a Maharam algebra; (ii) ^(91) is a Dedekind complete If-space with unit; (iii) the duality between L°°(9l) and Lx(9l), derived from the isomorphism between Lx(9l) and L°°(9l)x [52E], represents L°°(9l) as

Proof (a) (i) o (ii) We know that 91 is Dedekind complete iff L00 is Dedekind complete [43Da]. If 91 has a 1, then ^1 is the unit of 136

MAHARAM ALGEBRAS 00

[53

a

1/°°. Conversely if L has a unit e, then {a: x ^ e} is bounded above in 21 [43Bd], and its upper bound must be the 1 of 21. (b) (ii) => (Hi) We know that the canonical map from L°° to L°°xx takes L00 onto a solid linear subspace of LQ0XXJ because L00 is Dedekind complete [32C]. Now L°°xx = L#(2t)x is an If-space with unit [26C], and its unit is J, given by jf = ||/|| whenever/ ^ 0 in L # . But for/ ^ 0 in L#, Il/H = / ( # l ) [44Bb]. So J is the image of #1 in L°°xx, and the image of L00 is the whole of L°°xx. Also, we have seen in 52G that L # separates the points of L00, so L00 is isomorphic to LooXX as a Riesz space. At the same time, 52G shows that the norm of L00 is precisely that induced by its embedding in L # '; we recall from 26C that

Lf = L#x.

Thus L00 is identified with L # '. Since L # and L 1 are canonically isomorphic, we can identify L00 with L 1 '. (c) (iii) => (ii) Finally, the dual of L 1 is certainly a Dedekind complete M-space with unit [26C], so if it is isomorphic to L00 the condition (ii) must be satisfied. 53C Definition if pb\ < oo.

A measure algebra (91, fi) has finite magnitude

53D Lemma A Dedekind tr-complete measure algebra of finite magnitude is a Maharam algebra. Proof Of course any measure algebra (21, fi) of finite magnitude is semi-finite, as 21 = W. Now suppose that 0 <=: B \ in 21. Then C = {a\b : b e B, a is an upper bound for B} 10 [41Ha], So by 51Cb there is a sequence (cn)ne1x in C such that mfnelxcn = 0. If cn = aw\6w, where bneB and aw is an upper bound for B for each neN, then infneNaw = supweN6w must be the least upper bound of B. [Cf. 18B.] 53E Proposition ram algebra.

Any product of Maharam algebras is a Maha-

Proof Obviously, any product of Boolean algebras is a Boolean algebra. But also a product of semi-finite measure rings is semi-finite [51F], and a product of Dedekind complete Boolean rings is Dedekind 137

53]

MEASURE ALGEBRAS

complete [13Hb]. Thus each of the defining properties of a Maharam algebra is preserved in the product. 53F Exercises (a) A measure algebra (81, /*) has countable magnitude if there is a sequence <&w}weN in 21/ such that sup7leNa7l = 1. Show that any measure algebra of countable magnitude is semifinite, and that it is a Maharam algebra iff it is Dedekind o*-complete. (b) Let 91 be any Dedekind complete Boolean algebra. For each a G 91, let 9ta = {b: b e 91, b <= a} <\ 91. If A ^ 31 is a disjoint set such that sup-4 = 1, show that the map b \-^(b o a)aeA: SH-^ ]JaeA%a is a Boolean ring isomorphism between 91 and YlaeA^aNow let (91, fi) be a Maharam algebra. For as 91, let ju,a: 9la -> [0, oo] be the restriction of/* to 9la. Then if A satisfies the same conditions as before, (9t,/*) is isomorphic to YlaeA^a^a) a s measure ring. (c) Any Maharam algebra is isomorphic to a product of Maharam algebras of finite magnitude. [Hint: in (b) above, use Zorn's lemma to find a maximal disjoint set A ^ %f,~\ *(d) Let (81,/*) be any semi-finite measure ring. For each aeW, define/v 9 t x 9 t - > R b y Pa(b,c)=/i(a(b

+ c)) V b,CE91.

Then each pa is a pseudo-metric [cf. 5lGa]. The collection {pa: ae W} defines a Hausdorff uniformity on 91 with respect to which the ring operations on 91 and the map ju,: 91-> [0, oo] are all uniformly continuous ([0, oo] being given its usual compact uniformity). If 91 is a Maharam algebra, it is complete with respect to this uniformity [cf. Nakano's theorem, 23K]; otherwise, 91 may be completed, and, with the natural extensions of/* and the ring operations, the completion is a Maharam algebra. Notes and comments L00 and L1 now constitute a dual pair of perfect Riesz spaces; L00 is isomorphic to L l x and L 1 to LcoX. The main value of this result lies in the fact that L l x = L1', so that L00 can be regarded as a representation of the normed space dual of L1, or of S given the norm || || v From 53C-53Fa we see that this result will apply to any product of Dedekind cr-complete measure algebras offiniteor countable magnitude. The theorem 53B can be looked at from another direction. Suppose that 91 is a Dedekind complete Boolean algebra; under what conditions is L°°(9l) a perfect Riesz space? We have seen in 4XF that it need not 138

MAHARAM ALGEBRAS

[53

be. Now 53B shows that it is sufficient that there should exist a strictly positive semi-finite measure on 91. In fact it is possible to show that this condition is also necessary (cf. the remark at the end of 52G). There is an effective structure theory known for Maharam algebras. The first step is to express an arbitrary Maharam algebra as a product of Maharam algebras of finite magnitude [53Fb/c]. A much deeper argument, due to MAHABAM, can now be used to express a Maharam algebra of finite magnitude as a product of homogeneous algebras, and to classify completely homogeneous Maharam algebras of finite magnitude. (Maharam's theorem may also be found in SEMADENI [§§26.4-5]). This classification must not be confused with the much easier representation theorem of Kakutani [611]. 54 Measure-preserving ring homomorphisms When we come to examine homomorphisms between measure rings, we find that the most natural ones are the 'measure-preserving5 ring homomorphisms, that is, ring homomorphisms n such that net, and a always have the same measure. On the ideal of elements of finite measure, such homomorphisms have to be order-continuous [54B], so we obtain linear maps between the Lfl spaces as well as between the S and L00 spaces. Of course, a measure-preserving ring homomorphism has to be one-to-one, so in effect we are talking about subrings of measure rings. What is interesting is the fact that the map between the S spaces preserves the norm || \\l9 so that we find ourselves with a natural map between the L1 spaces. Now this canonical map between the L 1 spaces is in the opposite direction from the canonical map between the L# spaces; so the isomorphism between the L1 and L # spaces has complex effects. 54A Definition Let (91,/*) and (93, v) be measure rings. A ring homomorphism n: 91 -> 93 is measure-preserving if v(na) = /ia for every a e 91. A measure-preserving ring homomorphism must be one-to-one, for aj=0=>fia>0=>

v(na) > 0 => na 4= 0.

It need not be order-continuous, or even sequentially order-continuous; but we do have the following result. 54B Proposition Let (91,/*) and (93, *>) be measure rings, and n: 91 -> 93 a measure-preserving ring homomorphism. Then7r[9J/] ^ 93/ 139

54]

MEASURE ALGEBRAS

and TT:W->W is an order-continuous measure-preserving ring homomorphism. Proof Clearly n[W] c [51Cb], so iniaeAv(7ra) = proves that n:W->W measure-preserving ring

93'. Now if 0 c ^U 0 in 9t>, mfaeA/ia = 0 0. As v is strictly positive, inf n[A] = 0. This is order-continuous, and of course it is a homomorphism.

Remark The effect of this is that we can apply 54E below, which refers to order-continuous measure-preserving ring homomorphisms between semi-finite measure rings, to any measure-preserving ring homomorphism, by considering its restriction to the ideal of elements of finite measure. 54C Proposition Let (%/i) and (93, y) be measure rings, and n: 91 -> SB a measure-preserving ring homomorphism. Then n:W-^W induces a Biesz homomorphism n: 5(910 -> 5(930 [45B]. Now IITOHX = ll^Hi for every

Proof

xeS(W).

IfaeW, vn(xa) = vx{na) = v{na) = fia = fl(xa).

So 07r=/i: 5(910->R. Now

for every a; e S(W), using the definition of || || x [52E] and the fact that 7T is a Riesz homomorphism. 54D Corollary n: S(W) -> S(W) extends to a norm-preserving Riesz homomorphism n: Lx(9l) -> Proof By definition, L\%) and L x (8) are the completions of 5(810 and 5(930 respectively for the norms || l^. So n has an extension to a norm-preserving linear map from Lx(9l) to L1(93), which is a Riesz homomorphism because the lattice operations are continuous on both sides. 54E Proposition Let (9t,/0 and (93,^) be semi-finite measure rings. Let n: 91 -> 93 be an order-continuous measure-preserving ring homomorphism. Let 0: Lx(9l) -> L#(«) and ft: Lx(93) -> L#(93) be the 140

MEASURE.PRESERVING RING HOMOMORPHISMS

[54

canonical isomorphisms [52E]. Then by 54D and 45E we have a diagram:

Now this diagram commutes, i.e.

Proof

Suppose that ae 9l/. Then (TT'^TT) (#a) is given by:

= *>7r(a n 6) = /*(« n 6)

for every b e 31. So n'ft7rxa = ^ a i n ^#(9l)- But n'ljrn and ^5 are both linear, so they agree on 5(217); and they are both continuous, so they agree on L1(9l). 54F Corollary

Turning this diagram round, we can write

and see that Pnn is the identity on L1(9l). Thus if the norm-preserving Riesz homomorphism n is taken as embedding L1(3l) as a closed Riesz subspace of L1(S8), then Pn is a projection from L1(S8) onto L1(3l). i^ can be defined by saying that, for any 2/eL1(93), Pny is that member of such that (Pny, Xa) =

for every a G 21.

<

Notes and comments Since measure-preserving ring homomorphisms have to be one-to-one, we may regard all the work above as results about subrings of measure rings. If 21 is a subring of 93, then 141

54]

MEASURE ALGEBRAS

%f = % n ffi is a subring of ffi, and S(W) is a Riesz subspace of S(W) [45Da]. Now 54C shows that the norm || || x on S(W) agrees with the norm || || x on 5(93'), so that L\%) will be just the closure of S(W) in £!(»). Next, we see from 54B that the embedding of W in ffi is ordercontinuous; if 0 c: J. 10 in W, then ^4 10 in ffi also. So we have a natural map from L#(S3>) to L#(Sl>) [45E]. If we identify L#(»>) with £!(©/) = ^(SB) and L#(W) with L1{W) = L1{%), we get a map P : L^SB) -^L^Sl), which turns out to be a projection [54F]. These projections are particularly important in probability theory. The point is that a map n: 91 -> 33 leads naturally both to a map w i L 1 ^ ) - > £ ! ( » ) and to a map P^: L 1 ^ ) - > L 1 ^ ) . Thus £!(«), regarded as the completion of S(W), is a covariant construction; but identified with L#(W) it is contravariant.

5X Examples for Chapter 5 At this stage I restrict my examples to measure rings which are rings of sets. 5XB tries to show how, in a semi-finite measure ring (31, fi), L1(9l) depends on [i, even though it is always isomorphic to L#(3l), which does not depend on /JL. 5XC discusses a simple example of a measure-preserving ring homomorphism.

5XA Two simple types of measure ring (a) Let X be any set, and 31 any subring of &*X. Define /i: 31 -> [0, oo] by: fiE = n

HE has n members, where weN;

fiE = oo if E is infinite. Then (31,/^) is a measure ring. (b) If 31 is any Boolean ring, define/i: 31 -> [0, oo]by/^0 = 0,/ia == oo if a 4= 0. Then (31,/^) is a measure ring. 5XB L 1 and L# Let X be any set, and let 31 be a subring of 0>X containing all finite subsets of X, Let £: X -> [0, oo] be a function such that £(t) > 0 for every teX. Define /*: 31 -> [0, oo] by

142

EXAMPLES FOR CHAPTER 5 [5X Then pi is a measure on 21. (Note that, because \t}e% for every teX, the embedding 31 -> SPX is order-continuous. If A c % and sup A exists in 21, then sup A = \JA.) S(W) can be identified with a linear subspace of R x , as in 4XB. For xeS{W)

>

fnx) = Y

the sum being absolutely convergent if we employ the convention that

0.co = 0.So

1^ = 2^1^)1^)

V

XeS{W).

x

Now L (2)[) can be identified with \\x\\\\xx = using the same arguments as those which show that lx(X) is complete [2XB]. Let us regard L°°(2l) also as a Riesz subspace of Rx. Then the duality between L°°(2l) and S(W) [52A] is given by

<*,V> = St e x*Wy(0«0

V xeS(W), yeL«(a);

this formula is easily verified if # = #2? and y = ^JP7, and is extended to general x and y by linearity and continuity. Also, because every finite subset of X belongs to 21, L°°(2l) is order-dense in /°°(X). It follows from 17B and 2XA, or otherwise, that L#(2l) = L°°(2l)x can be identified with V-(X). So the canonical map from S(W) to L#(2l) is given by x\-+xx£, where (x x £) (t) = x(t) £(t) for every t e X. Clearly /i is semi-finite iff £(£) < oo for every t eX. In this case, the isomorphism from L1(2l) to L#(2l) [52E] is given by x h-> x x £for every # G L1, and this is now onto. This example shows how Lx(2l) is subtly dependent on the measure/£, even though (as long as /i is semi-finite) it is isomorphic to /1(X), which is entirely independent of /i. Since 21 contains all finite subsets of X, it will be a Dedekind complete Boolean algebra iff it is SPX. Now L°°(2l) can be identified with /°°(X) [4XCa], which is a perfect Riesz space [2XA]. Thus (%/i) is a Maharam algebra iff 21 = SPX and £(£) < oo for every teX. 5XC Let X be the set N x N; let 33 = 0>X; let <£mn>m,neN be a double sequence such that L

> 0 V m,neN,

2m,WGN£m7l < oo.

Let v be the measure on 33 given by

143

5X]

MEASURE ALGEBRAS

Now let 91 be the algebra 0>N, and define where r\m = 2 n 6 N £ m n for every m e N . Then (9t,/*) and (SB,j/) are Maharam algebras of the type described in 5XB; in fact they are of finite magnitude. If we write nA = AxN V i c N , then TT: 91 -> SB is a measure-preserving ring homomorphism. Identifying L°°(9l) and L°°(S3) with /°°(N) and /°°(NxN) respectively, n: L°°(9l) ->• L°°(93) is given by (nx) (m, n) = x(m) V m, n e N, # e /^(N). So, now identifying L#(9l) and L#(93) with P(N) and PfNxN) respectively, TT': L#(S3) -> L#(9l) is given by (TT^) (m) = SneM«(m, n) V m e N, 2 e lx(N x N). At the same time, 5XB shows how Lx(9l) and L1(93) can be identified

and

{«/:yeR***, 2 m , n6H ? mre |y(f»,n)| < oo}.

Clearly TT: L^Sl) -> L 1 ^ ) is still given by (nx) (m,n) — x(m)

V m,«eN,a;eL 1 (2l);

and Pn: Lx(58) -> L^Sl) is given by (P.y)(m) = Vm^n^mnV^n)

V meN,

Thus, for y eL^SB), (w-P,y) ("*, n) = ^SrtrfCmiy^.«) V TO, the projection nPv: L1(93) -> L1(S3) 'averages' y down each column.

144

6. Measure spaces The object of this chapter is to show how the concepts of ordinary measure theory may be joined to the more abstract work done so far. The first step is to describe the measure algebra of a measure space [6 ID]; then § 62 and § 63 show how the L00 and L1 spaces of this measure algebra can be identified with the usual function spaces. In view of the importance of Maharam measure algebras [§53], we are naturally concerned to identify the corresponding measure spaces; the search for useful criteria involves us in consideration of a number of special kinds of measure space, which are discussed in § 64. In §65 we examine an important class of function spaces, including the IP spaces, which exemplify the results of Chapters 2 and 3 on locally convex topological Riesz spaces. 61 Definitions and basic properties The most important idea here is the construction of the measure algebra of a measure space [6ID]. The rest of the section comprises a miscellany of definitions and results which will be used later. 61A Definition A measure space is a triple (X, S, ji), where X is a set, S is a er-subalgebra of SPX [i.e. a subalgebra closed under countable unions and intersections], and ji\ 2 -> [0, oo] is a function such that: (i) (10 = 0; (ii) /i(E [}F)= fiE+jiF if E, J F G S and E[]F = 0 ; (iii) if (Enyn€$ is an increasing sequence in 2,

61B Notes (a) For a note on the use of 'oo', see 51A. (b) Observe that I allow X = 0; in this case S = SPX — { 0 } and fi must be the zero function. (c) /i is called a measure on X, and the elements of 2 are called measurable sets. 145

61] MEASURE SPACES 61C Elementary results Let (X, 2, /i) be a measure space. (a) If E, F G S and E ^F, then /iE ^ /iF. (b) HE.FGS,

(c)

If

then/J(JE7 U J1)

<^i>neN ^ &ny Sequence in 2,

Proof (a) /«JP = ju,E+/i(F\E) ^ /iE. (b) /*(# U J?7) = fiE+/i(F\E) ^ /iE+fiF. (C)

61D The measure algebra (a) Let (X, S, /i) be a measure space. Write S o for {E:EeX,/u,E = 0}. Consider S as a Boolean ring. Then S o is an ideal of S, by 6lCa and 61Cb [see 41J], and we can form the quotient 21 = S/S o ; 91 is a Boolean algebra and the canonical map i?H-»ir:Z!-^2l is a ring homomorphism. The 1 of % is X\ We recall also that

Em c F' o E\FeJl0 o ji{E\F) = 0 [41J]. Now

W =F' o /^V*7) = MJ 1 ^) = 0 => fiE = /i(E (]F)= tiF.

So fi factors through 21, that is, there is a function fi : 21 -> [0, oo] such that /6JST = /i'E' for every . E G S . Henceforth I shall allow myself to write [i instead of [i for this function on 21. (b) Now the map E i-» E': S -> 21 is sequentially order-continuous. P Suppose that (En}neln j 0 in 2. Since HneN^ belongs to S and is a lower bound for {En:neN} in S, n^eN^n = ^ - ^et a be any lower bound for {E»:neN} in S. Let JS7eS be such that ^7# = a. Then so b 61 y ^a and 6lCc fiE ^ "Lnaii{E\En) = 0 because E' ^E'n for every neN. Thus ^ = 0 and a — 0. As a is arbitrary, <^> ne N 10 in 21. Because the map E\-^E' is a, ring homomorphism, this is enough to show that it is sequentially ordercontinuous. Q (c) It follows that (%/i) is a measure algebra. P 21 is a Boolean algebra and fi is a function from 21 to [0, oo]. (i) If a, 6 e 21 and a n b = 0, 146

DEFINITIONS AND BASIC PROPERTIES [61 let E, J^eS be such that E' = aandi^ = b.A$(E()Fy = E' o F' = 0, /i(E f)F) = O, so /i(a u 6) = ji((E [)F)')= = jiE+jiF

fi{E [) F) =

/iE+/i{F\E)

= ju,a+fib.

(ii) If ae% let S G S be such that E* = a. Then fia = 0 o [iE = 0 o i ? e 2 0 O a = 0. (iii) If neNf a in 21, choose for each neN an EneH such that E^ = an. Set # n = Ui<w-®i- Since the map E\-^E' is a lattice homomorphism, V

Let 0 = \Jn^Gn. By (b) above, 0' = sup n e H 0; = a. So Thus the three conditions of 51A are satisfied and (%/i) is a measure algebra. Q (d) It also follows from (b) that 91 is Dedekind cr-complete. P Let (ari)neK ^ e a n y sequence in 21. For each ne N, choose an EneH such that En = a n . Then H = n we N^n Gs > a n d ^ = i n f n e N ^ i n s - Because the map E\-+E* is a sequentially order-continuous lattice homomorphism, H' = infneNaw in 21. By 411, this is enough to show that 21 is Dedekind cr-complete. Q (e) Now (2t,/£) is the measure algebra of the measure space

*61E Inverse-measure-preserving functions Given two measure spaces (X,TI,JU>) and (Y,T,v), a natural class of functions to consider is given by the following definition. A function / : X -> Y is inverse-measure-preserving if f-1[F]e^ and j^f-^F] = vF for every FeT. Let 21 and 33 be the measure algebras of (X, S, jx) and (Y, T, J>) respectively. If/: X -> F is inverse-measure-preserving, then there is a sequentially order-continuous measure-preserving ring homomorphism n: 93 -> 21 given by 7r[i^'] = (/- 1 !/])* for every Fe T. P (i) We must verify first that n is well-defined. But this is a direct consequence of the fact that/" 1 : T -> S is a ring homomorphism such that / ^ [ - P J G S O for every .FeT 0 . (ii) The same argument shows that n is a ring homomorphism. Now suppose that (bn)ne^ 10 in 93. Choose Fn e T such that F'n = 6W for each 147

61]

MEASURE SPACES

n e N. Now the map J^i-^-F'iT^SBisa sequentially order-continuous lattice homomorphism, so

and v(Dn^K)

= 0- Similarly,

But /if^lCinenFJ = HHneM = °> »O infneN7T&w = 0. As <6n>weM is arbitrary, n is sequentially order-continuous. (iii) Finally, it is obvious that n is measure-preserving. Q [See also 61Jc, e.] 61F Further definitions

Let (X, 2, /i) be a measure space.

(a) Following 51Ba, write

2> ^{EiEe^/iE

< oo}.

If 91 is the measure algebra of (X, S,/0, then W = S>/So. (b) (X, S,/*) is semi-finite if its measure algebra is semi-finite [5lBb], that is, whenever EeTt and fiE = oo, there is an F c iJ such that - P G S and 0 < /iF < oo. *(c) A set .EG2 is purely infinite if /^JE7 = oo, and, for every FeZ such that F ^ E, fiF is either 0 or oo. Thus (X, 2,/0 is semi-finite iff there are no purely infinite sets in S. (d) A set A c X is negligible if there is an F e S such that A^F and /^F = 0. Now (i) any subset of a negligible set is negligible, (ii) a countable union of negligible sets is negligible [using 6lCc]. (e) Suppose that O is any property applicable to points of X. I shall write '(£) for almost every teX\ or '$(£) p.p. (t)\ or '® almost everywhere on X\ or < ' !> p.p.' to mean that X\{t: 0(0} is negligible. For instance, if #: X -» R is a function, then 'x = Op.p.' means that {t: x(t) 4= 0} is negligible. Or, if (xn)neJX is a sequence of real-valued functions on X, then <<^n(0)n6N ~* Op.p.(£)' means that {» is negligible. (f) (X,l,,fi) is called complete if, whenever /iF = 0 and E ^F, then J 5 G S (SO that /4.Z? = 0); i.e. if every negligible set is measurable. (g) (X, H,fi) is of finite magnitude, or totally finite, if its measure algebra is [530], i.e. if ju,X < oo. 148

DEFINITIONS AND BASIC PROPERTIES 61G Direct s u m of measure spaces indexed family of measure spaces. Let the 'disjoint union' of (Xt}ieI.

[61

Let <(X t ,S t ,^)> te7 be an

For E c X and tel write

El = {t:(t,i)eE}c:Xl. Let 2 = {E: E c X, iS?t e S t V *e/}, and define /*: S -> [0,oo] by It is easy to see that (X, £,/0 is a measure space. I shall call it the direct sum of <(X t ,S t ,/0X 6 / . In effect, the maps t h-> (t, t): X t -> X embed each Xt as a subset of X, and (XL}L€l now appears as a partition of X, A set E £ X belongs to S iff # t = .0 n XL e S4 = S n ^ X t for every 4 e /, and in this case if we identify each /£t with the restriction of fi to S t . [See also 64Ga.] 61H Measure algebra of the direct sum Clearly, in this construction, 2 is isomorphic to the Boolean ring product YlteI^L [41K]. It follows that if 21^ is the measure algebra of (Xp S t , fiL) for each tel, while 91 is the measure algebra of (X, £,/£), then 91 £ I L e i ^ r (For if , then /eJEr = Oiff^ t ^ = 0 for every tel.) And, given any So (91, /*) is the measure ring product of <(9lt, /OXei> a s defined in 51F. Consequently, the direct sum of semi-finite measure spaces is semifinite [61Fb, 51F]. *61I Representation of measure algebras Some of the work of this chapter will appear in sharper relief if I give a general representation theorem. We have seen in 6 ID that the measure algebra of a measure space is always Dedekind
See SIKORSKI, §29.1, or HALMOS L.B.A., p. 100, §23. 149

61]

MEASURE SPACES

Proposition Let (91, fi) be any Dedekind [0, oo] by vE = /i(ftE) for every E e 2. Then it is easy to see that (becauseftis sequentially order-continuous) (X, 2, v) is a measure space. Since

ftE = 0 o fi(ftE) = 0 o vE = 0, (9t,/£) is isomorphic to the measure algebra of (X, 2, v). 61J Exercises (a) Let (X, 2,/^) be a semi-finite measure space. Then jiE = sup{/iF .FeX^F ^ E} for every Ee2 [cf. 51Gc]. (b) Let (X, 2,/^) be a measure space. Then ji\ YJ -> R is countably additive [cf. 51E]. *(c) Let (X, 2,/0 be a measure space, Y any set, and/: X -> F any T = {F:F ^ Y, f-^F]eS}. function. Let Then T is a cr-subalgebra of & Y and there is a unique measure on T for which/is inverse-measure-preserving. (d) Let ((X t ,2 t ,/£ t )) t e J be an indexed family of measure spaces, with direct sum (X, 2 , / J ) . Then (X, 2,/0 is complete iff (X p 2 p /£ t ) is complete for every i e I. *(e) Let (X, S, /^) and (F, T, v) be measure spaces. Let/: X -> T be a function such that/[jE7] eT and vf[E] = ^ for every EGX. Then if 21 and 95 are the measure algebras of (X, S, /^) and (Y, T, y) respectively, there is a measure-preserving ring homomorphism n: W -> SKgiven by n(E') = (f[E])' for every EeU. Notes and comments In 6ID we have the first link between Chapter 5 and Chapter 6; others will appear in 62H and 63A; it is my purpose in this book to emphasize these connections. In every construction involving measure spaces, the effect on the measure algebras is one of the most important things to lookfor. Thus 61E, 61H and 6lJe are typical investigations. One of the oddities of measure theory is the lack of a completely convincing class of morphisms. The inverse-measure-|freserving maps of 6 IE are probably the most important ones; but many curious results [e.g. 61Je] attend other functions. Inverse-measure-preserving functions constitute a category for which the measure algebra is a contravariant construction; but for the functions of 61Je it is covariant. 150

MEASURABLE FUNCTIONS, L°

[62

62 Measurable functions; the space L° Given a measure space (X, 2, fi) ,we call a real-valued function on X measurable if the inverse image of an open set in R is always in 2 ; we have already encountered bounded measurable functions in 4XD. The space M(S) of all real-valued measurable functions on X is now a Riesz subspace of R x , and also a (T-sublattiee (i.e. it is closed under countable suprema and infima) and a subring (for ordinary pointwise multiplication in R x ) . If we say that two measurable functions are equivalent if they are equal almost everywhere, we find that the set L° of equivalence classes inherits most of the structure of M(2) to become a Dedekind cr-complete Riesz space with an order-continuous multiplication [62G, 621]. The main importance of the construction L° lies in its solid linear subspaces. If 91 is the measure algebra of (X, 2,/^), then L°°(2l) has a canonical representation as a solid linear subspace of L° [62H]. In the next section I shall show how L1(9l) can also be embedded naturally in L°, and § 65 deals with a large class of subspaces of which L1 and L00 are archetypes. L° is in itself a remarkable space. Although its most dramatic properties depend on certain conditions being imposed on (X, 2,/^), and will therefore be held back for the moment, 62K and 62Mh give a foretaste of what can be proved. 62A Definition Let (X, 2, /i) be a measure space and (Y, @) a topological space. A function/: X-> Y is measurable if /- 1 [6r]e2 for every Oe Y 'measurable' if j M p ^ e S for every FeT. This may not coincide with the definition above. Consider, for instance, the case X = Y = R, with @ the usual topology and S = T the class of Lebesgue measurable sets. Then there exist functions / : X -» Y which are (S, <S)-measurable, indeed continuous, but not (S, T)measurable. 62B Lemma Let (X, 2,/^) be a measure space, and (F, @) and (Z, %) topological spaces. If/: X -> Y is measurable and g: Y -> Z is continuous, then gf: X -> Z is measurable. 151

62]

MEASURE SPACES

Proof

For if G e % g-^G] e ©, so

62C Lemma Let (X, S,/*) be a measure space, and x: X -> R and 2/: X -> R measurable functions, where R is given its usual topology. Then the function ^ : z -> R2 t^ ^ is measurable. Proof

Let ( ? c R 2 be open. Then 6? can be expressed as where Jw and Jn are open subsets of R for each weN. (Infact, unless # = 0 = 0 x 0 , each Jw and Jn can be taken to be an open interval with rational end-points.) Now U W 6 N 4 X «4>

{t: (x(t), y{t))eGf) = U » 6 M M 4 ] n j 62D Lemma Let (X, £,/£) be a measure space. Then a function x: X -> R is measurable iff {t: x(t) > a} e £ for every a e R. Proof Obviously the condition is necessary. Conversely, suppose it is satisfied, and let G £ R be open. Then G can be expressed as for suitably chosen sequences neN and (fin}ne]x, and now ocn}\{t:x(t) > / 62E Proposition Let (X, 2,/^) be a measure space. Let M be the set of all measurable functions from X to R. Then Mis a Riesz subspace and a cr-sublattice of R x . Moreover, xxyeM for all x,yeM, where

(xxy)(t) = x(t)y(t) V teX. Proof If x,yeM, then x + y e M , by 62B, for + : R 2 - > R is continuous and ty-+(x(t),y(t)): X -» R2 is measurable by 62C. Similarly, a# and |#| belong to M for every a e R , because J3H*a/?: R->Rand | |: R -> R are continuous. Thus M is a Riesz subspace of R x . If A c M is countable and x = sup A exists in R x , then for every a e R . Using 62D in both directions, and the fact that S is closed under countable unions, we see that xeM. Equally, if inf^i exists in R*, then i n f A = _ s u p ( _ A) G M # Thus M is a cr-sublattice 152

MEASURABLE FUNCTIONS, L°

[62

Finally, M is closed under x because multiplication, like addition, is a continuous function from R2 to R. 62F Definitions (a) Let (Z,S,/*) be a measure space. I shall write M — M(2) for the space of real-valued measurable functions on X, described in 62E above. (b) Consider M o = {x:xeM, x = Op.p.}; that is,

XGM0

o fi{t:x(t) =f= 0} = 0.

Then Mo is a solid linear subspace of M, and also a cr-sublattice. P Suppose, for instance, that x, y e M0. Then {t: (x + y) (t) + 0} c {*:*(*) * 0} U {t:y(t) # 0}; so

* 0} ^ / $ : a ( 0 * 0 } + ^ : y ( 0 4= 0} = 0.

