No title

(3.16)


=

LL j=l er3j

aJxso (x) =

L

aerXs" (x),

er

where tilT = LjEer (fj. Let us denote by 6+(X) the class of simple functions that are 2: 0 on X It follows from (3.16) that such functions can be put in the form (3.13) with a J 2': O. It also follow~ from (3.16) that 6+(X) is the class of ~-measurable functions with range in a finite subset of [0,00).

29

3. Integration on Measure Spaces

Given rp E (5+(X), we define the integral

(3.17) This quantity is possibly +00, if /1,(S)) = 00 for some S). However, if a) then by convention we take ajfJ(S)) = 0 whether or not fJ(S)) is finite.

= 0,

Note that by (3.16)

(3.18) ) u3j

j

u

Note also that if the set of nonzero values taken on by rp is {b 1 , ... , bM}, then M

(3 19)

rp(x)

= I:>kXTk (x),

Tk

= {x

EX: rp(x)

= bd.

k=l

Each Tk is a union of sets Su in (3.16), and each corresponding coefficient au is equal to bk. It follows that

(320)

f

rp dfJ

=

L aufJ(Su) = L bkfJ(Tk) ,

.

k

u

the first identity by (3.18) and the second by additivity of fJ. This identity shows that J rp dfJ i:, well defined, independently of the particular representation of rp in (3.13) If rp. 'I/; E

(5 + (X).

one al&o veri fie:, immediately that

(3.21) and

(322) The following is also easy, though worthy of the status of a lemma. Lemma 3.3. If rp E (5+(X), then

(323)

..\(A)

=

J

rpdfJ

A

=

J

XA rpdfJ

30

1,8

3

a measure on

Integration on l\!easure Spaces

~.

Here, the last identity is a definition of fA'P dJL. Note that if 'P is given by (313), then N

>'(A) = 2:a,p(A n Sj).

(3.24)

J=1

The proof of Lemma :3.3 is an exercise Now let M+(X) denote the set of all measurablf' functions [0, ooJ. For .f E M+(X), we define the integral (3.25)

J.t

dJL = sup

{J

'PriJL' 0::; 'P::;

f.

.t . X

----)

'P E e+(X)}.

This is somewhat analogous to the definition of the lower Riemann integral 1(1). One difference is that the quantity defined by (3.25) might be 00. Our hypothesis that f is measurable makes it unnecessary to introduce also an analogue of the upper integral That (3.25) agrees with (3.17) for f = y E e+(X) follows from (3.22). Note that, for any f E M+(X), we can find a sequence (3.26) For example, set 'PJ(:r) = k/2J on {x EX: k2-) ::; f(x) < (k + 1)2- j } if k < j . 2}, and set y)(x) = ) on {.r EX: J ::; f(x)}. If K < ex and 0::; f(:r) ::; K on SeX, then 'PJ ----) f uniformly on S, as J ----) x. When (3.26) holds. one should expect .f YJ dp to converge to .f f dp. That this b true is a special case of the following central result, knmvn as the Monotone Convergence Theorem. Theorem 3.4. If.t) E M+(X) and .tj / , then fJ / (:~

f

E

M+(X), and

27)

Proof. Given that fJ(x) /, the limit certainly exists for each x, though it might be +x The measurability of the limit f followb fwm Proposition 3.2. Clearly J fn dl'" ::; .f f dJL for all n, and f in dJL is increasing. so
To get it, take c\' E (0.1). Let 'P E e+(X), 'P ::; f Let En = {x EX· fn(:r) 2: o:'P(x)}, Then {En} is an increasing sequence of measurabk sets w hose union is X, and (3.28)

J

f n dp2:

Ji En

n

dp2:

0:

J

'Pdp

E"

= o:>.(En).

3.

31

Integration on Measure Spaces

where' >.(En) is as in (3.23). By Lemma 3.3, together with the property (35), we have

(:3.29)

.I

lim.f i.p dl1, =

'P dj.l.

En

Hence (328) implies

J

\Ve next establish an additivity result Proposition 3.5. Iffn E M+(X)

Z8

a countable sequence and f = L:n~l fn,

then (330)

Proof. First coIlf:>ider the CabE' f = fl + h. As in (3.26), take ipnJ E e;+(X) f:>uch that i.pnj / ' f n as j ........ ex; Then ip J) + i.p2) / ' h + h, so, by the :!\Ionotone Convergence Theorem together with (3.21), we have

J

(lJ

+ h) dj.l =

(3.:n)

lim

= lim =

J J

J

+ ip2)) dj.l

('PIj

i.pJ)

fl dt),

dj.l

+

+ lim

J12

J

'P2) dt),

dj.l.

By induction,

I (2: In) N

(:3 :32)

.

/Ii

ellL

n=l

= 2: n=I

J

fn rij.l,

given fn E M+(X). for any N < oc, and applying the l\Ionotone Convergenc(' TheOIem again yields (3.30) The nt'xt result ib known as Fatou's Lemma. Proposition 3.6. If fn E M+(X), then

(3.:3:3 )

.I

(lim inf

In) dj.l

::; lim inf

.I

fn dj.l

32

3. Integration on Measure Spaces

Proof. For each k E Z+, j 2 k

==}-

infn~k

fn :s; /j, so

(3.34) Of course, infn>k fn / lim inf fn as k -----* 00. Thus, letting k and applying the Monotone Convergence Theorem, we have

-----* 00

in (3.25)

lim! (inf fn) dJ.L :s; lim inf! fj dJ.L, ! (lim inf fn) dJ.L = k-+oo n2:k

(3.35)

which gives (3.33). We now tackle the definition of f f dJ.L when f(x) is not necessarily 20 everywhere. For a general measurable function f : X -----* JR, we can write f = f+ - f-, where f+(x) = f(x) when f(x) 2 0, f+(x) = 0 otherwise, and f-(x) = - f(x) where f(x) :s; 0, f-(x) = 0 otherwise. Thus f+, f- E M+(X). We define the integral

(3.36) provided at least one of the terms on the right is finite. If they are both finite or, equivalently, if

! lfl

(3.37) we say

f

dJ.L <

00,

is mtegrable and write

(3.38) For short, we might write

f

E £1(X) or simply

f

E £1.

The following basic result is parallel in statement, though not in proof, to Proposition 1.1. Proposition 3.7. The set £1(X,J.L) is a linear space, and the map

(3.39) is a linear map.

J.

33

integration on Measure Spaces

Proof. From Ih + 121::::; Ihl + 112[, it easily follows that h E £J(X,J-t) =? h + 12 E C 1 (X, J-t). Also, clearly aj E ~ =? ajh E C1(X, J-t) in this case. Let us establish that, if h E C 1 (X, J-t), then

(3.40)

rt -

Indeed, if 9 = h + 12, then g+ - g- = fl g- + It + ft· Hence, by Proposition 3.5,

(3.41) =

J J

(g+

+ fl + f 2 ) dJ-t =

(g-

+ ft + ft) dJ-t =

+ ft -

J J

g+ dJ-t g- dJ-t

f2' so g+ + fl

J +J

+

fl dJ-t ft dJ-t

+ f2

=

J +J +

f2 dJ-t ft dJ-t.

Equating the right sides, we get (340). The other ingredient in linearity, (342)

J

af dJ-t = a

J

f dJ-t, for f E C1(X, J-t), a E

~,

is easy. We can also integrate a complex-valued function. If f : X ---+ te, write f = h + ih where fj are real valued. We say f is measurable if hand 12 are both measurable. If iI, 12 E C1(X, J-t), we set (3.43)

J Jh f dJ-t

=

dJ-t

J12

+i

dJ-t.

Often, we use the same notation, C1(X, J-t), to denote the set of such complexvalued integrable functions. The rest of the results of this section will deal specifically with real-valued functions, but the reader should be able to formulate extensions to complex-valued functions The next result is known as the Dominated Convergence Theorem. Theorem 3.8. Let fn E C1(X, p,), for n E

(3.44)

fn(x)

---+

z+.

Suppose f : X

f(x)

for all x E X, and suppose that there is a functwn 9 satisfying (3.45) Then (3.46)

9 E C 1 (X,J-t),

Ifnl:S g, for all n.

---+ ~

and

34

3. Integration on Measure Spaces

Proof. We know f is mt>af:,urable; of course then if I ::; 9 =} f E £} Now if 9 satisfies (345), then 9 + fn 2: O. Thu:", by Fatou's Lemma and (3.40),

.I

q dp

+

.I f

dp

(3.47)

=

.I

(g

+ f) dp

.I

::; lim inf

=

.I

grilL

(g

+ In) dp

+ lim inf

.I

In dfJ·

and hence

.I .t

(348)

dll ::; lim inf

Jf

n dll.

whenever the hypothese:" of Theolem 38 hold Then the :"ame argument applie~ with fn replaced by - In, so.f (-f) dll::; liminf J(-fn) dlL or. equivalently,

Jf

(3.49)

dJl 2: lim

~up.l fn dll.

The inequalities (348) (3.49) yield (3.46) The next result is known a~ Egoroff's Theorem. It hac, many uses, including an alternative approach to the proof of the Dominated Convergence Theorem (explored in some of the exercises below).

Proposition 3.9. Let (X, J. p) be a mea91pre 5pace 9/1,ch that 11(X) < :)0 Assume fJ are J-m,ea5UTable and f,(:l') -----> f(J;) faT all .1: E X Then, gwen any 8 > O. there eX1<;t.s B E J .such that (350)

Proof. For

(351)

/l(B) ::; 8 and fj E

> 0, FnE

----->

f umformllJ on X \ B.

n E Z+, set

= {.r

E X

for some J 2: n, IfJ(:I:) - I(x)l2: E}

Note that each 'ie't FnE belong& to J The' hypothesi& implie& for each E > 0, so /1(FnE ) -----> 0 as n -----> rx., In paltie-ular, given any 8 such that

(352)

> 0, if we

~ct

E

=

2- k ,

nn>

1 FOE

= 0.

-

there exist~ n

= n(k)

3.

35

Integration on Measure Spaces

Let B

= Uk 2':1 F n (k),2-k,

B E J. We ~ee that IL(B) ::; () and

x tic B, J 2': n(k)

(353)

=?

If)(x) - f(x)1 < 2- k .

This establishes (350). "Ve mention a point which is quite simple but important for applications. Namely. in all the convergeuce theorcms established aboVE'. thc hypothesis that a ~equC'nce of mea~urablc function~ fj converges for all :£ E X can be weakened to the hypothe~is thclt IJ converges ai-most everywhere, i.e., fJ converges for all x E XI. where Xl C X is a measurable subset such that fL(X \ Xd = 0 The proof is ea~y. just restrict attention to the measure spacE' (X 1 ,JI.ltd, where JI = {S E J' S c Xd = {S n Xl: S E J} and JL I i~ J1 restricted to J 1· To close this chapter. we relate the Riemann integral, considered in Chapter 1. to the Lebe~gue integraL which is the integral defined in this chapter whC'n fL i~ Lebe~gue measure The theory of the Lebesgue integral allows us to specify preci~ely which bounded functions on an interval I = [a, b] are Riemalln integrable. Indeed. let ..p : 1-+ lR be a bounded function, and recall the dehnitioll (1.;))-(1 6) ofl(
(142).

J1i

d.T

~ 1(
1

J

On the otll('r hand. (~)

./ 9, dx /I(
\Vf'

~ee

that

f · "'>

55)

j' > -

•J

~ '/-'.

(I). / ':J

9 <_

{(c r '•

where f and 9 are bounded and meac;umble. and by the ~Jonotone Convergence Theorem, together with (:3.54), we have

1(
(356)

J

is Riemann integrable if and only if I (f - g) dx = O. hence if and only if f(:r) = g(x) for almost all x E I. (See Exercbe 4 below.) On the other hane!. if E is the collection of endpoints of all the intervab in the partitions P J (a countable set, hence of Lebesgue measure zero), we have Con~equently rp

(3.57)

V x tic E,

I(x) = g(x) ~ rp is continuous at x

Thus we have the following proposition

36

3. Integration on Measure Spaces

Proposition 3.10. If

and only if the set of points x E I at which

Exercises 1 Suppose J is an algebra of sub~ets of X. Show that provided it is closed under countable disjoint unions.

J is a O"-algebra,

2. Establish the properties (3.3)-(3.6) of a measure. 3. Prove Lemma 33. Hznt. Show that, if f.L1, ... , f.LN are measures on X and aJ E [0,(0), then A = a1f.L1

+ ... + aNf.LN

is a measure Also show that, if f.L is a measure on X and E J measurable, then

C

X are

(3.58) are measures 4. Let (X,J,/-l) be a measure space and (3.59)

f =

0 f.L-a.E'

To ~ay that f = 0 f.L-a e

O}, /-leN) =

o.

¢::::::}

f

Jf

E M+(X) Show that df.L

=0

is to say that, for

N = {x

E X

f(x) f

5 A measure /-l on (X, M) is said to be complete provided

A E M, /-leA) = 0, SeA Suppo~e /-l

(3.60)

==}

S E M, and f.L(8) =

o.

is a measure on (X, J) that is not complete. Show that

J=

{E U 8: E E J, 8 c FE J, f.L(F) = O}

is a O"-algebra, that there is a unique measure Ii on (X, J) such that J.L on J, and that Ii is complete We call Ii the completion of J.L.

Ii =

3.

37

Integration on Measure Spaces

6. Let fJ, be a measure on (X, J'), as above, and /i its completion, on (X, J). If f : X ---> lR is J-measurable, show that there exists fo that is J'measurable, such that fo = f outside a set of /i-measure zero. Hint. First treat the case of simple functions, then positive functions, via the approximation (3.26). 7. Let

f

E

M+(X) and set, for A >"(A) =

(3.61)

E

J',

J

fdfJ, =

J

XAfdfJ,.

A

Show that >.. is a measure. Hint Use the Monotone Convergence Theorem. Note that this result is much stronger than Lemma 3.3.

8 Let f E .cJ(X,fJ,) be given. Show that, for every such that (3.62)

S E J', fJ,(S) <

(j ===?

J

[fl dfJ, <

E

> 0, there is a

(j

>0

E.

S

Hint. Pick cp E 6+(X) such that 0 ::; cp ::; Iff and c/2. Then A = sup cp < 00, so pick (j < c/2A.

J cp dfJ, 2 J If I dfJ, -

9. Give an alternative proof of the Dominated Convergence Theorem, de-ducing it from Egoroff's Theorem and the result of Exercise 8, rather than from Fatou's Lemma. Hznt. Given Ifni::; 9 E .cJ(X,j1), fn ---> f, c > 0, pick a measurable set Y C X such that

fJ,(Y) <

(3.63)

00,

J

gdfJ, < c,

X\Y

and pick a measurable S C Y, such that fJ,(S) < c and fn on Y\ S. 10. Let /j be measurable functions on (X, J', fJ,). We say /j provided fJ,(Eje:) ---> 0 as j ---> 00, for each E > 0, where

--->

--->

f uniformly

f in measure

Ejc = {x EX: [/j(x) - f(x)[ > c}. Show that, if fJ everywhere.

--->

f in measure, then some subsequence /jv

--->

f almost

38

3. Integration on Measure Space6

11. Show that the Dominated Convergence Theorem continues to hold if the hypothesis (3.44) that fn ---4 f pointwise is changed to fn ---4 f in measure. 12. Let (X,'J,f-L) be a measure space and assume f-L(X) < 00 Let M(X,'J) denote the space of measurable functions and let Af(X, 'J, I.L) denote the set of equivalence clasbes of measurable functions, where

f For measurable (3.64)

f

rv

9

<¢:=::?

f(x) = g(:1:), f-L-a e.

and g, set

d(f,g) =

J + if _ gl [f-g[

1

df-L.

x

Show that this is a distanC'e function, making Af(X, 'J, f-L) a metriC' bpaC'e. and that fn ---4 f in this metric if and only if fn ---4 f in measure, as defined in Exercise 10. 13. Show that, if f-L(X) < 00, then fn ---4 f, f-L-a.e. implies fn ---4 f measure. Give a counterexample for X = Z, with counting measure

In

In Exercises 14-18, we deal with Lebebgue measure on the line lR or on an interval I = [a,b]. A set S c lR is called an Fa set if S = UJ_ >1 FJ , FJ closed. S is called a G(j &et if S = nJ~l G j , G j open 14. If 0 c I ib open, show that there exist fj E C(I) such that 0 :S fj(x) /' xo(x) for all x E I. If K c I is closed, show that there' exist fJ E C(I) such that 1 2 fJ(x) ~ XK(X) for all x E I. 15. If S is an Fa set or a G(j set, show that there exi&t fJ E C(I) &uch that O:S fJ :S 1 and fJ(x) ---4 xs(x) for almost all x E I. Hint Tackle the question of convergence in mea&ure. 16. Let ScI be Lebesgue measurable. Show that there exist& an Fa set So and a G(j set S1 such that So eSc S1 and m(S1 \ So) = O. Deduce that there exist fJ E C(I) such that 0 :S fJ :S 1 and fj ---4 xs almost everywhere.

39

3. Integration on Measure Spaces 17. Let 1 E Cl(I,dx). Show that there exist fJ E C(I) 5uch that almost everywhere. Hint First approximate 1 by simple functions, as in (3.26).

1J

----+

1

18. Given 1 E Cl(I, dx) and 0 > 0, show that there is a Lebesgue measurable ScI such that m(S) < 0 and 111\S is continuous. Hint. Apply Egoroff's Theorem to the results of Exercise 17. This result is known as LUbin's Theorem 19. If (X, -J', f-L) is a measure space and that there exist Aj E -J' such that

1:X

----+

[0,00] is measurable, show

(3.65) Hint. Set Al

= {x

EX: 1(x)

2: 1} and, inductively, for k 2: 2,

20. Let (X,-J',J-l) be a mea5ure space. Given

1 E M+(X),

t E (0,00), set

(3.66) Show that (3.67)

Then use (3.26) and the l\lonotone Convergence Theorem. Show that, if i.{JJ / 1, then, for each

Hint. First verify this for simple functions

tE(O,oo), S
c

to deduce that JW
S
c ... /

Sf(t)

JWf(t)dt.

21. Let P : [0,1] ----+ [0,1] be the homeomorphism described in Exercise 10 of Chapter 2, satisfying P(L) = K,

m(L) > 0,

m(K) = 0.

Show that there is a subset S C K 5uch that p-l(S) is not Lebesgue measurable.

40

3. Integration on Measure Spaces

Hint. Apply Exercise 7 of Chapter 2 to X

=

L.

22. Let Un be measurable functions on (X,:J, f-L) Assume

Un

-----+

U pointwise,

Un

~ 0,

j Un df-L

-----+

j U df-L <

00.

Prove that

jlu-unldf-L-----+ O. Hint Write

IU - unl = 2(u - u n)+ - (u - un) and apply the Dominated Convergence Theorem to the first term on the right side. 23 Below we display the "guts" (though not the skeleton or the flesh) of the derivation of Fatou's Lemma and the Dominated Convergence Theorem (DCT) from the Monotone Convergence Theorem (MCT), as done in the proofs of Proposition 3.6 and Theorem 3 8. Familiarize yourself with the details of these arguments, to the point where you can look at the material displayed below and see the entire argument.

MCT

==}

Fatou

==}

DCT.

(Fatou)

j liminf fndf-L = lim j(inf fn)df-L:S liminfjfjdf-L, k~oo

n?:.k

T MCT

(DCT)

j (g

+ f) df-L :S lim inf j

(g

+ fn) df-L =

T Fatou (analOgue for fn replaced by - fn).

j 9 df-L + lim inf j fn df-L

Chapter

4

LP Spaces

Let (X, f-L) be a measure space. As in Chapter 3, we say a measurable function f belongs to £ 1 (X, f-L) provided

[[fIlL1 =

(4.1)

J

[f(x)[ df-L(x) <

00.

X

Elements of L 1 (X, f-L) consist of equivalence classes of elements of £ 1 (X, f-L), where we say

(4.2)

f

rv ]

{:}

f(x) = ](x) for f-L-almost every x.

With a slight abuse of notation, we denote by f both a measurable function in £l(X,f-L) and its equivalence class in Ll(X,f-L)' Also we say f, defined only almost everywhere on X, belongs to L1 (X, f-L), if there exists] E £1 (X, f-L), equal f-L-almost everywhere to f. The quantity [[fl[p defined by (4.1) is called the L1 norm of fIn general, a normed linear ~pace is a vector space equipped with a positive function IIv[1 having the properties

[[av[[ = la[ . Ilvll, for v (4.3)

IIv +wll

s

IIvll

E

V, a E C (or JR.),

+ [[wll,

!lv[[ > 0, unless v =

o.