Thus M o is closed under addition. And it is easy to see that it is solid and closed under scalar multiplication. If A c M o is countable and x = sup A exists in M, then x(t) = sup^^ y(t) for every t e X [see 62E]; so is negligible. Thus M o is a cr-sublattice of M. Q (c) Now let us write L° = L°(L,/i) for the Riesz space quotient M/Mo, described in 14G. (d) We observe that, for x,yeM,

x ^ y o {x-y)+eM0 o /i{t: x(t) > y(t)} = 0, where x* is the image of x in L°. Thus

x = y o x = y p.p. 62G Proposition Let (Z,S,/^) be any measure space. Then L° = L°(Z,/i), defined in 62F above, is a Dedekind cr-complete Riesz space, and the canonical map from M = M(S) to L° is sequentially order- continuous. Proof In the construction of quotient spaces in 14G it was remarked that the canonical map is always a Riesz homomorphism. Now suppose that (xn}neV 10 in M, and that u is a lower bound of {x'n: n e N} in L°. Let x e Mbe such that x" = u. Then x' < x'n, so (x — xn)+ e Mo for every 153

62]

MEASURE SPACES

neN [62Fd]; but ((x-xn)+)ne-sfx+, so x+eM0, as M o is a cr-sublattice of M [62Fb]. It follows that u = x ^ 0 in L°. As w is arbitrary, min^x'n = 0; as (%n}nexis arbitrary, the canonical map is sequentially order-continuous. [See 14Lb.] Now let (un)nelx be any sequence in L° bounded above by u e L°. For each neN, choose a,nxneMsuch that x'n = un, and also ernxeM such that x' = u. Set which exists in M because M is a cr-sublattice of R x , and is therefore Dedekind cr-complete. Because the canonical map from M to L° is a sequentially order-continuous Riesz homomorphism,

As (un}nel!t is arbitrary, L° is Dedekind cr-complete. 62H We come now to the identification of the L00 space of the measure algebra of (X, 2,/^). Theorem Let (X, 2, ju,) be any measure space, and 21 its measure algebra [6ID]. Then L°°(9l) can be identified, as Riesz space, with the super-order-dense solid linear subspace of L°(S,/^) generated by e = (^Z)*.ForeachJE?Gi;,thisembeddingidentifies^(JS7#)inLGO(9l)with Y in L°. In particular, e becomes the unit of the M-space L°°(2l). Proof In 4XB we have seen that L°°(S) can be identified with a space of real-valued functions on X, which in 4XD (using the fact that S is a cr-subalgebra of 0>X) were identified as those bounded functions which are measurable by the criterion of 62D. Thus L°°(S) can be thought of as the solid linear subspace of M(S) generated by %X. Now 21 is defined as the Boolean ring quotient S/So, where S o is the ideal of sets of measure 0 [61D]. So, by 45Dc, L°°(9t) £ Loo(S)/Loo(S0). But it was pointed out in 4XE that I/°°(S0) *s precisely {x:xeL™(Z),

{t:x(t) * 0}eSo},

because S o is a cr-idealof S. Thus L°°(S0) = L°°(S) n Mo, where Mo is the space of functions which are zero almost everywhere, as in 62F. It follows at once that Loo(S)/Loo(S0) can be identified, as Riesz space, with the image of L°°(S) in L° = M/Mo. And because L°°(S) is the solid linear subspace of M generated by ^X, its image in L° will be the solid linear subspace generated by e = 154

MEASURABLE FUNCTIONS, L°

[62

On examining this embedding, we see that, for any J57e2, L°°(8l) is identified with the image of %E in LCO(S)/L00(S0); that is, with (X^y. The unit of L°°(2t) is of course #(JL '), since X' is the identity of the algebra 21, and this is identified with (x^Y = e. Finally, suppose that u ^ 0 in L°. Then there is an x e M+ such that x = u; now (x A ^ ^ > n 6 N t a? in M, so <^ A we>neN f u in L°. Thus L°°(3l) is super-order-dense in L°. 621 The multiplication on L°. In 62E it was pointed out that M(S) always has a natural multiplicative structure. Since Mo, as defined in 62F, is clearly an ideal of M, we can define a multiplication on L° = M/Mo by writing x* xy' = {xxy)'

V x,yeM.

XX is the multiplicative identity of M, so e — (xX)' is the multiplicative identity of L°. If xeM+, the map y\-*xxy\ M - > M is a Riesz homomorphism. From this it follows easily that the map v \-+u x v: L°^L° is a Riesz homomorphism for every w e L o+ . In fact, this map is order-continuous. P Suppose that 0 c i j 0 in L°. Let w ^ 0 be a lower bound for {wxvive^}. Take some voe^L and let x,yo,zeM+ be such that x = u, y0' = v0 and z = ^ . Then if zt = ZA(XXy0) we shall have zx* = w A (u x v0) = w. And Define x2: X -> R by »!(*) = xtf)-1 = 0

if »(«) 4= 0, if x(t) = 0.

Using 62D, or otherwise, it is easy to see that xx e M. Now so u± x u < e, where ^ = #j, and 0<%x^^%xttxv^v

V VEA.

So 0 = % x w — (xx x zj'. But now

{t: zx{t) 4= 0} = {^: (xx x zx) (t) + 0},

so Z1EM0 and w = zj = 0. As w is arbitrary, infve^^ x v = 0; as A is arbitrary, the map v H-> ^ x v is order-continuous. Q (Note also the results in 62Mc, which will be used later.) 155

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62J The next theorem describes one of the most remarkable properties of the space L°. In its strongest form it requires the measure algebra to be Dedekind complete, and will appear in 64D. To begin with I give a sequential form which is always true. First we need a lemma. Lemma Let (X, 2, /i) be a measure space, and let (un}nelS( be a sequence in L°(S, /i)+. Then either {un: n e N} is bounded above or there is some v > 0 in L° such that Jcv = supneNwn A hv V i e N . Proof

For each weN, choose # n eM(2) such that x*n = wn. Set

(i) If /t(*W) = 0, exists in M. Now in L°, so {^ :weN} has an upper bound. (ii) Iifi(X\E) > 0, set y = x(x\E) a n d v = V- Since A % V * G N, A hv V J G N . As /i(X\E) > 0, v > 0 in L°, as required. 62K Theorem Let (X, S, fi) be a measure space and E an Archimedean Riesz space. Suppose that F is a super-order-dense Riesz subspace of £ and that T: F ~+ L°(L,/i) is an order-continuous Riesz homomorphism. Then T has a unique extension to an order-continuous Riesz homomorphism from E to L°(Z,fi). Proof

I mean to apply 17B. To do so, I must show that

has a least upper bound in L° for every x e £ + . (a) I show first that if {yn}n^ is a sequence in F+ which is bounded above by x e E, then {Tyn: n e N} is bounded above in L°. P ? For suppose otherwise. By the last lemma, there is a v > 0 in L° such that kv = SVLVnesTyn Akv V k6N. In particular, setting k = 1, there is an m e N such that Tym A V > 0. 156

MEASURABLE FUNCTIONS, L°

[62

Now, because E is Archimedean, in E [use 15D], So, because F is order-dense in E, ym = sup A, where A = {y:yeF,

3 a > 0, y ^ (2/m-az)+}.

(For any upper bound for A must be an upper bound for and therefore greater than or equal to y.) Now T: F-> L° is ordercontinuous and A\ym, so Tym = s u p ^ ^ T y and there is a y e A such „ that m w = 2ty A v > 0. Let a > 0 be such that«/ ^ (2/m~~aa0+Let us now examine, for k, n e N, x A % ^ ^ A k(ym - aa:)+ = x A ak(arxym - o;)+ < a-xi/m + (^ - a~1i/?w)+ A aA;(a-12/m - *) + using 14Kb and 14B1. So kw = kw Ahv = kw A supneyf Tyn A &# N

T^

A

kTy

But L° is also Archimedean, so this is impossible, since w > 0. X Q (b) Now let x e fi+. Then there is a sequence
Thus

u = sup{Ty:2/eJP, 0 < y < a;}.

(c) Now by 17B there is a unique extension of T to an ordercontinuous increasing linear map U: E -> L°, and by 17Ca ?7 is a Riesz homomorphism. *62L A corollary of this theorem throws an interesting light on the real nature of the construction L°. 157

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Corollary Let (X, 2,/^) and (Y,T,v) be measure spaces, with measure algebras 21 and 33 respectively. Let n: 21 -> 33 be a Boolean ring isomorphism. Then ft: ^(21) -> L°°(SB) is a Riesz space isomorphism which extends uniquely to a Riesz space isomorphism from Proof We know from 45C that n: L°°(9l) -> L°°(93) is onto and normpreserving; so of course it is a Riesz space isomorphism. Since L°°(SS) is solid in L°(T, v), fr: L°°(2l) -> L°(T, v) is order-continuous [17E]. Now L°°(9l) is super-order-dense in L°(2,/£) [62H], so we can apply the theorem to obtain an order-continuous Riesz homomorphism

( / 0 ( )

extending n. To show that U is the required Riesz space isomorphism, we can use the following device. 77—1:93 -» 21 is also an isomorphism, so (7r-1)A:L0O(93)->L0O(2l) has an extension to an order-continuous Riesz homomorphism F:L°(T,y)->L°(2,/*). Now VU: L°(S,/0 -> L°(S,/0 is an ordercontinuous increasing linear map extending (TT" 1 )^ : L°°(2t) -> L°°(93). By 45F, (TT-1)^ = (7r~17r)A, which is of course the identity on L°°(2l). Since L°°(2l) is order-dense in L°(L,JLL), VU is the identity on L°(Z,ju,). Similarly, UV is the identity on L°(T, v). Thus U is indeed an isomorphism. 62M Exercises (a) If (X, S, /i) is a measure space and (#n>neN ^s a sequence in M(S) such that (xn(t)}ne^ -> o;(^) for every ^ e l , then iu: X -> R is measurable. (b) (i) Let 21 be any Boolean ring, and Z its Stone space. Then 5(21), regarded as a linear subspace of R z , is closed under the multiplication in R z . Consequently L°°(2l) is also closed under multiplication, (ii) Let (X, 2,/^) be a measure space and 21 its measure algebra. Show that Lco(2l) is closed under the multiplication in L°(S, ju,), and that this is the unique norm-continuous bilinear multiplication on LG0(2l) such that ^ x ^ 6 = x(a n b) for all a,be21. Consequently it agrees with the multiplication on LG0(2l) defined in part (i) above. (c) Let (X, S, /i) be a measure space, (i) Show that \uxv\ = \u\x \v\ for all u,veL°(L,fi). (ii) Show that a set A ^ L° is solid i& uxveA whenever ueA and |v| ^ e. (iii) Show that uxv = Oiff|w| A \v\ = 0. *(iv) Show that an element u of L° has a multiplicative inverse iff it is a weak order unit. 158

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[62

*(d) Show that in 62L the Riesz homomorphism /)

is multiplicative. *(e) Let (X, Ti,ju,) be a measure space and E an Archimedean Riesz space. Let F be a super-order-dense Riesz subspace of E and a sequentially order-continuous Riesz homomorphism. Then T has a unique extension to a sequentially order-continuous Riesz homomorphism from E to L°. [Use 17Gc] *(f) Let (X, 2,/*) and (F, T, y) be measure spaces, with measure algebras 21 and 95 respectively. Let n: 21 -> 93 be a sequentially ordercontinuous ring homomorphism. Then TT: £(21) -> 5(93) is sequentially order-continuous [hint: use the technique of 42J/42O], so extends uniquely to a multiplicative sequentially order-continuous Riesz homomorphism from L°(Z,,ji) to I/°(T, v). [Use (e) above.] (g) Let (X, 2,/0 be a measure space. Suppose that 0 in R, {n(un - auo)+: w e N} is bounded above in L°. (h) Let (X, 2,/^) be a measure space. Let E be any Archimedean Riesz space and T: L°(2,/£) -> E an increasing linear map. Then T is sequentially order-continuous. [Use (g) above.] Notes and comments The most important result above is the representation of the L00 space of the measure algebra of a measure space as a solid linear subspace of the L° space. Of course it is this solid linear subspace that is normally thought of as the L00 space of a measure space; so from the ordinary point of view it is this result which justifies the emphasis I have laid on the construction L°°(2l), and also its name. If we look closer at the construction of L°, se see that it does not really depend on the measure. The original space M(2), of course, is defined by the pair (X, S). Now the solid linear subspace M o in 62P depends only on the cr-ideal S o of S, not on anything else about fi. Thus it is not surprising that L° = M/MQ is determined by the algebra 21 = S/So, without further reference to the measure [62L]. The same result is the reason for the notation I use. Since L° is determined by the algebra 2 and the function/^: S -> [0, oo], regardless of how 2 is embedded in 0>X, I write L° = L°(2,/0. Similarly, M is determined, up to a multiplicative Riesz space isomorphism, by the algebra 2. For it is easy to prove 62 J and 62K with M replacing L°, and 159

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therefore to prove an analogue of 62L showing that if 2 and T are cr-algebras of sets which are isomorphic as Boolean algebras, then M(2) ~ M(T). Indeed, every space M(2) can be regarded as an L°. For suppose that X is a set and 2 a cr-subalgebra of 8PX. Define fi: 2 -> [0, oo] by fiE — n if E has n members, where % G N ; = oo otherwise. Then (X,2,/^) is a measure space. Since 2 0 = {0}, 2/2 0 ~ 2 and L°(L,ju,) ~ M(2). Thus it is unsurprising that most of the results of this section [62G, 621, 62J] consist in showing that L° spaces have properties which are a matter of course for M(2) spaces. It is clear that the theorem 62K relies only on the lemma 62J and the Dedekind ^-completeness of L°. A more powerful result along the same lines is given in 64C/D. *Note that we have seen in 5XAb that any Dedekind cr-complete Boolean algebra 91 can be given a measure, so by 611 it is isomorphic to the measure algebra of a measure space. This gives us a construction of L°(9l), which by 62L is well-defined up to isomorphism. The functorial nature of the construction L° is given by 62Mf. 63 Integration The theory of integration on an arbitrary measure space (X, 2,/*) can be developed in many ways. All the customary approaches, however, proceed by defining a space 8H/0 of' integrable' real-valued functions, together with an 'integral', a sequentially order-continuous increasing linear functional J which extends the functional fi: 5(2*) -> R. On S1 a semi-norm || 1^ is defined by writing ||#||i = J|#| for every x efi1, and S 1 is found to be complete, in the sense that every Cauchy sequence has at least one limit. It can then be pointed out that

H = {x:xe21, 1x^ = 0} is a linear subspace of S1, so that the quotient St/H is a Banach space, in fact an L-space. In most systems, all integrable functions are measurable; so, in the language of §62 above, S 1 c M(2) and H = S 1 0 Mo. So the quotient QXIH can be identified with the image of S 1 in L°(2,/^). It is usually easy to show that 5(2*) is || || ^dense in 2 1 , so that its image (identified in 62H with 5(21*)) becomes dense in S1/!*; it follows at once that fiVH is isomorphic to L1(9l) as defined in § 52, where 91 is, as usual, the measure algebra of (X, 2,/*). 160

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In this book, however, I propose to use this work to demonstrate the power of the general theorems I have introduced so far. We know that S(W) is a super-order-dense Riesz subspace of L1(9l) [52E]; it is easy to see that the natural embedding of S(W) in L° [62H] is an ordercontinuous Riesz homomorphism. So by 62K it extends to an embedding of L1 in L°. I define &1 in reverse as

the characterization of members of S 1 requires care, but proceeds satisfactorily [63D]. We can accordingly regard the integral on S 1 as the linear functional defined by the integral on the i-space L1, i.e. jx = jx' = ||#'||i whenever

x ^ 0 in S1.

The famous convergence theorems are now consequences of the fact that L 1 is an L-space [26HJ-1, 63Ma-d]. I prove without difficulty that L1 is solid in L° [63E]; it follows that uxveL1 whenever ueL1 and veL00. Now, writing (u,v) = jux v9 we have a duality between L 1 and L00; this is of course the usual duality taken between these spaces; we have to check that it agrees with the identification of L1 with LG0X in 52E. On the basis of this, I can use previous results to establish a form of the Radon-Nikodym theorem [63J]. 63A Theorem Let (X,Tt,/i) be any measure space, with measure algebra (91, /i). The embedding of S(W) as a Riesz subspace of L°(S, JLL), described in 62H, extends uniquely to an order-continuous embedding of L1(9l) as a Riesz subspace of Proofs (a) It will help if we regard the embedding of S(W) in L° as induced by a one-to-one Riesz homomorphism T: S(W) -> L°. Since L°°(9l) is solid in L° [62H], L«>(W) is solid in L°°(8l) [45Db], and S(W) is order-dense in L°°(21O [43Bc], the embeddings S(W) <= L"(W) S L»(«) c L° are all order-continuous [17E], so T: S(W)->L° is order-continuous. (b) Now 62K shows at once that T has a unique order-continuous extension U: L1(9l) -> L°, because by 52E S(W) is super-order-dense in L1. We see also from 17Cb that U: L1 -> L° is a one-to-one Riesz homomorphism. So we can think of U as an actual embedding of L1 as a Riesz subspace of L°. FTR 6

161

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63B Integration of real-valued functions: the space S 1 We have now found a canonical representation of L 1 as a Riesz subspace of L°. Let us write £ \ or S 1 ^), for Then S 1 is a Riesz subspace of M(S). Since L 1 is an L-space, it has a natural integral J given by

Jw = ||~+||i—||«""||x

V UEL1

[see 26C]. This induces an increasing linear functional on S1, writing

jx = jx'. 1

Members of S are the integrable real-valued functions, and J x or J xd/ioY j x(t)/i{dt) is the integral of x with respect to /i. 63C In order to describe S 1 more adequately, we must first look again at the embedding of 5(21/) in L°. Lemma Let (X, 2, /i) be a measure space and 91 its measure algebra. (a) S(U) c $i(fi) and jxdju, = ju(x) for every x e 5(S'), where S 1 and J are defined in 63B, and ju,: Sffi) -> R is the linear functional associated with the additive functional /JL on S^. (b) If u e S(W), then there is an x e S(Lf) such that x = u. Proof (a) For suppose that EeU. Then KeW and x{E'), in S(W), is identified with (xEY i n L° [ 6 2 H 1- So ^ e S 1 and

fXE = HxEYli = ll^")lli = ^ =f*E = filxE). Now the result follows because S(2J) is the linear space generated by :EeYJ} and J and ju are both linear functionals. (b) follows directly from 45Cb, as W = {Em : 63D Proposition Let (X, S,/^) be a measure space, and let xeM(Z). Then xe&i/i) iff (i) a = sup$y:ye5(S>), 0 < y < |a;|} < oo; (ii) {t: t eX, x(t) + 0} has no purely infinite measurable subset [for definition, see 61Fc]. In this case, a = j\x\ djit = \\x'lv Proof

(a) Suppose that xeQ1, i.e. that x EL1, embedded in L° by

63A. Then j\x\ =j\x\ #

= \\xm\\v If yeS(J2)

0 ^ 2 / ' ^ |o; | inL°and

fiy = jy = jy' = \\y'\\i 162

and 0 ^ y ^ \x\, then

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[63

using 63Ca. So a ^ ||#"||i < oo. On the other hand, let Fe^L be any purely infinite set. Then ju,(F n E) = 0 for every i?G 2', b e c a u s e / ^ n •#) must be either 0 or oo. So (xF)' A (##)' = x(E OF)' = 0 for every J2eS>. It follows that (xF)' A|M| = 0 for every ueS(W) [using 63Cb]. Now S(W) is order-dense in L1, so A = {u:ueS(W), 1

O^u^

\x\}^\x\

1

in L ; as the embedding of L in L° is order-continuous, J. f \x'\ in L°; so by the distributive law [14D], (XF A \x\)m = # F ' A |a?'| = mipu€AxF'

Thus

hu = 0.

ju,{t:teF, \x(t)\ > 0} = 0, 7

and i' cannot be a subset of {t: \x(t)\ > 0}. Thus x satisfies both conditions. (b) Conversely, suppose that x e M(S)+ satisfies the conditions. Let A be the set r Then A f and HZ ' ^Hi ^ aforevery^G^ [using 63Ca again], so v = sup A exists in L 1 [because the Zrspace L1 is Levi and Dedekind complete, by 26B]. Again, the embedding of L 1 in L° being order-continuous, v = sup.4 in L°. Since u ^ x' for every usA, v ^ x' in L°. Let ZEM(?Z)+ be such that a;" = v. Now consider j» = ^ . ^ < ? Suppose, if possible, that/*# > 0. For yeQ, set since ^ = Ur6Q^> there is a yeQ such that /d!^, > 0. Now y > 0 as 2 ^ 0, so Fy c {^: o:(^) =)= 0}. By the condition (ii), Fy is not purely infinite, and there is a measurable set H ^ Fy such that 0 < fiH < oo. Let y — yxH. Then i/ e S(Lf) and 2/ ^ x, so ^/# G^4. Thus y' 0,

which is impossible. X So we see that /iF = 0 and x' ^ z'; thus #* = z' eL1 and Also, because the norm on the i-space L 1 is Fatou, ), 0 ^ y ^ x} = oc. Thus the result is true for x ^ 0. In general, it is clear that \x\ and 6-2

163

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x+ satisfy the conditions if x does, and now x = 2x+- \x\ eS 1 ,

while a HI K 111 == IK 11163E Corollaries (a) S 1 is solid in M(2) and L 1 is solid in L°. (b) For any x ^ 0 in fi1, / # = sup{Jy :ye5(S/), 0 ^ y < «}. Proof (a) It is obvious from the criterion in 63D that S 1 is solid in M, and it follows that L1 is solid in L°. (Or use 17Gd.) (b) This is just a restatement of the last sentence in 63D, identifying jy with fly by 63Ca. If (X, £, /i) is a semi-finite measure space, then L1 is

63F Lemma order-dense in

Proof For, if 91 is the measure algebra of (Z,S,/^) then L°°(9t) is order-dense in L° [62H], and 5(91) is order-dense in L°°(8t) [43Bc]. But, if (Z, S,/e) is semi-finite, then S(W) is order-dense in 5(91) [5lGd]. So [using 15E, or otherwise], 5(910 i s order-dense in L° and L1 is orderdense in L°. 63G The duality between L1 and L00 Let (X,S,/^) be a semifinite measure space, with measure algebra 91. In 62H and 63A we have identified both L°°(9t) and L\%) with subspaces of L°(L,/i). liueL1 00 and veL , then \v\ < IML^, where e = #X* [see 62H]. So | U X V | = \u\ X \v\ ^ \u\ X ||v||«x>C = ||v||oo|^|. 1

As L is solid [63E], uxveL1 and Now we already have a duality between L 1 and L00 given by the identification of L 1 with L°°x in 52E. This duality is continuous for the norms of L 1 and L00 and is defined by saying that (X

M

^, 6 G 91.

/

But, given a e 9l and b e 91, because %E x xF = #(# n J7) in M(S) for all E,FeX.

164

So

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But now it follows that (u, v} = juxv for every u e S(W) and v e S( 21), and therefore for every u e L1 and v e L00, by continuity. Thus the duality between L1 and L00, under which L 1 is a representation of L°°x, is given by (u,v} = \uxv

V w e L1, v eL00,

and this is the way in which I shall generally regard it from now on. *63H For any measure space (X, £,/£), we can identify L1(3l) with L°°(21OX [52F]. Exactly the same arguments show that the same formula (u,v) = juxv gives this duality also. 631 Notation Let (X, 2,/^) be a measure space. Suppose that

xe&(/i) and that # e 2 . Then \x x xE\ < \x\ soxx xEeZ1. We write

regarding x^ as a member of L°°(2l). 63J The Radon-Nikoctym theorem famous theorem into the light of day.

I can now bring this

First form Let (91, ji) be a measure ring. Let v: W -> R be bounded and completely additive. Then there is a unique ^eL 1 (9l) such that pa = (u, xa) for every a e W. Second form Let (X, 2,/£) be a measure space. Let v: 2 ' -> R be bounded, countably additive, and such that vE = 0 whenever ju,E = 0. Then there is an integrable function x: X -> R such that vE = \Ex for every EHJ Proof The 'first form' is just 44Bb (which finds u in L#(W)) and 52E/F (which is the isomorphism between Lx(9l) and L#(W)). The 'second form' follows. You see at once that the condition '/iE = 0 => vE = 0' is simply to ensure that p descends to a function p': W -> R given by p'(E') = PE for every Eeltf. (As usual, 91 is the measure algebra of (X, £,/£).) Then J/ is bounded and additive; in fact it is completely additive. P By 5 ID, it is enough to show that p' is countably additive. 165

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Suppose that
such

which is what we need. Q So there is a ueLx{%) such that (u,xE'} = v E = vE for every . Now let XG&^/I) be such that x = u. Then for every E e 2X, using the notation of 631 and the identification of the duality in 63G/H. *63K The topology on L° For any measure space (X, S, ju,), there is a 'natural 5 topology on L°(Ii,/i), which may be defined as follows. For each u ^ 0 in L1, define pu[v) = ju A \v\ V veL°. 1

Because L is solid in L°, pu:L° -> R + is well-defined. Now it is a Fatou pseudo-norm. P (i) Of course, if \v\ < \w\, then pu(v) ^ pu(w). (ii) If v and w belong to L°, ^ A |t? + w| R is order-continuous), (iv) Consequently, if 0 <= A \w in L° sopu(w) = suvveApu(v). (v) Also, if {a|w| :a > 0}|0 in so infa>opM(av) = inf a > o pja |v|) = 0. Q Since, given u, v e {L1)+9 Pu+v

the system {pu: ^ G (L1)4"} defines a Fatou topology % on L° which by (iii) above is also Lebesgue. When JLCX < oo, % (or the topology it induces on M(S)) is commonly called the topology of convergence in measure; see 63Mk. I regret that space does not allow me to discuss at length its remarkable properties, but some of them are in 63L, 63M j-k, 64E, 64Jc(v) and 64Je. For two examples see 6XAa and 6XBc. 166

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*63L Lemma If (X, 2,/^) is semi-finite, the topology % described above is Hausdorff. Proof Let v be a non-zero member of L° and let yeM(L) be such that y' = v. Then fi{t:y(t) 4= 0} 4= 0, so there is an neN such that fi{t: \y(t)\ ^ 2~n} =f= 0. Let H be a measurable subset of {t: \y(t)\ ^ 2~n} such that 0 < /iH < 00; such a set exists as (X, 2,/^) is semi-finite. Set

u=

(XH)'EL1.

Then

As v is arbitrary, this shows that % is HausdorfF. [See also 63Mj.] 63M Exercises (a) (B.Levi's theorem.) Let ( X , 2 , / J ) be a measure space and (xn}nBlS[ a sequence of integrable real-valued functions on X such that cc = s u p n e N J ^ < 00 and, for each n, xn ^ xn+1 p.p. Then (i) there is an xeM(L) such that x(t) = supn6H#TO(£) p.p. (t), (ii) any such x is integrable, with J # = a. [See 26Hj.] (b) Let (X, 2,/£) be a measure space. Let (x^)n^ be a sequence of integrable real-valued functions on X such that SweN J \xn\ < °°- Then there is an integrable function x such that x(t) = YineXxn(t) P-P- (0(c) (Fatou's lemma.) Let (X, S,/^) be a measure space, and (xnyne^ a sequence of non-negative integrable real-valued functions on X such thatsup neN Jo; n < cxj.Thenliminf^^oo^^) < oop.p. (t),andifi/:X->R is defined by y(t) = liminfw_>0Oa:w(^)

when this is finite, 0 otherwise,

then y is integrable and J y < liminf^ooja^. [See 26Hk.] (d) (Lebesgue's Dominated Convergence Theorem.) Let (X, S, /i) be a measure space and x a non-negative integrable real-valued function onX. Let (%nyn€Tn be a sequence of measurable real-valued functions on X such that (i) |#J ^ x for every n e N (ii) (xn(t))ne1x -> i/(^) for every teX. Then jy = l i m ^ ^ j xn. [See 26HL] (e) Let (X, 2, /^) be a measure space and x an integrable real-valued function on X. Let FeTt be such that /iF > 0 and #(£) > 0 for every teF. Then jFx > 0. (f) Let (X, 2,/^) be a measure space, and x: X -> R an integrable function, (i) The map jPh->JF#: S->R is countably additive, (ii) For every e > 0, there is a yeSi^J) such that J |o: — 2/| ^ e. (iii) For every e > 0 there is a set EeHf and a # > 0 such that \\Fx\ < e whenever i ^ S d ^ ^ ) 8 167

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(g) Let (X, 2,/^) be a measure space of finite magnitude. Then L°°(2) c £i(/£), and J: L°°(S) -> R is continuous for || \\n. (h) Let (X, 2,/^) be a measure space, and i c : I - > R a non-negative integrable function, (i) For every a > 0, /i{t: x(t) > a} < oo. (ii) There is a sequence (%n}ne$ in 5(2') such that (xn)nelx t x in M(S); i.e. S(Zf) is super-order-dense in S 1 ^). *(i) Let (X, 2,/^) and (Y, T, p) be measure spaces. Let/: X -» Y be an inverse-measure-preserving function [6IE], (i) If #: Y-> R is measurable, so is xf: X -> R. In this case, J #di> exists i&jxfdju, exists, and they are then equal, (ii) The map x\-+xf: M(T) -> M(S) induces a sequentially order-continuous Riesz homomorphism from L°(T, y) to L°(2,/£), which includes norm-preserving maps from L 1 to L 1 and from L00 to L00. These maps are precisely the maps n described in 54D and 45B, where n is the homomorphism between the measure algebras described in 61E. See also 62Mf. *(j) Let (X,li,/i) be a measure space. For EEILJ and e > 0, let U(E,e) = {x: xeM(Ii)9 p{t: teE, \x(t)\ ^ e} ^ e} c L°(S,/^). Then {U(E, e): EeU, e > 0} is a base of neighbourhoods of 0 for the topology % on L° described in 63K. Hence, or otherwise, show that % is HausdorffifF(X, 2,/^) is semi-finite. *(k) Let (X, 2,/£) be a measure space of finite magnitude, and suppose that (xn}neV is a sequence in M(S). (i) Show that (x'n}neI[ -> 0 in L°(Z,/i) for the topology £ of 63K iff (/i{t:\xn(t)\>e})n€jx->0

V e>0.

(ii) Let / : R -> R be continuous and bounded. Show that there is a function R given by 0(a;') = jfxd/i

V »eM(S),

and that is continuous for S. Notes and comments In Chapters 4 and 5, the relationships 5(910 £ ^(Sl) and 5(910 £ S(9i) £ L°°(9l) were established for any measure ring 91. Now the effect of 62H and 63A is to show that when 91 is the measure algebra of a measure space (X, 2, /i), then all these spaces can be embedded in L0(L,ji). This is, of course, the ordinary way of looking at L 1 and L°°. So the work of 63G/H is essential, since we must be sure not only that the abstract spaces of Chapters 4 and 5 are isomorphic to the usual function spaces, but also that they are in the familiar duality. 168

INTEGRATION

[63

Most of the results in this section are, like 63G, a matter of checking that we really are in the situation we expected. But we nowfindour labour beginning to bear fruit, in a quick proof of the Radon-Nikodym theorem [63J]. The 'first form' is just a restatement of earlier results; the 'second form', the conventional Radon-Nikodym theorem, is a straightforward corollary. It is instructive to take a conventional proof of the Radon-Nikodym theorem [e.g. WILLIAMSON p. 100, theorem 6.3d, or ROYDEN chap. 11, §5] for contrast with the proof here. The 'Jordan decomposition theorem' is 42N; the 'Hahn decomposition theorem' is 52Ha. The proof proceeds very much on the lines of 52D, reading HJ for 21 throughout. It finishes with an argument based essentially on the fact that the space of bounded countably additive functionals on U is an i-space, just as in 52E. You will see that the fact that 2 is a cr-algebra of sets is vital, for the sake of the Hahn decomposition; whereas no such condition is required in § 52. In fact it is used, not in this part of the argument, but in the embedding of L 1 in L° [62K/63A]. *It is clear from 63Mj that the topology of L° can be described without reference to the theory of integration; but this seems an unnecessary refinement. It has usually been studied for measure spaces of finite magnitude [see 63Mk], and various notions of convergence have been introduced in other cases. But I think there can be no doubt that the topology of 63K is the right one for ordinary purposes.

64 Maharam measure spaces Now that we have identified the function spaces L 1 and L00, the work of § 53 can be seen in a new light. I shall define aMaharam measure space to be one with Maharam measure algebra, so that by 53B the dual of L 1 is identified with L00. We find that, because L00 is Dedekind complete, so is L°, and that this leads to striking consequences [64D,64E]. The brief discussion in 53C-53E showed that a product of measure algebras of finite magnitude is Maharam; so a direct sum of measure spaces of finite magnitude is Maharam. Such a measure space I will call decomposable. All important Maharam measure spaces are of this kind. A useful sufficient condition for a measure space to be decomposable is in 641. 64A Definition A measure space is Maharam (sometimes called localizable) if its measure algebra is Maharam, i.e. semi-finite and Dedekind complete [53A]. 169

64] MEASURE SPACES

64B Theorem If (X,H,/i) is a semi-finite measure space, with measure algebra 31, then these are equivalent: (i) (X, 2, /i) is Maharam; (ii) L°°(2t) is Dedekind complete; (iii) L°(L,fi) is Dedekind complete; (iv) the natural duality between L1(3l) and LG0(9l) represents L00 as Proof In 53B it was shown that (i), (ii), and (iv) are equivalent. (Note that of course L°°(2l) has a unit, e = (xX)\) Obviously (iii) => (ii), as L00 is a solid linear subspace of L° [62H]. Conversely, suppose that L00 is Dedekind complete and that A c (L°)+ is non-empty and bounded above. Then for each n e N.

vw = svp{uAne:ueA} 00

exists in L , because L00 is Dedekind complete; as L00 is solid in L°, #w = sup{WAneiueA} in L°. Now sup^L = sup^eNv% exists in L° because L° is Dedekind cr-complete [62G]. Thus (ii) => (iii). 64C We can now prove a stronger version of 62K, which will be useful later, as well as being intrinsically remarkable. As before, I begin with a lemma. Lemma Let (X, 2,/£) be a Maharam measure space, and let A c: L°(£, /i)+. Then either A is bounded above or there is a v > 0 such &# = supwe^i6A&# V i e N . Proof

If ^4 is not bounded above, set un = svipUEAuAne

V neN,

#

where as usual e = (^^) ; these all exist because L° is Dedekind complete [64B]. Then {un: n e N} is not bounded above, so by 62J there is 8bV > 0 such that Jcv = supweNwn A kv < supMe^it Akv ^ Jcv for every jfc e N. 64D Theorem Let (X, 2,/^) be a Maharam measure space and £ an Archimedean Eiesz space. Suppose that F is an order-dense Riesz subspace of E and that T: F -> L°(S,/^) is an order-continuous Riesz 170

MAHARAM MEASURE SPACES [64 homomorphism. Then T has a unique extension to an order-continuous Kiesz homomorphism from E to L°. Proof

Just as in 62K, I use 17B. Suppose that x ^ 0 in E, and set G = {zizeF, 0 < z < x}.

? Suppose, if possible, that T[C] is not bounded above in L°. Then by 64C there is a v > 0 in L° such that kv = mp{hv ATzizeC}

V &eN.

In particular, there is a i/0 e C such that 0 A Ty0 > 0. Now, because E is Archimedean, ?/0 = sup A where A = {y:yeF+,

3 a > 0, y ^ (yo-ax)+}.

7

Asl is an order-continuous Riesz homomorphism, TyQ = sup T[A], and there is a ?/eJ. with w = vhTy > 0. Let a > 0 be such that V < (Vo-oix)+.

Now, just as in 62K, consider, for zeC and heN, zAky^xAky^xA k(yo-ax)+ — x A OLkiar^y^ — x)+ ^ a - 1 ^ + (x - a - ^ o ) ^ A otkict-^Q - x)+

So

kw = kw Akv = kw A supzeCkv A Tz kv ATZ ^ supzeCkTy A Tz

contradicting the fact that w > 0. X Since L° is Dedekind complete [64B], sup T[G] exists. By 17B, T has a unique extension to an order-continuous increasing linear map from E to L°, which by 17Ca is a Riesz homomorphism. *64E Theorem Let (X, S, /i) be a semi-finite measure space. Then it is Maharam iff the usual topology % on L°(S,/^) [63K] is complete, and in this case % is Levi. Proof By 63L, % is Hausdorff. We know it is Lebesgue and locally solid [63K], so if it is complete, L° must be Dedekind complete [24E], and (Z, lt,/i) is Maharam [64B]. 171

64]

MEASURE SPACES

Conversely, if (X, 2,/*) is Maharam, % is Levi. P Suppose that A <= (L°)+ is directed upwards and bounded for %. ? If A is not bounded above, let v > 0 in L° be such that lev = 8vpU€Au A lev V 1ceN

[64C] Then v = sup^^fe- 3 ^ At> for every & ^ 1. Let p be a continuous Fatou pseudo-norm on L° such that y = pv > 0 [23B or 63K]. Let& ^ 1 be such that k~xA c [u:pu ^ Jy}, i.e./)(&-%) < | y for every . Then -4 fc-%

Ati) =

using the fact that {k~xu f\v:ueA}\v. X Thus ^l is bounded above. As A is arbitrary, % is Levi. Q But we know that % is Fatou [63K] and that L° is Dedekind complete [64B], so by Nakano's theorem [23K], % is complete. 64F Proposition (a) A measure space of finite magnitude is Maharam. (b) A direct sum of Maharam measure spaces is Maharam. Proof (a) follows directly from 53D, since the measure algebra of any measure space is Dedekind o*-complete [61Dd]. Now (b) follows from 53E since the measure algebra of a direct sum is the product of the measure algebras [61H]. 64G Definition (a) The last proposition shows the importance of the following concept. A measure space (X, 2, JLL) is decomposable (sometimes called strictly localizable) if it is (isomorphic to) a direct sum of measure spaces of finite magnitude. From the discussion in 61G, it is clear that (X, S, fi) is decomposable iff there is a partition (Xt}LeI of X such that: (i) XL e 2/ for every i e I; (ii) given a set E c X, E e 2 iff E n XL e 2 for every i e I; (iii) for every (b) A special case of the above [see 64Ha] is the following. A measure space ( X , 2 , / J ) is o-finite or of countable magnitude if there is a sequence {En}neIS in 2* such that X = UneNEn. (c) A weaker property is important to us because of its use in 641 below. A measure space (X, 2, /JL) is locally determined if it is semifinite and, for any set E c X,

EoFeX 172

V

MAHARAM MEASURE SPACES

[64

64H Proposition (a) A measure space of countable magnitude is decomposable. (b) A decomposable measure space is Maharam and locally determined. Proof (a) Let (X, 2, [i) be a measure space of countable magnitude. Let (En}neji be a sequence in 2^ such that X = (JneH^ for each WGN. Then (X n ) n e N is a partition of X, and Xne 2 ' for every

neN. If E ^X, then

Finally, if .EG2, ^

H X,) = li

So <Xn>weN has the required properties, and (X, 2, ju,) is decomposable. (b) By 64F, a decomposable measure space is Maharam; it is therefore semi-finite. So it is obviously locally determined. *64I Proposition Let (X,S,/*) be a locally determined measure space which is complete in the sense of 6lFf. Suppose that there is a disjoint family s/ c 2 such that = 1 = JT in the measure algebra 91 of (X, 2,/^). Then (X, 2,/*) is decomposable. Proof (a) Observe first that if FeHf, there is a countable set <s/0 c j/such that ju,(F\\Jstf0) = 0. P Since ^ is disjoint, the set must be finite for each n e N. So F)

> 0}

is countable. Consequently U^o a n ( i -® = J P\U^o b^h. belong to 2 . Now /i(H (]E) = 0 for every Hesf, so U' n fl" = 0 in 91 for every HGS/. AS sup{IT \Hesf) = 1, JST = 0, i.e. ^ = 0, as required. Q (b) Now suppose that E c X and that E nHeH for every Hes/. 173

64]

MEASURE SPACES

Then 2?e2. P Let FeU. By (a) above, there is a countable set such that /i(F\\Jsio) = O. Now E (\H ftFel, for every , so E n F n U ^ O G S - O n the other hand, 0

J/OCJ/

which is of zero measure; so E n F\\J J / O G S as (X, 2,/^) is complete. Thus E CiFelZ. But i'7 was an arbitrary member of 2'; as (X, 2,/^) is locally determined, i? e 2. Q (c) Moreover, if i ? e 2 , then ^ = ^He^ME H # ) . P (i) Suppose first that EeHf. Then there is a countable J / 0 C si such that fi(E\\J J/ 0 ) = 0. Since J / 0 is countable, iiE = p{E n U ^o) < ^He^0ME oH)^ ZH^M®

n 5).

7

(ii) In general, we know that /^i? = sup {/iF: J? e 2^, F ^ E} because (X, 2,/^) is semi-finite [6lJa]. So IiE = s\ip{/iF:FeIS,

F ^ E}

^^,

e/finite}

because «s/ is disjoint. Q (d) So if we set Ho = X\ \J stf, 88 = si U {flj,} is a suitable partition of X. For by (b) above, # o e 2 , and by (c) /iH0 = 0. Now if JE/ C X and for every £Te^, then .S E 2 by (b) and n £T) = by (c). Thus (X, 2,/^) is decomposable. 64J Exercises (a) Let (X, 2,/^) be any measure space. Let / be the cr-ideal of negligible sets in 0>X [6lFd]. Show that 2' = { = {G:lE,Fe^

E c G c F,/i(F\E) = 0}

is a cr-subalgebra of SPX, and that there is a unique extension of /i to a measure/^'on 2'. Show that: (i) (X, 2', /i') is a complete measure space; (ii) (X, 2,/^) and (X, 2',/^') have isomorphic measure algebras; (iii) M(2') = {x:xeRx, 3 * / E M ( 2 ) , y = a; p.p.}; 174

MAHARAM MEASURE SPACES [64

(iv) £/>(£,/*) ~ L ° ( 2 > ' ) ; 3 ye&(/i), y = a; p.p.}; (v) S V ) = {x:xeRX, (vi) (X, 2',/^') is semi-finite or Maharam iff (X, 2,/^) is; (vii) if (X,S,/G) is decomposable, so is (X,2',/£'). (b) Let (X, 2,/^) be any measure space. Let

2' = {E:EnFeX

V.

Define/: S'-> [0,oo] by Show that: (i) (X, 2',/G') is a locally determined measure space; (ii) if 9t and S3 are the measure algebras of (X, 2,/^) and (X, S ' , / 0 respectively, then 917 £ 93^; (iii) if (X, 2,/J) is semi-finite, then / is an extension of fi; (iv) if (X, 2,/^) is Maharam, then 91 c^ 93, so that L°(S,/0 - L°(2',/0; (v) if (X,2,/*) is complete, so is ( X , 2 ' , / 0 . [Hint for (iv): if # e 2 ' , let J J e S be such that

*(c) Let (X, 2,/^) be a semi-finite measure space, with measure algebra 91. Show that these are equivalent: (i) (X, 2,/0 is of countable magnitude; (ii) 91 is of countable magnitude [53Fa]; (iii) Lc0(9l) has the countable sup property; (iv) L°(L,/i) has the countable sup property; (v) the topology % on L°(S, fi) [63K] is metrizable. (d) Let (X, 2,/^) be a measure space of countable magnitude, and v: 2 -> R a countably additive functional such that vE = 0 whenever fiE = 0. Then there is an integrable function x: X -» R such that jEx = vE for every EeH. *(e) Let (X, 2,/0 be a measure space of countable magnitude and (xn}nen a sequence in M(2). Show that <^>neN -> 0 for the usual topology on L°(2,/0 [63K] iff every subsequence (yn)neN of <»n>weH has in turn a subsequence 0 p.p. (t). (f) A direct sum of locally determined measure spaces is locally determined. A direct sum of decomposable measure spaces is decomposable. *(g) Any Maharam algebra is the measure algebra of a decomposable measure space. [Use 53Fc, 611.] *(h) Let (X, 2,ft)be a Maharam measure space such that whenever E c X is such that EnFel, and ju,(E ()F) = 0 for every Fe2>, then . Show that (X, 2,/J) is locally determined. 175

64]

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Notes and comments This section demonstrates the power of the abstract theory we have been studying. It shows that, for a semi-finite measure space, Dedekind completeness of the measure algebra is equivalent to a series of very important properties of the function spaces [64B]. There is an application of 64D in the next section [65A]. It is clear that 64C/D use only the Dedekind completeness of the measure algebra, not the fact that it is semi-finite. This is significant because we know that any Dedekind complete Boolean algebra can be expressed as the measure algebra of some measure space [see the note at the end of § 62]. In fact 64D is characteristic of Dedekind complete Riesz spaces satisfying 64C; these include the C^ spaces of VULIKH [chapter v, §2] or LUXEMBURG & ZAANEN R.S. [§47]. We can see from 64Ja and 64Jb that any semi-finite measure space can be converted into a complete measure space and thence into a complete locally determined space, by adding new measurable sets. In fact some important constructions for measure spaces can be adjusted so as to yield complete locally determined spaces directly [see, for example, 71A]. There do exist Maharam measure spaces which are not decomposable but which are complete and locally determined (and are therefore unaffected by the transformations in 64Ja/b), but these are all thoughly unnatural. This is why 641 is so important; it is clearly a necessary and sufficient condition for a complete locally determined measure space to be decomposable, and it is in practice nearly a necessary condition for a properly constructed measure space to be Maharam. Measures of countable magnitude, which include, for instance, all the ordinary Lebesgue-Stieltjes measures on finite-dimensional Euclidean spaces, are the simplest non-trivial decomposable measures. They have a number of special properties such as those in 64Jc-e.

65 Banach function spaces In §§ 62 and 63 we have identified two of the most important subspaces of L°. A great many other spaces have attracted interest, principally L2 and the other IP spaces. In this section, I show how an important class of normed spaces can be discussed in terms of the concepts I have introduced; specific examples are in 6XD-6XH. I begin with a powerful general result which enables us to identify the dual space Ex for a large proportion of naturally arising Riesz spaces. 176

BANAGH FUNCTION SPACES

[65

65A Theorem Let (X, S, /i) be a Maharam measure space. Let E be an order-dense solid linear subspace of L° (2,/j). Let

F = {w.weLQ^uxweL1

V ueE}.

Then F is a solid linear subspace of L°. Define a duality between E and F by writing.

(uyw} = juxw

\f ueE, weF.

Then this duality induces a Riesz space isomorphism between F and £ x . Proof (a) Because L 1 is a linear subspace of L° [63A] and x is bilinear, F is a linear subspace of L°. Because L 1 is solid [63Ea] and E is a Riesz subspace, F is solid. P Suppose that weF and that \v\ < |w|. Then for any ueE \uxv\ = \u\ x jv| < \u\ x \w\ = |^6 x w\ eLl [using 62Mc(i)]. So u x v e L1. As u is arbitrary, veF. Q (b) Clearly the duality given is bilinear, and therefore induces a linear map T: F -> JE*. Now T[F] c £x# p Suppose first that weF+ and that 0 cz J. 10 in £. Then {wxw;: us A] 10 in L 1 [621]. So inf^e^C?7^) (w) = inf we ^/^ x w = 0, because J is order-continuous on L1. Thus TweEx.

weF, Tw = Tw+-Tw~eEx. Q (c) In fact J7: F->£xisaRieszhomomorphism. T is increasing, as

Now, for any

P It is clear that

(Tw)(u) = Juxw ^ 0 V w,w ^ 0. So suppose that v A w = 0 in F, and that u e E+. Let x,y,ze M(S)+ be such that x' — u, y' — v and z' = w. Examine

in M(S). Clearly ^ ( 0 = x(t) if 2/(^) > 2(Q, and xx(t) = 0 otherwise. So x x i (2/""2;)+ = x x (2/~2;)+> while ^ x (z — y)+ = 0. Now set u± = x\ in L°. Then 0 < u± ^ ^ and % x v = %x(«;--w)+ = \xxx(y — z)+]' = [xx (y — z)+]# = wx (v — w)+ = uxv, while similarly u±xw = u±x (w — v)+ = [a^x (z-~2/)+]' = 0. 177

65]

MEASURE SPACES

So

(Tv A Tw) (u) < (Tw) (ux) + (Tv) (u - u±) = J %X W + J (U — Ux) X V = 0.

As u is arbitrary, Tv A Tw = 0. But this shows that T is a Biesz homomorphism [14Eb]. Q (d) It follows (because E is order-dense in L°) that T is one-to-one. P If weF and w =)= 0, then |w| > 0. So there is a weE such that 0 < w ^ |w|. Now u x |w| > 0, so \Tw\ (u) = T(\w\)(u)

= jux\w\>

0,

and Tw + 0. Q (e) Next, G = T[F] is order-dense in £ x . P Suppose that / > 0 in £ x . Let w0 > 0 in E be such that/& 0 > 0. Now since L00 is orderdense in L° [62H], we know that A = {u:ueL00, 0 0. Let ux = ||w||~1w, so that/% > 0 and ||ux\ oo = 1. Consider the functional g: L00 -> R given by g(v) =f(u1xv)

V veL™.

00

Since the maps V H ^ X V I L - ^ £ and/: £ -> R are both increasing and order-continuous, ge (L°°)x. So by 63G there is a weL1 with Jwxt? = <7(#) =f(u1xv)

V veL00;

and, setting v = e, j w = /(%) > 0, so w > 0. Now suppose that u e E+. Then once again B = {V:VGL00, 0 < v ^ w}f w,

so ButifveB,

{w;xv:?;G£}t^xw. jwxv=f(uxxv)

^f(v) ^f(u),

1

since % ^ e = (^^)*. As L is an i-space, {t^xvrveJS} is bounded above in L1, i.e. wxueL1, and jwxu = suj)veBjwxv

^f(u).

1

So for any %e£, \wx u\ = wx |u\ eL , and weF. As w > 0, I 7 ^ > 0 in Ex, by (d) above. Finally, we have seen that, for any ueE+,

(Tw)(u) = jwxu ^ f(u), so 0 < Tw < / . Thus [using 15E] T[F] is order-dense in £ x . 178

BANAGH FUNCTION SPACES

[65

(f) Thus T is a Riesz space isomorphism between F and the orderdense Riesz subspace G of Ex. Define 8: G -> L° by 8 = T~\ Now 8 is a one-to-one Riesz homomorphism, and 8[G] = F is solid in L° [(a) above], so 8 is order-continuous [17E]. Therefore, S extends to a one-to-one Riesz homomorphism U: Ex -> L° [64D, using 17Cb]. Now E7[EX] c F. P Suppose t h a t / ^ 0 in E x , and consider 4 = { V : V G F + , TV < / }

Then, because G is order-dense in E x and 8 is order-continuous, ^4 f Uf in L°. But suppose that ueE. Then {|w| xv:veA}\\u\ x £7/. JH

(2T)(H)/(H)

So once again |w x Uf\ = \u\ x UfeL1. As % is arbitrary, £7/ eF. Now for a n y / e £ x , 17/ = I7/+ - Uf-eF. Q (g) But now T: F-* Ex and £7: £ x -> F are both one-to-one Riesz homomorphisms, and UT is the identity on F. It follows at once that T is an isomorphism between F and Ex. *65B Corollary The same result applies for any order-dense linear subspace E of L°, whether solid or not. Sketch of proof Just as in part (a) of the proof above, F is certainly solid. Now let G be the solid hull of E in L°. Then it is easy to see that

F = {w:weL°, uxweL1

V ueG},

so that the duality between F and G sets up an isomorphism between F and G x . Now using 17B it is easy to see that every increasing ordercontinuous functional / : E -> R extends to an increasing ordercontinuous functional on G, and therefore corresponds to an element of F. The remainder of the verification that F is isomorphic to Ex is relatively straightforward [cf. 17Gb]. 65G Definition Let (X, £,/£) be a semi-finite measure space. An extended Fatou norm on L°(L,/i) is a function p: L° -> [0,00] such that: (i) p(u + v) ^p(u)+p(v) V u,veL°; (ii) p(ocu) = \a\p(u) V usL°, a e R ; 179

65] (iii) (iv) (v) (vi)

MEASURE SPACES if \u\ ^ \v\ inL°, then p(u) ^ p(v); if 0 <= A f v in L°, then /}(v) [see 65E below]. The conditions (v) and (vi) are designed to eliminate uninstructive complications. For examples of extended Fatou norms, see 6XE-6XG. 65D Theorem Let (X, 2,/^) be a semi-finite measure space and p an extended Fatou norm on L°(Lt/i). Then p has an associate 0, given by ^ = mv^uxw\\i:ueu>,p{u) ^ 1}. 6 is also an extended Fatou norm, and the associate of 6 is now p, i.e. p(u) = sup{||t& x t^||x: ^(t£;) ^ 1 } V ueL°. Note Here I am employing the convention of 6XE, that if v e L°\LX then lv\\x = 00. Proof I t is easy to verify, directly from the fact that || ||x is an extended Fatou norm, that 6 satisfies the conditions (i)-(iv) of 65C. If w > 0, then, because Z> is order-dense, there is a u e Z> such that 0 0. Thus 6 satisfies condition (vi) of 65C. Before showing that 6 satisfies 65C(v), I shall show that p is the associate of 6. Of course, if p(u0) ^ 1, then 1. Consider

U ^{u'.ueU^piu)

< 1}.

Then by conditions (i)—(iii) of 65C, U is solid and convex. Moreover, if 0 c i c ( 7 and A f v in L1, then by 65C (iv) It follows from 23L (or otherwise) that U is closed for the norm topology on L1. Also, since L 1 is order-dense in L° [63F],

180

BANACH FUNCTION SPACES

[65

and there is a ux eB such that p(u^ > 1, i.e. ux £ U, Now by the HahnBanach theorem there is a continuous linear functional/: L 1 -> R such that/(%) > 1 but \fu\ ^ 1 for every ue U. Recall that (L1)' = (L1)* [260]. Consider | / | in (L1)*. As ?7 is solid,

:H^H}^l

V ueU,

while of course | / | (%) ^ |/(%)| > 1. Now the natural duality between L 1 and L00 represents L 1 as (L°°)x [63G], so it must map L00 onto an order-dense Riesz subspace of (L1)x = (L1)' [32B]. Thus if we know that {ib: we (7} f | / | in (L1)x, so by 16Db there is a we C such that >s/ \ < c We now find that 6(w) < 1. P Suppose that t^GL0 and that yo(^) ^ 1. Once again, because L 1 is order-dense in L°, {vx w:veL1, 0 ^ v ^ \u\}^ \u\xw = \uxw\, so J v x ^ : V G L 1 , 0 ^ v ^ |^|}

= sup{u)(v): vGL1, 0

As t6 is arbitrary, this shows that 6(w) < 1. Q Consequently,

11% x Hli ^ / % x ^ > 1. Thus we see that, for w0 e L°, p(u0) < 1 o ^(ii0) < 1.

Since both /> and <j) satisfy the positive-homogeneity condition 65C(ii), p = 0 in L°, §5(t;) = yo(v) > 0; so there is a weL° such that #(w) < 1 and ||v x w\\t > 0. Then 0 < v A |^| < v and #(# A |^|) < oo, so 6 satisfies the condition (v) of 65C, and is an extended Fatou norm. 65E Proposition Let (X, 2,/^) be a Maharam measure space and p an extended Fatou norm on L°(L,ju,). Then p, restricted to LP, is a Fatou norm, and the topology it induced on LP is Levi and complete. 181

65]

MEASURE SPACES

Proof The conditions (i)-(iv) and (vi) of 65C make it clear that p is a Fatou norm on LP. If A is a non-empty set in (LP)+ which is bounded and directed upwards, then it must be bounded above in L°. P ? For otherwise, by 64C, there is a v > 0 in L° such that

{uAkv:ueA}fkv

V &eN.

Now it follows that, for any & e N, Jcp(v) = p(kv) < $wpueAp{u

A ho) ^ mpUEAp(u)

< oo,

so p(v) — 0; which contradicts 65C(vi). X Q As L° is Dedekind complete [64B] it follows that w0 = sup A exists in L°. Now A f w0, so p(w0) < &wpueAp(u) < oo, and WOELP. Thus A is bounded above in LP. AS A is arbitrary, the norm topology on LP is Levi. It follows by Nakano's theorem [23K], or otherwise [see 65la], that LP is complete. 65F Proposition Let (X, 2,/*) be a Maharam measure space and p an extended Fatou norm on L°(S, /i). Let 6 be the associate of p [65D]. Then Ld = {w:weLQ, wxueL1 VUELP}, so Le may be identified with (LP)X. Proof (a) HUELP and we Ld, let a, = p(u).Ifot = 0, then w = 0 and u x w = 0 G L1. If a > 0, then oo > #(w) ^ ||a"~

HJ

l l l l j

Thus uxweL1 and ||ttx w^ < p(u)6(w). (b) Conversely, suppose that weL° and that wxueL1 for every UELP. Then |w| x |w| = |wx w| el^for every UELP. Set JS = {v:veL^, 0 ^ v < |^|}. For every VEB, VE(LP)', where v(u) = juxv V UELP, ||$|| = sup{|t(u)\ :p(u) < 1} = sup{||^x vWi'.pfa) ^ 1}

and

using the fact that {u: p(u) < 1} is solid. Morever, for any u e LP, *VPveB\Hu)\ < SUVvesj \u\ X V ^ j \u\ X \w\ < 00.

So by the uniform boundedness theorem (since LP is a Banach space, 65E)

182

'

oo > sup vei? ||$|| = mVveBd(v)

= 0(\w\) = 0(w),

BANACH FUNCTION SPACES

[65

since Bf \w\ because U is order-dense in L°. Thus weLd. As w is arbitrary, this shows that

Le = {w.wxueL1

V us LP},

as required. (c) Now by 65A above the natural duality between LP and L$, writing (u, v) = J u x v, represents Le as (LP)X. 65G Corollary If (X, 2, ju,) is a Maharam measure space and p an extended Fatou norm on L°(2, /^), then LP is perfect in the sense of 33A. Proof For if d is the associate of/?, then p is the associate of 6 [65D], so (Z>)xx = (L*)x = £P. 65H Proposition Let (X, S,/^) be a Maharam measure space and p an extended Fatou norm on L°(L,JLL). Then (L^)' = (LP)~, and the following are equivalent: (i) the norm topology on LP is Lebesgue; (ii) whenever 0 c: A\ 0 in Z>, infweJ,yo(it) = 0; (iii) (Lpy = (L/>)X; (iv) whenever (un}nex is a disjoint sequence in (L°)+ and then neH -> 0. Proof (a) (L^)' = (Z>)~ by 25G, because by 65E I> is a Banach lattice. (b) Of course (i) and (ii) are equivalent by definition, and (i) and (iii) are equivalent by 25M. Because the norm topology on LP is Levi [65E], we see that, for a disjoint sequence (un)neJS in (Z>)+, sup neH />(Si< n ^) < oo iff

{un:neN} is bounded above in LP. SO (iv) is equivalent to 'whenever (un}n€}X is a disjoint sequence in (LP)+ which is bounded above, then which by 24J is equivalent to (i). 651 Exercises (a) Let (X, £, /i) be any semi-finite measure space, and p an extended Fatou norm on L°(£, /i). Show that (i) p is a Fatou 183

65]

MEASURE SPACES

norm on 2>; (ii) if (un}neV is a bounded increasing sequence in Z>, then supneHwn exists in L° and belongs to Z> [use 62 J]; (iii) Z> is complete [use 25Na]; (iv) if 0 is the associate of p, Ld = {W.WXUEL1

V UELP},

just as in 65F; (v) the whole of 65H is still true [use (ii) and (iii) above instead of 65E]. *(b) Let (X, 2, /£) be a semi-finite measure space, and/? an extended Fatou norm on L°(L,/i). Give L° its usual topology [63K]. Then the embedding L? £ L° is continuous. Notes and comments The theorem 65A/B gives a general method of describing order-continuous linear functionals on most of the usual function and sequence spaces [see 6XD]; it includes, for instance, the identifications (I00)* = l\ (P) x = I00, and (co)x = I1 in 2XA-2XC, as well as (L°°)x = L 1 [63G] and (L1)* = L00 [64B]. Moreover, it can be shown that if E is any Riesz space such that Ex separates the points of E, then E can be represented as an order-dense Riesz subspace of L°(£,/£)forsomeMaharammeasurespace(X,2,/£) [FREMLIN A.K.S. II]; so that 65B gives a picture of the duality between E and Ex which is in some sense complete. The examples 6XE-6XH give some idea of the kind of spaces that can be generated by extended Fatou norms. They include most of the solid normed function spaces that have attracted interest. In fact, using the representation theorem mentioned above and the method of 6XH, it is not hard to show that if E is any Dedekind complete Riesz space with a Fatou norm inducing a Levi topology, and if Ex separates the points of JE, then E is an Z> space. When the underlying measure space is Maharam, then 65E-65H give a description of these function spaces which in its own terms is fairly complete. They are all perfect and their topologies are Fatou and Levi. The question of when they are Lebesgue is tackled in 65H. For the sake of simplicity, the underlying measure space is assumed Maharam all through this work. But in 651a I list those results which can be extended to arbitrary semi-finite measure spaces. They give a good idea of the way in which, for metric spaces, it is often enough to be able to handle sequences. In 6X11 give a example of a class of spaces which is very similar to the function spaces of this section except that the defining functions are not norms and the topologies are consequently not locally convex. 184

EXAMPLES FOR CHAPTER 6

[6X

6X Examples for Chapter 6 I begin with simple examples of measure spaces; the most important one is of course Lebesgue measure [6XAb, 6XB]. Because Lebesgue measure is diffuse [6XBa], the associated L° space is not locally convex [6XBb]. The next paragraph [6XC] is an examination of a natural inverse-measure-preserving function, showing how the ideas of § 54 are linked to Fubini's theorem. The rest of the section is a discussion of some simple function spaces, based on the ideas of § 65, showing how the results of Chapters 2 and 3 can be applied. 6XA Measure spaces Define /i: 2 -> [0, oo] by fiE — n

(a) Let X be any set, and set 2 = SPX.

if E has n members, where neN;

= oo otherwise. Then (X, 2,/^) is a complete decomposable measure space. As pE = 0 iff E=

0,

its measure algebra 21 is isomorphic to 2, and L°(2,/^) ~ M(2) = R x . L°°(«) - 2>°(2) = t»(X) [4XCa], a n d L 1 ^ ) s £ V ) = P(X) [cf. 5XB]. The usual topology on L° [63K] is precisely the product topology on R* [cf. 1XD]. (b) Lebesgue measure For the construction of Lebesgue measure on R, I refer you to WIDOM, chapter 1, or BARTLE, p. 96, chapter 9; an alternative methodis suggestedin 7XA. Its basic property is that /i[<x,fi] =fi—0Lwhenever a < fi in R. It is complete and of countable magnitude. (c) For further examples, see MTJNROE [chapter in] and Chapter 7 below; also HALMOS M.T., chapter xi, §§57-60, for a description of Haar measure. 6XB Lebesgue measure on [0,1] Now let us take X to be the unit interval [0,1], and 2 the class of Lebesgue measurable subsets of X; let [i be the restriction of Lebesgue measure to 2. Then (X, 2, /i) is a complete measure space and fiX — 1. (a) Suppose that 2?e2 and that [iE > 0. Let n ^ 1 be such that a = rr1 < fiE. Since [0,1] = \Ji
0 < fi(E n [oci, a(t + 1)])

^oc
6X]

MEASURE SPACES

Thus (X, 2,/^) is diffuse, i.e. for every non-negligible EGS, there is an / e S such that F c E and neither F nor E\F is negligible. Another way of putting this is to say that the measure algebra of (X, 2,/0 has no minimal non-zero elements or 'atoms'. (b) Consider now L°(Z,/i). If/: L°->R is increasing and linear, / = 0. P ? Otherwise, let g: M(S) -> R be given by gx = f(x') for every xeM(2,). Let x0 > 0 in M(S) be such that gx0 > 0. For each n e N , observe that i, 2-^(i+ 1)]), n

so there must be an i < 2 such that where Now let

yn = xo x X(i2~~% 2"-(i+1)]). w,

2~n. If we set

so that g(zn) = 2»and/*{r- zn(y) * E == {y: sup ne

Y) = oo},

we see that

.t(y) 4= 0},

Egz

so that Let Then z' ^ zn in L°, so

fiE z

2-< = 0.