The second of these conditions is called the triangle inequality. Given a norm on V, setting d( u, v) = II u - v II defines a distance function on V, making it a metric space. It is easy to see that £l(X,f-L) is a vector space and that IIflip satisfies the first two conditions in (4.3). However, [If[[Ll = 0 if and only if f = 0

-

41

42

LP Spaces

4

almost everywhere. (Recall Exerci&e 4 of Chapter 3.) That is the reason we define L1 (X, fJ,) to consist of equivalence classes defined by (4.2), so L1 (X, fJ,) becomes a normed linear space. Generally speaking, a sequence (v]) in a normed linear space is said to be a Cauchy ~equence if [Iv] - 11k II ----+ 0 as ], k ----+ 00. If every Cauchy sequence ha~ a limit in V, then V is said to be complete, a complete normed linear space is called a Banach space. Theorem 4.1. L1 (X, /1,) is a Banach space.

The proof of completeness of L 1 (X, fJ,) makes use of the following two lemma~, which are essentially re~tatements of the Uonotone Convergence Theorem and the Dominated Convergence Theorem, respectively Lemma 4.2. If fj E £l(X,/L), 0 ::; hex) ::; h(x)::; ", and IlfJllu ::; C < 00, then lim fJ(x) = f(x), with f E L1(X,/L) and Ilfj - fllu ----+ 0 a5 J-'OO

J

----+ 00.

Proof. We know that f E M+(X). The Monotone Convergence Theorem implies J fJ dfJ, /' J f dfJ, Thus J f d/L ::; C. Since Ilf) - fliLl = J f d/L fj d/L in this case, the lemma follows.

J

Lemma 4.3. If fj E £l(X,fJ,),limf](x)

= f(x) /L-a.e., and

~f there is an

£l(X'/L) such that If)(x)1 ::; F(x) 11,-a.e., for all j, then f and Ilfj - fliLl ----+ O.

FE

Proof. Apply the Dominated Convergence Theorem to 9j a.e. Note that 19) I ::; 2F.

E

Ll(X,fJ,),

= IfJ - fl

----+

0

To show L1(X,/L) b complete, suppose (In) is Cauchy in L1. Passing to a subsequence, we can assume Ilfn+l - fnllLl ::; 2- n . Consider the infinite series 00

(4.4)

hex)

+ L[Jn+1(X) -

fn(x)].

n=l

N ow the partial sums are dominated by m

(4.5)

Ih(x)1 + L Ifn+1(x) - fn(x)1 = Ih(x)1 + Gm(x), n=l

and since 0 ::; G 1 ::; G2 ::; ... and llGmll v ::; E2- n ::; 1, we deduce from Lemma 4.2 that G m / ' G fJ,-a.e. and in Ll-norm. Hence the infinite series

4. LP Spaces

43

(4.4) is convergent a.e., to a limit f(x), and via Lemma 4.3 we deduce that fn ---+ f in L1-norm. This proves completeness. Continuing with a description of LP spaces, we define Loo(X, J1,) to consist of bounded measurable functions, L 00 (X, J1,) to consist of equivalence classes of such functions, via (42), and we define Ilfll£<>o to be the smallest sup of J", f. It i8 easy to show that Loo(X, J1,) is a Banach space. For p E (1. 00 ), we dt'fine LP (X, J1,) to com,ist of measurable functions f such that [/II(x)IP dJ1,(x)]

(46)

lip

x is finitt'. LP (X, It) consists of equivalence classes, via (4.2), and the LP- norm IIIIILP is given by (4.6). This time it takes a little work to verify the triangle inequality. That thi& holds is the content of Minkowski's inequality:

111+ gilLP :S

(4.7)

IIIIILP

+ IlglluJ.

One neat way to t'stablish this is by the following characterization of the LP-norm. Suppose p and q are related by

1 p

We claim that, if

(4.9)

1 q

-+-=1

(48)

I

IIIIILP

E

LP(X, JL),

= ~up{IIIhll[) : h E Lq(X,J1,),

IlhllLq

= 1}.

We can apply (4.9) to 1+ g, which belongs to LP(X, J1,) if I and 9 do, since II + glP :S 2P (IIIP + IgIP) Given this, (4.7) follows easily from the inequality II (f + g)hllv :S IIIhllLl + IIghllv· The identity (4.9) can be regarded as two inequalitit's The ":S" part can be proved by choosing h(x) to be an appropriate multiple Clf(x)IP-l. We leave this as an exercise. The convt'r&e inequality, "2':," is a consequence of Holder's mequality: 1 p

1 q

-+-=l.

(4.10)

Holder's inequality can be proved via the following inequality for positive numbers:

(4.11)

aP

ab < - -p

bq

+ -q

for a, b > 0,

4.

44

LP Spaces

assuming that p E (1, (0) and (4.8) holds. In fact, we claim that, given a,b > 0, 1jp+ 1jq = 1, aP p

bq

'P(t) = - tP + - C q

(4.12)

q

===}

inf 'P(t) = ab,

t>O

which implies (4.11) since the right side of (4.11) is 'P(1). As for (4 12), note that 'P( t) ---> (X) as t "" 0 and as t / ' 00, and the unique critical point occurs for aPtP = bqe q, i.e., for t = blip ja l / q, giving the desired conclusion. Applying (4.11) to the integrand in (4.10) gives

J

1 If(x)g(x)1 dJL(x) :::; -llfll~p

(4.13)

p

1

+ -llglllq· q

This looks weaker than (4.10), but now replace f by tf and g by elg, so that the left side of (4.13) is dominated by (4.14) for all t > O. Another application of (4.12) then gives Holder's inequality. Consequently (4.6) defines a norm on LP(X,{L) Completeness follows as in the p = 1 case discussed above. In detail, given Un) Cauchy in LP(X, {L), we can pass to the case Ilfn+l fnllLP :::; 2- n and define Gm as in (4.5). We have IIGmllLP :::; 1, and hence (via the Monotone Convergence Theorem) deduce that G m / ' G, {L-a.e , and in LP-norm Hence the series (4.4) converges, {L-a.e., to a limit f (x). Since If - fm+11 :::; G - G m , we have by the Dominated Convergence Theorem that

J

J

x

x

if - fm+ll P d{L:::;

as m

---> 00.

(G - Gm)P d{L

--->

0,

Hence LP(X, {L) is complete To summarize, we have

Theorem 4.4. For p E [1, (0), LP( X, {L), with norm given by (4.6), is a Banach space

It is frequently useful to show that a certain linear subspace L of a Banach space V is dense. We give an important case of this here; C(X) denotes the space of continuous functions on X Proposition 4.5. If {L is a finite Borel measure on a compact metric space X, then C(X) is dense in LP(X, JL) for each p E [1, (0).

4.

45

LP Spaces

Proof. First, let K be any compact subset of X. The functions

(4.15)

fK,n(X) = [1

+n

dist(x, K)] -1 E C(X)

are all ::; 1 and decrease monotonically to the characteristic function XK equal to 1 on K, 0 on X \ K. The Monotone Convergence Theorem gives fK,n ----+ XK in LP(X, J-L) for 1 ::; p < 00. Now let A c X be any measurable set. Any Borel measure on a compact metric space is regular, i.e., (416)

J-L(A) = sup{J-L(K) . K

c A, K compact}.

In case X = I = [a, b] and J-L = m is Lebesgue measure, this follows from (2.20) together with the consequence of Theorem 2.11, that all Borel sets in I are Lebesgue measurable. The general case follows from results that will be established in the next chapter; see (5.60). Thus there exists an increasing sequence K j of compact subsets of A such that J-L(A \ Uj K j ) = O. Again, the Monotone Convergence Theorem implies XK J ----+ XA in LP(X, J-L) for 1 ::; p < 00. Thus all simple functions on X are in the closure of C(X) in LP(X, J-L) for p E [1,00). Construction of LP(X, J-L) directly shows that each f E LP(X, J-L) is a norm limit of simple functions, so the result is proved. Using a cut-off, we can easily deduce the following. Let Coo(I~) denote the space of continuous functions on ~ with compact support. Corollary 4.6. For 1 ::; p

< 00, the space

Coo(~)

is dense in

LP(~).

The case L'2(X'll) is special. In addition to the L 2 -norm, there is an mner product, defined by (4.17)

(j, g)u =

.I

f(x)g(x) dJ-L(x).

x

This makes L'2(X, J-L) into a Hzlbert space. It is worthwhile to consider the general notion of Hilbert space in some detail. We devote the next few pages to this and then return to the specific consideration of L2(X, J-L). Generally, a Hilbert space H is a complete inner product space. That is to say, first the space H is a linear space provided with an inner product, denoted (u, v), for u and v in H, satisfying the following defining conditions:

(aul (4.18)

+ U2,V) = a(ul'v) + (U2,V), (u,v) = (v,u), (u, u) > 0 unless u = 0

4.

46

LP Spaces

To such an inner product there is assigned a norm, denoted by

Ilull = V(u,u)

(4.19)

To establish that the triangle inequality holds for Ilu + vii, we can expand jju + vl1 2 = (u + V,u + v) and deduce that this is ::; [[lull + Ilvll]2, as a consequence of Cauchy's inequahty: (4.20)

l(u,v)l::; Ilull·llvll,

a result that can be proved as follows. The fact that (u - v, u - v) 2': 0 implies 2 Re (u, v) ::; Ilul1 2 + Ilv11 2 ; replacing u by eiBu with eiB chosen so that eiB (u, v) is real and positive, we get (4.21) Now in (4.21) we can replace u by tu and v by C 1 v to get (4.22) Minimizing over t gives (4.20). This establishes Cauchy's inequality, so we can deduce the triangle inequality. Thus (4.19) defineb a norm on H. Note the parallel between this argument and the proof of (4.7), via (4.10). The completeness hypothesis on H is that, with thib norm, H is a Banach bpace. The nice properties of Hilbert spaces arise from their similarity with familiar Euclidean bpace, so a great deal of geometrical intuition ib available. For example, we say u and 'V are orthogonal and write u .1 v, provided (u, v) = O. Note that the Pythagorean Theorem holds on a general Hilbert space: (4.23) Thib follows directly from expanding (u

+ v, u + v).

Another useful identity is the following, called the "parallelogram law," valid for all u, v E H (4.24) This also follows directly by expanding (u+v, u+v)+ (u-v, u-v), observing some cancellations. One important application of this simple identity is to the following existence re&ult. Let K be any closed, convex subset of H. Convexity implies x, y E K ::::} (x + y)J2 E K. Given x E H, we define the distance from x to K to be (4.25)

d(x, K)

= inf {llx - yll : y

E K}.

4.

LP Spaces

'l{

Proposition 4.7. If K cHis a nonempty, closed. convex set in a Hilbert space H and if x E H, then there ~s a unique z E K such that d(x, K) =

Ilx - zll· Proof. We can pick Yn E K such that Ilx - Y'lii ---+ d = d(x, K). It will suffice to show that (Yn) mw"t be a Cauchy sequence Use (4.24) with u = Ym - x, 'U = X - Yn, to get

Since K b convex, (Yn

+ Ym)/2

lim sup llYn

E K, so

ilx -

(Yn

- Ymll'2 :s 2d'2 + 2d'2 -

+ Ym)/211 2': d. 4d'2

Therefore

:s 0,

mn-oo

which implies convergence. In particular, this result applies when K i& a closed linear subspace of H. In this case, for x E H, denote by PKX the point in K closest to x. We have (426) We claim that x - PK x belongs to the closed linear space K J.., called the orthogonal complement of K, defined by (4.27)

K J..

= {u

E H . (u, v)

= 0 for all 'U

E K}.

Indeed, take any v E K. Then ~(t) = =

Ilx Ilx -

+ t'Oll'2 2 PKxii + 2t PKx

Re (x - PKX, v)

+ t '2 11'011 '2

is minimal at t = 0, so ~'(O) = 0, i.e., Re (x - PKX, '/') = 0, for all v E K. Replacing v by iv shows that (x - PKX, v) al&o has vanishing imaginary part for any v E K, so our claim i& established The decomposition (4.26) gives (4.28) with Xl = PKx, X2 = X - PKX. Clearly such a decomposition is unique. This implies that H is an orthogonal direct sum of K and K J..; we write (429)

48

4.

£P Spaces

From this it is clear that (430) that (4.31) and that P K and PKl.. are lmear maps on H. We call PK the orthogonal projection of H on K. Note that PKx is uniquely charac-terized by the condition (4.32)

PKX E K, (PKX, v)

=

(x, v) for all v E K

We remark that if K is a linear subspace of H that is not closed, then K ~ coincides with K~, and (4.30) becomes (K ~ ) ~ = K. Using the orthogonal projection discussed above, we can establish the following result. Proposition 4.8. If H is a Hilbert space and

(4.33)


f) for

---+

C %s a continuous

all u E H.

Proof. Consider K = Ker

We note that the correspondence

f---+

f

gives a conjugate linear isomor-

H'-+H,

where H' denotes the space of all continuous linear maps

-+

C.

Recall that our interest in Hilbert spaces arises from our interest L2(X, /-L). Let us record the content of Proposition 4.8 in that case.

In

4. £P Spaces

49

Corollary 4.9. If
~

C is a continuous linear map, there exists

a unique f E L2(X,J-L) such that (4.35)


J

u(x)f(x) dJ-L(x) ,

Vu E L2(X,J-L).

We can use orthogonal projection operators to construct an orthonormal basis of a Hilbert space H. Let us assume H is separable, i.e., H has a countable dense subset S = {Vj : j 2: I}. See the exercises for results on separability. In such a case, pick a countable subset {Wj : J 2: I} of S such that each Wk is linearly independent of {w J : J < k} and such that the linear span of {Wj} is dense in H Let Lk = span {Wj : 1 ::; j ::; k}. We define an orthonormal set {uJ : j 2: I} inductively, as follows. Set U1 = wl/llw111. Suppose you have {uJ . 1 ::; j ::; k}, an orthonormal basis of Lk. Then set

(4.36) where Pk is the orthogonal projection onto L k . One does not need the construction involving Proposition 4.7 to get Pk here. We can simply set k

Pkf = ~(J,uJ)Uj.

(4.37)

j=l

The set {Uj : j 2: I} so constructed is orthonormal, i.e., (Uk, ue) = (he. Also its linear span is dense in H, since the linear span coincides with that of {Wj:j2: 1}. We claim that, for each that

f

E H,

Pkf

-t

f

as k

~ 00.

To see this, note

k

(4.38)

IIfl12 = IJPkfl1 2+ Ilf - Pkfl1 2 2: IJPkfl12 = ~ I(J, uj)12. j=l

W,

We also see that, if n 2: k, then IJPnf - Pk f 112 = I:k<j:::;n I(J, Uj and hence that (PkJ) is a Cauchy sequence in H, for each f. We claim that the limit is equal to f, hence that

(4.39)

f = L(J, Uj)Uj. J

Indeed, we now know that the right side of (4.39) defines an element of H; call it g. Then, f - 9 has inner product 0 with each u J ' hence with all

4.

50

V'Spaces

elements of the linear span of {Uj}, hence with all elements of the closure, i.e., with all elements of H, so f - 9 = O. In the case H = L2(1, Tn), with I = [-1f,1f] and Tn = dx/21f, an orthonormal basis is given by (440) See the exercises for a proof of this. In such a casE', (4.39) is the expansion of a function in a Fourier series. We use Corollary 4.9 to prove an important result known as the RadonTheorem. Let f.L and v be two finite measures on (X, J). Let

N~kodym

a = f.L

(4.41)

+ 2v,

W = 2f.L + v.

On the Hilbert space H = L2(X, a), consider the linear functional cp : H C given by

cpU) =

(4.42)

---t

J

f(x) dW(x).

X

Note that IcpU)1 :s: 2 f If I da :s: 2va(X)llfIIL2(x,Lt). By Corollary 4.9, there exists 9 E £2(X, a) such that, for any f E £2(X, (Ie) = £2(X, f.L) n £2(X, v), (4.43)

J

f(x) dw(x)

x

=

J

f(x)g(x) da(x).

X

In particular, this holds for any bounded measurable tity is equivalent to

(4.44)

J

f(2g - 1) dv =

f.

Note that thit. iden-

J

f(2 - g) df.L,

for all f E £2(X, a). If we let f be the characteristic function of SIR = {x E X . g(x) < 1/2 - 1/t'} or of S2£ = {x EX: g(x) > 2 + l/C}, we see that f.L(SJ£) = v(SJc) = O. As a consequence, we can arrange that 1/2 :s: g(x) :s: 2, for all x E X We also see that Z = {x EX: g(x) = 1/2} must have f.L-measure zero. (Similarly, {x: g(x) = 2} has v-measure zero.) Also, (4.44) holds for all f E M+(X), by the Monotone Convergence Theorem We say that v is absolutely continuous with respect to f.L and write v < < f.L, provided (445)

f.L(S)

= 0 ====} v(S) = O.

4.

LP Spaces

51

In such a case, we see that Z = {x EX: g(x) = 1/2} has v-measure zero. Given FE M+(X), we can set (4.46)

F(x) f(x) = 2g(x) - l'

h(x) = 2 - g(x) 2g(a:) - 1

and apply (4.44) to get (4.47)

.I

.I

F(x) dv(x) =

F(x)h(x) dp,(x)

x

X

for all positive measurable F Note that taking F = 1 gives h E Ll(X, p,). The result we have just obtained is known as the Radon-Nikodym Theorem. We record a formal statement Theorem 4.10. Let p, and v be two fimte measures on (X, ~). If v is absolutely continuous wzth respect to p" then (4.47) holds for some nonnegative hE L1 (X, p,) and every positive measurable F.

We mention that (4 47) abo holds for every bounded measurable F. H we do not assume that v < <

(4.48)

we can still consider

p"

h(x) = 2 - g(x) 2g(x)-1

if g(x)

i- 2'1

if g(x) =

0

1

2'

and we have (4.49)

.I

F dv

=

y

J

Fhdp,

y

for any positive measurable F, where (4.50)

Y = X \ Z =

{x EX. g(x) i-l}.

Recall that p,(X \ Y) = O. We can define the mea&ure A on (X,~) by (4.51)

A(E)

= v(Y n E).

Then we have (4.52)

.I X

FdA

=

.I x

Fhdp,

52

4.

IJ' Spaces

for all positive measurable F. Write

(4.53)

p(E)

= v(E \ Y) = v(E n Z),

so

(454)

v

= A + p.

Now the measure A i& supported on Y, i.e., A(X \ Y) = O. Similarly, p is supported on Z. Thu!'> A and p have disjoint supports. Generally, two measures with disjoint supports are said to be mutually smgular. When two measures A and p are mutually singular, we write A .1. p. We have the following result, known as the Lebesgue decomposition of v with respect to J-L. Theorem 4.11. If J-L and v are finite measures on (X, ~), then we can wnte

(455)

v

This decomposztion

tS

= A + p,

A < < J-L,

p ..1

J-L.

unique.

Proof. The measure& A and p are given by (4.51) and (452). The fact that A < < J-L i& contained in (4.52). As we have noted, J-L(X \ Y) = 0, so J-L is supported on y, which is disjoint from Z, on which p is supported, hence p .1. J-L.

-

-

-

If also A and pare mea!'>ures such that v = A + p, A < < J-L, p ..1 I)', we have Z E ~ such that p is supported on Z and J-L(i) = 0. Now J-L(ZUZ) = 0. so A(Z U Z) = 0 and ~(Z U Z) = 0, and, for E E M,

A(E) = A(E \ Z) = v(E \ (Z U Z)), ~(E) = ~(E \ Z) = v(E \ (Z U Z)) This give!'> uniqueness. We say a measure J-L on (X,~) i& (T-finite if we can write X as a countable union UJ~l Xj where Xj E ~ and J-L(Xj ) < 00. A paradigm case is Lebe&gue measure on X = ITt There are routine extensions of Theorems 4.10-4.11 to the case where I)' and v are (T-finite measureh, which we leave to the reader.

Exercises 1. Let V and W be normed linear &paces. Suppose we have linear trans-

formations

(4.56)

4.

LP Spaces

with C independent of j. (We say {Tj } is uniformly bounded.) Suppose also T : V --7 W satisfies this bound. Let L be a dense subspace of V. Then show that (4.57)

Tjv

--7

Tv, V vEL

==}

Tjv

--7

Tv, V v E V.

2. Define

(4.58)

Ts : LP(lR)

~

LP(lR),

Tsf(x)

= f(x - s).

Show that, for p E [1,00), (4.59)

f E LP(lR)

==}

Tsf

--7

f in LP- norm, as [,

--7

Hint. Apply Exercise 1, with V = W = LP(lR) , L Corollary 4.6. Note that IITsfliLl' = IlfllLl'.

O.

= Coo(lR),

as in

One says a metric space is separable if it has a countable dense subset. 3. If I = [a, b] C lR and a :S a < ;3 :S b, define

'Pa;3(x)

=

dist( x,I \ [a,;3]).