= sup^z

for every n e N, which is impossible. X Q *(c) Consequently, if % is the usual topology on L° [63K], then the dual of L° for X is {0} [22D]. In particular, % is not locally convex, though it is complete and metrizable [64E, 64Jc]. For further striking properties of X, see PRYCE U.S.

*6XC An inverse-measure-preserving function Let / be the unit interval [0,1], and let X = I2 c R2; let ju, be the restriction of Lebesgue planar measure to subsets of X, and 2 the domain of ft. Let (Y,T,v) be Lebesgue linear measure on / , as in 6XB above. Define / : X -> Y by writing f(oc,fi) = a for all ot,/3el. Then / is inversemeasure-preserving. Let 21 be the measure algebra of (X, 2,/^) and 95 that of (Y, T, v). Then the measure-preserving ring homomorphism n: S3 -> % of 6IE is given by n = {f_1[E]).

186

=

(Exiy

EXAMPLES FOR CHAPTER 6

[6X

So n: L°°(95) -» L°°(3l) is given by n(x') = (a® 1)' V aJGL°°(T), where 1 = # / and # (g) 2/ is given by (x ® y) (a, /?) = s(a) */(/?) V a, /? G J. Similarly, rr: L^SB) -> L^Sl) [54D] is given by 7r(aO = (s® 1)" V ze&(v). On the other hand, Pn: L\%) -> L^SB) is given by (Pnu, v) = O , TTV) V ^ G ^(31),

t; G L°°(93)

[54F], i.e. [using the representation of the duality between L1(3l) and L°°(9l) given in 63P]. Now Fubini's theorem [MUNROE, §28, or WILLIAMSON, p. 63, § 4.2] tells us that, given xe z(ot) = (x(a,ft)v{dfi) exists and that

^-p.p. (a),

J zdv = J #eZ/£.

So, applying this to x x (y ® 1), (P7Tx',y') = jxx(y®

l)d/i = jy(a)z(a)v(da) =

As this is true for every 2/GL G0 (T), Pnx' = z\ Thus Pn ' averages' members of L1(3l) over vertical lines, in a fashion perfectly analogous to 5XC. 6XD Sequence spaces Suppose, in 6XAa, that X = N. Then M(S), which is isomorphic to L°(S, /i), becomes RN, and we may call its subspaces 'sequence spaces'. A Riesz subspace E of RN is easily seen to be order-dense iff it includes s0, the space of sequences with only finitely many non-zero terms. In this case we may use 65B to identify Ex with {y.xxyel^N)

V xeE},

identifying I1 with S 1 - L1. Observe that (RN)X = s0 and (s0)* = RN. This result also includes the identifications of (/1)x, (/°°)x and (co)x given in 2XA-2XC. 187

6X]

MEASURE SPACES

6XE Extended Fatou norms For any semi-finite measure space (X, 2,/^), we may extend || ||x and || H^ to the whole of L°(2,/£) by writing ||^||1==oo for UEL°\L\ IKHoo = oo for ueL°\L<°. Because the norm topologies on L1 and L00 are Fatou and Levi, it is now clear (reversing part of the argument of 65E) that || ||x and || H*, are extended Fatou norms in the sense of 65C, and also that they are associates in the sense of 65D [using 52G and 63G]. 6XF Lp spaces (1 1 in R. Set q = (p—l^p, so that p ^ + q-1 = 1. Let (X,2,/£) be any semi-finite measure space. If x, y e M(2), x =y ox = y p.p. => \x\* = \y\P p.p. o (\x\*Y = (\y\*>)\ Thus we may define \x'\& to be equal to (|#| p )'. Now we write

M , = (II Mix)1'* for each u e L0(S, /i), where || || x is permitted to take the value oo, as in 6XE. Then || \\p is an extended Fatou norm and its associate is || || r Proof (i) Holder's inequality If a, J3 e R, then p-iaP + q-1/?*^ a/3 [HARDY, LITTLEWOOD & P6LYA,

§2.5]. It follows that if u,v ^ OinL0, |^|a ^ uxv.

Consequently ||t^ x v||x ^ 1 whenever \\u\\p < 1 and ||t>||ff ^ 1 (since p~1 + q~1= 1). So |[^x v||x < H^llpll^llg whenever H^H^ and \\v\\q are finite, for

\\au\\p = |a| \\u\\p V a e R ,

ueL°.

(ii) It follows that

IKIlp = TOp{lk for every ueL°.

xv

lli : llvll« < 0

P We have just seen that \\u\\p ^ suplWuxvliilvWe < 1}

for every u such that \\u\\p < oo, and therefore for every ueL°. If H^IIP = 0, then |^|^ = Osoi^ = 0, and the equality is certainly satisfied. 188

EXAMPLES FOR CHAPTER 6 [6X If 0 < \\u\\p < oo, set v = \u\*!*. Then

II^IU = (II l^i^I[x)a/^ = (II |^l^IU) a/ « = (1KIU)^« = ^ say. So \\fiv\\Q = 1, and using the fact that 1 -f (p/q) = p, so that p — (p/q) = 1. So in this case also the equality is satisfied. Finally, if H ^ = oo, we use the fact that (X, 2,/*) is semi-finite to see that oo = \\u\\p = a u p { \ \ w \ \ p : w e S ( W ) , O^w^ \u\} , 0 ^ w < \ u \ , \\v\\q < 1} (iii) It follows at once that || ||p satisfies the conditions (i)-(iv) of 65C, and it satisfies (v) and (vi) because || \\t does. Similarly || ||g is an extended Fatou norm, and is of course the associate of || ||p. (b) We normally write Lp for the Banach space

{u:\\u\\p)' = (LP)- = (Z>) x may be identified with L«, a t least when (X, S, fi) is Maharam. Thus, in this case, IP is reflexive. * Actually, Lp always has the countable sup property (because L1 does), so is necessarily Dedekind complete, even when (X, £ , / 0 is not Maharam. We can see that the norm topology of Z> is Levi because the norm topology of L 1 is Levi; thus IP is always complete [see 25Na]. The argument of 65A shows that Lq can be thought of as an order-dense Riesz subspace of (Lp)x, which b y 17F is solid. Now, because LQ is a Banach space and the norm on 2> is that induced b y the duality [part (ii) of the proof of (a) above], a %8{Lp, L 5 )-bounded set is || Unbounded; so Z> is Levi for S£S(L^,L«). I t follows from 33B that L* can be identified with (L«) x . Similarly, L« = (L^) x = (2>)', so again Z> is reflexive. 6XG L 2 spaces If in 6XF we set p = q = 2, we get the outstanding special case f Til ro T(\ 5 ^ L2 = {u:ueL°, uxueL1}. The identification between L2 and (L2)' corresponds of course to an inner product, writing j x v V u, veL2. 189

6X]

MEASURE SPACES

We know that L2 is complete; therefore it is a real Hilbert space. In view of its great importance, it is worth noting that many of the arguments of 6XF can be simplified in this special case. *6XH Orlicz spaces Let us consider the properties that the unit ball U of a Banach function space Z> defined by an extended Fatou norm p must have. It is a subset of an L° space which is solid and convex [65C(i)-(iii)], and JTJ £ U [65C(iv)]. If u > 0 in L°, there is a v G U such that 0 < v ^ u [65C(v)], and if cm e U for every a e R, then u = 0 [65C(vi)]. Clearly these properties are necessary and sufficient for U to be {u: p(u) < 1} for some extended Fatou norm p. We may regard 6XF as defining || \\p by Now the function a\-+ap: R+ -> R + is a continuous monotonic function which, for p ^ 1, is convex, i.e. (yot + (l-y)/3)*> ^

yav

+ (1 - y) ft*

whenever a, JS ^ 0 and 0 < y ^ 1. This is clearly the essential property to ensure that {u: J \u\p < 1} is convex. So now suppose that [0, oo] is any convex monotonic nondecreasing function satisfying the following: (i) O is continuous on the left, and O(0) = 0; (ii) there is an a > 0 such that O(a) < oo; (iii) (\x(t)\)dt < 1}. (Because O is permitted to take the value oo, we must be careful to choose x such that O(|a;(£)|) < oo for all t; if this is impossible, then u$U. Because O is monotonic, the function t h-» O( \x(t) |) will always be measurable, using the criterion of 62D.) It is immediate from the definition, because O is convex and monotonic, that U is solid and convex. It is a little harder to see that J U c JJ. But, for any u ^ 0 in U, we may define Qu e L1 by where x' = u. Now if 0 <= A t u0 in (L°)+ and A ^ U, in the unit ball of L1, and has a supremum v0 say. Using the leftcontinuity of O, we can show that ®w0 ^ v0, so that uoe U. Conditions 190

EXAMPLES FOR CHAPTER 6

[6X

(ii) and (iii) on O now ensure the other properties of U (since, as always, the measure space is supposed semi-finite). For a description of the way in which O and its ' complementary Young's function' can be used to generate an associated pair of Banach function spaces, I refer you now to ZAANEN, chapter 5, §§ 4-6. Apart from the obvious cases of Z>, 1 1,

we find that U is just the unit ball of L00. *6XI IP spaces (0 VEL°; now it is easy to see that || \\p: L° -> [0, oo] satisfies all the conditions of 65C except (ii). Instead, we see that \\ocu\\p = |a|* ||^||p V ueL°, a e R . However, this is enough to ensure that || \\p is a Fatou pseudo-norm

onL» = {u:\\u\\p
Just as in 6XF, Lp now has the countable sup property and is Dedekind complete, and its pseudo-norm topology is Levi; so IP is an example of a complete metrizable topological Biesz space which is not (except in trivial cases) locally convex. Further reading for Chapter 6 It has unfortunately been impossible to include more than a few of the concepts of abstract measure theory. Probably the best introduction to general modern measure theory is still HALMOS M.T. One of the most serious omissions above is the lack of any reference to product measures; HEWITT & STROMBERG [chapter vi] are very sound, though they restrict their attention to spaces of countable magnitude. An alternative look at the foundations, with some of the ideas of § 64, can be found in IONESCXJ TTJLCEA [chapter i]. Here also is a proof of the Lifting Theorem, one of the most remarkable results of pure measure theory. There are a great many important results in DUNFORD & SCHWARTZ; it is instructive to seek to refine their theorems and proofs in the light of the work above. Since many of the most fruitful developments of measure theory 191

6X]

MEASURE SPACES

have been stimulated by problems arising in the theory of probability, it is worth making an effort to cross the language barrier between the two subjects. A concise and energetic survey of measure theory from a probabilist's viewpoint is in MEYER. Here again it may be profitable to investigate whether the ideas of Chapters 4 and 5 above could be used to clarify some of the concepts.

192

7.

Representation of linear functionals

Suppose that we are given a set X, a Riesz subspace E of R x , and an increasing linear functional / : E -» R. My object in this chapter is to discuss conditions under which/is an 'integral', that is, when there is a measure / i o n l such that J xd/i exists and is equal to fx for every x e E. A necessary, and nearly sufficient, condition is that / should be 'sequentially smooth' [71B-71G]. Further conditions on/and E lead, of course, to stronger results [§§ 72, 73]. The outlines of the theory are not hard to appreciate. Unfortunately, the technical refinements needed for the strongest results are complex, and in their most general forms they are difficult to grasp intuitively. Each extra scrap of information costs us a good deal of hard work. I shall try to summarize the theory in a way which will show its essential structure and maintain reasonable simplicity in the theorems, though the proofs will inevitably be lengthy. The general approach of the first two sections of this chapter is close to that of TOPS0E T.M.

71 Sequentially smooth functionals The first representation theorem I give [71G] is the most natural, in the sense that the hypotheses and consequences are most directly related here. In order not to have to repeat them in the next section, I remove certain parts of the argument into separate lemmas. The first of these is a substantial result, giving a powerful method of constructing measure spaces. 71A Theorem Let X be a set, and Jf* c 0>X a sublattice containing 0 and closed under countable intersections. Let A: JT -> R+ be an increasing function such that: (i) A0 = 0 ;

(ii) ifH, KeJfa,ndH(]K= 0, A(H[)K) ^ XH + XK; (iii)

IfF^eJfandH^F, , K <= F\H};

FTR

7

193

71]

REPRESENTATION OF LINEAR FUNCTIONALS

(iv) if (Kn)n6ix is a decreasing sequence in Jf and C[ne^Kn = 0 , then infweNAJ5Tw = 0. Then A has an extension to a complete locally determined measure /A defined on a c-algebra £ =2 C%* such that (a) /*J0 for every i? e S ; (A) if JE? C X and E n # e 2 for every iTe Jf, then Proof

Define A*: 0>X -> [0, oo] by A^^L

for every i c j , Set and let ju, be the restriction of A* to S. The whole of the rest of the proof consists of a demonstration t h a t fi is the required measure. (a) Let us begin with two simple observations. If A, B c: X and A n J5 = 0 , then A^-4 U B) > \*A + A^S. P AHc^ + AH8B ^ sup{A(#[) < A*(il U B), using condition (ii). Q (b)

; K ^ A) sup {\+(K (because A* is obviously increasing) by (a) above. Q (c) Suppose that E,Fe"L. Then X\E and JE? u -F e S. IfEnF = 0 , then A*{E u F) = A*E + \*F. P It is immediate from the defimtion of 2 that if EeH then . NOW let KG JT. Then

194

SEQUENTIALLY SMOOTH FUNCTIONALS

[71

(because .Fe 2) = A*(K nF) + A*((K\F) (]E) + A*((K\F)\E) (because 2Je2, using (b) above with A = K\F) = A*(K n (E u F)) + A*(K\(E u F)) (because FeZ, using (b) above with A = K n (E U F)). As K is arbitrary, 2?u.Fe2. Finally, if E(\F = 0, then, again using (b), . Q (d) So we see that S is a subalgebra of ^ X If KeJf, ^iT = AiT, because A is increasing. So if H,

then

n F) + sup{AiT: As H is arbitrary, J ' G S . Thus X c S and /^ = A^ = A on «yf. (e) Now suppose that {En}neV is an increasing sequence in S and that i£ = UneH^n- T h e n E el> and fiE = ]imn_^ao/iEn. P I shall rely heavily on the fact that (by (c) above) 12 = {F:Fe'L, A*F < 00} is a subring of £PX and A^: U -> R is additive. Let HeX* and e > 0. For each neN, choose an Hnetf such that Hn c j f f ^ and AJ3^ ^ \*{H\En) - 2~ne. Now we know that, for each neN,Hn\jHn+1c:H\En;8o X{Hn U Hn+1) ^ and

A*(Hn+1\Hn)

= A ( ^ U Hn+1) - XHn ^ 2 - e .

So, if we set Kn = C[i^nHi for each n e N , 2e, and 1

=

\Hn+1-A*{Hn+1\Kn)

Let Z = RneH-^n ^ ^ W - W

ek n 0 W t h a t

A*(H\E) > l i n v ^ , A^Zf n En) + AK. To compute AK, consider \+(H\K). If i^G JT and F c Z ? ^ , then <^ n ^ > T I 6 N I ^ a n d neN -> 0 ; t n i s i s where we use the 7-2

195

71] REPRESENTATION OF LINEAR FUNGTIONALS condition (iv) of the theorem, and this is why the decreasing sequence had to be found. Now we see that AF = limn^A(F n As F is arbitrary, A*(H\K) ^ limn^A*(H\Kn), and XK > \imn^AKn > limn^ So

As e is arbitrary, \E

AH;

as H is arbitrary, Eel,. Moreover, if, above, H £ E, then K = 0; so by equation (*), < 3e- A s e i s arbitrary, lim,,^^ A*(H\En) = 0, so

Now, because H is arbitrary, as required. Q (f) Thus (X, S, fi) is a measure space, and clearly it has the property (a).If2?£2and/£i? > 0, there is a KeX such that K c jJandAif > 0; now fiK = AK < oo. Thus (X, 2,/0 is semi-finite. If 2? c X and # n iTeS for every KeJf, then A^^

n E) + A*

for every Ke2^,80 Eel,; thus S has the property (/?), and as X c £/, it follows at once that (X, 2,/j) is locally determined. HE ^ Fell and / ^ = 0, then for any KeJf, A*{K [)F) = AK,

so Eel. Thus (X, l,/i) is complete. 71B Definition Let X be a set and E a Riesz subspace of Rx. An increasing linear functional / : E -> R is sequentially smooth if (fxn}nen -> 0 whenever (xn)neji is a sequence in E such that 196

SEQUENTIALLY SMOOTH FUNGTIONALS

[71

Remarks In §72 I shall introduce 'smooth' functional. Note that the property of being sequentially smooth is not intrinsic to the functional/and the space E; it depends on the embedding of E in R x . A sequentially order-continuous increasing linear functional on E must be sequentially smooth, but the converse is false. For instance, it is not hard to show that there is no non-trivial sequentially order-continuous increasing linear functional on C([0,1]), but every increasing linear functional on C([0,1]) is smooth. Consequently we must, at several points below, distinguish between bounds in E and bounds in Rx. I shall use the phrase ' inf A = z in R x ' to mean that z(t) = infxeAx(t) for every teX; while 'supA = z in E' would mean that z was the least member of E which was an upper bound for A. 71C Definition Let X be a set and E a Riesz subspace of R x . I shall say that a functional / : E -> R is an integral if there is a measure /i on X such that J xd/i exists and is equal to fx for every xe E. In this case, I shall call/ 'the integral with respect to /i9. An integral is always linear, increasing and sequentially smooth [Lebesgue's theorem, 63Md]. To obtain a converse result, we need an extra condition. 71D Definition Let X be a set and E a Riesz subspace of R x . I shall call E truncated if x A ;yX e E for every x e E. 71E Lemma Let X be a set and E a truncated Riesz subspace of Rx. Let x e E. Then (a) x A a#X e E for every a > 0 (b) if H = {t: x(t) > 1}, then there is a sequence (%n)nejs in E such that (xn}nelj( | xH in R x . Proof

(a) For x A a#X = a(arxx A #X). (b) Set n

xn =

2n(xAXX-xA*nxX),

where otn = 1 - 2~ . 71F Lemma Let (X, £,/£) be a measure space and E a truncated Riesz subspace of Rx such that, for every xe E, Hx = {t:x{t) 197

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Suppose that / is a sequentially smooth increasing linear functional on E such that, for each x e E, y>

XHX}.

Then / is the integral with respect to ji. Proof

(a) Let x e E+, and for each n, i e N set

Set

ynr = 2i
then ynr(t) = 2~^ifthereisai ^ r such that 2~nlc < x(t) < 2~n(k + l), and 2~n(r+1) otherwise. Since XFni < 2nix A 2~n^ +1)XX~X ^m < 2»Lf(a; A 2 - ( i + 1) XX) -f(x A jynrd/i = Xi
So

x

Put i/w = supreNi/nr in R ; then by B. Levi's theorem [63Ma], j ynd/i exists and is not greater than/x. But yn(t) = 2-H whenever

2~H < x(t) < 2~n(i +1).

So (yn)neN t ^ in R^j and by B. Levi's theorem again, J x d[i < /#. (b) Again, let a; be any member of E+, and let e > 0. Then »B = a ; A 2 ^ I - a ; A 2 ^ I e E

V neN,

x

and (xn}nEl[ f x in R , i.e. <x — xn)nelSi j 0 in R x . So there is an n e N such thsbtfx < fxn + e, because/is sequentially smooth. Now set H = {t:x(t) ^ 2 - ^ } G S , n

and set y = 2 XH, so that xn ^ y. Let z e E be such that z ^ ^J? and

fz^fiH + 2~ne. Then /a: ^ c + X = e+f(2"z)-f(2*z-xn) ^ e + 2nfz-f{2nz-xn)d/i (applying (a) above to 2nz — xn) 2nju,H - f yd/i + j xndju, = 2e+jxnd/i ^ 2e+ (xd/i. 198

SEQUENTIALLY SMOOTH FUNGTIONALS

[71

As e is arbitrary, fx^j xd/i. Thus fx = J xdju, for every x e £ + . Clearly nowfx = j xdfi for every x e E, i.e. / is the integral with respect to fi. 71G Theorem Let X be a set and E a truncated Riesz subspace of Kx. Let/: £ -> R be a sequentially smooth increasing linear functional. Then there is a complete locally determined measure /ionX such that / is the integral with respect to [i. Proof (a) Let $f be the family of all those sets K £ X such that there exists a sequence (xn)nelx in E with xK = mfnelssxn in R x . Define A: J T - > R + b y AZ = i n f { f x : x e E , x ^

x K)

v

KeJlT.

I propose to prove that 3T and A satisfy the conditions of 71 A. (b) Obviously 0 eX*, setting xn = 0 for each WGN. If £T, I G JT, let nGN and (yn)nejsi be sequences in £ such that and #J5L = infneN2/w in R x ; then so H \J Ke Ctif. If (Kn}neJX is any sequence in JT, then choose for each neN a sequence <^X e u i n ^ such that #iTw = mfiEixxni; then so DneN^w G ^ - Thus CJT is a sublattice of ^ X , closed under countable intersections. (c) Of course A is increasing and A0 = 0. If H, Ketf* and H n K = 0, let <#n>n6N a n ( i (2/?I)WGN b e sequences in £ such that X# = inf^gu^ and #X = infweN2/n in Rx. Now let a: be any member of £ such that x ^ # ( # U K). For each ^ e N, set

Then neN 10 in R^, so (fzn)ne1x -> 0. But x ^ (x A inf i
/ x ^ /(x A i n f ^ ^ ) +/(a; A inf^^i/J

-f(zn)

taking the limit as n -> oo, fx ^ AZf + AiT. As x is arbitrary, as required by 7lA(ii). 199

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(d) Now suppose that F,He$T and that F c H. Fix on a sequence (xn)nen ^n E such that %F = inine1ssxn in R x . Let e > 0. Take an #e £ such that # ^ # # and/a; ^ Ai7 + e. Set

Then

JBT0 =

lt:teF, (xn-x)(t)

>j

^ V

= J1 n n»erf: (e-1+1) (*»-*) (0 > i}But { ^ ( e ^ + l ) ^ - « ) ( * ) > 1}GJT for every nsN, by 7lEb. So, because Cft* is closed under countable intersections [(b) above], . NOW suppose that z is any member of £ such that z

x ^ xF >

so

As 2J is arbitrary, XF ^ ( < (1 + e) (e + *H) + Bap{AK:KeJr,

K c

As e is arbitrary, AJ7 < AH + swp{AK:KeX',

K c

and 7lA(iii) is satisfied. (e) Finally, suppose that (Kn}n€n ^ 0 in Jf*. For each WGN, choose a sequence (xni)ieIS in £ such that ^i?^ = inf^eNx^ in R x . Set Then (yn)ne1x | and inf neN ^ = infneN^J5Tn = 0 in R^. So (fyn)nex -> 0. But yw > XKn, so neN -> 0. (f) Thus all the conditions of 71A are satisfied, and there is a complete locally determined measure /i on X extending A. But now we find [using 7lEb again] that ju, satisfies the conditions of 7IF, as all the sets Hx actually belong to X\ So/is the integral with respect to fi. 71H Exercises (a) Show that under the conditions of 71 A, there is only one extension of A to a complete locally determined measure with the property (a). (b) Suppose that X is a set and J f a sublattice of SPX containing 0 . Let A: JT -> R+ be an increasing functional satisfying the conditions (i)-(iii) only of 71 A. Then A has an extension to an additive functional on the subring of ^ X generated by CtiT. 200

SEQUENTIALLY SMOOTH FUNGTIONALS

[71

(c) Let (X, T, v) be any measure space. Let $T = 17 and let A be the restriction of v to T*. Show that CrfT and A satisfy the conditions of 71 A, and that the measure ju, produced by 71A is precisely the complete measure on X obtained by applying 64Ja and 64Jb to (X,T, v). (d) Let X be a set, 91 a subring of 0*X, and /i: 91 -> R an increasing additive functional. Show that fi has an extension to a measure on Xiff, whenever {An)neV is a decreasing sequence in 91 such that Oneit^n — & > then (ju>An}nex -> 0. [Hint: either show t h a t

is sequentially smooth, using the method of 42Jc, and apply 71G, or apply 71A directly to a suitable J f ^ 91.] *(e) Let X be a set and E a Riesz subspace of Rx containing %X, Let S o be the smallest c-subalgebra of SPX such that E £ M(S0), i.e. the cr-subalgebra of 0*X generated by the sets {t: x(t) > a} as x runs through E and a through R. Let / : E -> R be a sequentially smooth increasing linear functional. Show that there is a unique measure /£0 with domain 2 0 such that / is the integral with respect to /£0. [Hint: define $T, S and fi as in 71G. Show that JJL and ju,0 agree on Jf, and therefore on So.] Notes and comments I t is a slightly surprising fact that, in 71G, we cannot dispense with the condition that E should be truncated; there is a counterexample in 7XB. Otherwise, it is clear that this theorem goes about as far as it can, since if we have any measure space (X, T, v), we can set E = S1(^), the space of y-integrable functions on X, and that the theorem will then generate the ' complete locally determined extension' of v described in 64Ja-b [cf. 71Hc above]. Readers who have encountered the Caratheodory ' outer measure' construction for measure spaces will recognize that 71A is based on a similar' inner measure' method. Up to a point they are interchangeable (thus 7lHd can be used to tackle product measures, while an outer measure method will give 71G), but outer measures are probably more important for general measure theory. However, I shall not have occasion to use them in this book. The advantage of 71A is that it directly generates a locally determined measure space, which outer measures often fail to do. 71G can also be thought of as a way of extending a sequentially smooth functional to a whole Sx-space; the point about an S1-space being that it is closed under countable suprema and infima, at least of sets which are order-bounded in itself. Indeed there exist proofs 201

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which set out to make this extension directly [see LOOMIS, § 12, pp. 29 et seq.]; this approach has the advantage, when E is not truncated, that equivalents of S 1 and L1 can be constructed even though they may not correspond to any measure. I t is a remarkable fact that the range space R in this result can be replaced by a Dedekind cr-complete Riesz space F iff F is weakly cr-distributive [WEIGHT E.T.]. 72 Smooth functionals: quasi-Radon measure spaces A natural adaption of 7IB leads us to the concept of 'smooth' functional, in which the decreasing sequence of 7 IB is replaced by a directed set. In order to give a description of smooth functionals which will correspond to 71G, we need to consider a special type of measure space. Because its important applications are to functionals defined on spaces of continuous functions, I express the theory in terms of measures on topological spaces, the 'quasi-Radon' measures. 72A Definition A quasi-Radon measure space is a quadruple (X, %, 2, fi), where (X, 2, fi) is a measure space and % is a topology on X such that: (i) (X, 2, fi) is complete and locally determined; (ii) % £ 2 (i.e. every open set is measurable); (iii) if E e 2 and /iE > 0, there is a G e % such that JLLG < oo and fi(E ()O)>0; (iv) fiE = sup{jiF:F ^E,F closed} for every Ee2; (v) if 0 c ^ f in S, then

Remark This definition probably strikes you as undesirably long and technical. I hope that by the end of this section it will be apparent that there are good reasons for each feature. The following theorem is a reason for many of them. 72B Theorem composable.

A quasi-Radon measure space (X,£,2,/£) is de-

Proof I shall apply the criterion of 641. Let us call a set F e 2 supporting if fi(GftF) > 0 whenever G is an open set meeting F. Let iTbe the collection of all disjoint families of 202

SMOOTH FUNCTIONALS

[72

supporting sets of finite measure. By Zorn's lemma, 9£ has a maximal member s/. 1 Now suppose, if possible, that in the measure algebra 51 of (X, 2 , / J ) . Then there must be a non-zero ae% such that a n F' = 0 for every F es/; i.e. there is an i ? e 2 such that fiE > 0 but ju,(E nF) = 0 for every F e stf. (i) Let Goe% be such that /*(?0 < oo and /*((?0 n E) > 0 [72A(iii)]. Every Fes/ is supporting, so

^ = {i^e^^nG 0 * 0} = {F:Fes/,ji(FnG0)>0} which is countable, because /^(?0 < oo and s/ is disjoint, so that

{F:Fes/9/i{F(\G0)>2-^} cannot have more than 2n/iG0 members. So also, because /i(E (]F) = 0 for every Fes/0, fi{E n U^o) = °? a n ( i x = /i(O0 f]E) > 0. Now E1()F = 0 for every JP GS/. (ii) Let ^ = {G:(?G2, ffc^^nffjsO}. Then <9\\ let = U^. We know that = sup{[iG:Ge@} ^ /iG0 < oo [73A(v)]; so for any e > 0 there is a 6r e & such that /^G ^ /^J? — e. Now n ^ ) ^ /6(G n #i) +MH\G) ^ e. As e is arbitrary, /^(£T n E±) = 0. Set Fo = E±\H; then ^ = [iEx > 0, J i n J1 = 0 for every FGS/, FO C
/i(FQ n 0) +/JL{E1\F0) = /i(F0 n 0).