Show that the linear span over Q of {'Pa;3 : a,;3 E QnI} is dense in C(I), and deduce that C(1) i& separable. From the denseness and continuity of the inclusion L: C(l) ~ LP(I), prove that £P(l) is separable, for 1 :S p < 00. Then prove that LP(lR) is separable, for 1 :S p < 00. 4 Let X be a compact metric space; X has a countable dense subset {Zj : j ;::=: I}. Given 0 < p < (diam X)/2, set

1/;jp(x)

=

dist (x, X \ Bp(zj))

Show that the algebra generated by {1,Uj,p : j E Z+, P E Q+} and 1 is dense in C(X) and deduce that C(X) is separable Conclude, from Proposition 4.5, that VeX, fJ) i& separable, for p E [1,00), if f.L is a finite measure on the CT-algebra of Borel sets in X. 5. Let {Uj : j ;::=: I} be a countable orthonormal set in a Hilbert space H. Show that 00

(4.60)

~(f,Uj)Uj = Pvf, j=l

4. LP Spaces

54

where V is the closure of the linear span of {11,j}. The Stonc-Weierl:>tra&s Theorem states that, if X is a compact Hausdorff space and A an algcbra of functions in C]R(X) (the space of real-valued continuoul:> functions on X), such that 1 E A, and if A has the property of &eparating points, i.e., for any two distinct p, q E X, there cxists f E A such that f(p) i= f(q). then A i& dense in C]R(X). If A is an algebra in Cc(X) with these propeltie~, plus the property that f E A =} J E A, then A is dense in Cc(X). A proof is given in Appendix A. 6. Use the Stone-Weierstras~ TheOlem to show that, if ek(:c) = eikx , as in (4.40), then the linear span E of {ek : k E Z} is dense in C(Sl), where Sl = JR/27rZ; hence E is dense in LP(SI,dx/27r), for p E [1,00). Hence {ell: k E Z} is an orthonormal basis of L'2(Sl,dx/27r)

7. For f,11, E Crr(JR), the space of smooth in JR, set Kf11,(x) = f

(4.61)

* 11,(x) =

function~

with compact support

J

f(y)11,(x - y) dy

Show that, for 1 ::; p < 00, K f has a unique bounded exten&ion: (4.62) The operation in (461) il:> called convolutwn. H~nt. For f, 'U E Co(JR), if f il:> ~upported in [a, b], show that

f

* 11,(x) = n-+oo lim

b

n

- a n

L J=O

.

f(Lb n

+ (1 -

L)a) TJ'U,(X), n

where Tj11,(X) = 11,(x - Jb/n - (1 - J/n)a) U&e the triangle inequality (4.7) to estimate norml:>. 8. Show that there is a unique extenl:>ion from f E Co(JR) to fELl (JR) of Kf11" with (462) continuing to hold, giving a continuou& linear map K : L1(JR) --+ £(LP(JR»). 9 Let fJ be a sequence of nonnegative functions in L1 (JR) such that (4.63)

/ fJ dx = 1,

supp fJ

C

{:c

E

JR . Ixl < Ijj}

Show that, for 1 ::; p < 00, (4.64)

11, E LP (JR)

====}

fJ

* 11, --+ 11, in

LP norm, as J

--t

00

4.

LP Spaces

55

Derive the same conclusion, upon weakening the second hypothesis in (4.63) to

J

= (3j(c;)

fJ dx

-+

1 as

J -+ 00,

V c; >

o.

Ixl<€ Hmt. First verify the ('onclusion for

U

E Co(JR). Then use Exercise 1.

10. Given fELl (SI), 0 < r < 1, define 00

(465)

Prf(O) =

2.=

f(n) =

f(n)rlnleinB,

n=-oc

~ r27f f(O)e- inB dO. 27r

Jo

Show that (4.66)

Prf(O) = Pr

* f(O) = -1 127f P, (0 27r

0

cp)f(cp) dcp,

where 00

(4.67)

1 r2 P, (0) = '"' rlnleinB = 2· ~ 1 - 2r cos 0 + r n=-oo

Show that (4.68) 11 If f E U(Sl), 1 ::; P <

(4.69)

-

1

27r 00,

Pr f

l27f Pr(O) dO = l. .0

~how that -+

f,

as r /

1,

in LP- norm. If f E C(Sl), show that you have uniform convergence in (4.69). Thif> if> known as Abel ('onvergence of Fourier series. Hint. Use a variant of the analy~is needed for Exercbe 9 12. Show that Exercises 10-11 providE' an alternative proof of the conclusion of Exercise 6, that the linear span of ek (0) = eikB , k E Z, is dense in C(Sl) and in LP(SI), for 1 ::; P < 00, and hence that {ek : k E Z} is an orthonormal basis of L2(SI,dO/27r). 13. Suppose (X,~, Ji) is the completion of (X,~, M) and LP (X, Ji) are identicaL

Show that LP(X, M)

56

4. V Spaces

Hint. Consult Exercise 6 of Chapter 3.

14. Show that if (4.70)

f

E V(X, J-L) and p E [1,(0), then, for ,\ E (0,00), J-L( {x EX:

If(x)J > ,\}) ::;

,\-P IIfll~r

Deduce that if fk ~ 0 in LP-norm, for some p E [1, (0), then !k ~ 0 in measure, as defined in Exercise 10 of Chapter 3. Hint. Denote the set being measured in (4.70) by EA and note that Jfl P dJ-L ~ '\PJ-L(EA). The inequality (4.70) is called Tchebychev's inequality.

lEA

Chapter 5

The Caratheodory Construction of Measures

Recall how our construction of Lebesgue measure in Chapter 2 proceeded from an initial notion of the size of a very restricted class of subsets of lR, the intervals. Using this notion, we defined an "outer measure" by the process (2.1) and proceeded from there to obtain a measure. In general, an outer measure on a set X is a function /-L* : P(X) satisfying the following three conditions:

---+

[0,00]

/-L*(f/J) = 0,

(5.1)

(5.2)

A

(5.3)

Aj

C

c B

=====?

X countable

/-L*(A) ::; /-L*(B),

=====?

/-L*(UAj) ::; L/-L*(Aj). j

j

Recall that P(X) is the collection of all subsets of X. Parallel to (2.1), here is a way to construct outer measures. Let £ be some family of subsets of X, such that f/J E £ and X = Uj :2:1 Xj for some countable collection Xj E £. Suppose you have a function

(5.4) For any SeX, set

p: £

----+

[0,00],

p(f/J) = 0.

/-L*(S) = inf {LP(Ej ) : E j E £, S

(5.5)

j~l

Here {Ej

}

is a countable cover of S by sets in £.

c

UE

j }.

j~l

-

57

58

5.

The Caratheodory Construction of Measures

Proposition 5.1. Under the hypotheses above. J-L*, defined by (5.5), is an outer measure. Proof. Property (5.1) follow~ from p(f/J) = 0 and (5.2) from the fact that, when A c B, any countable cover of B by elements of E i& also a cover of A. The proof of (5.3) works the 8ame way a& the proof of Proposition 2.1; each Aj has a countable cover {Ejk : k 2: 1} by elements of E, &uch that J-L*(AJ ) 2: I.:k p(EJk) - 2- J E. Then {Ejk : ), k 2: 1} i~ a countable cover of Uy Aj by element8 of E, and WE' obtain (5.3) in the limit a8 E -----> O.

If we have an outer mea8ure J-L*, then, as sugge:;,ted when we 8tated Proposition 2.12, we ~a:v that a 8et A c X i8 J-L* -measurablE' if and only if (5.6)

J-L*(Y)

= J-L*(Y n A) + J-L*(Y \ A).

vY

c X.

(By (5.3), :S alwaY8 holds, 80 the condition to check is 2:.) DenotE' by M the class of J-L* -mea~urable sub8et8 of X The following. re~ult is known a.'> Caratheodory's Theorem. Theorem 5.2. If J-L* ts an outer measv,re on X, then the class M of J-L*measurable sets is a (J-algebra, and the restriction of J-L* to M is a measv,re. Proof. Clearly 0 E M. Also, if A E M, then, for all Y c X, YnAc = Y\A and Y \ AC = Y n A, so M i8 cl08ed under complement8.

Next, supp08e A J E M. We want to 8how that (5.6) holds with A = Al U A 2. Since J-L* is :;,ubadditive, it ~uffice:;, to e~tabli8h that

for all Y c X. Note that Al U A2 = A1 U (A2 n AI) i~ a di~joint union We see that the right :;,ide of (5.7) is

+ J-L*(Y n A2 n AJ:) + J-L*(Y n Al n A 2) = /-L*(Y n A l ) + J-L*(Y n AI) :S J-L*(Y n Ad

(5.8)

=/-L*(Y), where the la&t two identitie8 U8e A2 E M and Al E M, re&pedively Thi:;, yield8 (5.7) and ~how8 that M i8 an algebra, i e., M i8 closed under complement8 and finite union8. We next check additivity when A l , A2 E M are disjoint Indeed, if we 8et Y = Al U A 2 , A = Al in (5.6), we get

5

59

The Caratheodory Construction of Measures

Inductively, we have, for Aj EM, disjoint, (5.10) Taking N /

tL*

N

N

j=I

j=I

(U AJ) = L:tL*(AJ).

()() and using monotonicity of tL* yield tL*(U AJ) :::: L:tL*(AJ), j~l

J~I

and ~ince the reversE' inequality holds by (53), we have for a countable family A), (5 11)

tL*

(U Aj) = L j~[

tL*(AJ)'

A J EM, disjoint.

J~l

Compare the proof of Proposition 2.7, leading to Theorem 2.8 To finbh the proof of Theorem 5.2, we need to show that, if A J E M is a countable disjoint family, then A = UJ~l A J is tL* -measurable. Let En = Uj:S:n A J • The mea~urability of An implies tL*(Y

n En) = tL*(Y n En nAn) + tL*(Y n En n A~) = tL*(Y nAn) + tL*(Y n En-I),

for any Y C X. Inductively, we obtain n

(5.12)

IL*(Y

n En)

= LIL*(Y

n AJ )

j=I

We know M is an algebra.

(5.13)

~o

En E M, and hence, using (5.12), we have

n

: : L: tL*(Y n Aj) + Il*(Y n A

C

),

j=l

the last inequality holding because En C A. Taking n

--+ 00,

J~l

(5.14)

:::: tL*(U(YnA J )) +tL*(YnAr) j~l

= tL*(Y n A) + IL*(Y n A

C)

::::

tL*(Y).

we have

60

5.

The Caratheodory Construction of Measures

Thus A E M, and Theorem 5.2 is proved. While the construction (5.4)-(55) is very general, you need extra structure to relate /-L* (S) to p( S) when S E [.. The following is a convenient setting. Let A be an algebra of subsetI', of X, i.e, a nonempty collection of subsets of X, closed under finite unions and under complements, hence under finite intersections. A function /-Lo : A ---+ [0,00] is called a premeasure if it satisfie& the following two conditions:

/-Lo(0) = 0,

(515) (5.16) SJ

E

A countable, disjoint, USj = SEA ~ /-Lo(S) = L/-Lo(SJ)' j

j

As an example, let X = I = [a, b], and let A consist of fimte unions of intervals (open, closed, or half-open) in I If S = Uf=l Jk is a disjoint union of intervals, take /-Lo(S) = I:f=l £(Jk), so /-Lo is the restriction of Lebesgue (outer) measure to A. In this case, (5.16) can be demonstrated by an argument similar to that needed for Exercise 1 in Chapter 2.

If /-Lo is a premeasure on A, it induces an outer measure on X via the construction (5.4)-(5.5), i.e.,

(5.17)

/-L*(E)

= inf

{ L /-Lo(Aj) . Aj E A, E J~O

c

UAj} J~O

Proposition 5.3. If /-Lo is a premeasure on A and /-L* ~s defined by (5.17). then

(5.18)

SEA ~ /-L*(S) = /-Lo(S),

and every set in A is /-L* -measurable. Proof. To prove (5.18), first note that /-L*(S) ::; /-Lo(S) for SEA ~ince S covers itself. Suppose SEA, and S C Uj~l A J, A J E A Then let Bn = S n (An \ UJ
(519)

/-Lo(S)

=

L/-Lo(BJ)::; L/-Lo(Aj ). j~l

j~l

It follows that /-Lo(S) ::; /-L*(S). This proves (5.18).

5.

The Caratheodory Construction of Measures

61

To prove that each A E A is J.L* -measurable, if Y c X and c > 0, there is a sequence {Bj : j 2: 1} c A with Y c Uj 2:1 B J and LJ.Lo(Bj) ::; J.L*(Y) +c. Since J.Lo is additive on A, J.L*(Y) + c 2: ~ J.Lo(Bj n A) + ~ J.Lo(BJ n A C ) (5.20)

J2:1

2: IL*(Y n A) + J.L*(Y n A'), the latter inequality holding, e.g, because {BJ n A : j 2: I} is a cover of Y n A by elements of A Taking c ~ 0, we obtain (5.6), so any A E A is J.L* -measurable. When we combine the last result with Theorem 5.2, we obtain an extension of the premeasure J.Lo to a measure. Theorem 5.4. Let A C P(X) be an algebra, J.Lo a pr-emeasure on A, and M = a(A) the a-algebra generated by A. Then ther-e exists a measur-e J.L on M whose r-estr-iction to A ~s J.Lo, namely J.L* wher-e J.L* zs given by (5.17).

1M'

Proof. In fact, the class of J.L* -measurable sets is a a-algebra containing A; hence it contains a(A). There is nothing more to prove.

We next examine the extent to which the extension J.L of J.Lo to a(A) is unique. Suppose 1/ is a measure on M = a(A), which also agrees with J.Lo on A If E E M and E C Uj 2:1 A J , A J E A, then (521)

I/(E) ::;

I:: I/(Aj) = I:: J.Lo(A j2:1

J ),

J2:1

so, by the construction (5.17) of J.L*, (5.22)

I/(E) ::; J.L(E),

VE E a(A).

Also. by the property (3.5),

(523)

B j E A, B J /

B ~ I/(B) = lim I/(Bn) = lim J.L(Bn) = J.L(B). n---+oo

n~oo

Suppose now that E E M and J.L( E) < 00 Fix c > O. Then we can choose the cover A J in (5.17) such that LIL(A j ) < J.L(E) + c. Hence, with Bn =

UAj /

B =

UA

j,

j~n

we have J.L(B) < J.L(E)+c, so J.L(B\E) < c, and hence, by (5.22), I/(B\E) < c. Meanwhile, by (5.23), I/(B) = J.L(B), so

(5.24)

J.L(E) ::; J.L(B)

= I/(B) = I/(E) + I/(B \ E) ::; I/(E) + c.

Taking c ~ 0 gives J.L(E) ::; I/(E), as long as J.L(E) < (5.22), this yields the following.

00.

In concert with

62

5.

Tbe Caratbeodory Construction of Measures

Proposition 5.5. Let A be an algebra of subsets of X, generating the 0"algebra M = O"(A). Let t-to be a premeasure on A, and let J.i be the measure on M gwen by Theorem 5·4, extendinq t-to· If v is another measure on M whzch agrees unth t-to on A, then for all S E M, v(S) ::; t-t(S), and /1,(5)

(5.25)

<

00 ===}

J1,(5)

=

v(S).

Furthermore. zf there is a countable family A J such that

x

(5.26)

=

U Aj ,

A J EM.

t-t(A))

<

00,

)21

then t-t(S)

= v(S)

for all S E M.

Proof. The implication (5.25) was established above. If (5.26) holds, we can assume the Aj are disjoint. If 5 EM, we have the disjoint union 5 = US)' 5) = 5 n A). By (5.25), J1,(Sj) = v(Sj), and then peS) = v(S) follows by countable additivity.

If (5.26) holds, we bay (X, M, t-t) is a O"-finite measure space. See Exercbe 15 for another proof of Proposition 5.5. If Theorem 5 4 is applied to the example mentioned after (5.16), it yields the fact that finite unions of intervals in I = [a, b] are Lebesgue measurable and hence so are sets in the O"-algebra O"(A) generated by this algebra of subsets of I. In particular, all open sets in I are measurable, since they are countable unions of open intervals. Hence all closed sets in I are measurable Of course, this recaptures results obtained in Chapter 2.

Recall that, in our treatment of Lebesgue measure on an interval, we gave a definition of measurability different from (5.6) and then showed in Proposition 2.12 that the definition of measurability given there implied (56). We now give a result in counterpoint to Proposition 2 12. Proposition 5.6. Suppose we are gwen an algebra A of subsets of X and a premeasure t-to on A, with associated outer measure t-t*, defined by (5 17). Assume Z C X is t-t*-measurable and t-t*(Z) < 00. Then a set S C Z is t-t* -measurable zf and only if

(5.27)

t-t*(S)

+ J1,*(Z \

S)

= t-t*(Z).

Proof. We need to show that, if (5.27) holds, then, for any Y C X,

(5.28)

5.

The Caratheodory Construction of Measures

Only the "~" part needs to be established, so we can assume {L*(Y) < 00. Write Y = Yo U Y1 , a disjoint union, where Yo = Y n Z, Yl = Y n Using the measurability of Z, we have

ze.

(5.29) and

provided S

c Z. while

(531) Consequently it suffices to prove (5.28) for Y = Yo, i e., for Y we bring in a lemma

c

Z. For that,

Lemma 5.7. Let {L* arise from a pTemeasure on an algebra A. Let Aa conszst of countable unions of sets zn A. Then

(532)

SeX

==?

{L*(S)

=

inf{{L*(E) : E E A a , SeE}.

Proof. We leave thi& to the reader; see Exercise 3 at the end of this chapter.

We continue' the proof of Proposition 5 6. Given E > 0, there exists A E Aa such that A ::J Y and {L*(A) ::;: {L*(Y) + E. Set A = An Z Then A E M (the (J-algebra of It*-measurable ~ets), A ::J Y (if Y c Z), and It*(A) ::;: {L*(Y) + E. We claim that

(5.33) Suppose this is known Then we have (5.34) for all E > 0. In the limit E -+ 0, we obtain the ~ part of (5.28), and the problem is solved. Thus it remains to establish (5.32), given A E M, A c Z, {L*(Z) < 00, and given the hypothesis (5.27). Using the measurability of A, we have (535)

64

5.

The Caratheodory Construction of Measures

Now the left side of (5.35) is equal to J1,*(S) + J1,*(Z \ S), by hypothesis. By subadditivity of outer measure, the right side of (5.35) is

n S) + {t*(A n SC) + J1,* ((Z \ A) n S) + J1,* ((Z \ A) n SC) = {t*(S n A) + J1,*((Z \ S) n A) + J1,*(S \ A) + {t*((Z \ S) \ A) = {t*(S) + {t*(Z \ S), ~ {t*(A

(5.36)

the last identity following by grouping together the odd terms and the even terms on the second line of (536) and using measurability of A Since the bottom line of (5.36) is equal to the left side of (5.35), the ~ in (5.36) must be equality. That inequality arose from the sum of two inequalities, and so both of them must be equalities. One of them is the desired result (5.33). This finishes the proof of Proposition 5 6. We next describe an important class of outer measures on metric spaces, for which all open sets and all closed sets can be shown to be measurable. This result will play an important role in Chapter 12, on Hausdorff measures, and in Chapter 13, on Radon measures. Let X be a metric space, with distance function d(x, V). An outer measure J1,* on X is called a metric outer measure provided (5.37)

p(Sl,S2) = inf {d(Xl,X2): Xj =::::}

E

J1,*(Sl U S2) = {t*(Sl)

Sj} > 0

+ {t*(S2).

Note that the part (2.15)-(2.16) of Lemma 2.4 is the statement that Lebesgue outer measure is a metric outer measure on I = [a, b]. The following is another result of Caratheodory: Proposition 5.8. If {t* zs a metric outer measure on a metnc space X, then every closed subset of X is {t* -measurable. Proof. We must show that if Fe X is closed and Y C X satisfies J1,*(Y) < then

00,

(5.38)

+ J1,*(Y \

F).

Bn = {x E Y \ F : p(x, F) ::::

~}

{t*(Y) :::: J1,*(Y n F)

Consider

Note that Bn / (5.39)

Y \ F. Also p(Bn, F) ::::

lin,

so, by (5.37),

5.

'1 'he Caratbeodory Construction of Measures

Thus it will suffice to show that (5.40) Let C n = Bn+l \ Bn. Note that

Hence, for any N, one obtains inductively (using (5.37)) that N

L M*(C

2J )

N

= M*

(U C

j=l

(541)

2j )

:s: M*(Y),

j=l

N

N

LM*(C2J +l) = M*(U C2j+l) and consequently (5.42)

:s: M*(Y),

j=l

j=l

I:j 2:1 M*(CJ ) < 00

Now countable subadditivity implies

M*(Y \ F)

:s: M*(Bn) + L

M*(C]),

J2:n

so as n

-> 00,

the last sum tends to zero, and we obtain

(5.43) the last inequality by monotonicity. This implies (5.40) and finishes the proof. As one may have noticed, at several points in this chapter and previous chapters, we have considered a class of sets with certain desirable properties and have wanted to prove it was a o--algebra. The following result, called the Monotone Class Lemma, sometimes furnishes a convenient tool for doing this. We define a monotone class on a set X to be a collection C c P(X) having the properties (5.44)

Ej E C, E J

/'

E

=:}

E E C,

Ej E C, E J

\.

E

=:}

E E C.

The smallest monotone class containing a collection [; monotone class generated by [;.

c P(X) is called the

Proposition 5.9. If A is an algebra of subsets of X and C is the monotone class generated by A, then C = dA).

66

5.

The Caratheodory Construction of MeasUIes

Proof. Clearly C C o-(A) We will show that C is a o--algebra. If E E C, let

(545)

C(E)={FEC'E\F, F\E, EnFEC}.

Clearly 0, E E C(E), and C(E) is eahily verified to be a monotone class. Also, E E C(F) {:} F E C(E) Since A is an algebra, we have

E, FE A===} F

(5.46)

E

C(E)

In other words, if E E A, then A C C(E), hence C C C(E). Thus, if FE C, then FE C(E) for all E E A. Thih implies E E C(F),

(547) Thus A (5.48)

\I E E A. FE C.

c C(F) and hence C C C(F), for all F

E

C. In other words,

E, F E C ===} E \ F, E n F E C.

Since X E A c C, it follows that C is an algebra of sets. We finally note that C must be a o--algebra. Indeed, if E j E C, thE'n FJ = Un~j E J E C, and (5.44) implies Fj /' F = Uj~l E J E C. This finishes the proof. ThE' proof of Proposition 6.2 in the next chapter will provide a nice application of the Monotone Class Lemma.

Exercises 1. Let I = [a, b] and let A be the algc'brcl of subsets of I consisting of finite unions of intervals, as in the example after (5.16). Let

lR be a monotonically increasing function, i.e., assume that x < y ::::}
(5.49)


=

lim

=

lim
u,,"x

with the convention that lR+ as follow~. If 5 = U~=l .h is a union of mutually disjoint intervals, set 1),0(5) = I:~=l f-lO(Jk), where f-lO is defined on a single interval J C I as follows:

J),o([x,yJ) =
f-lo([x,y») =
5

67

The Caratheodory Construction of Measures

Also, set fJ,o(0) = O. Show that fJ,o is a premeasure. The measure then produced via Theorem 5.4 is called "Lebesgue-Stieltjes measure." Hmt. Say J is a closed interval and J = Uk>l ,h, with disjoint intervals J",. First show that L~=l fJ,O(Jk) ::; fJ,0(1) for each m, so fJ,o(J) 2': Lk>l fJ,O(Jk). For the reVE'rse inequality, one can argue as follows. Pick c >-0 Say the endpoints of Jk are ak and bj, so (ak' bk ) C Jk C [ak, bkJ. If ak E Jk, pick Uk < ak such that 0 ::; 'iJ-(ak) - 'iJ+(Uk) < c2- k . If 0,,,, Jk, let Uk = ak· Make an appropriate analogous choice of bk, and let Jk = (Uk, bk) Note that I-Io(1k) < fJ,O(Jk) + c2- k +1. Now J C Uk21 Jk, and this is an open cover. Take a finite subcover.

tt

For Exercises 2 6, suppo~e we are given an algebra A of subsets of X and a premeasure tlo on A, with associated outer measure fJ,*) defined by (5.17). Let ACT consist of countable unions of sets in A. 2 Show that

(5.51)

E E ACT

==?

fJ,*(E) = sup {Po(A) A E A, ACE}.

3 Show that

(5.52)

8 eX==? p*(5)

= inf {ri*(E) : E

E

ACT' 5

C

E}.

4. Let A8 C'onsbt of countable intersections of sets in A, and set (5 G3) Show that /),*(5) ::; p*(5). If 11,*(5) < 00. show that 5 i~ p*-measurable if and only if /1,*(8) = p*(5). Hmt. Show that, if 8 C Z E Aa. /1*(Z) < 00, then p*(Z) = p*(5) + p*(Z \ 5). Then use Proposition 5.6.

5. Given 8 C X, show that there exist 50 E

Ao-,j

and 51 E

A,jo-

such that

(554) Show that, if p*(5) < 00, then 5 is /h*-mea~urable if and only if p(51 \ 50) = O. Here, p is the measure on o-(A) given by Theorem 54. 6. Let fJ,# be the outer measure on X obtained via (5.17), with (A, po) replaced by (o-(A),fJ,). Equivalently,

(5.55)

fJ,#(5) = inf {fJ,(A) : 5 CAE o-(A)}.

68

5.

The Caratheodory Construction of Measures

Clearly f.L#(S) ::; f.L*(S) for all SeX. Show that f.L# = f.L* Hint. If SeA E (T(A), show that, for any E > 0, there exists BEAu such that A C Band f.L(B) ::; f.L(A) + E 7. If f.L* is an outer measure on X, then clearly (556) If f.L * arises from a premeasure, via (5 17), bhow that M = f.L * (S). Hmt. First note that An E cr(A), An /' A ==? f.L*(An) /' f.L*(A). Then show that, for any E > 0, there exist Zn E (T(A) such that Zn =:> Sn, Zn /" and f.L*(Zn) = f.L*(Sn). l\lake use of (5.54).

8. As in Exercbe 5 of Chapter 3, a measure f.L on (X, M) ib said to be c-omplete provided (557)

A E M, f.L(A) = 0, SeA

====?

S E M, and f.L(S) =

o.

Show that the measureb provided by Theorem 5 2, from outer mea&ures, are all complete. In particular, Lebesgue measure i& complete. 9 If f.L is a measure on (X, J) that is not complete, form the outer measure 11* by (517), with A replaced by J Let M be the (T-algebra of f.L*measurable setb Show that Theorem 5.2 produc-es a measure 7i on (X, M) which is complete and buch that J eM and 7i = f.L on J Show that M = J, given by Exercise 5 of Chapter 3, and that 7i coincides with the measure produced there, the "completion" of 11.

In Exercbes 10-14, suppose X is a compact metric space, 113 the (Talgebra of Borel subsetb of X, and f.L a finite measure on (X,Q3). Construct the outer measure It* via (5.17), with A replaced by 113. (Note that Propobition 5 3 applies in this case.) 10. Show that f.L* must be a metric outer measure. Hmt If SI, S2 satisfy (537), construct compact K C X such that S1 eKe X \ S2. Apply (5.6) with A = K, Y = Sl U S2. 11 We know that 113 = (T(A) where A is the algebra generated by the compact sets in X. Show that A consists of sets of the form N£

(558)

M£

Nl

Ml

un··· u n E

0:£=1 {3£= 1

0:1 =1 {31 =1

a {3,

5.

69

The Caratheodory Construction of Measures

where each

is either open or closed.

EOI.{3 C X

12. Show that (5.59)

SeX

===}-

j.t*(S) = inf {j.t*(0) : S C 0, 0 open}.

Hint By Exercise 6, 11*(S) is given by (552) or, equivalently, by (5.17) (with some changes in notation) Show that, given A J E A, £ > 0, there exist open sets OJ such that A J C OJ and j.t(Oj) :s: j.t(Aj) + 2- j £. It suffices to make such a construction for each EOI.{3 in (5.58), and we need worry only about the case when EOI.{3 is compact; i.e., we need to verify (5.59) when S is compact. For this, keep in mind that X is a metric space. 13. Show that (5.60)

S E 23

===}-

11( S)

= sup

{j.t( K) : K C S, K compact}.

14 If f E £1 (X, j.t) and £ > 0, show that there exists a compact K C X such that j.t( X \ K) < £ and f IK is continuous. Hmt. Using Proposition 45 (or otherwise), produce fv E C(X) such that fv ----7 f, j.t-a.e. Then use Egoroff's Theorem. Then use (5.60). This result is Lusin's Theorem. A special case was stated in Exercise 16 of Chapter 3. 15. Use the Monotone Cla~~ Lemma to give another proof of Proposition 5.5. Furthermore, establish the following variant of Proposition 5.5 Proposition 5.5A. Let A be an algebra of subsets of X, generating the a-algebra M = a(A). Let j.t and v be measures on M. Assume there

exist AJ such that X

=

UA

J,

A J E A,

j.t(AJ ) <

00.

j~l

If j.t = v on A, then j.t = v on M. How do these two propositions differ? 16. Suppose £ is a collection of subsets of X having the following properties: (1) The intersection of any two elements of £ belongs to £ (2) The collection of finite disjoint unions of elements of £ is an algebra

A.

o

5.

Let

ji,b :

The Caratheodory Construction of Measures

E --> [0, 001 satisfy

(a) /Lb(0) = 0,

(b) E j E E countable, disjoint, Uj E j = E E E::::} /Lb(E) = 'L- j /Lb(EJ ) Show that /Lb extends to a premeasure /Lo on A, satisfying K

El

, ... ,

EK E E disjoint ::::} ILo(El u· . u EK)

=

L

/Lb(Ej ).

j=l

Show that if

ji,*

/L*(E)

is defined by (5.17), then also, for E eX,

= inf{I:: /Lb(Aj) : A J j~

E

E, E c

UA

J }.

J~

Hint. Start with uniqueness; if also F l , ... , FL E E are disjoint and U~'=l Fc = U~l Ek, write this set as Uk,e(Ek n Fe) to show ILO is well defined. Note. An example is X = [0,1]' E = the collection of intervals in [0,1]' /Lb(J) = t(J), the length. Another family of examples arises at the beginning of Chapter 6.

17. Suppose E is a collection of subsets of X satisfying property (1) Exercise 16 and also satisfying (2') E E E ::::} X \ E is a finite disjoint union of elements of E. Show that E satisfies property (2) in Exercise 16

III

Chapter 6

Product Measures

If (X, M, ft) and (Y,N, v) are measure spaces, we construct a a-algebra M ® N of subsets of X x Y and a product measure ft x v on M ® N, as

follows. We say that a rectangle is a set of the form A x E c X x Y, with A E M, E EN. We define M ®N to be the a-algebra generated by the collection R of rectangles. Note that

(6.1)

(A x E)n(B x F) (A x E) U (B x F)

= (AnB) x (EnF), = A x (E \ F) U (A U B) x (E n F) U

B x (F\E),

the latter decomposition being a disjoint union of rectangles. Also,

(62) Hence the collection of finite disjoint unions of rectangles is an algebra, which we denote M f2J.N, and M ®N is the a-algebra generated by M f2J.N. We define the set function

7r . M

f2J. N ---+

N

(6.3)

[0,00] by

N

7r(U(A

J X

EJ ))

=

j=l

L ft(Aj)v(Ej), j=l

when the rectangles R j = A J x E j are disjoint. We claim on M f2J. N. We need to check that (6.4)

t-t(A)v(E)

7r is a premeasure

= Lft(AJ)v(Ej ) j2:1

when A x E is a countable disjoint union of rectangles Aj x E j . Establishing (6.4) will also show that (6.3) is well defined for an element of M f2J. N

-

71

72

6.

Product Measures

which has several different representations as disjoint unions of rectangles. To prove (6.4), write

(6.5)

XA(X)xE(Y) = XAXE(X,y) = LXAjXEj(X,y) = LXAj(X)xEj(Y) j2:1

Integrating with respect to x and applying Proposition 3 5 yield

M(A)XE(Y) = L

XEj (y)

(6.6) =

J

XAj(X) dM(X)

LIL(Aj)XEj(Y),

VY E Y

Integrating this with respect to Y and using a parallel argument, we get (6.4). We mention that this setup is a special case of that treated in Exercise 16 of Chapter 5. Here [; = n, the collection of rectangles. and A = M ~ N. Using the construction (5.17), we obtain Note that, for SeX X Y,

(6.7) 7l"*(S)

= inf{L M(Ak)V(Ek) : Ak

eW

outer measure

EM, Ek EN, S

k2:1

c

7l"*

on X x Y

U (AkX Ek) } k2:1

By Theorem 5.4, the re~triction of 7l"* to M ®N is a measure, extending the pre measure on M ~ N defined above. We call this measure on M ® N the prod uct measure M x v We note that, if M and v are a-finitc, so is M x v, and, by Proposition 5.5, in this case M x v is the umque measure on M ® N with the property that (IL x v)(A x E) = M(A)v(E), A E M, E EN. If f is a measurablc function on (X x Y, M ® N), we want to see when the iterated mtegral

(68)

.f[.f f(x,y)dv(y)] dM(X)

is well defined and equal to the "product integral" of f with respect to the product measure 11, x v To begin, we define the x-section fx and y-section fY of a function f on X x Y by

(6.9)

fr(Y)

=

r(x) = f(x, y).

Related concepts are those of the x-section Ex and y-section EY of a set E c X x Y, defined by (6.10)

Note that (6.11)

Ex = {y E Y : (x, y) E E},

EY

= {x EX: (x, y)

E E}.

6.

73

Product Measures

Lemma 6.1. If E E M®N, then Ex EN for all x E X and EY EM for all y E Y. More generally, if f is M ®N-measurable, then fx ~s N-measurable for all x E X and fY is M -measurable for all y E Y. Proof. Let

F = {E c X x Y

Ex E N for all x and EY E M for all y}.

Then every rectangle is certainly in F Furthermore,

(U EJ).r = U (EJ)x' j?:l

and (Exr

= (EC)x'

j?:l

Thus F is a cr-algebra, 50 F::J M ® N. Since (jx)-l(S) the lemma follows.

= (j-l(S))x and (jyr1(S) = (j-l(S))Y, the rest of

The next result i5 a basic special case of the forthcoming Theorem 6.3, when f i5 a characteristic function. Proposition 6.2. Assume l.l and v are cr-finite. and take E E M ® N.

Then, cp(x) = v(Ex) and 1.f;(y) = p,(EY) are measurable,

(612)

and (6.13)

(Jt x v)(E) =

J

v(Ex) dJ.l(x)

=

J

p,(EY) dv(y).

Proof. First we assume l.l and v are finite. Let C consist of sets E E M rg;N

such that (6.12) and (6.13) hold, our goal if:> to prove that C = M ®N. We will do this by applying the l\lonotone Cla58 Lemma, Propof:>ition 5.9. Thus, to start, we see that rectanglef:> A x B clearly belong to C, and so do finite disjoint union5 of rectangles, f:>0 M rgj N c C It remains to show that C is a monotone clasf:> Let E j E C, E J / E. Then Ej / EY, so 'Ij}J(Y) = 1.l(E%) / 'tj;(y) = J.l(EY), and the Monotone Convergence Theorem implies (6.14)

J

p,(EY) dv(y) = lim

J

'ljJj(Y) dv(y)

= lim (p,xv)(EJ ) = (p,xv)(E).

Similarly we obain the other identity in (6.13), 50 E E C. On the other hand, if Ej E C and E J ~ E, then EY ~ EY, and hence 'ljJj(Y) ~ 'IjJ(y), provided p, is finite. As long as v is also finite, we can apply

74

6.

Product Measures

the Dominated Convergence Theorem and again get (6.14). Similarly we get the rest of (6.12)-(6.13) for such E, so C has been shown to be a monotone dass, and the proposition is proved, at least for finite measures JI and v. The extem,ion to the CT-finite case is routine. The next two results, known as Tonelli's Theorem and Fubini's Theorem, respectively, are the major result::, of thb chapteI Theorem 6.3. Assume (X, M. jJ,) and (Y,N, v) are CT-finite measure spaces. If f E M+(X x Y), then the funetzons

g(x) =

(615)

J

fx dv,

hey)

=

J

fY dJI

are measurable, hence elements of M+(X) and M+(Y). r'espectwel1j. and

J

(616)

f(x, y) d(II x v)

=

J

g(x) dJL(x)

=

J

hey) dv(y).

Proof. Take a sequence of simple functions fj E 6+(X x Y), fj /' f, such as comMucted in (:320). The results above, for f replaced by fj, are immediate from Proposition 6.2 and additivity of the illtegraL Note that, for each x E X, fJX /' fx and, for each y E y, fi /' p. Thus, with obvious notation, gj /' 9 and h J / ' h, by the l\Ionotone Convergence Theorem. Then (6.16) also follows by the l\Ionotone Convergence Theorem Theorem 6.4. Let X and Y be as m Theorem 6.S. If j E £1 (X X Y, II x v), then

(6.17)

JY

fI E £1 (Y.v) for JL-a e. x E X.

E £1 (X. 11) for v-a.e. y E y,

so g and h, given by (6 15), are defined almost everywhere. We have

(6 18) and (6 16) holds

Proof. If we write f = f+ - f- with f± E M+(X x Y)n£l(X x Y,II x v), then Theorem 6.3 applies to f+ and to f-, yielding g± E M+ (X) and h± E M+(Y) such that

(6.19)

I

f±(x, y) d(JL x v)

=

J

g± dJL

=

J

h± dv

6.

Product Measures

75

The finiteness of the first integral in (6.19) implies that g± and h± are finite a.e. and belong to L1 (X, J.t) and L1 (Y, 1/). respectively. Thus 9 = g+ - gand h = h+ - h- satisfy (6.18), and (6.16) also follows from (6.19). Let us write (616) in thE' form /

f(x, y) d(ll, x 1/) = / [ / f(x. y) dl/(Y)] dp,(x)

(6.20)

= / [/ f(x. y) dtt(x)]

dl/(y).

The first intq!;ral is called a produ("t integral and the other two quantities in the triple identitv (620) are called iterated integrals It is u~eful to have ~ome sufficient ("onditionfl for a function 9 . X x Y to be M ®N-measurable Clearly if 7fx : X x Y --t X and 7fy : X x Y are the standard projectionfl. then (621)

f :X 9 :Y

--4 --4

lR M-measurable

===}

f 0 7fx M ® N-measurable,

N-mea~urable

===}

go 7fy M ® N-measurable

lR

--t --t

lR Y

Also, by Propo~ition 3.1, the algebra 2l ("onsbting of finite sums of products of functions of the type (6.21) are M ® }J-measurable. Also recall that pointwifle limit~ of measurablE' functiom, are measurable. Using these facts together with the StonE'-\Veierstra~s Theorem. we ("an prove the following: Proposition 6.5. Let X and Y be compact Hausdorff spaces and Il3x, ll3y their (J -algebras of Borel sets. Then an'/) contmuous f : X x Y --4 lR is 23 x ® 'By -measurable Proof. The observation~ above imply that f is Il3x ® ll3 y -measurable if f belongs to the algebra C of finite ~um~ of fun("tions of the form g(x)h(y), with 9 and h continuou~. Abo, f is 'B x ® ll3y-mea~urable if f belongs to the clo~ure C of C in C(X x Y). However, the Stone-Weierstrass Theorem implies C = C(X x Y), so we have the re~mlt. If X and Yare compact metric

space~,

we ("an

~ay

more.

Proposition 6.6. If X and Yare compact metric spaces, then any compact K c X x Y belongs to 'Bx ® ll3y. Hence

(6.22)

Il3x x Y = 'Bx ® 'By

6.

76

Product Measures

Proof. In this case, X x Y is a metric space, and, if we denote its distance function by d, then

'lJK(Z) = d(z, K)

(6.23)

is a continuous function on X x Y such that K = {z E X x Y . 'lJK(Z) = O}. According to Proposition 6.5, 'IJ K is SB X ® SBy-measurable, &0 each compact K c X x Y belongs to SBx ® SBy. Thi& implies SBxxy C SBx ® SBy. For the reverse containment, we argue as follows Let Ax = {A eX' A x E E SBxxy, Vcompact E

c Y}.

Then Ax is a cr-algebra, containing all compact A eX, so Ax Hence A E SBx, E c Y compact ====? A x E E SBxxy

~

SB x.

Now, let £y

= {E c Y : A

x E E SBxxy, V A E SBx}

We see that £y is a cr-algebra, containing all compact E ment above, so £y ~ SBy. Hence A E SBx, E E SBy

====?

C

Y, by the argu-

A x E E SBxxy.

It follows that SB x x Y ~ SB x ® SBy. Note that this latter containment holds whenever X and Yare compact Hausdorff 8paces; we do not need them to

be metric spaces. The following i& an immediate consequence of Propo8ition 6.6. Corollary 6.7. If X and Yare cr-compact metric spaces. then (6.22) holds

The product measure space (X x Y, M ® N, ft x v) is almost never complete, but one can construct its completion (X x Y, M ® N, ft x v) via the proces& (3.44). The following i~ a version of the Fubini-Tonelli Theorem for M ® N-measurable function8. Proposition 6.8. Assume (X, M, ft) and (Y,N, v) are complete, cr-finite measure spaces. Let f : X x Y -----t lR be M ® N -measurable and satisfy

(a) f 2: 0

(b) f

or

E

£: 1 (X

X

Y, M ® N)

Then fx is N-measurable . .tor t--t-a.e. x,

(6.24)

fY zs M-measurable, for v-a.e. y.

In case (b), fx and fY are integrable for a e. x (resp., a.e. y). Also, g and h, defined a.e by (6.15), are measurable, and in case (b), integrable, and (6.25)

J

f d(t--t x v)

=

J

g(x) dt--t(x)

=

J

h(Y) dv(y).

6.