As (? is arbitrary, we see that Fo is supporting. (iv) I t follows that Fo is a supporting set of finite measure, not 0 , such that Fo 0 F = 0 for every Fes/; which contradicts the supposed maximality of si. X 203

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Since (X, 2,/^) is complete and locally determined [72A (i)], it is decomposable [641]. 72C Definition Let X be a set and E a Riesz subspace of Rx. An increasing linear functional/: E -> R is smooth if mfxeAfx = 0 whenever A is a non-empty set in E such that A j 0 in R x [i.e. .4 is directed downwards and mixeAx(t) = 0 for every teX]. 72D Proposition Let (X,%,H,/i) be a quasi-Radon measure space. Let £ be a Riesz subspace of C(X) such that fx = jxd/i exists for every xe E. Then/: E -> R is smooth. Proof Suppose that 0 c= A c E and that A \ 0 in R x . Fix Let e > 0. Let 8 > 0 be such that

xoeA.

and let 6r0 = {£: #0(£) > 8}. Then 6r0 is open and ju,G0 < oo. If ju,G0 = 0, set y = x0; then fy = jx0 ^ e. Otherwise, let neN be such that and set rj = 2~ne. Then for any -E e £ such that /^i? < r/,

<

/

^ 2e.

For xeA, set

Zf, = {t: te G0,x(t) < ( ^ o ) " 1 elThen {Hx:xeA}>[G0, so by 73A(v) there is an xeA such that ju,Hx ^ /^(?0 — 7j, i.e. /i(G0\Hx) < T/. Now let i/ G J. be such that y ^ x Ax0. Then X(GO\HX)

+ yx

X(X\G0)

+ xQ x X(GO\HX) + XOA 8XX.

So in this case also fy = jy ^ e + 2e + e = 4te. As e is arbitrary, miyeAfy = 0; as J. is arbitrary,/is smooth. 72E Theorem Let X be a set and E a truncated Riesz subspace of R x [definition: 7ID]. Let / : £->R be a smooth increasing linear functional. Let % be the coarsest topology on X such that every member of E is continuous. Then there is a measure ju, on X, quasiRadon for %, such t h a t / is the integral with respect to /i. 204

SMOOTH FUNGTIONALS

[72

Proof Let Jf* be the collection of all !£-closed sets K c X for which there is some xeE with x ^ ;\;JT. Define A: J f -> R + by V Just as in 71G, I propose to show that JT and A satisfy the conditions of 71 A. (a) It will help to begin with a brief examination of %. For each x G E, let Hx be {t: sc(J) > 0}, and set Since Hx(]Hy = HXAy for all x,yeE, {Hx: a; e £} is a base for a topology Xo on 7 . Let 3^ = {0: # c X, On Y e St0}; then %x is a topology on X. Suppose that x e £ + . For any a ^ 0, where Z — X — XK ocxXeE [using 71Ea]. Similarly, {t:teY,x(t) 0} e %0 for each y e E. So {«:fer, s(0 < a}GS:o and {^:x(t) < a}e%v Thus every member of E+, and therefore every member of E, is continuous for 2^. So % c ^ j . This shows that, if 6? G % and ^eGn 7, then there is an x e £ such that teHx = {u:x(u) > 0 } g f t (b) I t follows that for any K e ^T, ^1 = { X : X G £ , a; ^ ^ Z } | ^ Z

in R^.

P Of coursed j , and by the definition of JT, ^4 4= 0.Fixo; o G^[, and let t be any point of X. IfteK, then x0 A ^ X G ^ 4 , SO inf^g^a;^) = 1. If « G ! \ 7 , thena;0(^) = 0, soinf^^x^) = 0. If te Y\K, then by (a) above there is a z e E such that te{u:z(u) > 0 } c I \ I +

Now (xo — nz) eA for every neN, so inf^^x^) < inf^XQ-nz)+(t) = 0. Thus in all cases infxe^#(£) = (x%) (t), as required. Q (c) Now we can proceed along the same lines as in 71G. Clearly, 0 GJf, as 0 is closed and #0 ^ 0 e £. If if, iTG^, and if # # ^ x and y where x and 2/ belong to £, then J? U ^K" is closed and X(H[)K) ^x + yeE; 205

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so H U K G X. If neN is any sequence in Jf, let a; e E be such that XKO ^ #; then J5T = C\nE^Kn is closed and x% ^x,aoKeJf. Thus J f is a sublattice of ^.X, closed under countable intersections and containing 0 . (d) Of course A is increasing, and A 0 = 0. If H and K belong to Ctif and H [)K = 0, then by (b) above

#}!#

in

So {a; A ?/: a; G ^4, i/ G JS} 10 in R x . I t follows, because / is smooth, that for any e > 0 there exist x±e A and yxeB such t h a t / (a^ A 2/x) ^ e. There is also a zx e E such that z± ^ ^(if U Jf) and

fz±-e «! A yx) -f{z±

Ax1Ay1)-e

As e is arbitrary, A(Zf u K) ^ A£f + AiT, and 7lA(ii) is satisfied. (e) Now suppose that F,HGX and that F ^ H. Let e > 0. Take an xeE such that
Then iT0 is closed and Ko c f , so Ko e X. Let »e E be such that z > xKo- Since 2 + (1 + e) x ^ xF>

AF ^ (l+e)fx+fz. As z is arbitrary, AF As e is arbitrary, AF ^ AH + tmv{AK:Ke3r,

K

as required by 7lA(iii). (f) Finally, suppose that si c jf* is non-empty, and that si \ 0 in 0>X (i.e. J / | and 0 ^ = 0 ) . Then in£Kej,\K = 0. P Set

^4 = {x:xeE, 3 Kes/,x ^ Then J. | in £, and inf ^ = i n f ^ ^ Z = 0 206

SMOOTH FUNGTIONALS

[72

in Kx [using (b) above]. So 0 = in£xeAfx = m£K6J,AK.

Q

In particular, 7lA(iv) is satisfied. (g) Accordingly, there is a complete locally determined measure /u, extending A; and if 2 is the domain of fi,

(a) /lE = sup{Aif :Ke3f,K^E}

for every Eel,,

(/?) If E c X and E n if G S for every if G JT, then Eel*. From 7 IF we see again t h a t / i s the integral with respect to JLC. (h) We must now check that (X, %, 2,/J) is quasi-Radon; let us go through the conditions of 72A in order. (i) We know that (X, 2,/^) is complete and locally determined. (ii) I f ( ? G X a n d i T e J r , t h e n i r \ ( ? G j r c S ; s o i r n G G S . By{g)(fi) above, ffeS. (iii) If EeH and /iE > 0, there is a K e Ctif such that K^E and > 0. Now there is an x e E such that a: ^ ^Z". So if 6? = {t: o;(^) > | } , ? ^ 2/x < oo, while

/i{G (\E)> p(G

(\K)=/iK>0.

(iv) For every Ee S, < sup{/^i7:F <^E,Fclosed} using (g) (a). (v) If 0 c f f (?O in X, let if G X be such that if c 0O. Then if \G G JT for every ^ e ^ , and {if \ 0 : G e &} | 0 . So by (f) above, inf {/*(if \£): Ge&} = inf {A(if \£): Ge&} = 0. So

/eJC < suv{fi(KnG):Ge&}

^ sup{/e0:Ge9).

As K is arbitrary, /£# 0 ^ su

Thus the last condition of 73A is satisfied and the proof is complete. *72F Lemma Let (X,%,S,/^) be a quasi-Radon measure space, (a) Let V = {G:Ge%,fiG< oo}. Then (i) /iE = sup{/i(E nG):Ge%f} for every EeS; (ii) if JE? c Z and JS7 n » G S for every GeW, then JS?GS. 207

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(b) If v is another quasi-Radon measure on (X, %), and if /u,G = vG for every GeX, then /i = v. Proof (a) (i) Let <x, = awp{ji(G<\E):GeV}. Then OL^/IE, so if a = oo then certainly a = /dS1. On the other hand, if a < oo, we can choose a sequence (Gn}neI[ in 2/ such that

p(E(\Gn)& a-2-» V neN. Now set ^ = E\DneVGn. * I f ^ i > °> there is a ( ? G ^ such that /i(G (]E1)> 0 [72A(iii)], and an ne N such that /i{G 0 Ex) > 2~n. But nowG[)Gne%f, so

which is impossible. X Thus /iE± = 0, and

(ii) Now suppose that E c X is such that ^7 n # e S for every G e £', and suppose that FeU. Then by (i) there is a sequence {Gn}neV in 2 / such that sup^^/A^ n Gn) = /^J7, so that /e(JI\UneH®n) = °- S o E n J 7 \U 7 l eN^ G ^ because (Z, 2,/J) is complete, while

by hypothesis. So ^7 n FeS. As -P is arbitrary, Ee2t, because (X, 2,/JL) is locally determined. (b) The basic point we have to check is that [i and v have the same domains of measurable sets. Let T be the domain of v. Observe first that if F c X is closed, n G) \QeV). : Ge %*}

using (a) (i) above and observing that W = {G: G e %, /iG < ex)} = {G: G e %, vG < oo}.

So /^ and i^ agree on the closed sets. Now suppose that EeT. Let f. By 72A(iv), applied to v, there exist increasing sequences 208

SMOOTH FUNCTIONALS

[72

^>weN and (Hn)ne1S( of closed subsets ofGnE and G\E respectively, such that ^VnexvFn = v(G n E)y suVneNvHn = v(G\E). Set

F^\JneMFn^G(\E9

H=(JneIiHn

then F and H both belong to S, and u #)) < infneN/*(£\(i^ U Hn)) = =

infne}Xv(G\(Fn[)Hn)) vG-v(GnE)-v(G\E)

= 0. Now (G 0 E)\F ^ G\{F [) H); so, because (X,S,/0 is complete, (G n E)\Fei: and # n # e 2 . AS £ is arbitrary, .ff e 2 [(a) (ii) above]. So we see that T ^ 2. Similarly, S £ T. Now it is obvious from 72A(iv) that fi and y are equal, since we have seen that they agree on the closed sets. 72G Exercises (a) Let (X, %, 2,/^) be a quasi-Radon measure space, and Ee2. Then inf{/iG :Ge%, G ^ E}i& either oo or /JLE. (b) Let (X, %) be a topological space, and let /i and y be two quasiRadon measures on (X, %) which agree on the closed sets. Then they are equal. (c) Suppose that, in 72E, Y = X, i.e. for every teX there is an x e E such that x(t) + 0. Then there is only one measure fi which is quasi-Radon for (X, %) and such that/is the integral with respect to /i. (d) Let (X, %, 2,/^) be a quasi-Radon measure space and suppose that 0 cz <S f Go in %. Then /i(E n Go) = s u p { / ^ n G) :Ge&} for every JE/GS. Notes and comments The correspondence between 71G and 72E is made clear by part (b) of the proof of 72E; in 71G the family Jf* consists of those sets K such that ^J5T is the infimum of a sequence in E, while in 72E we take sets K such that x& is ^he infimum of an arbitrary family in E. The effect of introducing the topology is to reduce the rather obscure relationship between a measure and a space of functions on which it acts as a smooth functional to relatively straightforward 209

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REPRESENTATION OF LINEAR FUNGTIONALS

relationships between both and the topology. For instance, putting 72E and 72B together, we have the result that a smooth functional is the integral with respect to a decomposable measure; it is not clear that there is any easy direct proof of this. Quasi-Radon measure spaces have many remarkable properties; but the majority belong to pure measure theory and we shall have to pass them by. My aim has been to characterize smooth linear functionals in a way which will be an effective step towards the classical theorems of the next section. Note carefully the technical result 72Fb. A quasiRadon measure is determined by its values on the open sets, and can therefore for some purposes be identified with its restriction to the Borel sets.

73 Radon measures and Riesz' theorem Much the most important applications of the work of the last section are those in 73D and 73E below. In these results we find ourselves considering locally compact Hausdorff quasi-Radon measure spaces in which compact sets are of finite measure. These are the Radon measure spaces of twenty years ago; but in the last decade it has been shown that a rather wider class can be studied with the same methods. Again, I shall have to pass by the most interesting properties of these spaces, and concentrate on the results most closely related to those we have seen already. 73A Definition A Radon measure space is a quadruple (X, %, 2,/^), where (X, 2,/*) is a measure space and % is a Hausdorff topology on X such that: (i) (X, 2,/^) is complete and locally determined; (ii) fcsS; (iii) if t E X, there is a G e % such that t e 0 and fiG < oo; (iv) if # 6 2, fiE = sup{/iF :F c E, F compact}. (Note that because % is Hausdorff, every compact set is closed and therefore measurable.) 73B Because of the rather different forms of the definitions, I had better set out in detail the following result: Proposition A Radon measure space (X, %, £,/£) is quasi-Radon. 210

RADON MEASURES AND RIESZ' THEOREM

[73

Proof We know that 72A(i) and 72(Aii) are satisfied. If EeTi and jiE > 0, there is a compact F c E such that jiF > 0 [73A (iv)]. We know that and \J@ = Xby 73A (iii). So, as F is compact, there must be a Ge & such that F e g , Now /iG < oo and /i(SJ nff)>/«f > 0 . Thus 72A (iii) is satisfied. 72A(iv) is obvious from 73A(iv), because every compact set is closed. If 9 is a non-empty set in % and ^ f (?0, then fiG0 = sup {/^J7 :F ^ Go, F compact}. But if F c 6?0 and i'7 is compact, there must be a 6? e 2? such that G 2 -P. So /£#0 ^ sup {/*
/*(# n J7) > [i{H f)F)^ /i(O 0 F) -fi(G\H) > a.

But H(\ F is compact and H(\F <^ F ^ E. As a is arbitrary, [iE ^ suip{/iK:K ^ E,K compact} < /iE, and 73A(iv) is satisfied. 211

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REPRESENTATION OF LINEAR FUNGTIONALS

73D Riesz' theorem: first form Let (X,%) be a locally compact Hausdorff topological space. Let K = K(X) be the space of continuous functions on X with compact support [A2D]. Let/: K -> R be an increasing linear functional. Then there exists a unique Radon measure pbonX such t h a t / i s the integral with respect to fi. Proof (a) I show first that/is smooth. P Suppose that 0 a A c K and that A | 0 in Rx. Fix xoeA. Let O = {t: xo(t) > 0}; then 0 is compact, so there is a n ^ e K such that xx ^ %G in Rx [A2Eb]. Let e > 0. Let 8 > 0 be such that #/(#!) ^ e. For each x e .4, set

Since J.10 in R-^, {J^: ^ eA} f JT. Because G is compact, there is an XGA such that fi^ 2 G. Now let 2/GJ. be such that i/ ^ a; A#o. Then 2/ ^ Sxvsofy ^ 5 / ^ ) < e. As e is arbitrary, infyeAfy = 0; as J. is arbitrary,/is smooth. Q (b) Because 2 is completely regular, it is the coarsest topology on X such that every member of K is continuous [A2Ea]. And of course K is truncated. So by 72E there is a measure /i on X, quasi-Radon for X, such that / is the integral with respect to JLL. If F c; X is compact, there is an xeK such that x ^ ^ # [A2Eb]; so /iF ^fx
P We know that

liG = sup{/£j?: J7 c G, F compact}. But if F is a compact subset of G, there is an #elf such that

[A2Eb]. 80 fiF ^ jxd/i =fx. Thus. jiG 6r = sup {/a: :xeK, x for every Ge%. So v = /i by 72Fb. Thus /^ is unique. 212

RADON MEASURES AND RIESZ' THEOREM

[73

73E Riesz' theorem: second form Let (X, %) be a locally compact Hausdorff topological space. Let Co be the space of continuous functions on X converging to 0 at infinity [A2C], Let/: Co -> R be an increasing linear functional. Then there is a unique Radon measure /i on X such t h a t / i s the integral with respect to fi, and ju,X < oo. Proof Consider the restriction of / to K = K(X). This is still an increasing linear functional, so by 73D there is a unique Radon measure fionX such t h a t / r = jxdju, for every xeK. Now, because Co is complete for || ||oo,/is continuous [25E], so H/ll = sup{|/a:| :xeC0, \\x\\w ^ 1} < oo. But if F c X is compact, there is an x e K such that %F ^ x ^ #X, and now pF ^fx^ \\f\\. So fiX = anp{/iF :F ^X,F compact} < ||/|| < oo. If now x is any continuous function from X to R, x is certainly measurable, since every open set in X is measurable; so any bounded continuous function is integrable, because the integrable functions are solid in the measurable functions [63Ea], and the constant functions are integrable. Thus ju(x) — \xdp exists for every x e Co, and \jxd/i\ ^ j \x\ d/i ^ j | | # | | 0 0 ^ ^ = Halloo/kX"So ju: Co -> R is a continuous linear functional, agreeing with/ on K; as K is dense in Co, p, agrees with / on Co, i.e. / is the integral with respect to ji. ^Remark The original theorem of RIESZ referred to the special case X = [0,1] (in which case the two forms coincide) and showed that an arbitrary member of C([0,1])' could be expressed as a RiemannStieltjes integral. *73F It will be helpful later to be able to refer to an argument I have already used in 73D. Lemma Let (X, %, 2,/^) be a completely regular Radon measure space. Then (a) for any compact set K £ X, jiK = inf{jxdju,:xeC(X), x& < x (b) for any open set G c: X, [iO = sup{jxd/u,:xeC(X), 0 ^ x 213

73]

REPRESENTATION OF LINEAR FUNCTIONALS

Proof (a) Let & = {G:Ge%,/iG < oo}. Then ^ f X by 73A(iii). So there must be a G e & such that G ^ K. Let e > 0. Let .F c 6?\j£ be a compact set such that jiF ^ /i(G\K) — e [73A(iv)j. Set H = G\F; then H is open, H ^ K, and /dff ^/iK + e. Because !£ is completely regular and K is compact, there is an x e C(X) such that Now x is certainly measurable, so j xd/i exists and J xd/Jb ^ /^jff As e is arbitrary, inf {Jo:^:o;eC, (b) If a < /^ff, then there is a compact set K c (? such that /£.K' ^ a [73A(iv)]. Let H be an open set of finite measure including K, as in (a) above. Let x eC be such that x% < x ^ ^((? n £T). Then jx ^ e x i s t s and fiK ^ a.

Thus

/^© < sup{Jxdju,:xeC, 0 < x

73G Exercises (a) Let (X, %) be a Hausdorff topological space. Let /i and v be two Radon measures on (X, %) agreeing on the compact sets. Then they are equal. (b) Let (X, %, £,/£) be a Radon measure space. Let E c X be such that E n F e S for every compact set F c X. Then 2? e S. (c) Let X, £ and/: K(X) -> R satisfy the hypotheses of 73D. Let J^ be the set of compact subsets of X, and define A: CrfT -> R by XF = inf{/£:a;eX:, x Show that CriC and A satisfy the conditions of 71 A, and so prove 7 3D without using 72E. (d) Let X, % and/: C0(X) -> R satisfy the hypotheses of 73E. Show t h a t / i s smooth, and so prove 73E without using 7 3D. (e) (Prokhorof's theorem) Let (X, %) be a completely regular Hausdorff topological space, and / : Cb(X) -> R an increasing linear functional. Show that/is the integral with respect to a Radon measure on Xiff it is tight, i.e. for every e > 0 there is a compact set F £ X such that/a; < e whenever x 214

RADON MEASURES AND RIESZ' THEOREM

[73

[Hint: use 72E to find a quasi-Radon measure /i; now show that ju, satisfies 73A(iv).] *(f) Let (X, p) be a complete metric space. Show that every smooth increasing linear functional on Cb(X) is tight. *(g) Let (X,p) be a separable metric space. Show that every sequentially smooth increasing linear functional on Cb(X) is smooth. (h) Let (X, %) be a Hausdorff topological space and JT the set of compact subsets of X. Let A: CtiT ->- R be an increasing functional such that (i) A0 = 0, (ii) A(H \j K) = AH + AK whenever H, KeX and HnK= 0, (iii) if F, HeJT and F 2 H, then

AF = AH + s\xp{AK:KeJr, K c (iv) for every £eJT there is an open set G containing t such that sup {AK: KeJf, if <= (?} < oo. Then A has a unique extension to a Radon measure on X. (i) Let (X, X) be a Hausdorff topological space and 93 the family of Borel subsets of X (i.e. SB is the cr-subalgebra of 2PX generated by %). Then a measure / t o n l with domain 93 can be extended to a Radon measure on Ziff (i) it is regular (sometimes called inner regular) [iE = sup{[iF:F ^E,Fcompact}

V EeS8;

(ii) it is locally finite, i.e. every point of X has a neighbourhood of finite measure. In this case the extension of ju, is unique, and is precisely that obtained by applying the processes of 64Ja and 64Jb to (X, 93, fi). *(j) Using (f) and (g) above and 7lHe, or otherwise, show that if (X, p) is a complete separable metric space, then any measure/£ defined on the Borel sets of X, such that /iX < oo, must be inner regular. Notes and comments The theorems above have innumerable consequences in many branches of mathematics, and a classification of their applications wouldfilla book by itself; I shall make no attempt to describe them here. Observe that the 'sequential' theorem 71G is inadequate for the applications above. If we try to apply 71G to the situation of 73D or 73E, we shall of course find a measure /i0, since the increasing linear functional / i s certainly sequentially smooth. But the domain S o of /i0 may be too small. For we know only that functions in K(X) or C0(X) are measurable; with a little ingenuity (using 71A (/?)), we can prove that

{«:*(«) > 0 } e S 0 215

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REPRESENTATION OF LINEAR FUNCTIONALS

whenever x e C(X). But the cr-subalgebra generated by these sets, the (r-algebra of Baire sets, may fail to contain all compact sets [7XC]. Of course, we know t h a t / corresponds to some Radon measure fi on X, and it is not hard to show that fi is the unique Radon measure extending /i0. TOPS0E T.M. [§ 5] attempts to analyse the relationship between ji a n d ju,0. For further information on the remarkable properties of Radon measures, I refer you to SCHWARTZ. *I ought to point out that his definition of Radon measure does not quite coincide with mine. His ' Radon measures' are my' locally finite inner regular Borel measures', described in 73Gi. These need not be decomposable or even Maharam. But in view of the one-to-one correspondence described in 73Gi, the difference is only technical. 7X Examples for Chapter 7 The simplest example of the theorems of this chapter is of course the development of the Lebesgue integral from the Riemann integral [7XA]. 7XB shows that the conditions of 71G are all necessary for the result. 7XC shows how the procedures of 71G and 72E can produce different measures when applied to the same linear functional. 7XA Lebesgue measure on Rm There are various ways of defining a Riemann-type integral on continuous functions of compact support on Rm; for instance, given xeK(Rm),

exists, because x is uniformly continuous. Now / : JK"(Rm) -> R is an increasing linear functional, so by 7 3D there is a Radon measure /JL on Rm such that fx = J xd/i for every xeK(Rm). And of course JJL is Lebesgue measure on Rm. In this case, because every compact set in Rm is a Gs set, it does not matter whether we use 71G or 72E; the basic family Jf* is the set of all compact subsets of Rm in either case. 7XB A non-truncated function space Let X be the unit interval [0,1], and let / be the cr-ideal of meagre subsets of X, i.e. the o*-ideal of &X generated by the nowhere dense closed sets. Let E be the set of all those functions x: X -> R such that there exists some a G R with

216

EXAMPLES FOR CHAPTER 7 [7X In this case, write f(x) = a. Because X$I, f: E -> R is well-defined. Then we see that E is an order-dense Riesz subspace of Kx and that / : E -> R is a Riesz homomorphism. Moreover, because X is not a countable union of members of / , / is sequentially smooth. However, / is not an integral. P ? Suppose, if possible, that fi is a measure on X, with domain 2, such that jxd/i = fx for every xeE. Since the function e given by e(jS)= 1+/3 V peX belongs to E, fiX < f(e) = 1; also, because e is measurable, every open set in [0,1] belongs to S. Let S? be the algebra of regular open sets in X [4XF]. If weN^ 0 in S?, then int(n weN ^n) i s a regular open set included in every Gn, so must be empty; thus P[nE^GneI a n ( i Thus if v is the restriction of fi to 8?, J> : 5? -> R is an increasing countably additive functional, and vX = fiX ^ %f(e) = i- But this is impossible, because X is separable and Hausdorff and without isolated points [4XFh]. X Q So we see that the condition that E be truncated in 71G is essential. I do not know whether it is still needed in 72E. 7XC A Baire measure which is not a Borel measure Let o)x be the first uncountable ordinal, and let X be the set of all ordinals not greater than cov Then X is a well-ordered set, therefore Dedekind complete, and it has greatest and least members, so in its order topology it is compact [IXC]. For E c X, ^ is a cluster point of Eiff E is uncountable. If x: X -> R is a continuous function, then for every

{t:teX, |a(*) must be countable, so {t: t eX9 x(t) 4= x(a)i)} is countable. Prom this we see easily that the Baire sets of X are precisely the countable sets not containing o)v and their complements. Now suppose we define / : C(X) -> R by fx = x{o)-^) for every x e C(X). If we follow the procedure of 71G, we get a measure fi0 defined on the Baire sets by ji0E = 0 if E is countable,

1 otherwise.

In particular, {wj is not measurable. While of course the procedure of 72E produces a measure /i defined on all subsets of X by 0 if OJ^E,

[iE=l

if

O)XEE.

217

7X] REPRESENTATION OF LINEAR FUNGTIONALS [i is of course a Radon measure, the unique Radon measure extending /i0, and is also produced by applying 73D. * Actually, there is another Borel measure extending /i0. The point is that, given any Borel set E ^ X, either there is an uncountable closed set F such that F^a)-^ ^ E or there is an uncountable closed set F such that F\{X.) In the first case set /ixE = 1; otherwise set /ixE = 0. Then /ix is a countably additive functional extending ju,0; it is a Borel measure which is not regular [73Gi]. Further reading for Chapter 7 A great deal of work has been done on measures on topological spaces. Once again, the best starting point is probably HALMOS M.T., chapter x [§§50-6]. A more elaborate treatment is in BOTTRBAKI VI, where Riesz' theorem is made the basis of the whole theory, and a 'measure5 is regarded primarily as a linear functional on K( X). Many of the same ideas may be found in SCHWARTZ, who takes a line rather closer to that of §§72-3 above. An important direction of generalization for the work of this chapter is towards a theory of vector-valued measures and integrals. Measures taking values in normed spaces are discussed by DUNFORD & SCHWARTZ, [chapter ni]. Perhaps closer to the ideas of this book is the notion of a measure which takes values in an ordered linear space; the elementary techniques are exhaustively investigated by MCSHANE, while some more sophisticated results have been given recently in a series of papers by WRIGHT.

218

S. Weak compactness Weakly compact sets have proved fascinating objects of study in the theory of linear topological spaces. In addition to the rich harvest of general theory, many results have been proved for special spaces, which are often topological Riesz spaces. It is my object here to collect these together, and to give some new theorems. As we shall be considering pairs of Riesz spaces in duality, this chapter will carry on from Chapter 3. At a number of points I shall call on the more advanced results in the general theory of linear topological spaces laid out in the appendix [§ Al], 81 Weak compactness in E~ This section is a foundation for the rest. It will be helpful to have some results concerning weakly compact sets in E~ safely established before we go on to weakly compact sets in £ x . It turns out that the 'normal' topology on E~9 %lsl(E~, E), the coarsest locally solid topology finer than %S(E~, £), is useful for technical reasons. The results below are generally straightforward, with the exception of 8 lF(v) and 81H, where some relatively deep ideas come to the surface. 81A Definitions Let E and F be Riesz spaces. A bilinear functional < , >: E x F -> R is positive if (x, y) > 0 whenever x e E+ and yeF+; in this case we say that E and F are 'in positive duality'. In this case, %\S\(F, E) is that linear space topology on F which has for basic neighbourhoods of 0 the sets

as x runs through E+. This is locally solid and locally convex; it is the coarsest locally solid linear space topology finer than %S(F,E). It is defined by the Riesz seminorms of the form as x runs through E+. If F is a Riesz subspace of E~, then %\8{{F9 E) is the topology of uniform convergence on order-bounded sets of E [16Ea]; the same is 219

81]

WEAK COMPACTNESS

true if £ is a solid linear subspace (or any locally order-dense Riesz subspace) of F~ [31D, 3lEa]. 81B Proposition Let £ and F be Riesz spaces in positive duality, and let E± be the image of £ in F* under the canonical map x\->& givenby % ) = <*,*/> V yeF.xeE. Then E± c F~ and the dual F' of F under %lsl(F, E) is precisely the solid hull of Ex in F~. Proof Clearly, if x e £+, then £ e F~+. As £ = £+ - £+, £ x c F~. I t also follows that the solid hull of £ x is

{g:3xeE+, \g\ ^ £} = G say. Suppose geG; then there is an xe £ + such that \g\ ^ &, and so g is continuous. Conversely, if g e F', there is an x e £+ such that \gy\^<x,\y\} = %(\y\) so gGF~ and |gr| ^ x; thus geG.

v

yeF,

81C Proposition Let £ be any Riesz space. Then !£W(JE~, £) is a complete Levi Hausdorff Lebesgue Fatou topology. Proof 16Db gives us 'Fatou'; 23M gives us 'Levi 5 ; 16Ec gives us 'Lebesgue'. It is Hausdorff because £ separates the points of E~. Now completeness follows either from Nakano's theorem [23K] or by applying Grothendieck's criterion [AID], because £~ is precisely the set of linear functionals on £ which are bounded on the order-bounded sets, and %ls](E~, E) is the topology of uniform convergence on these. 81D Corollary Let £ be any Riesz space. Then %ls](Ex, E) is a complete Levi Hausdorff Lebesgue Fatou topology. Proof Observe that £|S,(EX, £) is just the topology on £ x induced by S£,s!(£~, £). Now £ x is a band in £~ [16H]; its properties therefore follow at once from those of E~ (recalling that by 22Ec £ x is closed). 81E Lemma Let £ be a Riesz space and A a subset of £~. Let C be the solid hull of A in £~. Then, for any x e £,

220

WEAK COMPACTNESS IN E~

Proof

[81

For we see from 3 ID that, given any fe E~ and xeE,

81F Now we can describe the convex relatively weakly compact sets in £~. The only really surprising condition is (v). Proposition Let £ be a Riesz space and A a subset of E~. Then these are equivalent: (i) The solid convex hull of A in E~ is relatively compact for

%JLEr,E). (ii) The convex hull of A is relatively compact for %S(E~, E). (iii) The solid hull of A in E~ is bounded for %S{E~, E). (iv) A is bounded for %l8l(E~, £). *(v) A is bounded for %S(E~, E) and sup weN sup /e ^|/# n | < oo whenever (xn}neIS is a disjoint sequence in £+ which is bounded above in E+. Proof (a) (i) => (ii) and (i) => (iii) o (iv) => (v) are elementary, in view of 81E. I shall prove that (ii) => (iv) => (i) and that (v) => (iv). (b) (ii) => (iv) Let B be the %S(E~, £)-closed convex hull of A in £~.ThenJ3is convex and %S(E~, £)-compact. So, if C c Eis%(E, Er)~ bounded, sup{|/x| :feB, xeG} < oo by the Banach-Mackey theorem [All]. In particular, consider the case C = [ — x, x], where x e E+; then co>8ixp{\fy\:feB,yeC} = suV{\f$x):fEB} by 16Ea. As x is arbitrary, B, and therefore A, is %l8\(E~, £)-bounded. (c) (iv)=>(i) For x e E, set p(x) = s\xpfeA\f\(\x$ (iv) For x G £ . define p(o;) = sup /e ^ \fx\ < oo. Let F b e

{y:yeE, sup{/>(#): |x| ^ |y|} < oo}. 221

81]

WEAK COMPACTNESS

Then F is a solid linear subspace of E. belong to F, then

P Clearly F is solid. If yv y2

m${p(x): \x\ < \y1 + y2\} ^ sup{p(a?): \x\ ^ \y±\ + \yj} = sup{/>(«?!+ a?a): 1^1 ^ |2/i|, |a?a| < |y a |} [using 14Jb]

: |a| ^ |ya|} < 00. Thus F is closed under addition, and is therefore a solid linear subspace. Q ? Now suppose, if possible, that F + E. Then there is an x0 e E+\F. Let J30 be the solid linear subspace of £ generated by #0, and let Fo = F f| £0> s o that F o I s a solid linear subspace of Eo not containing x0. We now find that there is a Riesz homomorphism/: £ 0 -> R such that/a; = 0 for every xeF0 but/x 0 =(= 0. P Let SE be the set of all those solid linear subspaces of EQ which include F o but do not contain x0. By Zorn's lemma, 2C has a maximal member H say. Any solid linear subspace of Eo properly including H must contain x0, and therefore must be £ 0 itself. Thus H is a maximal proper solid linear subspace of £ 0 , and the Riesz space quotient Eo/H has no non-trivial solid linear subspace. As H + JS0, £ 0 /H 4= {0}, and £0/flT - R [15G(v)]. Now let / be the composition of this isomorphism with the canonical map from Eo to EQ/H; then/is a Riesz homomorphism and the kernel of/ is H, which includes F o but does not contain x0. Q Let fx0 = a. Now construct sequences (zn)ne1j[ and (wn)n€li as follows. Set z0 = x0. Having chosen zn such that 0 < zn ^ x0 and /2 w = a, we know that zn ^ F o , so there is a w such that |u| ^ zn and /o(u) > 4n+p(zn). Set

t; = ^ — a" 1 /^) »n.