77

Product Measures

Proof. Using Exercise 6 of Chapter 3, write f = fo + !I, where fo is M 0Nmeasurable and satisfies either (a) or (b), while !I = 0, f.1 x v-a.e. Theorems 6.3-6.4 apply to fo, so it suffices to show that

(6.26)

!Ix = 0 v-a.e.,

for f.1-a .e x,

fi

= 0 f.1-a.e., for v-a.e.

y.

As in the proof of Theorems 6.3 6.4, it suffices to show that, for E E M 0 (f.1 x v)(E)

(6.27)

= 0 =?

= 0 for f.1-ae. x and f.1(EY) = 0 for v-a.e. v(Ex)

Now E c F with (f.1 x v)(F) = 0; hence Ex C Fx and EY is a consequence of the statement that, for F E M 0 N, (f.1 x v)(F)

(6.28)

N,

C

= 0 =? v(Fx) = 0 for f.1-a.e. x and f.1(FY) = 0 for v-a e

y.

FY. Thus (6.27)

y.

This in turn follows from (6.12)-(6.13), together with (3.59), so the proposition is proved. One can also consider products of K measure spaces (Xj ,Mj ,f.1j), constructing a (I-algebra ~ = Ml 0 ... 0 MK of subsets of Z = TI~l X j , generated by rectangles of the form R = Al X ... x AK, A J E M J. The product measure v = f.11 X ... x f.1K satisfies (6.29) K

v(S)

= inf {Lf.11(Aj1 )· f.1K(Aj K) : AJk

E

Mk, S C

U(II A jk ) }, j2':l k=l

J2':l

for S E ~. Analogues of the results discussed above follow without difficulty; we leave the details to the reader. There is also a notion of product measure for a countable infinite product of measure spaces, (Xj , M J , f.1j), j E Z+, provided f.1j(XJ ) = 1 for all j, i.e, each f.1J is a probability measure. Then one takes rectangles of the form (6.30)

R =

II A j ,

Aj E Mj,

A J = Xj except for finitely many j,

J2':l

and ~ is the (I-algebra generated by such sets. The product measure v is a probability measure on Z = TI j 2':l X j , satisfying

(6.31)

v(S)

= inf

{L: II f.1k(A jk ) : II Ajk E n, S c U(II A jk ) }, j2':l k2':l

k2':l

j2':l k2':l

6

78

Product Measures

for S E ~, where R denotes the class of rectangles of the form (630). Note that, for each "infinite product" Ih>l/1k(A j k) in (6.31), all but finitely many factors are 1 The proof that :"etting vCR) = I1J 2':l/1j(A J ), for sets R as in (6.30), yields a premeasure on the algebra of finite disjoint unions of such sets can be given parallel to the argument involving (6.5)-(66). Thus, buppobe A J E M j, Ej£ E M J' and we have a countable disjoint union Al

X

.

X

Ak

X

X k+ 1

X·

.

=

U Elf x

.. x EfC(f),f

X XfC(£)H

x .

£2':1

so XAr(xd· 'XAk(Xk)

= LXEl£(X1)"

XE"U)f(XK(f»)'

£2':1

One can integrate, first over Xl, then over X2, etc., obtaining, at the kth step, /11(A 1 )

..

/1k(A k )

= LMl(Elf)' . /1k(Ek£)XEk+l f(XkH)'"

XE",U)

f(XK(£))'

£2':1

Continue to integrate over X k + 1 , etc Any given term in the sum above will stabilize eventually, and we obtain

MI(AI) .. /1k(A k )

= Lf1l(Elf)

.. /1K(£) (EfC(f),f) ,

£2':1

as desired. With a little more effort, one can define a product measure on an uncountable product of probability spaces. For a treatment of thi:", we refer to

[HS].

Exercises Let A and B be algebras of bubbets of X and Y, respectively, and let /10 and Vo be pre measures on these algebras, giving rise to measures /1 and v, on M = (l(A) and N = (l(B), respectively. Let us denote by A ~ B the algebra of :"ubbets of X x Y generated by rectangles of the form (6.32)

R

=A

x E,

A E A, E E B.

u.

~

1 UU U{;L

IVH:::i;j,/jUrel>

1. Show that A IZl B consists of finite disjoint unions of rectangles of the form (6.32).

2 Show that o-(A IZl B) = M ® N. Hint. Show that o-(AIZlB) :) MIZlN, the latter being the algebra defined after (6.2). 3. Define a set function N

fO .

A IZl B

---+

[O,ooJ by N

fo(U(A j x E J )) = LMO(A.J)vo(Ej ), J=1

)=1

when the rectangles R j = AJ x BJ E A IZI B are disjoint. Show that is a well-defined premeasure on A IZI B.

fO

4. Show that the measure') on o-(AIZIB) obtained from 'Yo via Theorem 5.4 coincides with the product measure f.L x v, provided f.L and v are (T-finite. Consequently, for S E M ® N,

(6

:~3)

(f.L x v)(S) = inf { L t-to(Aj)vo(Ej) . A J E A B) E B, S

c

j~l

More generally, for any SeX x Y, the outer measure the right side of (6.33). (Compare (6.7) )

U(A

J

x Ej

)}.

j~l

7[*

(S) is given by

In ExerciE>es 5-6, aSf>urne (X, M, II,) and (Y,N, v) are o--finite measure spaces 5 Let Sn con&i&t of simple functions on (X x Y, M ® N) of the form f = I:~:1 aJXRJ , where each Rj is a rectangle, Rj = A) XE J , A J E M, E j E N, such that (f.L x v)(Ry) < 00. Show that Sn is dense in LP(X x Y,M ®N), for 1::; p < 00. 6. If {uJ • j 2: I} is an orthonormal basis of L 2(X,p,) and {Vk : k 2: I} if> an orthonormal basis of L2(y, v), show that {'UJ(X)Vk(Y) . j, k 2: I} is an orthonormal basiE> of the Hilbert space L 2 (X X Y, f.L xv) Hint. Show that the closure of the linear span of thi& orthonormal set contains Sn.

6.

80

Product Measures

In Exercises 7-10, let 00

z=II{O,I}.

(6.34)

k=l

We put on each factor {O, I} the probability measure assigning the measure ~ to {O} and to {I} On Z we put the product iT-alge bra and the product measure, which we denote p. Given a = (aI, a2, a3, .. ), define 00

(635)

P: Z

-----t

P(a) =

[0,1]'

I: ak2-k k=l

Thus P(a) is a number with dyadic expansion 0.ala2a3···.

7. Show that P is measurable Hint. Let a dyadic interval be one with endpoints of the form m2-€ and (m + 1)2-£, with m, f E Z+, m + 1 ::; 2£; a dyadic interval mayor may not contain its endpoints. Show that the iT-algebra generated by the collection of dyadic intervals in I is equal to the Borel iT-algebra 'B 8. Show that P is onto Show that P-1(x) consists of one point unless x has a finite dyadic expansion, in which case p- 1 (x) consists of 2 points. Hint. L~21 2- k = 2- 1

9. Show that S E 'B =? p(p-l(S)) = m(S), where m is Lebesgue measue on I. Hmt. Start with S E Ao, the algebra of finite (disjoint) unions of dyadic intervals. 10. Show that you can take a set Zo C Z of measure zero such that P : Z \ Zo ---+ I is one-to-one and onto and measure preserving. 11 For this exercise, make a slight change of notation; set ex)

Z=II{0,2}. k=l

We use the same sort of product measure as before. Consider the map G . Z ---+ [0, 1] given by 00

G(a)

= Lak3-k. k=l

6.

81

Product Measures

Show that C is a homeomorphism of Z onto the Cantor middle third set K. Show that the measure /1K on [0,1], given by /1K(S) = /1(C-I(S)), coincides with the Lebesgue-Stieltjes measure on [0, 1J given by Exercise 1 of Chapter 5, using the following function cpo On the middle third of [0,1], set cp = 1/2. On the middle third sets of the two intervals remaining, set cp = 1/4 and 3/4, respectively. Continue in the appropriate fashion, to define a continuous monotone function cp : [0, 1J -+ [0, 1J.

12. Let (X,J,/1) be a meabure space and let j ' X Consider the set U Show that U E that

= ((x,t) E X

-+

[0,00) be measurable.

x [0,00): j(x) < t}.

J 0113, where 113 is the class of Borel subsets of lR, and (/1 x

m)(U) =

J

j d/1,

x where m denotes Lebesgue measure on R Compare this result with Exercise 18 of Chapter 3. Hint. First verify this for simple functions, and then consider simple Cpj / ' j, as in (3.26) As an alternative, what can you deduce from Proposition 6.27 13 Here is a variant of Exercise 6. For j E N, let (Xj ,MJ ,/1j) be measure spaces satisfying /1J (XJ ) = 1. Form the infinite product measure space (Z, J, v) as in (6.30)-(6.31), with Z = IT j :2:1 X j . Suppose that for each J E N, {llJk . kEN} is an orthonormal basis of L2(Xj,/1j) and lljl = 1 Consider the collection offunctions on Z, 00

va(z) =

II llkak (Xk),

Z = (Xl, X2, X3, .. ),

k=l

where Q = (QI' Q2, Q3,·· ) runs over the set A of elements of IT~l N such that Qk =I- 1 for only finitely many k. Show that {Va:

Q

E

A}

is an orthonormal basis of L2(Z, v). Hint. For the denseness of the linear span, consult the hint for Exercise 6.

Chapter 7

Lebesgue Measure on ~n and on Manifolds

We constructed Lebm;gue measure on ]R in Chapter 2. The (I-algebra £1 of Lebesgue measurable subsets of]R contains the (I-algebra ~1 of Borel subsets of R The product construction of Chapter 6 applies to ]Rn = ]R x ... x R We have the (I-algebra

(7.1) which, by Corollary 6.7, is the (I-algebra of Borel subsets of ]Rn. We have the larger (I-algebra £1 Q9 .• Q9 £1. As indicated in Chapter 6, the product mea8ure rn x· . x rn (which we will also denote rn) is not complete on this product (I-algebra. Let (]Rn, £n, rn) denote the completion We will also denote the extended meat>ure rn on £n simply by rn. It is easy to &how that thi8 measure space is also the completion of (]Rn, ~n, rn). Frequently, we will omit the subt>cripts from simply ~ and £. respectively.

~n

and

£n,

denoting them

Recall that ~ 1 is the (I-algebra generated by the algebra Al of finite (disjoint) unions of intervals. It follow8 that ~n i8 the (I -algebra generated by the algebra Al ~ . ~ AI, consisting of finite (disjoint) unions of sets in ]Rn which are products of intervals; we call such sets cells. Products of intervals of identical length will be called cubes. It is often convenient to use the formula (6.33) specialized to Lebesgue measure.

(72)

rn(S)

= inf {L: rn(Qj) . QJ j~I

cells, S c

U QJ }, J~l

-

83

84

7.

Lebesgue Measure on ~n and on Manifolds

for S E IB, where, for a cell Q,

(73) In fact, (7.2) hold:;, for all S (7.4)

m(S)

E

£, since (via Exercise 9 of Chapter 5) we have

= inf {rn(E) : E

E IB, E :::) S},

for S E £.

Now ~n has an important structure in addition to its product structure, namely a linear structure The next re:;,ult shows how the measure of a Borel set changes under a linear tramJormation and, more generally, how the Lebesgue integral on ~n behaves under such a transformation. Here, Gl(n,JR.) denoteb the group of invertible linear transformations on JR.n, and det A denotes the determinant of A Proposition 7.1. If A E Gl(n, JR.). and %f f E M+(JR.n) or f E £1 (JR.n,dx), then

(75)

J

f(:r) dx

= Idet AI

J

f(Ax) dx.

Proof. Let 9 be the set of elements A E GI(n,JR.) for which (7.5) is true. Clearly I E g. U:sing det A-I = (det A)-I and det AB = (det A)(det B), it is ea:;,y to show that 9 is a subgroup of Gl(n, JR.). Thus, to prove the propo:;,itioll, it suffices to show that 9 contains all clements of the following three fOllllS. since it is an exerci:;,e in linear algebra to show that these elernmts generate Gl(n,JR.) Here, {ej : 1 ::; j ::; n} denotes the standard ba:;,is of JR.n and (j a permutation of {L. ,n}.

(7.6)

A 1 ej = ea ()), A 2 ej = c)e J ,

A 3e l

c)

= ('1 + C(;2,

i- 0, A 3 (') = c, for J 2: 2

For the cabe A = Al we can work directly If f = XR ib the characteribtic function of a rectangle El x ... x En. the re:;,ult is obvious By the con:;,truction of product meawre the identity (75) then holds for Xs, the characteristic function of a Borel set Then (7.5) follows for f E S+(JR.n ), by additivity, and for f E M+(JR. n ), by (3.25). A similar argument shows that (7.5) holds for transformations of the form A 2 . In such a case, a cell of size f(h), .. ,f(In) is mapped by A2 to a cdl of size !cllf(h), ... , !cnlf(In) , bO it:;, measure is mUltiplied by leI ... cnl = Idet A21· Thus, (75) ho1d& for f = Xs, S E IB, by (7.2)-(7.3), and hmce, as before, for morE' general f

7

Lebesgue Measure on

]Rn

85

and on Manifolds

We will show that (7.5) holds for each of the transformations of the type

A 3 , using the Fubini-Tonelli Theorem. We note that the identity

J

(7.7)

f(x) dx

~

=

J

f(x

+ c) dx,

f

E

~

is elementary. To keep the formulas short, take n we have

J

f(Xl

(7.8)

M+(lR),

J[J = J[J

+ CX2, X2) dx =

f(Xl

= 2. For the ca:;,e A = A 3 ,

+ CX2, X2) dX1]

dX2

f(xl, X2) dXl] dX2,

the second identity by (7.7) Again we get (7.5). It is clear how to extend the analysis of As to n 2: 3, so the proposition is proved.

REMARK 1. One consequence of Propoi:>ition 7 1 is the invariance of Lebesgue measure under rotations. REMARK 2. At the end of this chapter there are exercise sets on determinants and on row reduction, deriving the rei:>ults on matrices used in the proof of Proposition 7.1.

Proposition 7 1 generalizE's to a change of variables formula, which we establish nE'xt, an E'xtE'n&ion to Lebesgue measure of one of the fundamental results of multi-variable calculus. Let F : 0 ---+ n be a C 1 diffeomorphism, where 0 and 0 are open in lRn. Let G = F- 1 , and form the absolute values of the Jacobian dE'terminants:

(7.9)

J(X) = Idet DF(x)1,

K(y) = Idet DG(y)[.

Theorem 7.2. Let F . 0 ---+ 0 be a C l dtfJeomorphism. If either u E M+(O) or u E .c1(O,dx), then

(7.10)

J

u(x) dx

[/

=

J

u(F(x) )J(x) dx.

0

Proof. Take a Borel set Sec O. Fix family of cells {Q j} such that

(711)

E.

> 0, and cover S with a countable

86

7.

Lebesgue Measure on

]Rn

and on Manifolds

Subdividing each of the cells if necessary, we can assume that, if center of Qj and Yj = F(xj),

Xj

is the

(712) and

(7.13) where we set Q~ that

= Qj -

Xj,

centered at 0. From Prop05ition 7.1, we deduce

(7.14) It follows that

(7.15)

Taking

(716)

E --->

0, we have rn(S) 2:

J K(Y)xF(S)(Y) dy or, equivalently,

J

J

(')

[1

cp(x)dx 2:

cp(G(y))K(y)dy,

= XS. By additivity, we have (7.16) for any cp E e+(K), any compact K c O. The l\Ionotone Convergence Theorem then gives (7.16) f01 any cp E M+(O). Setting cp(x) = 'l/J(x)J(x), we have .r~')(x)J(x) dx 2: J IjJ(G(y)) dy or, equivalently, setting v(y) = 'Ip( G(y)), for cp

(7.17)

J

J

[1

(')

1'(Y) dy:S;

v(F(x))J(x) dx,

for all v E M+(n). Reversing the roles of 0 and 0, we hence have (7.10), for U E M+(O), from which the identity for u E £J(O,dx) follows. There are various more refined change of variables formular" some of which we will discuss below. First, we apply the results obtained so far to show how to integrate functions on a Riemannian manifold, i.e., a smooth manifold equipped with a metric tensor. Background material on manifoldr, and metric tensors is given in Appendix B.

7.

Lebesgue Measure on lRn and on Manifolds

87

Let M be a C 1 manifold of dimension n. A continuous metric tensor on M gives a continuous inner product on tangent vectors to M. In a local coordinate system (Xl, ... , x n ), idcnti(ying an open subset of M with an open set 0 C lRn , the metric tensor is given by a positive definite n x n matrix G(x) = (g)k(X)), and the inner product of vectors U and V is given by (718)

(U, V) = U· G(x)V

=

L9)k(X)U)(X)Vk(X), ),k

where U = 2::) uj(x)ej, V = 2::k vk(x)ek, and {el,' ., en} is the standard ba8i8 of }Rn. If we change coordinates by a C 1 diffeomorphism F : 0 ----'> 0, the metric tensor H(y) = (hJk(Y)) in the coordinate system y = F(x) is related to G (.r) by (7.19)

at y

DF(x)U H(y)DF(x)V = (U, V) = U . G(x)V,

= F(x), i.e., G(x) = DF(x)t H(y)DF(x),

(7.20)

or

Now, for a Borel measurable function the integral is given by (7.21)

J

udV =

U

J

u(x)J9dx,

8upported on a coordinate patch,

g(x) = dctG(x).

(One often use~ dS instead of dV when dim Ai = 2 or when one is integrating over the boundary of a manifold of dimension (n + 1).) To see that (7.21) is well defined, note that under the change of coordinates y = F(x) we have by (7.20) that det G(x) = (det DF(x))2 det H(y). Hence

Vh = I det DFI- 1 J9,

(7.22)

h = det H.

so, by (7.10),

J

u(y)Vhdy =

(7.23) =

J

u(F(x)) I det DFI- 1 J91 det DFI dx

J

u(F(x))J9dx.

7.

88

Lebesgue Measure on ]Rn and on Manifolds

More generally f M U dV is defined by writing u as a sum of terms supported on coordinate charts. We see that we have a well-defined measure on the (I-algebra of Borel subsets of a C 1 manifold with a CO metric tensor. In case M is an n-dimensional submanifold of]Rm and a local coordinate chart arises via a C 1 map

(7.24) using the dot product on ]Rm. An example is the graph of a real-valued C 1 function u on 0 C ]RTi. If M i& the graph of z = u(x), then At is an n-dimensional C 1 manifold in ]Rn+ 1 . The map
(7.25) If u i& C 1 , we see that the metric tensor is ('0 To calculate 9 = Idet(gJk) I at a given point Xo E n, if V'u(xo) i= 0, rotate coordinates so V'u(xo) is parallel to the xl-axis. We get

(7.26) In particular, the n-dimensional measure of the surface !vI is given by

V(M) =

J J

Vfjdx =

dV =

lV'u(x) 12) 1/2 dx.

n

D

Ai

J(1 +

Particularly important examples of Riemannian manifolds are the unit &pheres sn-1 in ]Rn,

sn-l = {x E]Rn : Ixl = I}. In ca&e n = 2, we have S1, the unit circle. Since the time of the ancient Greek geometers, the length of the unit circle has been used to define 11"; 11" = half the length of S1. Spherical polar coordinate& on ]Rn are defined in terms of a smooth diffeomorphism

(7.27)

R: (0, (0)

X

sn-l

~]Rn

\ 0,

R(r,w) = rw.

7.

Lebesgue Measure on

]Rn

and on Manifolds

89

Let (hem) denote the metric tensor on sn-l (induced from its inclusion in ]Rn) with respect to some coordinate chart cp : 0 cUe sn-l Then, with respect to the coordinate chart cp : (0,00) x 0 --> U C ]Rn given by cp(r, y) = rcp(y), the Euclidean metric tensor can be written

(f'Jd =

(7.28)

(1

r

2}

1,fm

).

(Compare (724).) To see that the' blank terms vanish, i e., orCP . ox] cp = 0, note that cp(x) cp(x) == 1 ::::} o;rJcp(.r) cp(:r) = 0 Now (7.28) yields (729) We therefore have the following ret>ult for integrating a function in spherical polar coordinates Let us denote by dS the measure on sn-l produced via its Riemannian metric. Proposition 7.3. If f E M+(Il{rt) or f E £l(JRn,dx), then

(730)

J

J [foOO

f(x) dx =

Rn

f(rw)r n - 1 dT] dS(w).

sn-l

See the exercises for applications of Proposition 7.3.

We return to the change of variable formula (7.10), approaching it from another point of view. which yields more general results. In fact, we start with a very general t>etting. Let (0, J', M) and (n, lB, v) be measure spaces and F : 0 --> n a measurable map, i e., S E lB ::::} p-l (S) E J'. If u : n --> R if> measurable and 2: 0, we have

J

J

o

n

u(F(x)) dM(X) =

(7.31)

U(T) dll F- I(;r),

where, for S E lB, (732) Now, if we have the property

v(S) = 0

(7.33)

=::::}

M(p-l(S)) = 0,

we can apply the Radon-Nikodym Theorem, Theorem 4.10, to conclude that there is a lB-measurable function J F-l 2: 0 t>uc-h that (7.34)

J

J

o

n

u(F(x)) dM(X) =

u(X)JF-l (.r) dV(x).

Now we specialize to 0, [2 open in Rn, J', lB the Borel setf> , M = v = Lebesgue measure. ASf>ume P : 0 --> n is continuous, hence' measurable. Suppose P is one-to-one and onto. The property (7.33) certainly holdf> if p-l is Lipf>chitz. If we interchange the roles of 0 and n and of P and p-l, we have the following.

90

7. Lebesgue Measure on

Proposition 7.4. Let F : 0

~n

and on ]Hanifolds

0 be both a Lipschitz map and a homeomorphism. There exists a Borel function JF ~ 0 such that, for any u E M+(O), -+

J

u(x)dx =

(7.35)

o

J

u(F(X))JF(X)dx.

(')

In Theorem 7.2, this result was established for a C 1 diffeomorphism F, and the factor JF(X) was identified with the abbolute value of the Jacobian determinant:

JF(X) = [detDF(x)[.

(7.36)

We will now establish this identity for a larger dass of maps F Note that, if Q ib a cell in 0, then (7.37)

m(F(Q)) =

J

J

o

Q

XF(Q)(x)dx =

Let us assume that 0 is bounded, so m(O) < Extend JF to be 0 on ~n \ 0. For c > 0, set (7.38)

Jc:(x) =

J

m(Q~(x))

,h(x)dx.

00,

JF(Y) dy =

and hence JF E L1(0,dx).

J

'Pc: (x - y)JF(Y) dy,

Q,,(x)

where Qc: is the cube of edge c, centered at x, so 'Pc:(x) have

= c-nXQe;(O)(x). We

(7.39) See Exercise 10 in the first exercise set below. Consequently, by Exercise 11 of Chapter 3, there is a sequence Cv -+ 0 such that

(7.40) Consequently, by (7.37), we have

(7.41)

m(F(Qc:(x))) m(Qc:(x))

----?

JF(X) for a.e. x E 0,

c

= Cv

-+

0

We next establish a lemma which will take the place of (7.14) in our extension of the range of validity of (7.10).

7.

91

Lebesgue Measure on IR n and on Manifolds

Lemma 7.5. If F : 0 Xo E 0, then

---+

n

is a homeomorphism that is differentiable at

(742)

Proof. First absume DF(xo) is invertible. Composing F with a linear tranbformation (and recalling Proposition 71), we can assume DF(xo) = I Thus

(7.43)

F(xo

+ y) = F(xo) + y + R(y),

Fix 6 E (0,1/2). and pkk EO so small that as in (7.13), we have, for E ::; EO,

IR(y)1 = o(IYI).

Iyl ::;

EO ::::::}

IR(y)1 ::;

61yl.

Then,

(7.44) Hence

(7.45) This implie& that the lim sup of the left side of (7.42) is ::; \det DF(xo)1 in this case. We now need to establish the reverse inequality for the lim inf. This is not hard, if one uses borne fundamental results from topology. The same reasoning that gives (744) &hows that the boundary 8Qc(xo) of Qc(xo) is mapped by F onto a set I;, contained in o

(7.46)

C = {x E 0

dist(x,8Qc(xo))::; &} = Q(l+o)c(xo) \ Q(l-o)c(xo).

Furthermore, for each T E [0,1]' we can define FT(xo+Y) = f(xo)+Y+TR(y), and each map FT maps 8Qc('£0) into C Recall that F is assumed to be a o

homeomorphibm. Thus, by degree theory, F(Qc(xo)) contains a point Xl if and only if the map FI&Qe(xo) has degree 1 about Xl. However, the maps o

FT [aQc(xo) all have the same degree about any point Xl E Q(1-o)c(xo), for T E [0,1], so we see that FlaQc(xol has degree 1 about each such point, since Fo obviously has this property. Thus we conclude that (7.47)

hence (7.48)

92

7.

Lebesgue Measure on 1R.n and on Manifolds

This establishes the lemma when DF(xo) is invertible. It remains to treat the case when det DF(xo)

= 0, i.e., A = DF(xo) is

not invertible. In this case, we have

(7.49)

F(xo

+ y) =

F(xo)

+ Ay + R(y),

Consequently, if we fix 0 > 0 and pick olyl, we have

(7.50)

F(Qc:(xo)) C {x E 1R.n

.

EO

IR(y)1

= o(IYI)·

so small that Iyl ::;

EO

:::=}

IR(y)1 ::;

dist(x - F(xo), AQc:(O)) ::; EO},

for E ::; EO. Now AQc:(O) is contained in a set {x E V Ixl::; KE}, where V = R(A) is a linear subspace of 1R.n of dimension::; n - 1. Hence, in this case,

(7.51) so the lim sup of the left side of (7 42) is ::; K 0 in this case. Letting 0 we finish the proof.

---+

0,

Using (7.41), we see that, if F : 0 ---+ n is as in Proposition 7.4 and if also F is differentiable for almost all x E 0, then the identity (7.36) holds for almost all x. We have the following extension of Theorem 7.2. Proposition 7.6. If F : 0 ---+ n is both a Lzpsch%tz map and a homeomorphism and zf F zs differentiable for almost all x E 0, then the conclusion of Theorem 7 2 holds It is a result of Rademacher that every Lipschitz function on 0 it:> differentiable almost everywhere. We will prove this in Chapter 11. To end this chapter, we show that Lebesgue measure on ]Rn is uniquely characterized by translation invariance, together with a simple finiteness condition. Proposition 7.7. Let). be a measure on ~n. Assume that). is translatwn invariant:

(7.52)

8 E

~n ===?

),(8) = ),(8 + y),

Also assume there exists a bounded open U

(7.53) Then).

0< )'(U)

<

C ]Rn

\;j

yE

]Rn.

such that

00.

= crn for some constant c > 0, where rn is Lebesgue measure on ]Rn.

7.

Lebesgue Measure on

93

and on Manifolds

]Rn

Proof. Let 8 be any bounded Borel set in ]Rn. By (7.52), for each y E ]Rn,

J

xs(x) d'\{x) =

(754)

J

xs{x - y) d'\(x).

Integrate this identity over a ball B with respect to dm(y). Since'\ is finite, one can apply Fubini's Theorem to write (7.55)

m(B)'(S)

~ J~ xs (x - y) dm(Y)]

(T-

d)'(x).

Now m(8) = 0 => the right side of (7.55) is zero, so ,\ «m Hence by the Radon-Nikodym Theorem we have f E M+(]Rn) such that

(756)

'\(8) =

J

f(x) dm(x),

V 8 E \)3n,

s and translation invariance (7.52) implies (7.57)

J

f(x

+ y) dm(x) =

J

f(x) dm(x),

s Note that (7.53) implies Iu f dm <

Vy E ]Rn, 8 E \)3n.

s

00,

so

f

is integrable on bounded subsets

of ]Rn. Consider the positive number

(7.58)

>'(U) = m(U) 1 c = m(U)

J

J

u

u

1 f(x) dm(x) = m(U)

f(x

+ y) dm(x).

J"

If we write (7.56)-(7.57) as >'(8) = f(x + y) dm(x) and integrate over U with respect to dm(y), we have by Fubini's Theorem that, for each 8 E \)3n, (7.59)

m(U)>'(8) =

JJ

f(x

8

+ y) dm(y) dm(x) = cm(U)m(8),

u

the last identity by (7.58). This proves that>. = cm.

Exercises 1. Let a

= I~ e- x2 dx. This is called a Gaussian integral. Show that an =

J

e- 1x12 dx.

IR n

94

7.

Lebesgue Measure on

2. Show that a 2 = 27r

fa= e-

r2

~n

and on Manifolds

r dr = 7r.

Hint. Apply Proposition 7.3 and use the fact that 27r is defined to be the length of Sl.

3. Let An denote the n-dimensional measure of the unit sphere sn. Show that

and show that 27rn/2

(7.60)

An - 1 =

f(~) ,

J:'

where f(z) e-ss z - 1 ds, for z > O. The function f(z) is called Euler's gamma function. Hint. Apply Proposition 7.3. 4. Show that zf(z) = f(z + 1). Hint. Integrate by parts. 5 Show that r(1)

=

1 and f(1/2)

6. Let B n be the unit ball in

Hint. Apply (7.30) to

f =

~n,

=

7r 1 / 2 . Deduce that, for k E Z+,

B n = {x E ~n .

Ixl <

I}. Show that

XBn.

7. Show that the arc of the circle lying above 0 ::; x ::; 1/2 has 1/12 the length of the entire circle Sl. Hint. Find the equilateral triangles in Figure 7.l. 8. Deduce from (7.26) that (7.61)

7r

[1/2

"6 = Jo

dx

00

an

(1)2n+1

Vf=X2 = n=O L 2n + 1:2

'

Lebesgue Measure on

]Rn

95

and on Manifolds

1

Figure 7.1

where ao = L an+l = an (2n + 1)/(2n + 2). Show that

U sing a calculator, sum the series over 0 :S n :S 20 to approximate 13 digits after the decimal point.

9. Consider a 2D surface !vI C IR3 , with a coordinate chart 'P : 0 M. For integrable f : AI -+ IR supported on U, show that

Jf M

dS

=

Jf

0

-+

1T

to

U c

'P(x) 10j'P x 02'P1 dXl d X2.

0

Hint. Apply (7.21). Note that 2 2 2 01'P' 01'P 01'P' 02'P) 9 = det ( 02'P 01'P 02'P' 02'P = 101'P1 102'P1 - (01'P' 02'P) .

Use lu x vi = lui' Ivl . I sinOI and u . v = lui' Ivl cosO to show that ju x vl 2 = juj2jvl 2 - (u . v?, and hence 9 = 101'P x 02'Pj2.

Lebesgue Measure on JRn and on Manifolds

7.

96

Exercises 10-11 extend the results of Exercises 7-8 of Chapter 4 to JR n .

KfU(X) = f

(7.62)

Show that, for 1 :::; p <

* u(x)

=

J

f(y)u(x - y) dy.

00,

(7.63) Show that there is a unique extension from f E CO'(JRn) to f E L1(JR n ) of Kfu, with (763) continuing to hold, giving a continuous linear map K : L1(JRn ) ----+ £(LP(JR n )). The operator (7.62) is called convolution. 11. Let fJ be a sequence of nonnegative functions in Ll(JRn ) such that

J

fJ dx = 1,

(7.64)

Show that, for 1 :::; p < (7.65)

u

E

suppfJ C {x E JRn: Ixl < 1/J}. 00,

LP(JRn ) ===} fj

* u ----+ u

in LP norm, as J

----+ 00.

Hint. See the hint for Exercise 9 of Chapter 4. The sequence (fJ) is called an approxzmate identzty. 12. Given f E £1 (JR n , dx) and 6 > 0, show that there is a Lebesgue measurable S C ~n such that m(S) < 6 and such that f[lRn\s is continuous. This is Lusin's Theorem for ll~n; compare Exercise 16 of Chapter 3 and Exercise 14 of Chapter 5. 13. Let u be a C 1 function on JRn with compact support, we write u E CJ(JRn) Show that, for 1 :::; j :::; n,

Jau

(7.66)

aX

J (x) dx = 0.

lR n

(7.67)

J

av

u(x)~

lR n

UXj

dx

=-

Jau

-v(x) dx. aXj

lR n

7.

Lebesgue Measure on

=

Hint. If, say, j

/ IR,,-l

~n

97

a.nd on Ma.nifolds

1, write the left side of (7.66) as the iterated integral

( / :X~ (Xl, . . , Xn) dXl) dX2"

dxn.

IR

Use the fact that the inner Lebesgue integral coincides with the Riemann integral, in combination with the Fundamental Theorem of Calculus, as established in Chapter 1 14. Consult a topology text for material about degree theory, used to establish (7.47). Sufficient material can be found in [Dug] or [Spa].

Exercises relating the Riemann and Lebesgue integrals on

~n

Let R = h x . x In be a ('ell in ~n, with Iv = [a v , bv ] intervals in R The notion of the Riemann integral of a bounded function J : R --+ ~ is defined similarly to the one-dimensional case treated in Chapter 1. If one has a partition of each Iv into J vl U .. U Jv,N(v), then a partition P of R consists of the cell::;

(768) where 0

~

Gv

~

N(v). Each cell has n-dimensional volume

(769) Then we set

(7.70)

11'(1)

=

LW! J(x)V{R

a ).

a

Note that 11'(1) ~ 11'(1). As in Chapter 1, we pass to limits via finer partition&. To be precise, if P and Q are two partitions of R, we say P refines Q and write P >- Q, if each partition of each interval factor Iv of R involved in the definition of Q is further refined in order to produce the partitions of the factor& Iv, u&ed to define P, via (7.68). It is an exercise to show that any two partitions of R have a common refinement. Note also that, as in (1.4),

(771) Consequently, if Pj are any two partitions of R and Q is a common refinement, we have

(7.72)

7.

98

Lebesgue Measure on lRn and on Manifolds

Now, whenever defined:

1 : R ---+ lR is

(7.73)

=

7(1)

bounded, the following quantities are well

7p (1) ,

inf

1(1) = sup

PElleR)

1p(1) ,

PElleR)

where ll(R) is the set of all partitiont; of R, as defined above. Clearly, by (7.72), 1(1) ::; 7(1). We then say that 1 it; Riemann integrable (on R) provided 7(1) = 1(1), and in such a case, we write 1 E R(R) and bet

(7.74)

1(1)

= 7(1) = 1(1).

The following exercises deal with the Riemann integral and its relation to the Lebesgue integraL 1. Show that

1 E C(R)

=?

j E R(R)

2. Show that j, 9 E R(R) =? j + 9 E R(R) and 1(1 + g) Hint Consult the proof of Proposition 1.1. 3. Show that

1,g E R(R)

=?

= 1(1) + 1(g)

19 E R(R)

4 Given a partition P of R, define maxsize (P) Extend Corollary 1.4 and (1.42) to the multi-diment;ional case. 5. Extend the proof of Propot;ition 3.10 to the following result.

Proposition 7.S. Let R be a cell in JR n , and let cp : R ---+ JR be bounded. Then cp it; Riemann integrable if and only if the set of pointt; x E R at which i.p is discontinuous has Lebet;gue measure zero. If thit; condition holdt;, then the Riemann integral and the Lebesgue integral of i.p coincide. 6. For a bounded set S

c JR n ,

with characterit;tic function Xs, set

as in (1.23), and define S to "have content" if and only if cont+(S) = cont - (S) Show that S hab content if and only if the set of boundary points of S hat; Lebesgue measure zero.

7.

Lebesgue Measure on IRn and on Manifolds

When doing Exercises 7-8, look back at Exercises 10-14 of Chapter 5. 7. If 0

c JRn is open and bounded, show that its Lebesgue measure is given

by m(O) = cont-(O). If K

c JRn is compact, show that its Lebesgue measure is given by m(K)

= cont+(K).

8. If S c JRn is a bounded set, show that its Lebe8gue outer measure is given by m*(S) = inf {m(O) . S C 0, open}.

Remove the hypothesis that S be bounded. Exercises on determinants

Let M (n, JR) denote the space of n x n real matrices. These exercises investigate the existence, uniqueness, and basic properties of the determinant, det : M(n,JR)

(775)

----+

JR,

which will be uniquely specified as a function 1J . }\,f(n, JR)

----+

JR satisfying

(a) 1J is linear in each column a) of A, (b) 1J(A) = -1J(A) if A i8 obtained from A by interchanging two columns, (c) 1J(I) = 1. 1 Let A = (al .... ,(J,n), where a) are column vectors; aj = (al), ... ,anj)t. Show that, if (a) holds, we have the expan8ion detA= La)1 det(ej,a2, ... ,an ) )

(7.76)

= .. =

ajll' . a)"n det (e Jl , e12 ,···, ej,,),

L Jr,

,j"

where {el' .. , en} is the standard basis of JR n,. 2. Show that, if (b) and (c) also hold, then (7.77)

det A

=

L o-ESn

(sgn a-)

ao-(l)1 ao-(2)2

. ao-(n)n,

100

7. Lebesgue Measure on

]Rn

and on Manifolds

where Sn is the set of permutations of {1, ... ,n} and (7.78)

sgn a = det (e u(l),

. .. ,

eu(n») = ±l.

To define sgn a, the "sign" of a permutation a, we note that every permutation a can be written as a product of transpositions: a = 'T1 ••. 'Tv, where a transposition of {1, ... ,n} interchanges two elements and leaves the rest fixed. We say sgn a = 1 if v is even and sgn a = -1 if v is odd. It is necessary to show that sgn a is independent of the choice of such a product representation. (Referring to (7.78) begs the question until we know that det is well defined.) 3. Let a E Sn act on a function of n variables by (7.79) Let P be the polynomial (7.80)

P(X1, ... , xn)

=

II

(Xj - Xk).

l:::;;j
Show that (7.81)

(aP)(x) = (sgn a) P(x)

and that this implies that sgn a is well defined. 4. Deduce that there is a unique determinant satisfying (a)-(c) and that it is given by (7.77). If (c) is replaced by 'I9(I) = r, show that 'I9(A) = r det A. 5. Show that, if (a)-(c) hold, it follows that (d) 'I9(A) = 'I9(A) if A is obtained from A by adding cal to ak, for some C E JR, where aI, ... ,an are the columns of A. 6. Show that (7.82)

Hint. For fixed A

det(AB) = (detA)(detB).

M(n,JR), apply Exercise 4 to 'l9 1(B) = det (AB). Note that if we represent B by its columns, as B = (bI, ... , bn ), then AB = (Ab1, ... , Abn ). To verify rule (b) for 'l91(B), note that if B = E

7.

101

Lebesgue Measure on IRn and on Manifolds

(bu(l), ... , bu(n») is obtained from B by permuting its columns, then AB has columns (Abu(l), ... ' Abu(n»), obtained by permuting the columns of

AB. 7. Show that (7.77) implies detA = detA t ,

(7.83)

where At denotes the transpose of A, i.e., A = (ajk) =? At = (akj). Conclude that one can replace columns by rows in the characterization (a)-(d) of determinants. Hint. au(j)j = a£T(£) with £ = (T(j), T = (T-I. Also, sgn (T = sgn T. 8. Show that

(7.84) det

(r

a12 a22

a21n n

a:

an2

)

= det

r 0 : 0

ann

0 a22 an2

a2n o)

:

= detA l1

ann

Hint. Do the first identity using (d), from Exercise 5. Then exploit uniqueness for det on M(n - 1, JR). . 1

9. Deduce that det(ej,a2, ... ,an ) = (-l)J- detA 1j where A kj is formed by deleting the kth column and the jth row from A.

10. Deduce from the first sum in (7.76) that n

(7.85)

~

. 1

det A = L..,..( -l)J- ajl det A 1j . j=l

More generally, for any k E {I, ... , n}, n

(7.86)

detA = L(-l)j-k ajk detA kj . j=l

This is called an expansion of det A by minors, down the kth column. 11. Let Ckj = (-l)j-k det Akj. Show that n

(7.87)

L j=l

aj£Ckj = 0,

if £ -=f k.

102

7.

Lebesgue Measure on lRn and on Manifolds

Deduce that C = (Cjk) satisfies

CA

(7.88)

=

(detA)I.

Hint. Reason as in Exercises 8-10 that the left side of (7.87) is equal to

with ae in the kth column as well as in the £th column. The identity (7.88) is known as Cramer's formula. 12. Show that

au ln

a2n ) a

det (

(7.89)

:

=

aU a 22 ... ann·

ann

Hint. Use (7.84) and induction.

Exercises on row reduction and matrix products We consider the following three types of row operations on an n x n matrix A = (ajk). If (T is a permutation of {1, ... ,n}, let

If

C

=

(Cl' ... , Cj)

Finally, if

C

and all

E lR and f.-t

Cj

i- v,

are nonzero, set

define

We relate these operations to left multiplication by matrices PO" Me, and Ej.lvc, defined by the following actions on the standard basis {el' ... ,en} of lR n : (7.90) and (7.91)

7.

Lebesgue Measure o~"lRn'and on Manifolds

.LUoJ

1. Show that

and that det Pu

=

3. Show that, if /.L

i=

sgn a,

det Me

v, then EJ-Lve

= q ...

en,

det EJ-Lve

= 1.

= p;;1 E 21e Pu, for some permutation a.

4. If B = Pu(A) and C = /.Le(B), show that A = PuMeC. Generalize this to other cases where a matrix C is obtained from a matrix A via a sequence of row operations. 5. If A is an invertible, real n x n matrix (i.e., A E Cl(n, JR.)), then the rows of A form a basis of JR. n . Use this to show that A can be transformed to the identity matrix via a sequence of row operations. Deduce that any A E Cl(n, JR.) can be written as a finite product of matrices of the form Pu , Me and EJ-Lve, hence as a finite product of matrices of the form listed in (7.6). Further exercises on Cl(n, JR.)