Then/*; = 0; also, as \fu\ =f(\u\) ^f(zn)

= a,

p(v) ^ pi^-a-^fulpizj ^ p(u)-p{zn) > 4M. + So max (p{v ), p(v~)) > 2n; set w = \v+ orw = \v~ in such a way that yo(w) > n. Again because la" 1 /^ ^ 1, -2zn ^ t? ^ 2zw, so 0 < i^ ^ zn; and /M; = 0 because /y = 0. Let geA be such that \g(w)\ ^ n, and let J3 = \g\ (zn). Set Then 0 < ^ w ^ ^ < zn, and

222

WEAK COMPACTNESS IN E~

[81

= (zn-J3w)+. Then = O and /(«»+i) =/(«») = a.

Continue. Now, because oo; which is impossible. X Thus F = £, and (x) = = sup{p(y):|y| < x) < oo for every xe JS+; i.e. ^4 is %lsl(E~, £)-bounded. 81G Lemma Let £ be a uniformly complete Archimedean Riesz space. If A c E~isboundedfor2; s (£^£),itisboundedfor2:, sl (£^,£). Proaf Let x e £ + . Let Ex be the solid linear subspace of E generated by x, and let || || x be the norm on Ex derived from its order unit x [25H]. Consider A as a set of functions on Ex. Since A c E~, each member of ^4 is bounded on the unit ball of Ex, i.e. is continuous for || || x. Also, A is bounded for %S(E~, Ex). By the uniform boundedness theorem (this is where we use the fact that E is uniformly complete, so that Ex is complete under || \\x), A is equicontinuous for || ||x, i.e. sup{\fy\ :feA, \y\ ^ x} < oo. But this is just As x is arbitrary, A is £W(IS~, £)-bounded, as stated. *81H The following lemma will be used later to give a criterion for compactness in the topology %S(E~, E~x) under suitable conditions. Lemma Let E be an Archimedean Riesz space and F a Riesz subspace of E~. Let i c f b e a non-empty %S(F, £)-bounded set such that, whenever (xnynejx is a disjoint sequence in E+ which is bounded above > Then for every x0 e E+ and e > 0, there is an / 0 e F+ such that

223

81]

WEAK COMPACTNESS

Proof £

"

(a) By 81F(v), A is bounded for £ |s| (E~,E). Define p:

> R + b

^

p(x) = mV{\f$\x$:feA};

then p is a Riesz seminorm on E. Let % be the topology on E defined by p. Then X satisfies the conditions of 24H. P Of course orderbounded sets are bounded, as X is locally solid. Now suppose that (xn)neix *s a disjoint sequence in £+, bounded above by x. Choose, for each neN, &nfneA such that \fn\ (\xn\) ^ p(xn)-2~n, and a yneE such that \yn\ ^xn and \fnyn\ > p(xn)-2.2-". Then (yt)neix and (yn)nen a r e both disjoint sequences in £+, bounded above by x, so P(*»)<|/»y»|+2.2-» ^ WVfeA \M\+ ™VfeA \fVn \ + 2. 2"^ -> 0 as n-+oo. Thus <^n)nGN -> 0? a » required. Q (b) ? Now suppose, if possible, that the result is false. Then there is an xQ G E+ and an e > 0 such that, for every g e F+, sup / e ^(|/| -g)+(x0) = sup / e ^(|/| - | / | A0) (x0) > 3e. So we may choose a sequence (fn}ne}$ in ^4 such that Next choose, for each n e N, an element 2/w of £ such that 0 ^ yn ^ x0 and

(|/.|-2»S*<

It follows that where

a = sup{|/| (a:0) :/G^4} < oo;

at the same time, \fn\ (yn) ^ 3e. Consider now, for m,neN, For each meN, (^mn)weN *s a n increasing sequence, bounded above by x0. So by 24H it is Cauchy for %, and there is an r e N such that Setn;m = zmr. Now if m 

WEAK COMPACTNESS IN £T

[81

Now set vm = infi<mwi for each m e N. Since (vm)meJj( is a decreasing sequence, bounded below, it is also Cauchy for X, and However, examine \fm\ (vm — vm+1), for any m e N . We know that I/ml (2/m-^m)+ < Si<m I/ml (2/m~^) + < S ^ 2e. Si<m So |/m| frm) > I/ml (^m A « m ) = |/ m | (^m)" l/m On the other hand, |/m| (*

So

p(vm-vm+1)

>\fm\(vm-vm+1)

> e-2-a

V meN,

and 0. X This contradiction shows that the lemma is true. 811 Exercises (a) Let E be an L-space. Then %lsl(E,Ef) is the norm topology on E. (b) Let E and F be Riesz spaces in positive separating duality. Then S|S|(F, E) is the topology on F induced by its embedding as a Riesz subspace of the product of a suitable family of L-spaces. *(c) Let E and F be Riesz spaces in positive duality. Suppose that %\S\(F, E) is Hausdorff and complete. Show that it is Lebesgue and Levi, and that F is perfect. [Hint: 26B.] (d) Let E be a perfect Riesz space. Then E is complete for %b(E, Ex), which is Fatou and Levi. (e) Let E be a perfect Riesz space and % a compatible complete metrizable linear space topology on E. Then % = %b(E, Ex). (f) Let E be any Riesz space. Then the Mackey topology %k(E, E~) is locally solid, and is the finest locally convex topology on E for which order-bounded sets are bounded. (g) Let £ be a uniformly complete Archimedean Riesz space. Then %k(E, Er) = %b(E, E~). *(h) Let £ be a uniformly complete Archimedean Riesz space. Then E~ is complete for £fc(£~, £). FTR

8

225

81J

WEAK COMPACTNESS

Notes and comments From 811a and 8lib we can see that these £|S|-topologies are a simple generalization of Z-space topologies. It is from this point of view that they are treated in KOTHE [§ 30.4 et seq.]. However, we shall be interested in them merely because they give efficient characterizations of the convex weakly compact sets in E~, in 81E(iv) and 81G. In part (d) of the proof of 8 IF we saw that, given a Riesz space EQ with an order-unit x0, and a proper solid subspace Fo of £ 0 , there is a Riesz homomorphism/: Eo -> R such that fy = O for every y e F o but fx0 4= 0. This can be regarded as a first step towards the representation theorem for Archimedean Riesz spaces with order units, which asserts that any such space is isomorphic to a norm-dense Riesz subspace of C(X) for some compact space X [LUXEMBURG & ZAANEN R.S., Theorem 45.3; see also the note concerning M-spaces with units at the end of § 26 above]. 8 IF (v) is a striking result. Here it is used only in the proof of 81H, which has important applications in §§ 82 and 83. 82 Weak compactness in Ex In this section I run through the principal characteristics of those sets in Ex which are compact for XS(EX, E). The most powerful results obtain when E is uniformly complete or Dedekind cr-complete; but it seems worth while to express the basic theorem 82E in terms of arbitrary Archimedean Riesz spaces. From these theorems we can deduce, as simple corollaries, some results concerning locally convex Lebesgue topologies. I have starred the paragraphs 82A-82D because, although they are interesting in themselves, they will not be used again. *82A Lemma Let E be an Archimedean Riesz space. Let (xn)ne$ be a decreasing sequence in E+. Then

is closed for %S(EX, £). Proof

Let

G = {y:yeE, 0 ^ y ^ xn V Then, just as in 15C,

{xn-y:neN, 226

WEAK COMPACTNESS IN E* [82 x

in £. So, for every fe £ , w£{\f\(xn-y):neN9yeC} = 0. Consequently,

A = {f:feEx,\f\(y) = 0 V yeC} = { / : / e £ x , / z = 0 whenever \z\ 0 there is an xeB such that x ^ x0 and M (#) ^ e- K is therefore possible to choose inductively sequences <0n>neH a

n d

<^>neN SUCh that:

2/0 is an arbitrary member of B; gneA and | ^ ^ - / ^ | < 2~n V

i^n

(choosing gn, using the fact t h a t / i s in the weak closure of A); yn+1eB,

yn+1 < yn

and (Zi
(choosing yn+1, using the facts that B10 and Si
D = {h:heEx,

V ieN.

l i m ^ | A | (^) = 0}

is %8(EX, £)-closed [82A], and contains every gn9 it contains g. But now

\fy\ ^ infi

8-2

227

82]

WEAK COMPACTNESS

As B is arbitrary, fe £ x by 32D. As / is arbitrary, A c £X; but if is %S(E*, £)-compact, i.e. %(EX, £)-compact, and A is therefore relatively %S(EX, £)-compact. *82C Proposition Let E be an Archimedean Riesz space, and A a subset of £ x . Then these are equivalent: (i) The convex hull of A is relatively compact for XS(EX, E); (ii) A is relatively compact for ZS(EX, £)andboundedforX, sl (£ x , £). Proof (a) (i) => (ii) follows from 8 IF (ii) => (iv). (b) (ii) => (i) By 82B, it is enough to show that every sequence in the convex hull T(A) of A has a cluster point in Ex. But given a sequence T(^4), there must be a sequence (fn)nGn in A such that for every neN, where .40 = {fn:neN}. Since ^ 0 is %lsl(Ex, £)-bounded, its %S(E~, £)-closed convex hull 0 is %8{Er, £)compact [8lP(iv)=> (ii)]. Once again, I shall use 32D to show that Cc Ex. Suppose that 0 c: B10 in E. As in 82B above, we may choose a decreasing sequence <^) 7l€N in B such that (2i<.l/<|)(y»X2-»

V neN.

Define T: E~ -+ /°°(N) by

Then T is linear and is continuous for £ s (£~, £) and £s(/°°, ^0), where 50 is the space of those sequences in R which have only finitely many non-zero terms. Consider D = {f:fe Ex, <|/| (yn))nelx -> 0}. Then D is %S(EX, E)-closed by 82A, and D ^ Ao, so D ^ AQ, which is XS(EX, £)-compact because A is supposed relatively %S(EX, £)-compact. Also T[D] c co(N). So yf^ol ^s a ^S(COJ <50)"comPac^ subset of c0. But T[A0] is || ||^-bounded, because Ao is S|S|(£X, £)-bounded, and {yn: 7i G N} is order-bounded in £. So %s{c0, s0) agrees with %s(c0,11) on T[AQ], because s0 is || || r dense in I1 = P(N), and T[J 0 ] is ^(Co,/ 1 )compact. Now by Krein's theorem its Xs(c0, P)-closed convex hull K is %s{c0, /^-compact, therefore ^(Z00, ^0)-closed. So T-^K] is a convex S^s(£^, £)-closed set, and includes (7. Thus T[C] ^K^c0. So if geC, MyeB\gy\ 228

^ MneJX\gyn\ = mfnelsl\(Tg) (n)\ = 0.

WEAK C O M P A C T N E S S IN E><

[82

As B is arbitrary, g e E x . Thus C c Ex. But (<7n>^eN has a cluster point in C, so has a cluster point in Ex; by 8IB, this shows that A is relatively compact for %S(EX, E). *82D Corollary If £ is a uniformly complete Archimedean Riesz space and A c Ex is relatively compact for £ s (£ x , E), so is Proof For by 81G any %(EX, £)-bounded set is £ |sl (E x , £)bounded. 82E Theorem Let E be an Archimedean Riesz space, and A a non-empty subset of Ex. Then these are equivalent: (i) The solid convex hull of A is relatively compact for %S(EX, E). (ii) {|/| :feA} is relatively countably compact for XS(EX, E). (iii) A is bounded for %S(EX, E) and, whenever 0 c: B \ 0 in E, infXeBmpfGA\fx\

= 0.

(iv) For any x e E+ and e > 0, there is an/ 0 e JSX+ such that (|/|-|/|A/0)(x)<e

V/ei.

X

(v) J. is bounded for £S(E , £) and, whenever (x^)ne^ is a disjoint sequence in E+ which is bounded above in E,

Proof

(a) (i) => (ii) is trivial. I shall prove that (ii) =>(v) =>(iv)=> (in) =>(i).

(b) (ii) => (v) Suppose that A satisfies (ii). In this case, SU

P/ G ^|/^| < S U P/e^|/| (N) < 00 V XGE, so A is certainly bounded for %S(EX, E). ? Suppose, if possible, that (xn)ne& ^s a disjoint sequence in £+, bounded above, such that neN -|>0. In this case, there is an e > 0 and a strictly increasing sequence {n(k))keN in N such that *uVf€A\fxn{k)\ >e

VkN,

and now we may choose a sequence (/^)^eN hi ^ such that

|AI (*«(*)) >|A(*«oo)|>e

VieN.

L e t / b e a cluster point of <|A|)*eN in £. Set B = {supi: A;eN}, 229

82] WEAK COMPACTNESS and let C be the set of upper bounds~of B in E. Then B f and C | , so C - 5 1 , and 0 - 2?10 [15C]. So there is a 2 e 0 - B such that/2; < e, i.e. there is a y e(7 and a J e N such that However, if j > &, a^H-S^av**) < V> s o As/is a cluster point of <|/y|>J€N3 f(y-Hi^kxn(i)) ^ e5 contradicting the choice of 1/ and &. X (c) (v)=>(iv)is81H. (d) (iv) => (iii) Suppose that A satisfies the condition (iv). (a) A is bounded. P For every x e E, there is an/ 0 e Ex+ such that (i/l - I/I A/o) (H) < 1 for every feA; but now sup /e ^|/x| ^ l+/ 0 (H) 0 and x0 e B. Then there is an / O G £x+ such that (|/| - |/| A/ 0 ) (XQ) < e for every feA, and an xe B such that # ^ x0 a,ndfox ^ e. But now |/*| < I/I (*) < ( I / H / I A/o)(*)+/o* < (1/1 - I/I A/o) («b)+c < 2e for every feA. As e is arbitrary, A satisfies (iii). (e) (iii)=> (i) Suppose A satisfies the condition (iii). Let C be the solid convex hull of A in Ex. (a) C also satisfies the condition (iii). P If xe E+ and aeR + , {/:/££*, I/I (x) < a} = {/:/££*, |y| < *=> |M < a} is a solid convex set in Ex; so sup /€C |/a| = sup{|/i/| :feA, \y\ ^ x}. Now define p: E -> R+ by /?(#) = sup/e^ |/r| for every xe £. Then p is a seminorm; let % be the linear space topology it generates on £. The condition (iii) on A asserts that % is Lebesgue. Accordingly orderbounded sets are bounded [24Db]; so, for any xe £+, BupteolM = *vp{pte): \y\ < ^} < °°> and (7 is £S(E , £)-bounded. Also, if 0 c: B\ 0 in E, mf*ei?sup/6C|/a;| = inf^^Bup^^p^) = 0 by 24Bc. Q X

230

WEAK COMPACTNESS IN E*

[82

((S) G is relatively compact for %8{EX, E). P As G is bounded, it is relatively compact in £* for %8(E*, E). Let
so ^ e E by 32D. As 0 is arbitrary, C £ £ x ; so C is relatively compact for %(EX,E). Q 82F Corollary Let E be an Archimedean Eiesz space. Then the Mackey topology %k(E, Ex) is Lebesgue iff it is locally solid. Proof (a) Suppose that % = %k(E, Ex) is Lebesgue. Let U be a neighbourhood of 0 for %. Then U° c £x has the property (iii) of 82E. So its solid convex hull C is relatively %S{EX, £)-compact, and C° is a neighbourhood of 0 for %. Now C° is a solid neighbourhood of 0 for %, and is included in U. Thus % is locally solid. (b) The converse follows from 24G. 82G Theorem Let £ be a Dedekind cr-complete Riesz space, and A a non-empty subset of £ x . Then the following are equivalent: (i) A is relatively countably compact for %S(EX, £). (ii) The solid convex hull of A is relatively compact for %S(EX, E). (iii) A is bounded for XS(EX, E) and, whenever (xn)nelSi j 0 in E,

Proof (a) (ii) => (i) is trivial; and (ii) => (iii) is a special case of 82E(i) => (iii). I shall prove that (i) => (ii) and that (iii) => (ii), in both cases using the criterion 82E(v). (b) (i) => (ii) Suppose that (xn}neJX is a disjoint sequence in E+ which is bounded above. Then, because E is Dedekind (r-complete, x = supneNa;n exists in E, and now X = S U p w e N S i < n ^ = 2ieN^>

the sum being for %S(E, Ex). Similarly, any subsequence of (xn}nels is summable for %8(E, Ex). So by AlHb,

andweseefrom82E(v) => (i) that the solid convex hull of ^4 is relatively compact. 231

82]

WEAK COMPACTNESS

(c) (iii) => (ii) Again suppose that (xn)ne1jf is a disjoint sequence in E which is bounded above. Let x = supneNa;w. Then (x — S$<w#$)neN 10 in E, so

aw = sup /e ^ | / ( » - S i < n ^ ) | -> °

as

a

aS

a

^ ~> °°>

and SWfeA \fan\ < | »+l| + \ n\ "> ° n-^ CO. 3S As (^>^eN arbitrary, the solid convex hull of J. is relatively compact. 82H Corollary If E is a Dedekind (r-complete Riesz space, any locally convex linear space topology % on E for which E' c £x j s Lebesgue. In particular, the Mackey topology %k(E, Ex) is Lebesgue and therefore [by 82F] locally solid. Proof If U is a closed absolutely convex neighbourhood of 0 for %9 then U° is %S(E\ JE)-compact, and therefore %S(EX, E)-compact. So by 82G its convex solid hull is relatively %S(EX, E)-compact. Now we see from 82E(iii) that if 0 cz B j 0 in E, B meets U00 = U. As B and U are arbitrary, % is Lebesgue. 821 Corollary Let E be an Archimedean Riesz space. Let F be the solid linear subspace of E x x generated by the canonical image of E in E x x [32B]. Then a subset A of E x is relatively compact for %S(EX, F) iff the solid convex hull of A is relatively compact for SS(EX, E). Proof The point is that 2 S (E X ,F), being finer than %(EX,E), is Levi. So it follows from 33B that the canonical map from E x to Fx is one-to-one and onto, and that we may identify E x with F x . Now if A c £x = Fx is relatively compact for %S(EX, F), so is its solid convex hull, by 82G(i)=> (ii), because F is certainly Dedekind (r-complete. As %S(EX, E) is coarser than %S(EX, F), the solid convex hull of A is relatively compact for %S(EX, E). Conversely, suppose that the solid convex hull of A is relatively %S(EX, E)-compact. Let ^ e F+. Then by the definition of F there is an xeE+ such that & ^ (p. Now, given e > 0, there is an/ 0 e E x + such that

e>(\f\-\f\*fo)(x)X\f\-\f\*Ut> for every / e A, using 82E (i) => (iv). Now, using 82E(iv) => (i), we find that A is relatively compact for %S{EX, F). 82 J Corollary Let E be a Dedekind cr-complete Riesz space. Then E x is sequentially complete for %(EX, E). 232

WEAK COMPACTNESS IN E* x

[82 x

Proof Let (fn)nelx be a sequence in E which is Cauchy for %s( E , E). I shall use 82E(v) to show that {fn: n e N} is relatively compact in JEX. Suppose that (xn}n€jx is a disjoint sequence in £+ which is bounded above. ? Suppose, if possible, that

In this case, there is an e > 0 and a subsequence (?/&)&GN °f such that Construct a sequence (^r)reN a s follows. We know that (?/&)/<; eN -^ 0 for XS(E, Ex). So, given r e N , there is a &(r) such that k(r) ^ r and ))] < e for every i ^ r. Now let m(r) be such that |/m(r)(^(r))| ^ 2e. As certainly m(r) > r for every r e N , (gr)re^ -> 0 for £S(22X, £). So N i s relatively compact for %S(EX, E); by 82G(i)=>(ii) and 82E(i) => (v), But i(j) ^ j and |^-(yA;(^))| ^ e f° r every J G N ; which entails a contradiction. X As x ( n)neTiis arbitrary, (fn)neln is relatively compact for £S(EX, £), by 82E(v)=> (i). So it has a cluster point feEx. But now, of course, (fn)neTS ~^/f° r ^s(^x> ^)- Thus every Cauchy sequence has a limit, and Ex is sequentially complete for %S(EX, £). Remark This result is true for all bands in E~; see SCHAEFER W.C., where there are further remarks on sequential convergence in E~. 82K Corollary Let £ be a perfect Riesz space, (a) If A c E is relatively compact for %S(E, Ex), so is the solid convex hull of A. (b) E is sequentially complete for %8(E, Ex). Proof For E may be regarded as (£ x ) x , and Ex is certainly Dedekind complete; so the result is immediate from 82G and 82J. 82L Exercises (a) Let E be a Dedekind complete Riesz space. Then [x, y] is compact for %S(E, Ex) for every x, y in E. (b) Let E be an Archimedean Riesz space and A a subset of £ x . Then the solid convex hull of A is relatively compact for %S(EX, E) iff, whenever (xn}nEN is an order-bounded increasing sequence in E and Un>neK ^ a sequence in A, (fn(xn+1 - xn))ne^ -* 0. 233

82]

WEAK COMPACTNESS

*(c) Find an L-space E and a sequence (xn}ne1x in E+ which converges for %8(E, Ex) but has no order-bounded subsequence. Notes and comments The proofs of the main theorems 82E and 82G above are relatively elementary because substantial parts of the argument have been removed to §§81 and Al. For a more direct approach to these results, see FREMLIN A.K.S.I. We may think of 82E(iv) and 82E(v) as being extremes, in that 82E(v) is the easiest of the five conditions to establish [see 82G, 82 J], while 82E(iv) is the easiest to draw conclusions from [see 821, 82Lb]. Thus it is unsurprising that the argument required to pass from (v) to (iv), using 24H, 82F(v) and 82H, is the hardest part of the work. It is perhaps worth observing that useful simplifications are possible if we allow ourselves to use the fact that A c £x. For Dedekind cr-complete spaces, the conditions in 82E and 82G form a very powerful list of equivalences. Their most natural applications are to perfect Riesz spaces, in which they can be used to discuss weak compactness for %S(E, Ex), as in 82K. This result has long been known for P and Z> spaces [6XF]; the generalization here is due essentially to DIEUDONNE. *The propositions 82B-82D are immediately reminiscent of the theorems of Eberlein and Krein. It is therefore significant that the example 8XC shows that they cannot be deduced from these theorems. Other examples in § 8X show how the results proved here for uniformly complete or Dedekind cr-complete spaces cannot be extended to the next wider class. But I do not know whether 82B-82F are true for non-Archimedean spaces. 83 Weak compactness in L-spaces The theorems of the last two sections have particularly direct applications to problems involving the identification of weakly compact sets in .L-spaces. Since i-spaces are perfect [33F], we can immediately apply 82E and 82G to obtain various criteria, as in 83A. Moreover, the most important .L-spaces are the L1-spaces, and the identification of these with L#-spaces in §52 sets them up for investigation by the methods developed above, as in 83F. The same arguments enable us to cope with .L-spaces which are defined as the duals of M-spaces [see 83B, 83J]. One of our criteria for weak compactness in L1-spaces [83F] can be generalized to characterize weak compactness in other perfect function spaces [831]. 234

WEAK COMPACTNESS IN L-SPACES

[83

83 A Lemma Let E be an i-space and A a subset of E. Then these are equivalent: (i) A is relatively compact for %8(E, Er); (ii) the solid convex hull of A is relatively compact for %8(E, E')\ (iii) for every e > 0 there is an x0 e E+ such that || H - \x\ A xo\\ < e V xeA. Proof Recall that an i-space is perfect [33F], so that E may be identified with £ x x = £ ' x [26C]. Now E' is certainly Dedekind complete, so (i) and (ii) are equivalent by 82G. Next, if the solid convex hull of A is relatively compact, and if e > 0, then by 82E there is an x0 e E+ such that

I N - M A &o|| = J ( N - M Aa0) <e V xeA, since J e E' [26C]. On the other hand, if this condition is satisfied, and if fe E' and e > 0, then there is an x0 e E+ such that

I/I (\x\-\x\ AX0) ^ I/I || 1*1-1*1 A*o|| < e V xeA, so A is relatively compact by 82E again. 83B Theorem Let G be a Riesz space with a Fatou norm || || such that (a) \\x wy\\ = max(||a;||, \\y\\) whenever x, y ^ 0 in G (/?) for any z e G + a n d a > 0, s u p { y : 0 ^ y ^ x> M <. a } exists in G. Let J. be a non-empty subset of G'. Then A is relatively compact for %S(G', G")ift (i) it is bounded for £ S (G', G) (ii) whenever (xn}ne^ is a norm-bounded disjoint sequence in G + ,

Proof We know that G' is an i-space [26D], so we may use the criterion of 83A. (a) If A is relatively compact for %S(G', G"), then of course it is bounded for %S(G\ G). Now suppose that (xnyneJX is a norm-bounded disjoint sequence in G+; let a = supn6H||a;w||. Let e > 0. Then there is a n / o e G'+ such that

Ill/I-I/I A/o|| < c

VfeA

[83A]. So, for any »eN,
83]

WEAK COMPACTNESS

So nGN ~^ 0 and there is an w o eN such tha,tfoxn n ^ n0. Now, for any TI ^ w0 a n d / e J.,

^ e for every

1X1 < I/I (*J < (I/I - I/I A/o) {xn) +foxn 0. (a) There is a wQ e G+ such that \f(w-wAWO)\

^ 2e

+

whenever/eA, we G and \\w\\ < 1. P ? Suppose otherwise. Choose sequences <#n>neN a n d <2/n>7ieNil1 G+ and weN in ^4 inductively, as follows. xQ = 0. Given xn, let yneG+ &ndfnEA be such that ||i/n|| < 1 and l/,(y»-y»A*j|>2e. 1 Set a;m+1 = zm V e- U/n| yn. Continue. Now, for each » e N , set <xn = el/JI" 1 , and let Bn = {w: 0 < w < yn, w\ By the hypothesis (a), Bn f; by the hypothesis (/?), z n = sup S m exists, and as || || is a Fatou norm, \zn\ ^ <xn. Of course 0 < zn ^ ?/„. Set

W-{yn-xn)+\\ so

^\\zn\ ^an,

I/ n « n | > \fn(yn-xn)+\ -ajfnl>2e-e

= e.

But examine umt\ un, where m < n. Actually, «*• A amw» ^ (ym - zm) A am(2/re - xn)+ < tom- 2/m A am2/ra) A a j y n - a - V m ) + (because 0^ymA xm+1 > a£ym) = (Vm ~ <xmyn)+ A (<xmyn - 2/OT)+ = 0. So by 14Kb umAun 236

= 0.

WEAK COMPACTNESS IN L-SPACES

[83

+

Thus (un}nels is a disjoint sequence in G , and \\un\\ ^ \\yn\\ ^ 1 for every neN. But

sup/eJ.KI > \fnun\ >e V neN, contradicting the hypothesis (ii). X Q (P) We now find that \f\(\w\-\w\Awo)^4e whenever fe A and \\w\\ < 1 in G.

P Suppose that

0 ^ u < \w\ — \w\ AW0. Set

v = u+ \w\ Aw0;

then \w\ Aw0 ^ v < \w\9 so ||v|| ^ 1 and |w| A wQ < v A wQ ^ |w| A i^0. Thus v — v A wQ = v — \w\ A w0 = u. Now \fu\ = \f(v-vAwo)\

^2e,

by part (a) above. As u is arbitrary, f+(\w\-\w\AwQ)

^ 2e.