The set Cl(n, JR.) is the subset of the space M(n, JR.) of n x n real matrices consisting of invertible matrices, where we say A E M(n, JR.) is invertible if and only if there exists B E M(n, JR.) such that AB = BA = I. (Then we write A-I = B.) 1. Show that (7.92)

Cl(n, JR.) = {A

E

M(n, JR.) : det A

i= O}.

Hint. Use (7.82) and (7.88).

2. Show that (7.93)

Cl(n, JR.) is open in M(n, JR.).

Hint. Use Exercise 1.

104

7.

Lebesgue Measure on

~n

and on Manifolds

Note that Gl(n,~) = Gl+(n,~) U GL(n, ~), where Gl±(n,~)

(7.94)

= {A

E M(n,~) :

± det A> O}.

The goal of the next six exercises is to show that

(7.95) 3. Show that (7.95) is a consequence of the following three assertions, regarding the matrices Ep,ve, Me, and P a , defined by (7.90)-(7.91): (a) There is a path from 1 to Ep,ve in Gl+(n, ~). (b) If el ... en > 0, there is a path from 1 to Me in Gl+(n, ~). (e) If sgnO" > 0, there is a path from 1 to Pa in Gl+(n,~). Hint. Use Exercise 5 from the previous exercise set. 4. Show that assertion (a) of Exercise 3 holds. Hint. Evaluate det Ep,v(te). 5. Show that assertion (b) of Exercise 3 holds. Hint. Reduce to the case where each ej = ±1. The number of -l's is even. There is a path from 1 to -1 in Gl+(2,~) given by (

That is, -1 E

COS

t

- sin t )

sin t

Gl+(2,~)

cos t

'

0::; t ::;

is a rotation by

7L

7L

6. Reduce the proof of assertion (c) in Exercise 3 to the following assertion: There is a path from 1 to PaT in Gl+(n, JR) whenever positions in Sn.

0"

and

T

are trans-

In turn, reduce the proof of this assertion to the following two assertions: (a) There is a path from

1=

C

1

J

to A =

G ~) 0 0 1

in Gl+(3, JR).

(b) There is a path from

C

1

1=

1

J

to

1 0

B=

C

0 1

J

in Gl+(4, JR).

7.

Lebesgue Measure on

jRn and

on NlamtOlQS

7. Prove assertion (a) of Exercise 6. Hint. Show that A is a rotation by

21T

-'-vv

/3 about the axis through (1,1, l)t.

8. Prove assertion (b) of Exercise 6. Hint. Show that B is the identity on the linear span of (1,1,0, O)t and (0,0,1, l)t and that it is rotation by 1T on the 2D orthogonal complement of this space.

Exercises on matrix integrals Let M(n, JR.) denote the space of n x n real matrices, M(n, C) the space of n x n complex matrices. Set

(7.96)

O(n)

=

{A

M(n, JR.) : A* A

=

(7.97)

U(n)

=

{A E M(n,C): A*A

=

I}, I},

=

I},

=

I}.

E

where Also set

SO(n) SU(n)

(7.98) (7.99)

= =

{A E O(n) : detA {A E U(n) : detA

1. Use Proposition B.7 from Appendix B to show that each of the four ma-

trix groups (7.96)-(7.99) is a smooth, compact submanifold of M(n, JR.) or M(n, C). 2. For each group G of the form (7.96)-(7.99), define, for 9 E G,

Rg, Lg : G

---7

G,

Rg(x)

=

xg,

Lg(x)

=

gx.

Show that, for each 9 E G, Rg and Lg preserve the metric tensor on G induced from M(n, JR.) or M(n, C). Hint. Show that R g, Lg : M (n, IF) ---7 M (n, IF) are isometries, with IF = JR. or C, as appropriate. 3. Let dV denote the volume element on G associated to the metric tensor discussed above, via (7.21). Show that for each J E Ll(G, dV), (7.100)

J

J

G

G

J(x) dV(x) =

J(gx) dV(x)

=

J

J(xg) dV(x),

G

V9

E

G.

106

7.

Lebesgue Measure on

]Rn

and on Manifolds

We set dg = V(G)-l dV and call this Haar measure on G.

J

4. Show that

f(g-l) dg

J

f(g) dg.

=

G

G

5. Let V be a finite-dimensional Hilbert space and assume 7r : G -----+ GI(V) is a continuous homomorphism such that 7r(g) : V -----+ V is unitary for each g E G. (One says 7r is a unitary representation of G on V.) For v E V, set

Pv =

J

7r(g)v dg.

G

Show that P is the orthogonal projection of V onto

Va

=

Hint. Show that 7r(g)v = v V g

p 2v =

{v

E

==}

J

V: 7r(g)v

Pv = v,

7r(h)Pv dh = Pv,

=

v, Vg

E

G}.

7r(h)Pv = Pv, P*v =

J

V hE G,

7r(g-l)v dg

= Pv.

6. Also let ,\ be a unitary representation of G on a finite-dimensional Hilbert space W. Give £(W, V) the Hilbert-Schmidt inner product, (A, B) = Tr B* A, and define a unitary representation v of G on £(W, V) by v(g)A = 7r(g)A,\(g-l). Define

Q : £(W, V) QA =

Show that

-----+

£(W, V) by

J

J

G

G

v(g)Adg =

7r(g)A,\(g-l) dg.

Q is the orthogonal projection of £(W, V) onto

I(7r,'\) = {A E £(W, V) : 7r(g)A = A'\(g), Vg E G}. 7. Assume the representations 7r and ,\ are irreducible, i.e., V and W have no nontrivial G-invariant linear subspaces. Schur's Lemma (cf. [TIl, Appendix B, §§7-8) implies that I(7r,'\) is either zero- or one-dimensional. Using Exercise 6, what can you conclude about

J

7rij(g)'\kC(g) dg?

G

Here 7rij (g) are the matrix entries of 7r (g) in an orthonormal basis of V, and ,\k£(g) are similarly defined. The resulting formulas are called the Weyl orthogonality relations.

Chapter 8

Signed Measures and Complex Measures

As opposed to the measures we have considered so far, a signed measure is allowed to take on both positive and negative values. To be precise, if M is a IT-algebra of subsets of X, a signed measure von M is a function

(S.l)

v:

M

-----+

[-00,00],

allowed to assume at most one of the values ±oo and satisfying

v(0) = 0

(S.2) and (S.3)

Ej EM disjoint sequence ===> v(UEj ) = 2:v(Ej), j

j

the series L: j v(Ej) being absolutely convergent. A signed measure taking values in [0,00] is what we have dealt with in Chapters 2-7; sometimes we call this a positive measure. If /11 and /12 are positive measures and one of them is finite, then /11 - /12 is a signed measure. The following result is easy to prove but useful. Proposition 8.1. If v is a signed measure on (X, M), then for a sequence

{Ej

} C

(S.4)

M, Ej /

E ===> v(E)

=

lim v(Ej), J->OO

and (S.5)

E j ". E, v(Ed finite ===> v(E) = lim v(Ej ). J->OO

-

107

108

8.

Signed Measures and Complex Measures

Proof. Exercise. If v is a signed measure on (X, M) and E E M, we say v is E-positive if v(F) 2: 0 for every F E M, FeE. Similarly we say v is E-negative if v(F) :S 0 for every such F. We say v is null on E if v(F) = 0 for all such F. Note that, if Pj is a sequence of sets in M, then

(8.6)

v Pj-positive

::::=::}

v P-positive, for P =

UPj. j

To see this, let Qn = Pn \ Uj:Sn-l Pj. Then Qn C Pn , so v is Qn-positive. Now, if FeU Pj, then v(F) = 2: v(F n Qn) 2: 0, as desired. The following key result is known as the Hahn Decomposition Theorem.

Theorem 8.2. If v is a signed measure on (X, M), then there exist P, N E = X, P n N = 0.

M such that v is P-positive and N -negative and PUN Let us assume that v does not take on the value be proved via the following:

+00.

The theorem can

Lemma 8.3. If v(A) > -00, then there exists a measurable PeA such that v is P-positive and v(P) 2: v(A). Given Lemma 8.3, we can prove Theorem 8.2 as follows. Let

s = sup {v(A) : A EM}. By (8.2), s 2: 0. Take Aj EM such that v(Aj) / s, v(Aj) > -00. By the lemma, v is Pj-positive on a sequence of sets Pj C A j , such that v(Pj ) -----) s. Set P = U Pj. By (8.6), v is P-positive. Then we deduce that v(P) = s. In particular, under our hypothesis, s < 00. Set N = X \ P. We claim that v is N-negative. If not, there is a measurable SeN with v(S) > 0, and then v(PUS) = s+v(S) > s, which is not possible. Thus we have the desired partition of X into P and N. To prove Lemma 8.3, it is convenient to start with a weaker result:

Lemma 8.4. If v(A) > -00, then for all c > 0, there exist measurable Be A such that v(B) 2: v(A) and v(E) 2: -c, for all measurable E C B. We show how Lemma 8.4 yields Lemma 8.3. We define a sequence of sets Aj EM inductively. Let Al = A. Given Aj, j :S n-l, take An C A n- 1 such that v(An) 2: v(An-d and v(E) 2: -lin, for all measurable E cAn. Lemma 8.4 says you can do this. Then let P = Aj. Clearly v is P-positive,

n

8.

109

Signed Measures and Complex Measures

and (8.5) implies that I/(P) = liml/(Aj) ~ I/(A).

It remains to prove Lemma 8.4. We use a proof by contradiction. If Lemma 8.4 is false, then (considering B = A), we see that, for some E > 0, there is a measurable El C A such that I/(El) -E. Thus I/(A \ E l ) ~ I/(A) + E. Next, considering B = A \ E l , we have a measurable E2 C A \ El such that I/(E2):S -E, so I/(A\(El nE2)) ~ I/(A)+2E. Continue, producing a disjoint sequence of measurable E j C A with I/(Ej) -E. Then Aj = A \ (El U· .. U E j ) has I/(Aj) 2: I/(A) + jE, and hence, by (8.5), we must have

:s

:s

Aj "" F,

I/(F)

= +00.

But we are working under the hypothesis that 1/ does not take on the value +00, so this gives a contradiction, and the proof is complete.

1/.

We call the pair P, N produced by Theorem 8.2 a "Hahn partition" for It is essentially unique, as the following result shows.

Proposition 8.5. If 1/, P and N are as in Theorem 8.2 and if also P, N is a Hahn partition for 1/, then 1/ is null on P /:;

P= N

/:;

N = (P n N) U (N n P).

Proof. Let E E M, E c P /:; P. Write E = EoUEl, a disjoint union, where we take Eo = En (P n N), El = En (N n P). Then I/(E) = I/(Eo) + I/(El)' But each I/(Ej) is simultaneously 2: and 0, so we have I/(Ej) = 0.

°

:s

If 1/ is a signed measure on (X, M) and we have a Hahn partition of X into PUN, we define two positive measures 1/+ and 1/_ by

(8.7)

I/+(E)

=

I/(P n E),

I/_(E)

= -I/(N n E),

for E E M. We have

(8.8) Now 1/+ is supported on P, i.e., on N, so

(8.9)

1/+

and

1/+

1/_

is null on X \ P. Similarly

1/_

is supported

have disjoint supports.

This is called the Hahn decomposition of the signed measure 1/. By Proposition 8.5, it is unique. We can also form the positive measure

(8.10)

110

8.

Signed Measures and Complex Measures

called the total variation measure of v. Given a signed measure v, we can define J f dv for a suitable class of measurable functions f· We can use the Hahn decomposition (8.8) to do this, setting (8.11) when v satisfies (8.8). In particular this is well defined for

f

E Ll(X,

Ivl).

Let f-l be a positive measure on (X, M). We say that a signed measure von (X, M) is absolutely continuous with respect to f-l (and write v < < f-l) provided that, for E E M, (8.12)

= 0 ==?

f-l(E)

veE)

=

o.

This definition was made in (4.45) in the case when v is also a positive measure. It is useful to have the following. Proposition 8.6. If f-l is a positive measure and v a signed measure on (X, M), with Hahn decomposition (8.8), then (8.13)

v «f-l

==?

v+

«

f-l and v_

«

f-l.

Proof. If E E M and f-l(E) = 0, then f-l(E n P) = f-l(E v+(E) = veE n P) = 0 and v_(E) = -veE n N) = O.

n N)

0, so

We can therefore apply Theorem 4.10 to v+ and v_ to obtain the following extension of the Radon-Nikodym Theorem. Theorem 8.1. Let f-l be a positive finite measure and v a finite signed measure on (X, M), such that v < < f-l. Then there exists h E c8x, f-l) such that

J

(8.14)

Fdv

x for all F E .cl(X,

=

J

Fhdf-l,

x

Ivl).

Proof. Note that v+(X) = v(P) < 00 and v_eX) = -v(N) < 00. Apply Theorem 4.10 to v±, obtaining h± E .cl(X, f-l), and let h = h+ - h_. The only difference between (4.47) and (8.14) is that this time h need not be ~ o.

8.

111

Signed Measures and Complex Measures

It is also useful to consider the concept of a complex measure on (X, M), which is defined to be a function

(8.15)

p :

M

C,

----+

satisfying

p(0) = 0

(8.16) and (8.17)

E EM disjoint sequence j

===}

p(UE

j )

=

j

LP(E

j ).

j

In this case, we do not allow p to assume any "infinite" values. It is clear that we can set

vo(S) = Re p(S),

(8.18)

VI

(S) = 1m p(S),

and Vj are signed measures, which do not take on either value. We have

p(S)

(8.19)

=

vo(S)

+00

or

-00

as a

+ iVl (S),

and we can define

J

(8.20)

F dp =

J

F dvo

+i

J

F dVl,

IF

where in turn dVj are defined as in (8.11). There is an obvious extension of the Radon-Nikodym Theorem: if p, is a finite positive measure and v a complex measure on (X,M), such that v « 11" then (8.14) continues to hold for some complex-valued h.

Exercises 1. 'Write down a proof of Proposition 8.1. In Exercises 2-3, let p, and v be finite positive measures on (X, J). For [0,(0), set i.pT = V - T p,. For each such T, i.pT is a signed measure.

T E

112

8.

Signed Measures and Complex Measures

2. Show that there exists a family P T and a family NT of elements of J such that Po = X, for each T E [0,00), PTlNT is a partition of X,

<

T1

T2 =====?

~

PT1

PT2 ,

and <(JT

is PT-positive and NT-negative, V T E [0,00).

Hint. To get the nesting property, make a preliminary Hahn decomposition

Pa, Na

= Q n [0,00), and set

for a E Q+

PT = 3. For each k E Z+

n

~

{Pa

:

= {O, 1,2, ... },

+ a E Q ,a:S T}. set

hk(X) = sup {T E Tk ·Z+: x E PT

}.

h E M+(X) and

Show that hk /

v

= h/1- + p,

p -1 /1-,

obtaining another proof of the Lebesgue-Radon-Nikodym Theorems 4.104.11. 4. Suppose /1-1 and /1-2 are finite positive measures on (X, M), and form the signed measure v = /1-1 - /1-2. Show that Ivl(E) :S /1-1 (E)

+ /1-2 (E) ,

VEE M.

Hint. First look at E c P and E c N, with P and N as in Theorem 8.2. 5. Take v as in Exercise 4. With fELl (X, /1-1 + /1-2)'

J

f dv =

Hint. v+

+ /1-2

=

v_

J f dv

J

f d/1-1 -

defined by (8.11), show that for

J

f d/1-2.

+ /1-1.

6. Let 9J1:(X, J) denote the set of finite signed measures on (X, J). Show that this is a linear space and, in fact, a Banach space, with norm (8.21)

Ilvll

= Ivl(X).

7. Suppose Vj are finite signed measures on (Xj, Jj). Show that VI x V2 is a well-defined, finite signed measure on (Xl x X2, J1 0 J2), and Ilvl x v211

=

Ilv111·llv211·

8. Let E be a Banach space. Define the notion of an E-valued measure on (X, M), and develop some basic properties.

Chapter 9

LP Spaces, II

In Chapter 4 we defined £P spaces and studied some of their basic properties as Banach spaces. We also studied special properties of L2 spaces, as Hilbert spaces. In particular, we characterized continuous linear functionals 'P : L2(X, p,) -+
w :V

-----+


(w : V -+ lR if V is a real vector space). Elements w E V' are called linear functionals on V. Sometimes one finds the following notation for the action of w E V' on v E V : (9.2)

(v,w) = w(v).

If V has norm II II, the condition for the map (9.1) to be continuous is the following: the set ofv E V such that Iw(v)1 ::::; 1 must be a neighborhood of 0 E V. Thus this set must contain a ball BR = {v E V : Ilvll ::::; R}, for some R > O. With C = 1/ R, it follows that w must satisfy

(9.3)

Iw(v)1 ::::; Cllvll

-

113

9. V Spaces, II

114

for some C < 00. The infimum of the C's for which this holds is defined to be Ilwll; equivalently, (9.4)

Ilwll = sup{lw(v)1 : Ilvll :S 1}.

It is easy to verify that V', with this norm, is also a Banach space. As shown in Chapter 4, if H is a Hilbert space, the inner product produces a conjugate linear isomorphism of H' with H; see (4.34). In particular, as noted in Corollary 4.9, L2(X, j.,l) is its own dual. We next identify the dual of LP(X, j.,l), for general p E [1,00), using the Radon-Nikodym Theorem. Proposition 9.1. Let (X, j.,l) be a (T-finite measure space. Let 1 :S p < 00. Then the dual space LP(X, j.,l)', with norm given by (9.4), is naturally isomorphic to Lq(X, j.,l), with l/p + l/q = l.

Note that Holder's inequality and its refinement (4.9) show that there is a natural inclusion i : Lq(X, j.,l) --+ V(X, j.,l)', which is an isometry. It remains to show that i is surjective. We give a proof in the case when j.,l(X) is finite, from which the general case is easily deduced. If w E LP(X, j.,l)', define a set function 1/ on measurable sets E c X by l/(E) = (XE, w), where XE is the characteristic function of E; 1/ is readily verified to be countably additive, as long as p < 00. Thus 1/ is a complex measure. We have

Jf

dl/ = (j,w),

first for f = XE, then for simple f. Furthermore, 1/ annihilates sets of j.,lmeasure zero, so the Radon-Nikodym Theorem, in the form of Theorem 8.7, implies

for some measurable function w. We hence have (9.5)

(j,w) =

J

fwdj.,l,

for simple f. A variant of the proof of (4.9) gives w E U(X,j.,l), IlwllLq Ilwll, and (9.5) holds for all f E V(X, j.,l). We mention that countable additivity of 1/ fails for p = 00; in fact LOO (X, j.,l)' is identified with the space of finitely additive bounded set functions on the (T-algebra of j.,l-measurable sets that annihilate sets of j.,l-measure zero; see [Yo] for a proof. Given a Banach space V, since V' is a Banach space, one can construct its dual, V". Note that the action (9.2) produces a natural linear map

(9.6)

/'\, : V

----7

V",

9.

LP Spaces, II

.L.LV

and it is obvious that 1I~(v)11 ::; Ilvll. In fact, 11~(v)11 = Ilvll, i.e., ~ is an isometry. More precisely, for any v E V, there exists w E V', Ilwll = 1, such that w(v) = Ilvll. One can check directly that this property holds when V = LP(X, /-l), for a O"-finite measure space, with 1 ::; p < 00. More generally, this property is a consequence of the Hahn-Banach Theorem; see, e.g., Chapter 3 of [RS] or §4 in Appendix A of [TI]. Sometimes ~ in (9.6) is surjective, so it gives an isometric isomorphism of V with V". In this case we say V is reflexive. Clearly any Hilbert space is reflexive. Also, in view of Proposition 9.1, we see that LP(X, /-l) is reflexive provided 1 < p < 00. On the other hand, Ll (X, /-l) is not reflexive; VXJ(X, /-l)' is strictly larger than Ll (X, /-l), except for the trivial cases where Ll(X, /-l) is finite dimensional. One of the values of identifying a certain Banach space Y as the dual V' of another Banach space V is the use of the "weak* topology" on V', in which a neighborhood basis of an element w E V' consists of sets of the form

In particular, for a sequence Wv E V', we say Wv -----> W in the weak* topology provided (v, w v ) -----> (v, w) for each v E V We have the following important result called Alaoglu's Theorem:

Proposition 9.2. If V is a Banach space, then the closed unit ball B is compact in the weak* topology.