Similarly,/~(|^| — \w\ AwQ) ^ 2e, so \f\(\w\-\w\Awo)^4e. Q (y) Observe next that A certainly satisfies the conditions of 81H, with F = G', E = G. So there is a n / o e G /+ such that

Now suppose that fe A and ||w|| < 1. Then

^ I/I (1^1-HI A ^ 4e + e = 5e. As w is arbitrary,

ll|/H/|A/oH 5e V/ei. (8) As e is arbitrary, A is relatively compact by the criterion of 83A. 237

83]

WEAK COMPACTNESS

83C Theorem Let % be a Boolean ring, and let A be a non-empty subset of £ = 5(91)' = L°°(9{)' = L°°(8t)~. Then A is relatively compact for £fi(E, E')iff(i)sup /e ^|/(;\;a)| < oo for every aeSt, (ii) whenever (aw)neN *s a disjoint sequence in 91,

Proof (a) Recall that by 25G L00' = L°°~. Because 5 is a normdense subspace of L00, we may identify 5 ' with L00' as normed linear space; because the positive cone of L°° is just the norm-closure of the positive cone of 5, the positive cones of 5 ' and L00' coincide, so we may identify 5 ' and L00' as Riesz spaces. (b) The space 5 ,with the norm || || «>, satisfies the conditions of 83B. P (a) is obvious, as 5 is defined to be a Riesz subspace of an /°° space. If x G 5+ and a > 0, then we know by 42Eb that x is expressible as Yii where (ai)i
On the other hand, suppose that condition (ii) of this theorem is satisfied, and that (xn}neIl is a disjoint sequence in S+ such that y = sup^guH^Jloo < oo. For any n (unless xn = 0), xn can be expressed as IijKmfaxh where ^j<m^ = |K||oo ^ 7, and each fa > 0 [42Ec], so Now let an be equal to one of the bj and such that

we know that there is an ccn > 0 such that otnxan < xn. (If xn = 0, set an = 0, ocn = 1; then the same statements are valid.) Since, for m =)= n, «>nXan A <XmXam < xn A Xm = 0,

238

WEAK COMPACTNESS IN L-SPACES

[83

Ti is a* disjoint sequence in 81. So <sup/e^ \f(xan)\}ne* -+ ° a n d ^ |/a?n|)neN ~> ®> a s required. Q Thus the conditions above are satisfied iff those of 83B are satisfied, i.e. iff A is relatively compact. 83D Corollary Let 91 be a Boolean ring, and A a non-empty subset of V* = L#(2t). Then A is relatively compact for XJJLftJJ*') iff the following two conditions hold: (i) sup / e ^|/(^a)| < oo for every ae8t; (ii) weH ->0 whenever n€N is a disjoint sequence in 21. Proof The point is that L# = L°°(3l)x is a band in E = L°°(2t)~. Consequently it is closed for the norm on E [22Ec], and is therefore closed for the weak topology %s( E, E'). So a subset A of L # is relatively compact for %S{L#, £'), which coincides with %S(L#, L #/ ), iff it is relatively compact for %S(E, E'). The result now follows at once from 83C. 83E Corollary Let 21 be a Boolean algebra, and A a subset of Ift = L#(2t); let C be the solid hull of A in L#. Then these are equivalent: (i) A is relatively compact for %s{lft,Lfl')\ (ii) G is relatively compact for %S(L#, L00); (iii) C is relatively compact for %S{L#, S), where S = 5(21), L™ = L°°(2l). Proof (a) If A is relatively compact for ZS(L#, L#f) so is G [83A], so C is relatively compact for £ S (L # , L00). Thus (i)=>(ii). Of course (b) (iii) => (i) Recall that in 44Bb L # is identified with the space of bounded completely additive functionals on 21, while in 42L Sx is identified with the space of locally bounded completely additive functionals. But if 21 has a 1, these are the same. So we may think of L# as5x. Now if C is relatively compact for %S(SX, S), then from 82E we see that for any e > 0 there is an / 0 e Sx+ such that lll/H/|A/o|| = ( | / | - | / | A/ 0 )(*l)<e V / 6 4 . So A is relatively compact for %S(L#,L#') by 83A. [Cf. 821.] 83F Proposition Let (21, /i) be a semi-finite measure ring and A a non-empty subset of L1 = Lx(2l). Then A is relatively compact for 239

83]

WEAK COMPACTNESS 1

^.(L , Llf)iff (i)supMe^|<w, xa)\ < oo for every a e 21 (ii) for every e > 0 there is an a e W and a S > 0 such that e

whenever

b e21, /^(a n 6) < S.

1

Proof Recall that we think of L as identified with Lf (21) [52E]. (a) If the conditions hold, then let (anynex be any disjoint sequence in 21. Given e > 0, find aeW and 8 > 0 as in (ii). Since (an)ne^ is disjoint, S ^ N M ^ n an) ^ ^ < oo. So there is an w0 such that /^(a n an) ^ 5 for all n ^ TI0. NOW As e is arbitrary, l i m ^ ^ s u p ^ ^ K ^ , ^ ^ ) ! = 0, and A is relatively ^(L 1 , L1/)-compact by the criterion in 83D. (b) Conversely, if A is relatively £ S (L\ L1')-compact, then certainly , x<x>)\ < °o for every ae2l. Also, given e > 0, there is a u0 e L1+ such that \u\ — |^| A^old < e

[83A]. Now there is an xoeS(W)+ such that ||ffo~~^olli ^ e, as 5 is dense in L1, and there is an asW and an a > 0 such that x0 [42Ec]. Accordingly for every ueA. Now set ^ = e/a. If 6G21 and/^(a n &) < ^, then

for every ^ e^4. As e is arbitrary, this establishes the second condition. 83G Lemma Let 21 be a Dedekind cr-complete Boolean algebra. If A c L#(2l) is bounded for %S(L#, S), it is bounded for the norm of L#. Proof I shall use 8 IF, though of course more direct approaches are possible. Let C be the convex hull of A in Lft. As in 83E, regard Lfl as identified with Sx. Since C is %S(SX, S^-bounded, it is certainly relatively £S(S*, 5)-compact. Let/ 0 belong to the closure of C in 5*. ? Iffo$S~, thenfx: 21-> R is not locally bounded [42H], so by 421 there is a disjoint sequence (pn}neJSi = <&n\#n+i>neN such that fo(xbn) > n+ 1 for every weN. Now however

240

WEAK COMPACTNESS IN L-SPACES

[83

exists for %S(S, L # ) for every strictly increasing sequence (n(i)}ie1jS. So by AlHd {xbn :n e N} is %b{S, L#)-bounded and °0 = ^VneMxK)

< SUp n6N SUp /eC |/(#6j| < 00,

which is impossible. X Thus/ 0 e £~. As/ 0 is arbitrary, C is relatively %S(S~, S)-compact, so by 81F A is £|8,(5~, S)-bounded, i.e.

83H Proposition Let 91 be a Dedekind cr-complete Boolean algebra and A a subset of L# = L#(9t). Then these are equivalent: (i) A is relatively compact for %8(L#9 L # '); (ii) A is relatively compact for %S(L#, L00); (iii) A is relatively compact for %8(L#, S)9 where S = 5(91), 00 L = L°°(9l). Proof (a) (i) o (ii) Of course (i) => (ii). Conversely, if A is relatively compact for %S(L#, L00), then by 82G the solid hull of A is relatively compact for %8{L#, L00), since by 43Db L00 is Dedekind cr-complete. Now it follows from 83E that A is relatively %S(L#, L#/)-compact. (b) (ii)o(iii) Of course (ii)=>(iii). On the other hand, if A is relatively compact for %S(L#, S), then by 83G its S£S(L#, 5)-closure A is norm-bounded. So %S(L#, S) and %8(L#, L00) agree on A, which is therefore %8{L#, L°°)-compact. 831 The same approach as in 83F gives a general criterion for weak compactness in Banach function spaces. Theorem Let (X,H,/i) be a Maharam measure space and E an order-dense solid linear subspace of L° = L°(L,/i). Let

F = {w.weLPtWxueL1

V ueE},

so that F can be identified with Ex as in 65A. Let A be a non-empty subset of F. Then A is relatively compact for %8(F, E) iff it is bounded and, for every e > 0 and u e £+, there is an H e 2 ' and a S > 0 such that

^ e V weA whenever iTeS and ju,(H (]K) < & (Notation: If Z G S and veL1,1 write J^v = Jv x ^ Z ' ; i.e. jKx' = for any x efi 1 ^)-) FTR

9

241

83]

WEAK COMPACTNESS

Proof (a) E, being solid in L°, is Dedekind complete; so if A is relatively compact, the solid hull of A is relatively compact [82G], and for any UEE+ and e > 0 there is a w0 e F+ such that (\w\ — \w\ Awo,u} ^ e V we A. Now w0 XUEL1, so there is a voeS(W)+ such that ||#0~^oXM\\i ^ e> and now there is an ae 31/ and an a > 0 such that #0 ^ a^a. Let He^ be such that H' = a, and let 5 = e/a. Now suppose that K e 2 and that *. Then for any we A, JK\uxw\ ^ jKuxw0+jux

(\w\ — |

= aju,(H n K) + 2e < 3e.

Thus the condition is satisfied. (b) On the other hand, suppose that the condition is satisfied. Suppose that <^n>neN is any disjoint sequence in £ + , bounded above by u. Let e > 0. Let He^ and 5 > 0 be such that fg-l^x w| ^ e whenever WE A, KEH and /i(H f]K) < 8. For each UEN, choose an xnEM(L)+ such that x'n = ^w, and let Kn = {t: xn(t) > 0}. Then fi(Km n ^ n ) = 0 whenever m + n, SO

S m 6 l / * ( * m H H) ^ flH < CO,

and there is an n0 such that /^(^% fl £T) ^ 8 for every ^ ^ n0. Now, for 'W ^ 'W

|<^w,ti?>| <
WEAK COMPACTNESS IN L-SPACES

[83

(ii) A is bounded for %S(E,K) and, whenever (xn}nelS( is a normbounded disjoint sequence in K+, (iii) A is bounded for %S(E,K) and, whenever {Gn}neti is a disjoint sequence of open sets in X and (g(n)}neJS[ is a sequence in B, (iv) A is bounded for £ S (E,K) and, whenever (Fn)neIi is & disjoint sequence of compact sets in X and (g(n))neN is a sequence in JS, Proof (a) Observe that K and || || ^ obviously satisfy the conditions of 83B. So (ii) => (i) is immediate. (b) (i) => (iv) Of course A is bounded for %8(E, K). Let neN be a disjoint sequence of compact sets in X, let {g(n)}neIS be a sequence in B, and let e > 0. By 83A there is an h e E+ such that

I I/I-I/I A A|| < e

WfeA.

Now for any compact set F ^ X, there certainly exists an x eK such that xF ^ ^ , a s l is locally compact [A2Eb]. So we see from 73F that for a n y / e £ +

lifF = inf{fx:xeK,

XF

<«

Nowif/ejB, ^ = inf {fx\xeK, x^<: x ^

e + inf {(/A A) («) :xeK 5 XF < x < Xx}

[since ||/—/A A|| < e, so/x < e + (/A h)x whenever 0 < x

< e + inf{hx:xeK, xF < « But we can see that, because (J^>neN is disjoint, 2»eN/^

[73E].

So there is an n0 such that fihFn ^ e for every w ^ n0. Now however

As e is arbitrary, (c) (iv) => (iii) For each WGN, choose a compact set JPW C Gn such that ^ ( ^ - ^ ^ ^(n)^n- 2 ~ w 5 tm " s i s possible because /^ (n) is a Radon 9-2

243

83]

WEAK COMPACTNESS

measure and /i^n)Gn ^ ju>g(n)X < oo. Now (Fn)nelS[ is disjoint, so (pg(n)Fn)nex -> 0, and it follows at once that <Jigin)Gn)ne^ -> 0. (d) (iii) => (ii) If (%n}neii is disjoint and norm-bounded in K+, set <* = sup neN \\zjv. For each neN, let Gw = {*: a^p) > 0}; then <w6N is a disjoint sequence of open sets in X. For each n e N, choose an/ w e .4 such that | |

and set g(n) = |/ n | eB. Then

But

(^gi^Gn)nejx ~> 0 by hypothesis, so <sup/eJ[ |/^ n |> neN -> 0.

*83K Sequentially order-continuous linear functionals and countably additive functionals Let E be any Archimedean Riesz space. Let £ 7 be the space of linear functionals / : E -> R such that (fxn}ne1S[ -> 0 whenever (xn}ne1x^ 0 in E. In the results below, which can be taken as a set of exercises, I show how the methods of this book may be used to establish many properties of E~, along the same lines as those already developed for £ x . (a) Let us call a linear space topology % on E sequentially Lebesgue if (xn)ne1sl -> 0 whenever (xn)neI[ 10 in E. (i) In this case, if (xn}nelSi 10 and U is any neighbourhood of 0, there is an n e N such that [0,#w] ^ JJ [cf. 24Bc]. (ii) Because E is Archimedean, order-bounded sets are bounded [cf. 24Db]. (b) (i) £ 7 c £~ [use (a) (ii) above]; (ii) E~ is a Riesz subspace of £~ [use (a) (i)]; (iii) E~ is a solid linear subspace of £~; (iv) £^ is a band in £~ [cf. 16H]; (v) £^ is the set of those linear functionals on £ which are expressible as the difference of two sequentially order-continuous increasing linear functionals. [See LUXEMBURG & ZAANEN B.F.S., note vi, §20.] (c) A subset of £ ^ which is relatively countably compact for S£s(£~, £) is relatively compact for %a(E?, £) [cf. 82B]. (d) If A c £7, then Y(A) is relatively compact for %(E~, E) iSA is relatively S£s(£~, £)-compact and £(S|(£~, £)-bounded [cf. 82C]. (e) If A is a non-empty subset of £^, then these are equivalent: (i) the solid convex hull of A is relatively compact for %S(E~, £); 244

WEAK COMPACTNESS IN L-SPACES

[83

(ii){|/| :fsA) is relatively countably compact for %S(E~, E); (iii) A is bounded for %s( E~, E) and, whenever

N ->0

[cf. 82E; see also 8XC]. (f) If E is Dedekind cr-complete and A is a non-empty subset of E~, then these are equivalent: (i) A is relatively countably compact for %S(E~, E); (ii) the solid convex hull of A is relatively compact for %S(E~, JE); (iii) A is bounded for %8(E?, E) and, whenever (xn)neV is a disjoint sequence in E+ which is bounded above in E, (iv) for any x e E+ and e > 0, there is an / 0 e E~+ such that for every feA [cf. 82G]. (g) If J57 is Dedekind cr-complete, then E~ is sequentially complete for£ s (E~ E)[tf.82J]. (h) If 91 is a Boolean algebra, then 5(9l)~ may be identified with the space of bounded countably additive functionals on 9t, and every member of 5(91)^ extends uniquely to a member of L°°(9l)~, so that iS(9t)~ can be identified with L°°(9l)~. With its natural norm, 5(9l)~ is an £-space [cf. 44B, 42P, 83E proof, part (b)]. (i) Let 91 be a Dedekind c-complete Boolean algebra, let S = 5(91), and let H = S~ = L°°(9l)~, the space of all countably additive functionals on 91 [(h) above, 42Q]. (i) If A c His bounded for %8(H, 5), it is bounded for the norm of H [cf. 83G]. (ii) If A s= H is relatively compact for %S(H, S), it is relatively compact for %8{H, H) [cf. 83H]. *(j) Let (X, S, fi) be the measure space consisting of R with Lebesgue measure. Then S 1 ^)^ is isomorphic, as Riesz space, to s 0 © I/°°(9l), where s0 is the space of functions on X zero except at finitely many points, and 91 is the measure algebra of (X, 2,/^). While S1(/^)x £ s0. 83L Exercises (a) Let X be any set, and let s be the space of functions in Kx taking only finitely many values [cf. 4XCa]. Let A be a subset of I1 = V-{X). Show that A is relatively countably compact for %8(P-,s)iSA is relatively compact for the norm on P(X). [Hint: use 83H and 83A.] *(b) Use the argument of KOTHE, 22.4.2-3, to prove (a) above when X — N. Now use this to prove Orlicz' theorem [A1H]. 245

83]

WEAK COMPACTNESS

*(c) Let (G, || ||) be a normed Riesz space satisfying the conditions of 83B. Show that, whenever A c G+ is non-empty and bounded above, where B is the set of upper bounds of A. Show that Theorem 83B applies to any Riesz space with a Riesz norm satisfying this condition. [See NAKANO C.A.C.R.]

*(d) Let (G, || ||) be a normed Riesz satisfying the conditions of 83B. (i) Show that the linear space G © R can be ordered by saying that x + ae = (x,oc) ^ 0 o \\x~\\ < a, to produce an Archimedean Riesz space Gx in which e is an order unit, and that the norm on G is now that induced by the norm || ||e on Gv (ii) Show that if X is the set of all Riesz homomorphisms/: G -> R such that 11/11 = 1, then G is isomorphic, as normed Riesz space, to a truncated Riesz subspace of Notes and comments The rather curious conditions of the basic theorem 83B are an attempt to find a common abstraction of the spaces 5(91) and K(X), when 21 does not have a 1 or X is not compact, respectively. Another is offered in 83Lc. Just as in Chapter 7, but apparently for different reasons, we find ourselves dealing with truncated Riesz subspaces of M-spaces with units [see 83Ld]. In effect, 83B is a version of 81H in a situation in which we want to discuss disjoint sequences in G+ which are order-bounded, not in G itself, but in G". It would also be possible to prove 83B by adapting the argument of81F/81H. It is now easy, using the technical device in part (c) of the proof of 83C, to establish the important remaining criteria that have been given for weak compactness in Z-spaces [83F, 83H, 83J]. Note that 83H can be proved by rather more direct arguments, since the ring 21 is here an algebra, and 83B tells us no more than 81H does. Indeed, it can be derived from the work of § 82, since we are examining the dual pair (Sx, S) rather than (S~, S). Note that for ^-spaces we have an equivalence between weak and norm compactness [83La]. KOTHE [§30.6] generalizes this to other perfect sequence spaces. Another generalization of 83H is given by MOORE & REBER. 83J, however, seems to lie deeper. The substantial link in the chain is (ii) => (i); this is due essentially to GROTHENDIECK A.L. and is the progenitor of 83B, 81H and even 24H. A number of authors have regarded this theorem as a criterion for weak compactness in spaces of 246

WEAK COMPACTNESS IN L-SPACES

[83

regular measures, divorcing them from spaces of continuous functions, and thereby obtaining generalizations to spaces X which are not locally compact; see, for instance, GANSSLER. 831 is a version of a result due to DIETJDONNE, who points out that a set A in F ~ Ex is relatively weakly compact i&{wxu:weA} is relatively weakly compact in L1 for every ue E+; this is because F is complete for %\sl{F, E) [81D], and therefore can be thought of as a closed subspace of a power of L 1 [cf. 8lib]. *In 83K I try to show how the same arguments that I have used to study Ex can be applied to E~. The point about E~ is that if E is actually a subspace of R x , then E~ may be more interesting than Ex [see 83Kj]. In view of the persistent appearance of sequences in § 82, it is not surprising that many of these results have forms applying to E~ 8X Examples for Chapter 8 The following three examples show how relatively familiar spaces afford examples of several phenomena relevant to the results of § 82. In 8XA we have a Riesz space E which is not uniformly complete; consequently there can be a set i c Ex such that A is relatively compact for %8(EX, E), but its convex hull is not. In 8XB we have a Riesz space E which is uniformly complete but not Dedekind (T-complete, so there can be a set A c: E such that A is relatively compact for %S(EX, E) but its solid hull is not. Finally, in 8XC, we have a space E such that the topology %k(Ex, E) is not complete; this shows that 82B and 82C cannot be deduced directly from Eberlein's and Krein's theorems. *At the same time we find that weakly compact sets in E~ behave a little differently from those in Ex. 8XA Let E be the subspace of /°°(N) consisting of sequences which are eventually constant. Then £ is a Riesz subspace of /^(N); it has an order unit e = ^N and || ||e = || H^ on E. E is isomorphic to 5(31) where

« = {/:/ c N, / is finite or N \ / is finite}.

We may identify Ex with /X(N) [use 65B or 4XCb]. Now let us define (fn}ne$ in Ex by writing

fnx = 2n+1(x(n)-x(n+l))

V xeE, neN.

X

Then n6H -> 0 for %S(E , E), so A = {fn:neN} is relatively compact for £ S (E X ,E). But if, for each rceN, gn = Si<w2"(<+1)/i» t h e n gnx = x(0) — x(n) for every x e E, so {gn}nejn has no cluster point in Ex; 247

8X]

WEAK COMPACTNESS

thus the convex hull of A is not relatively compact for %S(EX, E). We note that, of course, A is not bounded for %]8](EX, E) [82C]. This can happen only because E is not uniformly complete [81G]. 8XB Let E be the subspace of J°°(N) consisting of convergent sequences. Then E is a || || ^-closed Riesz subspace of /^(N); with || || ^ it is an ikf-space with unit, isomorphic to C(Y), where Y is the onepoint compactification of the discrete space N. Ex can be identified with P — P(N) (because c0 is order-dense in E, and (co)x ~ P; or use 65B). Define a sequence n(=N in Ex by fnx = x(2n)-x(2n+l) V xeE,neN. Set A = {fn: n e N}. Then the convex hull of A is relatively compact for %S(EX, E). P We can see that neN -> 0 for %S(EX, E). Also, SneNan/n exists for %S(EX, E) for every sequence {ocn}neV in P. So if we define?7: P - > £ b y

7 is a linear map continuous for %8{l1, cQ) and %S(EX, E). But the unit ball B of I 1 is compact for %8(l\ c0), so T[B] is compact for £S(EX, E); now clearly I7[JB] is the closed convex hull of A. Q Thus we see that A is equicontinuous for %k(E, Ex). Now, for each neN, define xne E by #w(i) = 0 if i ^ 2%, ^ ( i ) = 1 if i > 2w. Then (xn}ne^ 0in £ but/ n x n = - 1 for every nsN. Thus £*(£, £x) is not Lebesgue. We can also see that, for each neN, f^ is given by /+(#) = x(2n) for every O;GE; SO we see that {fn}neTH has no cluster point in Ex; thus the solid hull of A is not relatively compact for %S(EX, £), and %k(E, Ex) is not locally solid, therefore not Lebesgue [82P]. We have here a linear space topology for which E' c Ex but which is not Lebesgue. We can see also that A satisfies the condition (i) of 82G, but neither of the others. 8XC Let X be an uncountable set and let E be the set of functions x: X -> R which are ' convergent at infinity', i.e. for which there exists a real number / 0 x such that, for every e > 0,

248

EXAMPLES FOR CHAPTER 8

[8X

is finite. Then E is a || H^-closed Riesz subspace of l^iX); it is an M-space with unit, isomorphic to C(Y), where Y is the one-point compactification of the discrete space X. K « W l 0 in E, then <||a?n|«>>»eH -> 0. P Let e > 0. For each neN, {t:xn(t) 4= foxn} is countable, so there is a teX such that xn(t) = foxn for every neN. Thus neN10 a n ( i there is an m e N such that / 0 # m < e. Now X o = {£: #m(£) > e} is finite, so there is an n ^ m such that #n(£) ^ e for every teX0. Clearly ||a^||oo ^ e for every i ^ n. As e is arbitrary, <||#Joo>7ieN -> 0. Q As in 8XB above, E x can be identified with l\X). The function f0: E -> R is a Riesz homomorphism, and E~ = E' = E x ® Hn{/0}, since E ^ co(X) © R as Banach space. If A = {fifeE*, I/I < 1}, then A satisfies (iii) of 82H but neither (i) nor (ii), since / 0 is in the closure of A for %S(E~, JS). -4 is closed for %S(EX, E), and for every xe E there is a goeA such that

since x must attain its bounds on X. But J[ is not compact for %S(EX, E). By James' theorem [A1F], E x cannot be complete for ^ ( E x , E). *Observe that E~ = E~, since the norm topology on E is' sequentially Lebesgue' [83Ka]. The unit ball of E~ is therefore compact for %S(E~, E), but it does not satisfy either (iii) or (iv) of 83Kf; thus these cannot be added to the three conditions of 83Ke. Further reading for Chapter 8 The earliest version of 82E/82G in an abstract context which I have been able to trace seems to have been in NAKANO M.S.O.L.S.; in the abridged version [NAKANO L.L., theorems 28.7-28.10] I find (a) a proof of 82G(i) o 82G(iii) when Eis a perfect Riesz space and E x has the countable sup property (b) a proof that, if E is perfect, then 82E(iii) is a necessary and sufficient condition for relative weak compactness, AMEMIYA O.T.L.S. proves 82G for Dedekind complete spaces E, and gives some further references. Extensions of these ideas may be found in FREMLIN A.K.S.I. and MOORE & REBER. Many of the important Riesz spaces (notably the Lp and Orlicz spaces [6XF-6XH]) have symmetry properties which mean that we can say much more about their weakly compact sets; see GARLING s.s.s. and FREMLIN S.S. ; also GARLING I.O. and PHUONGc l c for yet further elaborations of these ideas. For extensions of Grothendieck's theorem [83 J], see GROTHENDIECK A.L., SEEVER, WELLS, TOPS0E C.S.M., GANSSLER, a n d SCHAEFER W.C.

249

Appendix Al Linear topological spaces This book has been written on the assumption that readers will be acquainted with the elementary theory of linear topological spaces. But it may be helpful if I set out explicitly some of the results I use, though the following list is far from complete. In general I give references rather than proofs. As in the rest of the book, all linear spaces are over the field R of real numbers. A1A Notation Let E and F be real linear spaces in duality, i.e. endowed with a bilinear functional < , ) : £ x F - > R . The fundamental topologies on E associated with this duality I denote as follows: %S{E, F) is the weak topology, the topology of uniform convergence on finite subsets of F; %C(E, F) is the Mackey topology, the topology of uniform convergence on convex %S(F, E)-compact sets; %b(E, F) is the strong topology, the topology of uniform convergence on %S(F, £)-bounded sets. A1B Mackey's theorem Let E and F be real linear spaces in duality, such that E separates the points of F. Let % be a locally convex linear space topology on E. Then the dual of E for % can be identified with FiS %S(E, F ) c X c %k(E, F). Proof

See BOTTKBAKI V, chapter iv, § 2, no. 3, Theorem 2; KELLEY & 18.8; KOTHE 21.4.3; ROBERTSON & ROBERTSON, chapter in, §7, theorem 7; or SCHAEFER T.V.S. IV, 3.2. NAMIOKA

A1C Closed graph theorem (a) Let E and F be complete metrizable linear topological spaces. Let T: £-> F be a linear map such that the graph of T is closed in E x F. Then T is continuous. (b) Let £ be a linear space, and © and % complete metrizable linear space topologies on E such that © ^ %. Then © = %. Proof

BOURBAKI v, Chapter i, §3, no. 3; KELLEY SCHAEFER T.V.S., chapter in, §2; KOTHE §15.12.

250

& NAMIOKA §11;

LINEAR TOPOLOGIGAL SPACES

[Al

Remark I have stated this theorem without any assumption of local convexity, and this is the form in which I shall use it. But the reader who prefers to restrict himself to locally convex spaces will have no difficulty in seeing where to interpolate this hypothesis. AID Grothendieck's criterion Let E and F be real linear spaces in duality. Let 501 be a family of subsets of F, directed upwards and covering F, such that each member of 9Ji is balanced, convex and %S(F, £)-bounded. Let % be the topology on E of uniform convergence on members of 501. Then E is complete under % iff every geF* which is %8(F, E)-continuous on each member of 3K is represented by a point of E. PrOOf

KELLEY

& NAMIOKA

1 6 . 9 ; KOTHE

21.9.2;

ROBERTSON

&

chapter vi §1, Cor. 1 toTh. 3 (p. 107); SCHAEFERT.V.S., IV.6.2, Cor. 2.

ROBERTSON,

Remark I think that we need only the easier half of this result; that E is complete if the criterion is satisfied. A1E Corollary Let E be a Banach space. Then its dual E' is complete for the Mackey topology %k{E\ E). Proof %k(E\ E) is the topology of uniform convergence on the family 50i of balanced convex weakly compact sets in E. So suppose that g e 22* is weakly continuous on each member of 9ft. Let (xn}neI( -> 0 for the strong (norm) topology on E. Then A = {0}\j {xn:neN} is compact for the strong topology; because E is complete in the strong topology, the closed balanced convex hull B of A is compact. But now Beffll, so g is weakly continuous on B. As certainly (xn)nE^ -> 0 for %S(E, £'), (g%n}neK -> 0. But as E is metrizable this is enough to show that g is continuous, i.e. that geE'. By Grothendieck's criterion, E' must be complete. Remark

This result is of course true for all Frechet spaces E.

A1F James' theorem Let £ be a complete locally convex Hausdorff linear topological space. Suppose that B ^ E is closed for %S(E, Er) and has the property that every/e E' is bounded and attains its bounds on B, Then B is compact for %S(E, £'). Proof

PRYCE

w.c. 251

Al]

APPENDIX

A1G Krein's theorem Let E be a complete locally convex Hausdorff linear topological space. Suppose that A c: E is relatively compact for %8{E, £'). Then so is its convex hull T(A). Proof Let B be the closure of T(A) for %8(E, Ef). Let fe E'. Then because A, the %S(E, Enclosure of A, is %(E, E')-compact, / is bounded and attains its bounds on A. But clearly the bounds of/ on A are the same as its bounds on B, and A c B. Thus / i s bounded and attains its bounds on B. By James' theorem [AlP], B is weakly compact, i.e. F(A) is relatively weakly compact. Alternative proof

KOTHE

24.5.4.

A1H Proposition Let E and F be linear spaces in duality. Let {xn}neN be a sequence in E such that every subsequence is summable for %S(E, F). Then: (a) If A £ F is relatively countably compact for %8(F, E) (i.e. if every sequence in A has a cluster point in F for %8(F, E)), then infn6H8up ye ^|| = 0 .

(b) In fact,

l i m , ^ , sup yeA \ (xn, y) | = 0.

(o) <*n>n€H -* 0 for the Mackey topology £*(£, F). (d) {xn: nGN} is bounded for the strong topology %b(E, F). *(e) ('Orlicz' theorem') (xn}neN is summable for the topology X of uniform convergence on those sets of F which are relatively countably compact for %8(F9 E), and therefore is summable for %k(E, F). Proof

(a)

? Suppose otherwise. Then e = J i n f ^ s u p ^ \(xn,y}\ > 0.

Por each neN, choose a,yneA such that \(xn,yn)\ ^ e. Let y be any cluster point of <2/JneN for %a(F, E). Obtain a strictly increasing sequence (n(i)}ie$ in N inductively, as follows. Set n(0) = 0. Given n(k), let m e N be such that for any r > m, |<*r,y>| ^ 2-^+D,

\(xr,yi)\

< 2-(^D

V i ^ n(k),

which exists as certainly (xr}re^ -> 0 for %S(E, F). Let

n(k+l) > ma,x (m,n(k)) be such t h a t

|<^,2/ n ( fc+1 )-2/>| ^ 2~<*+1> V i ^ n(k),

which exists as y is a cluster point of (yr}re^> Continue. 252

LINEAR TOPOLOGIGAL SPACES

[Al

We see now that (n(i))ie$ is a strictly increasing sequence such that K*ntt),y>l < 2~*>

\<*n(kh y»«>>| < 2-*,

| < ^ ) ? 2/n(&)-2/>| ^ 2-*,

whenever 0 y}\ -

So, if i/' is any cluster point of (yn(k)}k^ for %8{E', E), \(z,y'-y)\>e. However, since, for any i e N, we must have which is impossible. X (b) This follows at once, because we can apply (a) to any subsequence of <#n>n6N(c) This follows trivially from (b). (d) ? Otherwise, let B c F be a %S{F, £)-bounded set such that su PneNsuP2/esK^n?2/)l = °°- For each^eN, choose a yneB such that Then sup neN a n = oo. Let (fin)nejn be a sequence converging to 0 in R such that neN does not converge to 0. Then (j3nyn)ne1S( -> 0 for %S(F, E), so G = {/ffnyn: neN} is relatively compact for %8(F, E). But SUpy6O|<»n»y>| > \finVn>\ = \*nPn\>

so by (b) above (ocnfin)nElil -> 0, which contradicts the choice of *(e) Leta: = 2WGN#nfor!£s(E, F), and for each WGN let 2n = Sifl sup y6 ^|| > 0. 253

Al]

APPENDIX

Let (n(k)}k€lx be a strictly increasing sequence in N such that svpyeA\<x-zn{k),y)\ >e

VieN,

and for each JceN choose a yke A such that Now let (Jc(i)}ieja be a strictly increasing sequence in N such that \
>e

V

^ N ,

which exists because {zn(r)}relx -> x for X8(E, F), and set w

i

=

for each i e N. Then every subsequence of (w^^ is summable, being merely a collection of terms of a subsequence of (Xj}jejz. So we may apply (a) above to the sequence (w^^ and find that 0 = infieHSupy^K^y)! ^ inf; e N |<X,^)>| > e, which is impossible. X Thus i n f H s u p s u p ^ | < a ; - 2 w , y > | = 0, and x is a limit of (zn)neN f° r ^^e topology of uniform convergence on A. As A is arbitrary, nen -> ^ for 2 , as required. Remark Orlicz' theorem is not used in this book. But it is very close to some of the results in Chapter 8 and it has other applications in neighbouring parts of advanced measure theory. A l l Corollary (the 'Banach-Mackey theorem') Let E and F be linear spaces in duality. Let A be a subset of E such that SieN^^i exists for %s(E, F) whenever (x^^ is a sequence in A and iGN is a non-negative sequence in R such that 2ieNa£ = *• (F° r instance, suppose that A is convex, bounded and complete for some linear space topology on £ finer than %a(E, F).) Then A is bounded for %b{E, F). Proof Observe that in fact SieN^^i exists for %S(E, F) whenever (xi)ieTi i s a sequence in A and ^eN is a non-negative summable sequence. So if <^X £ N is a sequence in A, every subsequence of <2~^> ieN is £ s (E,F)-summable. By AlHd, {2~ixi:ieN} is bounded for %b(E, F). As <^X6N is arbitrary, A is bounded for %b(E, F). Alternative proof 254

See SCHAEFER T.V.S. II. 8.5 or KOTHE 20.11.3.

LINEAR TOPOLOGIGAL SPACES

[Al

*A1J Exercise Use A l l to show (i) that if £ is any locally convex linear topological space, then a %8(E, £')-bounded set is %k(E, £')bounded, (ii) that if £ is a complete locally convex linear topological space, then a %S(E, £')-bounded set is %b(E, £')-bounded. A2 Spaces of continuous functions Some of the most important topological Riesz spaces are spaces of continuous real-valued functions. In this section I give some definitions and state a few basic properties in the language I have used in this book. As a supplement I show how a result that is normally proved using measure theory is amenable to more primitive techniques. A2A Definition Let X be a topological space. I shall write C(X) for the space of all continuous real-valued functions on X. C(X) is a Riesz subspace of Rx [1XD]. It is a uniformly complete Archimedean Riesz space [25Hb]. A2B Definition Let X be a topological space. I shall write Cb(X) for the space of all bounded continuous functions from X to R. It is an order-dense solid linear subspace of C(X) [A2A], and a norm-closed Riesz subspace of /°°(Z) [2XA]. Under the norm || || w it is an M-space with unit #X [26Ad]. A2C Definition Let X be a topological space. Then C0(X) will be the space of those functions x e C(X) ' convergent to 0 at oo', i.e. such that

{t:\x(t)\>e}

is compact for every e > 0. C0(X) is a norm-closed solid linear subspace of Cb(X), and in its own right is an M-space [26Ac]. A2D Definition Let X be a topological space. Then K(X) is the space of continuous functions' of compact support \ i.e. those functions x G C(X) such that {t:x{t)*O} is compact in X. K(X) is a solid linear subspace of CQ(X), so it is uniformly complete [25Ni]. Also, K(X) is dense in C0(X) for || H^. P Let x e C0(X) and e > 0. Set V = (xThen it is easy to verify that yeK(X) and that \y — x\(X) ^ e. Q 255

Al]

APPENDIX

*A2E Lemma Let X be a locally compact Hausdorff topological space. (a) Suppose that teX and that G is a neighbourhood of t. Then there is an xeK(X) such that x(t) = 1 and 0 < x ^ %G in R x . (b) If G ^ X is open and F c C? is compact, then there is an # e K(X) such that xF ^ a? ^ ##. Proof (a) Let £T be a neighbourhood of t such that i? is compact. Then i7is compact and HausdorfF, therefore normal, and by Urysohn's lemma there is a continuous function y: H -> R such that ?/(£) = 1 and 2/(w) = 0 for every u e i7\int (GftH). Now if we set 2/(w) = 0 for every ueX\H, we see that yeK(X) and that x = y+/\xX satisfies the conditions. (b) Let A = {y :yeK{X)9 0 ^ y ^ xG}. For each y sA, let By (a) above, we know that \JyeAGy = #, and of course {©^: ye A) f because ^. f. As J7 is compact, there is a 2/ 6 A such that 6?^ 2 F. Now set x = 2y A *A2F How to avoid the Riesz representation theorem One of the most important applications of Riesz' theorem [73D/E] in functional analysis is to the following lemma. I think it is therefore of interest that the result can be proved without any call on measure theory. Lemma Let I be a compact HausdorfF topological space. Let (xn)neTx ^e a sequence in C = C(X) which is bounded for || 1^ and such that (xn(t))nex -> 0 for every teX. Then (xn)nelSi-+0 for First proof By 25G, C = Cr. Now if feC~+, then by Riesz' theorem [72D/E] there is a Radon measure jionX such that fx = J xd/i

V xeC,

and [iX < oo. Now {xn}ne^ is bounded above and below in Sx(/^) by constant functions, and is convergent to 0 at each point, so by Lebesgue's theorem [63Md], A s / i s arbitrary, (xn)neI[ -> 0 for %S(C, C~) = %8(C, C). 256

SPACES OF CONTINUOUS FUNCTIONS

[A2

Second proof By 25G, C" = C~. L e t / G C ~ + and e > 0. For each m e N, consider the sequence ccm-2-™e. Now, for any k Set wr = infm^ra;m for every reN. Then <^r)reNl- Moreover, for any teX, wr(t) < zr(t) ^ sup i>r 1^(01 V reN, so (wr(t))reN^0. By Dini's theorem, there is an reN such that M l * =^ e, so that/(^ y ) ^ e||/||. Now for any i ^ r,

(as (1^1 - wr)+ =

swpm
^ e | | / | | + S m < r 2 - e < 6(||/||+2). As e is arbitrary, we see that ieN -> 0. As/ is arbitrary,
References The following list comprises only works referred to in the text. For a systematic bibliography of the literature on Riesz spaces, see LUXEMBURG & ZAANEN R.S. Amemiya, I. [G.S.T.] * A general spectral theory in semi-ordered linear spaces.' J. Fac. Sci. Hokkaido Univ. (1) 12 (1953) 111-56. [O.L.T.S.] 'On ordered topological linear spaces'. In Proceedings of the International Symposium on Linear Spaces, Israel Academy of the Sciences and Humanities, Jerusalem, 1961, pp. 14-23. Baker, K. A. 'Free vector lattices'. Ganad. J. Math. 20 (1968) 58-66. Bartle, R. G. The Elements of Integration. Wiley, 1966. Birkhoff, G. Lattice Theory. American Mathematical Society, 1967. Bourbaki, N. Elements de Mathematique. Hermann (Actualites Scientifiques et Industrielles). [i] Vol. I, Theorie des Ensembles, A.S.I. §1212 (chaps. 1-2) (1966); §1243 (chap. 3) (1963). Translated into English as Theory of Sets, Hermann & Addison-Wesley, 1968. [v] Vol. v, Espaces Vectoriels Topologiques, A.S.I. §1189 (chaps. 1-2) (1966); §1229 (chaps. 3-5) (1964). [vi] Vol. vi, Integration, A.S.I. §1175 (chaps. 1-4) (1965); §1244 (chap. 5) (1967); §1281 (chap. 6) (1959); §1306 (chaps. 7-8) (1963); §1343 (chap. 9) (1969). Chacon, R. V. & Krengel, U. 'Linear modulus of a linear operator'. Proc. Amer. Math. Soc. 15 (1964) 553-9. Dieudonne, J. 'Sur les espaces de Kothe'. J. Analyse Math. 1 (1951) 81-115. Dunford, N. & Schwartz, J. T. Linear Operators I. Interscience, 1958. Fremlin, D. H. [A.K.S. I] 'Abstract Kothe spaces, i \ Proc. Cambridge Philos. 63 (1967) 653-60. [A.K.S. II] 'Abstract Kothe spaces, n ' . Proc. Cambridge Philos. Soc. 63 (1967) 951-6. [A.K.S. in] 'Abstract Kothe spaces, in'. Proc. Cambridge Philos. Soc. 63 (1967) 957-62. [s.s.] 'Stable subspaces of L1 + Lco\ Proc. Cambridge Philos. Soc. 64 (1968) 625-43. Ganssler, P. 'Compactness and sequential compactness in spaces of measures'. Z. Wahrscheinlichkeitstheorie und Verw. Gebiete 17 (1971) 124-46. 'A convergence theorem for measures in regular Hausdorff spaces'. Math. Scand. 29 (1971) 237-44. Garling, D. J. H. [s.s.s.] 'On symmetric sequence spaces'. Proc. London Math. Soc. (3) 16 (1966) 85-106. [i.o.] 'On ideals of operators in Hilbert space'. Proc. London Math. Soc. (3) 17 (1967) 115-38. Grothendieck, A. [c.c] 'Criteres de compacite* dans les espaces fonctionnels generaux'. Amer. J. Math. 74 (1952) 168-86. [A.L.] 'Sur les applications line*aires faiblement compactes d'espaces du type C{K)\ Canad. J. Math. 5 (1953) 129-73. 258

REFERENCES Halmos, P. R. [M.T.] Measure Theory. Van Nostrand, 1950. [N.S.T.] Naive Set Theory. Van Nostrand, 1960. [L.B.A.] Lectures on Boolean Algebras. Van Nostrand, 1963. Hardy, G. H., Littlewood, J. E. & Polya, G. Inequalities. C.U.P., 1934. Hewitt, E. & Stromberg, K. Real and Abstract Analysis. Springer, 1969. Ionescu Tulcea, A. & Ionescu Tulcea, C. Topics in the Theory of Lifting. Springer, 1969. Jameson, G. Ordered Linear Spaces. Springer, 1970. Kakutani, S. * Concrete representation of abstract X-spaces and the mean ergodic theorem'. Ann. of Math. 42 (1941) 523-37. Kelley, J. L. General Topology. Van Nostrand, 1955. Kelley, J. L., Namioka, I. et al. Linear Topological Spaces. Van Nostrand, 1963. Kothe, G. Topologische Lineare Relume I. Springer, 1960. Translated into English as Topological Vector Spaces I, Springer, 1969. Krengel, U. 'Uber den Absolutbetrag stetiger linear Operatoren und seine Anwendung auf ergodische Zerlegunger'. Math. Scand. 13 (1963) 151-87. Loomis, L. H. An Introduction to Harmonic Analysis. Van Nostrand, 1953. Luxemburg, W. A. J. 'Is every integral normal?' Bull. Amer. Math. Soc. 73 (1967) 685-8. Luxemburg, W. A. J. & Zaanen, A. C. [B.F.S.] 'Notes on Banach function spaces'. Nederl. Acad. Wetensch. Proc. (A): notes VI-VII, 66 (1963) 655-81; notes viii-xm, 67 (1964) 104-19, 360-76, 493-543; notes XIV-XVI, 68 (1965) 229-48, 415-46, 646-67. Also printed in Indag.Math. 25-7 (1963-5). [R.S.] Riesz Spaces I. North-Holland, 1971. Maharam, D. 'On homogeneous measure algebras'. Proc. Nat. Acad. Sci. U.S.A. 28 (1942) 108-11. McShane, E. J. ' Order-preserving maps and integration processes.' Ann. of Math. Studies 31, Princeton U.P., 1953. Meyer, P. A. Probability and Potentials. Blaisdell, 1966. Moore, L. C. & Reber, J. C. 'Mackey topologies which are locally convex Riesz topologies'. Duke Math. J. 39 (1972) 105-19. Munroe, M. E. Introduction to Measure and Integration. Addison-Wesley, 1953. Nakano, H. [C.A.C.B.] 'Uber die Charakterisierung des allgemeinen (7-Raumes'. Proc. Imperial Acad. Japan 17 (1941) 301-7. Reprinted in NAKANO S.O.L.S. [M.S.OX.S.] Modulared Semi-Ordered Linear Spaces. Tokyo, 1950. Reprinted in an abridged form as NAKANO L.L. [L.T.] 'Linear topologies on semi-ordered linear spaces'. J. Fac. Sci. Hokkaido Univ. (I) 12 (1953) 87-104. [S.O.L.S.] Semi-Ordered Linear Spaces. Maruzen, Tokyo, 1955. [L.L.] Linear Lattices. Wayne State University Press, 1966. Peressini, A. L. Ordered Topological Vector Spaces. Harper & Row, 1967. Phuong-Cac, N. 'Generalized Kothe function spaces, i \ Proc. CambridgePhilos. Soc. 65 (1969) 601-11. Pryce, J. D. [w. c ] 'Weak compactness in locally convex spaces'. Proc. Amer. Math. Soc. 17 (1966) 148-55. [U.S.] ' An unpleasant set in a non-locally-convex vector lattice'. Proc. Edinburgh Math. Soc. (2) 18 (1973) 229-33. Riesz, F. 'Sur les operations fonctionnelles lineaires'. C. R. Acad. Sci. Paris 149 (1909) 974-9. Robertson, A. P. & Robertson, W. Topological Vector Spaces. C.U.P., 1964. Royden, H. L. Real Analysis. Macmillan, 1963. 259

REFERENCES Schaefer, H. H. [T.V.S.] Topological Vector Spaces. Springer, 1966. (Paperback edition 1971.) [w.c] 'Weak convergence of measures'. Math. Ann. 193 (1971) 57-64. Schwartz, L. Radon Measures. To appear. Seever, G. L. 'Measures on 2^-spaces'. Trans. Amer. Math. Soc. 133 (1968) 267-80. Semadeni, Z. Banach Spaces of Continuous Functions I. Polish Scientific Publishers, 1971. Sikorski, R. Boolean Algebras. Springer, 1964. Taylor, A. E. Introduction to Functional Analysis. Wiley, 1958. Topsoe, F. [T.M.] Topology and Measure. Springer, 1970. [C.S.M.] ' Compactness in spaces of measures \ Studia Math. 36 (1970) 195-212. Vulikh, B. Z. Introduction to the Theory of Partially Ordered Spaces. WoltersNoordhoff, 1967. Wells, B. B. 'Weak compactness of measures'. Proc. Amer. Math. Soc. 20 (1969) 124-30. Widom, H. Lectures on Measure and Integration. Van Nostrand Reinhold, 1969. Williamson, J. H. Lebesgue Integration. Holt, Rinehart & Winston, 1962. Wright, J. D. M. 'Stone-algebra-valued measures and integrals'. Proc. London Math. Soc. (3) 19 (1969) 107-22. 'The measure extension problem for vector lattices'. Ann. Inst. Fourier (Grenoble) 21 (1971) 65-85. 'Vector lattice measures on locally compact spaces'. Math. Z. 120 (1971) 193-203. [E.T.] 'An extension theorem'. To appear in J. London Math. Soc. Zaanen, A. C. Linear Analysis. North-Holland, 1953.

260

Index of special symbols References are to section numbers; those in bold type indicate definitions, those in italic passing references b c

d

f

i k

I

see strong topology on a linear space (A1A) C 8XB, 8XC, A2A, A2F-G 2XC, 2XG, 4XCb 4XF1, A2B cb 43Ec, §43 notes, 73E, Co 83J, A2C E~ see dual spaces (83K) 3) HE W §51 (51Ba) If 61Fa ^ see filters of sections (11C) J HE K 73D, 83J, A2D, A2E %k see Mackey topologies (A1A) I1 2XB, 2XC, 2XF, 2XG, 4XC, 83La I2 2XE 2XA, 2XB, 2XF, 4XCa 1 L §52 (52E), 53B, 54D-F, 5XB-C,§63,64B,6XC5 83F

L2 6XG Lm §43 (43A), 44Ca, §45, §4Z, 51Gd, 51Ge, 52G, 53B, 62H, 62Mb, 63G-H, 64B Lv 6XF, 6X1 LP §65 (65C), §6X L° chap. 6 (62F)

m P

r s

t

45Ge, 4XC, 52^1, 52E-F, 52G, 52Hc, 54E, 5XB, 5XC, §83 L~ §16(16B), 18Eb,25Ld, 26E et seq., 33Ga9 42Rd Lx 16G, 16H, 18Ec, 25Ld, 26G, 33Ga, 42Rd S1 §63 (63B) M §62 (62Fa) Pn 54F, 5XC, 6XC see power set (4XA) 4XCb A2Ga R 12A;R^1XD, 6XAa9 6XD *o 4XOb, 6XD S §42 (42B), 43Bb, 43Ea, §45, 4XB, 4XC, 51Gd, 51Ge, 52A, 8XA % see weak topologies on linear spaces (A1A) I see under % (T see main index, after s see strong topology on a b linear space (A1A) 3: fc see Mackey topologies (A1A) A2Ga % see weak topologies on linear spaces (A1A) §81 (81A), A2Gb %8\

%l

x

0 see L°; c 0 ; C o ;

261

INDEX OF SPECIAL SYMBOLS 1 (in a Boolean algebra) 41G

see L1; P; S 1 ; || ||x

2 566 L 2 ; / 2 0 0 51A see also /°°, L00, || ^ < , ^ , > , ^ 11A | |

(\x\) 14A

(|/|, |T|) 16Ea, 16F (%\s\) see under %

26Ad, 2XA I 1 (||(|| p|| e) 25Ha, 2XA, 4£Bc, 42i?6, 4356, 43C, 52G, 6XE V

1/ v^JC/, u i u j

D.A.jQi

(|| I,) 6XF, 6X1 + (x+) 14A (/+, T+) 16Da (v+) 42H - (*-) 14A j 26C, 44Ce, 63B (/ E ) 631, 831 t (i?t, £fa) 11C (^(Bi)) see filters of sections

262

(11C) ]r see f above [ , ] 11F <] 41J A 4XA * (£*) see dual spaces x (£ x ) see dual spaces (Lx) <se6 under L ~ (E~, E'Z) see dual spaces (L~) see under L ' (£') see dual spaces (£') 45E et seq., 4XH, 52^c, 5XC, 6XC A

(EA) 566 completion of a linear topological space (ft) see evaluation map (31Ae) (0) §42 (42D) (n) §45 (45B), 4XH, §54, 5XC, 62L, 62Mf, 6XC <=, u, 2 5 n, \ 41F-G

Index References are to section numbers; those in bold type indicate definitions, those in italic passing references absolutely continuous additive functional, 42Rh, 42Ri, 63Mf additive function on a Boolean ring, §42 (42A), 44Ca algebra, see Boolean algebra almost everywhere, 61 Fe Amemiya's representation theorem, §14 notes Archimedean Riesz spaces, §15 (15A), 21Ba, 25Nh associate of an extended Fatou norm, 65D, 65F, 6XE, 6XF atom in a Boolean ring, 6XBa

metrizable, §25, §26, §2X, 6XBc, 6X1, 81Ie, A1C, A1E complete measure spaces, 61Ff, 61Jd, 641, 64Ja, 71A, 72A, 73A completely additive functionals on Boolean rings, 42K, 42L, 42Rj, 44B, 51D, 51E, 52C, 63J completely regular topological spaces, 73F completion of a linear topological space, 22F, 24Lf convergence in measure, 63K, 63L, 63Mj, 63Mk, 64E, 64Jc, 64Je, 6XAa, 6XBc countable magnitude Baire measure, 7XC measure algebras, 53Fa, 64Jc Baire sets in a topological space, §73 measure spaces, 64Gb, 64Ha, 64Jc-e countable sup property for Riesz spaces, notes Banach function spaces, § 65, 6XD et seq. §18 (18A), 24Lc, 25Lc, 2XA, 51Oe, Banach lattices, §26 (26Aa), §2X, 65E et 64Jc countably additive functionals on Boolean seq. Banach-Mackey theorem, A l l rings, 42M et seq., 44Cb, 4XG, 51D, band in a Riesz space, 14F, 15F, 22Ec 52Ha, 61Jb, 63J, 64Jd Boolean countably compact, see relatively countalgebra, 41G, 43Ea ably compact sets (82B) ring, chap. 4 (41 A), chap. 5, §83 Borel decomposable measure spaces, 64Ga et seq., 72B measure, 73Gi, 7XC set, 73Gi decreasing sequence, 11D bounded sets in linear topological spaces, Dedekind complete 81F, 810, 830, 83Ki, All, A1J Boolean ring, 43Da, §43 notes, 4XA, see also locally bounded additive 53A, 64A lattice, 13Ba, 13C functionals Riesz space, §16, 1XD, 23K, 24E, boundedly order complete linear space topology, §23 notes 43Da, §43 notes, 53B, 64B boundedness property for ordered linear see also Dedekind (7-complete below Dedekind er-complete spaces, 23Ne, 23Nf Boolean ring, 411, 42Q, 43Db, 52Ha, closed graph theorem, A1C 53D, 611, 830, 83H, 83Ki lattice, 13Bb, 13Ca compact sets in linear topological spaces, Riesz space, 15B, 25J, 43Db, 62G, 21Be, 26Hf, 2XG, 8IF, §82, §83, A1F, A1G 82G, 82H, 82J, 83Kf, 83Kg compatible linear space topologies, §21 determined, see locally determined measure spaces (64Gc) (21A), 22Ea complete linear space topologies, 23K, diffuse measure space, 6XBa directed set, 11G 81C, 81D, A1D-A1G

263

INDEX direct sum of measure spaces, 61G, 61H, 61Jd, 64Fb, 64Ga, 64Jf disjoint set and function, 13G distributive lattice, 13D, 14D, 41La dominated convergence, see Lebesgue's Dominated Convergence theorem dual spaces £*, 31Aa E~, 16B, 23M, 25E, 25G, 25M, 26G, 2XB, 2XC, §31, 33D, 42H, 65H, 6XBb, §81 E', 22D, 22Ga, 23Nh, 24F, 25E, 25G, 25M, 25Nd, 26C, 26D, 2XB, 2XC, 31Ad, 65H, 6XBc, 81B E x , 16G, 17Gb9 23M, 25M, 26C, 2XA-C, §§32-33, 42L, 44A, 4XFj, 52Hb, 65A-B, 65F, 6XD, 81D, §82, 831, §8X E~, 42P, 44Cb, 83K, 8XG evaluation map, chap. 3 (31Ae) extended Fatounorm, §65 (65C), 6XE-H extension of measures, 64Ja, 64Jb extremally disconnected topological spaces, §43 notes Fatou's lemma, 26Hk, 63Mc Fatou norm, 23A, §26, 42Re, 43Cf 65E, 651a pseudo-norm, §23 (23A), 63K, 6X1 semi norm, 23A, 23Na, 25Ha topology, §23 (23A), 24C, 24J see also extended Fatou norm filters, 23E of sections, 11G, 21Bd, 21Be, 24Bd, 24Da, 24H finite magnitude measure algebra, 53C, 53D measure space, 61Fg, 63Mg9 63Mkt 64Fa, 64Ga Fubini's theorem, 6XC Grothendieck's criterion for completeness, AID Hahn decomposition theorem, 52Ha, §63 notes Holder's inequality, 6XF ideals of Boolean rings, 41J of Reisz spaces, see solid linear subspaces increasing additive function, 42F function, 11D linear map, 14Ea, 25D, 25Nc, 25Ne inf, infimum, 11B

264

inner regular, see regular integrable functions, §63 (63B) integral, 26C, §63 (63B), chap. 7 (71G) integration, §63 inverse -measure -preserving functions, 61E, 61Jc, 63Mi, 6XC James' theorem, A IF Kakutani's theorems on representation of L-spaces, §26 notes on representation of measure algebras, 611 Krein's theorem, A1G, A2G lattice homomorphisms, 13E, 13Hc, 14Eb, 45A lattice operations, 13A, 22Bb, §22 notes, 25Nf lattices, §13 (13A), UA9 41F Lebesgue measure, 6XAb, 6XB, 6XG, 7XA Lebesgue's Dominated Convergence Theorem, 24Lg, 26H1, 63Md Lefcesgue topologies, §24 (24A), 25L, 25M, 26B, 26Ha, 26Hh, 2XB, 2XC, 2XF, 2XG, 33E, 63K, 65H, 651a, 6XF, 6X1, 81C, 81D, 82F, 82H, 8XB see also sequentially Lebesgue linear space topologies Levi's theorem, 26Hj, 63Ma Levi topologies, §23 (231), 25Ha, 25Nb, 25Nc, §26, §2X, §33, 64E, 65E, 6X1, 81C, 8ID, 81Id lexicographic ordering on R2, 1XF linear maps, 14E, §16, 42C linear topological spaces, chap. 2, §A1 localizable measure spaces, see Maharam measure space (64A); decomposable measure spaces (64Ga) locally bounded additive functionals, §42 (42G) locally compact topological spaces, 73D, 73E, A2E locally convex linear space topologies, 22Gc, 23Na, 24G, 24K, 25G, 25M, §26, §A1 locally determined measure spaces, §64 (64Gc), 71 A, 71G, 72A, 73A locally order-complete linear space topology, §23 notes locally order-dense Riesz sub space, 14F, §17 locally solid topologies, §22 (22A), 23A, 23Nh, 24C, 24Ld, 24Lf, 25F, 25G, 25K, 26E, 82F

INDEX JD-spaces, §26 (26Ab), 2XB, 33F, 44B, 811a, 811b, 82Lc, §83 see also L1, L# Mackey's theorem, A IB Maokey topologies, 81If, 82F, 82H, 8XB, 8XC, A1A, A1B, A1E, A1H magnitude, see countable magnitude; finite magnitude Maharam algebra, §53 (53A), 64A, 64Jg measure space, §64 (64A), §65 measurable function, §62 (62A) set, 61Bc measure, 51A, 61Bc algebra, 53A, 61De, 61H -preserving ring homomorphisms, §54 (54A), 5XC, 61E, 61Je, 6XC rings, chap. 5 (51A), 83F spaces, chap. 6 (61 A), 71A see also convergence in measure; extension of measures; quasi-Radon measure space; Radon measure space metrizable linear space topologies, 22Oe, 24Lc, §25 see also under complete linear space topologies M-spaces, §26 (26Ac), 2XC, 43A, 83B, A2C with unit, §26 (26Ad), 2XA, 53B, A2B multiplication in L°, 621, 62Mb, 62Mc Nakano's theorem, 23K, §23 notes, 53Fd, 64E negligible set, 61Fd normal subspace of a Riesz space, see band (14F)

order topology, HHa, IXC, 7X0 order unit, 141, 25Ha, 26Ad, 43Ea see also weak order unit Orlicz spaces, 6XH Orlicz' theorem, 83Lb, AIHe partially ordered linear spaces, §12 (12A) sets, §11 (HA) perfect Riesz spaces, 2XA, 2XBf §33 (33A), 65G, 81Ic-e, 82K positive cone, 12B duality, 81 A, 8 IB linear map, 14Ea power set, 4XA, 4XCa p.p. ('presque partoutf), 61Fe products of Boolean rings, 4IK, 45Ge of lattices, 13H of Maharam algebras, 53E, 53Fb, 53Fc of measure algebras, 61H of measure rings, 5IF of partially ordered sets, 11G of Riesz spaces, 14H, 15Ha, 1XD Prokhorof's theorem, 73Ge purely infinite measurable sets, 61 Fc, 63D quasi-Radon measure space, §72 (72A), 73B, 73C quotient Boolean rings, 41J, 41Ld9 45Dc, 45Gb, 4XE, 4XF, 61D Riesz spaces, 14G, 14Lb, 1XE, 25Nh, 25Nk, 45Dc, 62F Radon measure space, §73 (73A) see also quasi-Radon measure space Radon-Nikodym theorem, 63 J, §63 notes, 64Jd regular measures, 73Gi, 7XC open sets, §43 notes, 4XF see also completely regular topological spaces regularly embedded subspace, 17A relatively countably compact sets, 82B, 82E, 820, 83K, 83La, A1H relatively uniformly closed sets, 25Nj*

Ogasawara's representation theorem, § 15 notes order-bounded set, 11F order-closed set, HE, 11H, IXC order-complete, see Dedekind complete (13Ba); locally order-complete order-continuous increasing function, 11D, HHb increasing linear map, 14Ec, 17B—C lattice homomorphism, 13E et seq. on the left, 11D Riemann integral, 7XA Riesz homomorphism 17C-E, 62K, 64D Riesz homomorphism, 14Eb, 14F, 14G ring homomorphism, 45E et seq.; Riesz norm, 22A, 22Ga, 23Nh, 26'A measure-preserving, 54B Riesz pseudo-norm, 22A, 22C order-dense Riesz subspace, 14F, 15E, § 17 Riesz seminorm, 22A, 22Gc see also locally order-dense; super- Riesz space, 14A order-dense Riesz subspace, 14F, 14La

265

INDEX Riesz' theorem, 73D-E, A2F ring, see Boolean ring ring homomorphism, §45 (45A), 4XH, 62L, 62Mf see also measure-preserving ring homomorphisms sections, see filters of sections semi-finite measure ring, §51 (51Bb), 52E, 52G, 53Fa, 53Fd, 61Fb measure space, 61Fb, 61Ja, 63F, 63G, 63L, §64, 65D, 651a sequence spaces, 6XD sequentially Dedekind complete, see Dedekind or-complete (13Bb) sequentially Lebesgue linear space topologies, 83Ka, 8XC sequentially order-continuous increasing function, 11D linear map, 170c, 420, 62Mh, 83K Riesz homomorphism, 62Me ring homomorphism, 62Mf sequentially smooth functionals, §71 (71B), 73Gg, 7XB smooth functionals, §72 (72G), §73 solid hulls, 14G, 22Bc linear subspaces, 14F, 14G sets 14G, 14Jd, 22Be, 23L see also locally solid topologies Stone representation, 41E, 42Rf, 43Ec space, 41E, 41Le, §43 notes see also 611

Stone's theorem, 4ID strictly localizable, see decomposable (64Ga)

266

strong order unit, see order unit (141) strong topology on a linear space, 23Nh, 811, A1A, AlHd subalgebra of a Boolean algebra, 41G sublattice, 13F sup, 11B super-order-dense Riesz subspace, 14F, 17Gc, 43Bc, 62H, 62Me supremum, 11B cr-complete, see Dedekind er-complete (13Bb) cr-finite, see countable magnitude (64Gb) o--ideal, 41G (T-subalgebra, 41G, 4XD c-sublattice, 13F or-subring, 41G tight functional, 73Ge, 73Gf totally finite, seefinitemagnitude (61Fg) totally ordered set, IXC truncated Riesz subspace, §71 (71D), 72E, 83Ld uniformly closed, see relatively uniformly closed sets uniformly complete Riesz spaces, §25 (25Hb), 81G, 81Ig, 81Ih, 82D, 8XA, A2A-D unit of an ikf-space, 26Ad see also order unit (141); weak order unit (141) vector lattice, see Riesz space (14A) weak order unit, 141 weak topologies on linear spaces, 21C, 23M, 33B, chap. 8,§ Al (A1A)

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