C

V'

This result is readily deduced from the following fundamental result in topology:

Tychonov's Theorem. If {Xa : ex E A} is any family of compact Hausdorff spaces, then the Cartesian product TIaEA X a , with the product topology, is a compact Hausdorff space. For a proof of Tychonov's Theorem, see [Dug] or [RS]. To see how this yields Proposition 9.2, note that B E V' above, with the weak* topology, is homeomorphic to a closed subset of the Cartesian product TI{Xv : v E V}, where Xv = {z E C : Izl ::; Ilvll} and ~ : B -----> TI Xv is given by ~(w) = {w(v) : v E V}. If V is separable, we can say a little more. Let {Vj : j E Z+} be a dense subset of B 1 , and, for w, 0" E B c V', set

(9.8)

d(w, 0")

=

L2- j l(vj,w -

0")1.

j?O

This defines a metric on B, and it is clear that the identity map on B gives a continuous map id : (B, w*) -----> (B, d), from B with the weak* topology to B

116

9. V Spaces, II

with the metric topology. But a bijective continuous map from a compact Hausdorff space onto a Hausdorff space must be a homeomorphism, as is shown in Appendix A, so the two topologies must coincide. Thus we have Corollary 9.3. If V is a separable Banach space, then the unit ball B in V' is a compact metric space, with the weak* topology.

In case V is separable, one has /'i, : B -----t TI X Vj , mapping (B, w*) homeomorphically onto a closed subset of TI X Vj , for a countable dense subset {Vj} of V. The compactness of TI XVj follows from the special case of Tychonov's Theorem proven in Appendix A, Proposition A.17. REMARK.

Applying Proposition 9.2 and Corollary 9.3 to Proposition 9.1, we have Proposition 9.4. If (X, f-l) is a O"-jinite measure space, then, for 1 < p :S 00, the closed unit ball B in V(X, f-l) is compact in the weak* topology, relative to U(X, f-l), where p-I + q-I = 1. If Lq(X, f-l) is separable, then B is a compact metric space with this topology.

We turn now to more general mappings. If V and Ware two Banach spaces, we denote by £(V, W) the space of continuous linear transformations from V to W. We set £(V) = £(V, V). As in the derivation of (9.3), it is easy to see that, when V and Ware Banach spaces, a linear map T : V -----t W is continuous if and only if there exists a constant C < 00 such that (9.9)

IITvl1 :S Cllvll

for all v E V. Thus we call T a bounded operator. The infimum of all the C's for which this holds is defined to be IITII; equivalently, (9.10)

IITII

= sup {IITvll : Ilvll :S 1}.

It is clear that £(V, W) is a linear space. If V and Ware Banach spaces and T j E £(V, W), then IITI + T211 :S IITIII + IIT211; completeness is also easy to establish in this case, so £(V, W) is also a Banach space. If X is a third Banach space and S E £(W, X), it is clear that ST E £(V, X) and

(9.11)

IISTII :S IISII·IITII·

If V and Ware Banach spaces and T E £(V, W), then the adjoint T' E £(W', V') is uniquely defined to satisfy (9.12)

(Tv, w) = (v, T'w),

v

E

V,

wE

W'.

9.

LP Spaces, II

Granted that (9.13)

K,

117

is an isometry in (9.6), it is easy to see that

IITII = IIT'II = sup{I(Tv,w)l: IIvll = 1, IIwll = 1}.

When V and Ware reflexive, it is clear that Til = T. In case V and Ware Hilbert spaces, T E LeV, W), then we also have an adjoint T* E LeW, V), given by (Tv, w)

(9.14)

= (v, T*w),

v E V, w E W,

using the inner products on Wand V, respectively. As in (9.13) we have IITII = IIT*II· Also it is clear that T** = T. We consider some simple examples of bounded linear operators. If (X, J-t) is a measure space, f E V)Q(X, J-t), then the multiplication operator M f , defined by Mfu = fu, is bounded on LP(X, J-t) for each p E [1,00], with IIMfll = IIfIlL=. If X is a compact Hausdorff space and f E G(X), then Mf E £(G(X)), with IIMfll = IIflislip. Another class of examples, a little more elaborate than above, is given by integral operators of the form Ku(x) =

(9.15)

J

k(x, y) u(y) dv(y).

y

We have the following result: Proposition 9.5. Let (X, M, J-t) and (Y,N, v) be a-finite measure spaces. Suppose k is measurable on X x Y and (9.16)

J

Ik(x, y)1 dJ-t(x) :::; Gl

,

J

Ik(x, y)1 dV(y) :::; G2 ,

y

X

for all y and for all x, respectively. Then (9.15) defines K as a bounded operator from LP(Y, v) to LP(X, J-t), for each p E [1,00]' with 1

1

-+-=1. p q

(9.17)

Proof. For p E (1, 00 ), we estimate (9.18)

1/ / X

y

k(x, y)f(y)g(x) dJ-t(x) dv(y)1

9.

118

£P Spaces, II

via the estimate ab :::; aP /p+b q /q of (4.11), used to prove Holder's inequality. Apply this to If(y)g(x)l. Then (9.18) is dominated by (9.19) provided (9.16) holds. Replacing f, 9 by tf, Clg, we see that (9.18) is dominated by (Clt P /p)llfll~p + (C2/qt q )llglllq, and minimizing over t E (0, (0), via elementary calculus, we see that (9.18) is dominated by (9.20) proving the result. handled.

The exceptional cases p = 1 and p

00

are easily

We call k(x, y) the integral kernel of K. Note that K' is an integral operator, with kernel k'(x, y) = k(y, x). In the case of the Hilbert space L 2 (X,p,), K* is an integral operator, with kernel k*(x,y) = k(y,x). We next take a look at an important class of bounded operators on Hilbert spaces, the Hilbert-Schmidt operators, defined as follows. Let HI and H 2 be separable Hilbert spaces and A E £ (HI, H 2)' Let {Uj} be an orthonormal basis of HI. We say A is a Hilbert-Schmidt operator, or an HS operator for short, provided

L IIAujl12 <

(9.21)

00,

j

equivalently, if (9.22) j.k

where {vd is an orthonormal basis of H 2 . The class of HS operators in £(H1 ,H2) will be denoted HS(Hl,H2)' If HI = H2 = H, we denote the class HS(H). We define the Hilbert-Schmidt norm of an HS operator: (9.23) j,k

j

By the second identity in (9.23), we see that A* E £(H2' HI) is HS if A is and that (9.24)

IIA*IIHS

=

IIAIIHS.

IIAIIHS is clearly independent of the choice of orthonormal basis of H 2 , IIA*IIHS is independent of the choice of orthonormal basis of HI and hence so is IIAIIHS' Since

9.

£P Spaces, II

119

The first identity in (9.23) makes it clear that IIBAIIHS :::; IIBII . IIAIIHS if A is HS and B is bounded on H 2 • Since AB = (B* A*)*, we deduce via (9.24) that an analogous bound holds on IIABIIHS when B is bounded on HI. In other words, (9.25) if B j E £(Hj). In particular, recapitulating the observation made following (9.24), we have (9.26) if U j E £(Hj) are unitary transformations, i.e., invertible isometries of H j onto H j . We note that, if A is a Hilbert-Schmidt operator, then (9.27)

IIAII:::; IIAIIHS.

To see this, pick unit UI such that an orthonormal basis.

IIAuIl1

~

IIAII -

E:

and make that part of

The following classical result might be called the Hilbert-Schmidt Kernel Theorem. Proposition 9.6. 1fT: L2(XI,lLd -----t L 2(X 2,1L2) is Hilbert-Schmidt, then there exists K E L2(XI X X 2, ILl x 1L2) such that

(9.28) Proof. Pick orthonormal bases {fj} for L 2 (X d and {9k} for L 2 (X 2) and set

(9.29)

K(Xl' X2) =

L ajkfj(xd9k(x2) j,k

where ajk (Tfj,9k)· The hypothesis that T is HS is precisely what is necessary to guarantee that K E L2(XI x X 2), and then (9.28) is obvious. It is also clear that (9.30)

Also of interest is the converse, proved simply by reversing the argument and noting that {fj9k} is an orthonormal basis of L2(XI x X 2, ILl x 1L2) (see Exercise 6 of Chapter 6).

120

9.

Proposition 9.7. If K E L2(X1 operator T, satisfying (9.30).

X

£P Spaces, II

X 2, 11-1 x 11-2)' then (9.28) defines an HS

We now look at a particular linear operator of singular importance, the Fourier transform, defined by

Ff(~) = J(~) =

(9.31) when

f

(27r)-n/2

J

f(x)e- ix -€ dx

E L1(JRn). It is clear that

(9.32) It is convenient to bring in the Schwartz space of rapidly decreasing func-

tions:

(9.33)

Xfl ...

where xf3 = x~n, DOl = Dfl ... D';,n, with D j = -ia / ax j . Using an argument involving integration by parts (cf. (7.67)), it is easy to verify that (9.34)

and

(9.35) We define F* by (9.36) which differs from (9.31) only in the sign of the exponent. It is clear that F* satisfies the mapping properties (9.32), (9.34), and (9.37)

(Fu, v)

=

(u, F*v)

for u, v E S(JR n ), where (u, v) denotes the usual L2-inner product, (u, v) = fIRn u(x) vex) dx. The first major result is the Fourier inversion formula. The following is our first version. Proposition 9.8. We have the inversion formula

(9.38)

F* F = FF* = I

on S(JR n ).

121

9. £P Spaces, II

We will sneak up on the inversion formula by throwing in a convergence factor which will allow interchange of orders of integration. So, let us write, for f E S(lR n ),

F*Ff(x) = (27r)-n / [ / f(y)e- iyoe dy]e ixoe de (9.39)

= (27r)-n lim

c:~O

Jf

f(y) e-c:leI 2 ei(x-y)oe dy de.

We can interchange the order of integration on the right for any c > 0, to obtain (9.40)

F* Ff(x) = lim / f(y)p(c, x - y) dy, c:~o

where (9.41) Note that (9.42) where q(x) = p(l, x). In a moment we will show that (9.43) The derivation of this identity will also show that

J

q(x) dx = 1.

(9.44)

lR n

From this, it follows, as in the proof of (4.64) and (4.69), that (9.45)

lim / f(y)p(c, x - y) dy = f(x)

c:~o

for any f E S(lR n ), even for f bounded and continuous, so we have proved F* F = I on S(lRn); the proof that FF* = I on S(lRn) is identical. It remains to verify (9.43). We observe that p(c, x), defined by (9.41), is an entire holomorphic function of x E for any c > o. It is convenient to verify that

en,

(9.46)

9.

122

LP Spaces, II

from which (9.43) follows by analytic continuation. Now

p(c, ix) = (27r)-n (9.47)

=

J

e- x·e-eleI 2 dE,

(27r)-n e 1xI2 /4e

J

e-lx/2v1c+vIceI2 dE,

= (27r)-nc-n/2elxI2/4e

J

e- 1e12 dE,.

]R.n

To prove (9.46), it remains to show that

J

e -lel 2 dE, = 7rn/2.

(9.48)

]R.n

Indeed, for this see Exercises 1-2 in the first exercise set of Chapter 7. This completes the proof of the identity (9.46) and hence of (9.43). REMARK. See Exercises 7-8 for another demonstration of (9.43), not using analytic continuation.

In light of (9.37) and the Fourier inversion formula (9.38), we see that, for u, v E S(ffi.n), (9.49)

(Fu, Fv) = (u, v) = (F*u, F*v).

Thus F and F* extend uniquely from S(ffi.n) to isometries on L2(ffi.n), and are inverses to each other, i.e., they are unitary. Thus we have the Plancherel Theorem: Proposition 9.9. The Fourier transform

(9.50)

is unitary, with inverse F*. We now make some comments on the relation between the Fourier transform and convolutions. The convolution u * v of two functions on ffi.n is defined by

J =J

u * v(x) = (9.51)

u(y)v(x - y) dy u(x - y)v(y) dy.

9.

£P Spaces, II

123

Note that u * v = v * u. If u, v E S(JRn), so is u and also as a special case of Proposition 9.5,

* v.

As in (4.62) and (7.63)

(9.52) so the convolution has a unique continuous extension to a bilinear map (9.53) for 1 :s:: p < 00; one can directly perceive this also works for p = 00. Note that the right side of (9.40), for any c > 0, is an example of a convolution. Computing the Fourier transform of (9.51) leads immediately to the formula (9.54) Via Proposition 9.9, we have the estimate (9.55) which is stronger than the p = 2 case of (9.52). The estimate (9.55) does not hold if L2 is replaced by LP for p =I- 2. Useful strengthenings of (9.52) for other values of p E (1,00) lead one to the important topic of singular integral operators. See [Stl for a treatment of this and also Chapter 13 of

[TIl· It is useful to extend the Fourier transform from Ll(JRn ) -+ LOO(JRn ) to 9J1(JRn) -+ Loo(JRn), where 9J1(JRn ) denotes the space of finite (signed or, more generally, complex) Borel measures on JR n , and to extend the convolution product from Ll(JRn ) x Ll(JRn) -+ Ll(JR.n ) to 9J1(JR. n ) x 9J1(JR. n ) -+ 9J1(JR. n ). The formula (9.31) for J(~) makes it natural to set

(9.56)

p,(~)

= (27r)-n/2

Je-ix.~

dp,(x) ,

p, E 9J1(JR. n ).

It readily follows that 1p,(~)1 :s:: (27r)-n/211p,11, the norm on 9J1(JR. n ) defined by (8.21). Also the Dominated Convergence Theorem implies fi is continuous. Note that if u E S(JR. n ), then, by Fubini's Theorem,

Ju(~)p,(~) d~ = (9.57)

=

(27r)-n/2

J

JJu(~)e-ix.~ dp,(x) d~

u(x) dp,(x).

Applying Proposition 9.8, we deduce that, given p, E 9J1(JR n ), (9.58)

J

v(x) dp,(x)

=

Jv(~)p,(~) d~,

\Iv E S(JR n ).

9.

124

£P Spaces, II

A natural way to define the convolution J-L * v of two finite measures is suggested by the following computation. If f, 9 E Ll(JRn) and u E LOO(JR n ),

1

u(x)(f*g)(x)dx =

(9.59)

= Thus it is natural to define J-L

J-L * v(S)

(9.60) when S

c

=

*v

11 11

u(x)f(x-y)g(y)dydx u(x

+ y)f(x)g(y) dx dy.

by

11

xs(x

+ y) dJ-L(x) dv(y),

ffin is a Borel set and J-L, v E 9J1(JR n ). An equivalent formula is

It readily follows that J-L

* v E 9J1(lRn)

and

IIJ-L * vii:; IIJ-LII . Ilvll·

(9.62)

Note also that if u is a bounded Borel function on JR n ,

1

(9.63)

ud(J-L*v)

=

11

u(x+y)dJ-L(x)dv(y).

In fact, (9.63) follows by linearity from (9.60) if u is a finite linear combination of XS j ' Sj E ~ (JR n ), and then by the Dominated Convergence Theorem for a general bounded Borel function u. A particular consequence of (9.63) is the following analogue of (9.54):

(J-L (9.64)

* vr(~) =

(27f)-n/2

1e-i(x+y).~

dJ-L(x) dv(y)

= (27ft/2ji(OV(~).

Exercises 1. Extend the p = 2 case of Proposition 9.5 to the following result of Schur. Let (X, J-L) and (Y, v) be measure spaces, and let k(x, y) be measurable on (X x Y, J-Lx v). Assume that there are measurable functions p(x), q(y), positive a.e. on X and Y, respectively, such that

(9.65)

1

1

x

y

Ik(x, y)lp(x) dJ-L(x) :; C1q(y),

Ik(x, y)lq(y) dv(y) :; C2P(X).

V Spaces, II

Show that Ku(x)

=

Jy k(x, y)u(y) dv(y) defines a bounded operator

K : L2(y, v)

---+

L2(X, f-L),

IIKI12::; C 1 C2.

Give an appropriate modification of the hypothesis (9.65) in order to obtain an operator bound K : LP(Y, v) --+ V(X, f-L). Hint. To estimate (9.18), set

B(x) I IA(y) I If(y)g(x)1 = I A(y) f(y) . B(x)g(x), and apply the estimate ab ::; (a P I p)

+ (b q I q)

to this.

2. Show that k(x, y) is the integral kernel of a bounded map K : L2(JRn ) L2 (JR+.) provided it satisfies the estimate 9.66) Ik(x, y)1 ::; C ( lx' - y'1 2

+ xi + yi )

-n/2

,

--+

x = (Xl, x'), y = (Yb y').

Hint. Construct p(x) and q(y) so that (9.65) holds. Look for p(x) as a function of Xl alone. 3. Given u E LP(JRn), 1::; p < 00, and i.p E Cgo(JR n ), set v = i.p * u. (a) Show that v is continuous on JR n . Hint. Apply dominated convergence. (b) Show that h- l [v(x + hej) - v(x)] --+ (oi.ploxj) * u as h --+ 0, and conclude that v E CI(JR n ). (c) Show that v E Coo (JR n ) and Dav = (Dai.p) * u. 4. If u E LP(JRn) and w E Lpl (JRn), lip continuous on JR n .

+ lip'

= 1, show that u

*w

is

In Exercises 5-6, let (X, J, f-L) be a measure space and A an algebra of subsets of X such that O"(A) = J. Let A = span {XA : A

E

A}.

5. If f E LI(X,f-L) and IAfdf-L = 0 for all A E A, show that f = O. Hint. Show that {B E J : IB f df-L = O} is a monotone class. Apply Proposition 5.9. 6. If f-L(X) < 00, show that A is dense in L2(X, f-L). Hint. Show that the orthogonal complement A-.l of A in L2(X, f-L) is O.

126

9. V Spaces, II

Deduce that A is dense in VeX, J.L), for 1 :S p :S 2. Note. There is a general result known as the Hahn-Banach Theorem, which implies that if V is a Banach space and L a linear subspace which is not dense in V, then there exists a nonzero w E V' such that w( v) = 0 for all vEL. (See [RS], [TI], [Yo].) Using this, show that, when J.L(X) < 00, A is dense in LP(X, J.L) for all p E [1, (0).

Exercises 7-8 give another approach to the identity (9.43), i.e.,

(9.67)

(27r)-n

J

e-cf.~12+ix.~ d~ =

(47rE)-n/2 e - 1x I2 /4e,

one that does not use analytic continuation.

7. Show that to demonstrate (9.67) it suffices to treat the case n = l. Furthermore, it suffices to show that (9.68) 8. Note that g E S(JR.) and g satisfies the differential equation

(d~ + x )g(x) = o. g E S(JR.) and

Use (9.34)-(9.35) to deduce that

(~ + ~ )g(~) = O. Then establish (9.68). First show

g(O = Ce-e/ 2 ; then evaluate C.

9. Let pet, x) be given by (9.43) with t =

E,

i.e.,

(9.69) Show that

(9.70)

F-le-tl~12 Ff(x) =

J

f(y)p(t, x - y) dy.

J

Show that, if f E LP(JR.n ), 1 :S p < 00, then u(t, x) = f(y)p(t, x-y) dy satisfies the "diffusion equation" (also called the "heat equation"):

(9.71)

au

a2

a2

- = .6.u, .6.= -+"'+-2' at axi aXn

9.

127

LP Spaces, II

for t

> 0, and that

Hint. Use (9.35). 10. Verify that the map"" in (9.6) is an isometry when V = LP(X, f-l), 1 :s: p < 00, by showing directly that, given v E LP(X, f-l), there exists w E Lpi (X, f-l) such that IlwllLpl = 1 and I vw df-l = IlvIILP. 11. Assume f E C 2 ([0, 1]), f(O) = f(l) = O. Show that

f(x) where

12. If

f

G(x, y)

=

=

11

G(x, y)f"(y) dy,

:s: y :s: 1, O:S: y :s: x :s: 1.

-x(1 - y)

for O:S: x

-(1 - x)y

for

is as in Exercise 11, use Proposition 9.5 to show that

1I 1

(9.72)

o

f (x ) I dx

III If" (

:s: -

8

x ) I dx.

0

Exercises 13-14 deal with the trapezoidal 'rule, defined as follows. Let an interval I = [a, b] be divided into subintervals, all of length h. Given f E C(1), II f dx is approximated by (9.73)

where

h

=

[aj,

bj ] are the intervals into which I is subdivided, with bj

- aj

=

(b - a)/N.

13. Show that if I = [0, 1], then (9.74)

110 1 f

Hint. Subtract from and 1.

dx -

Tl fl :s: ~ 10 1 If"(x)1 dx.

f the linear function agreeing with f (x) at x

= 0

128

9. V Spaces, II

14. Show that, if I = [a, b] and

1 E C 2 (I),

1/ 1dx _7Jh 11 ~ ~2

(9.75)

I

then

/

11"(x)1 dx.

I

Hint. Derive (9.75) from (9.74) by scaling when I = [0, h]. Then sum this result over the intervals of length h into which I = [a, b] is subdivided.

1 * f..l(x) = / l(x - y) df..l(y). Show that

1 * f..l

E V(JR n )

and

Show that if also v E 9J1(JRn ), then

16. Suppose P is a bounded linear operator on a Hilbert space H satisfying p2 = P ,

P* = P.

Let K = Range P. Show that K = Ker (I - P) and K ~ = Ker P = Range (I - P). Deduce that K is closed and that P is the orthogonal projection of H onto K. 17. Let H be a Hilbert space and let T E ri. Show that T is an isometry if and only if T*T = I. Show that if T is an isometry, then TT* is the orthogonal projection of H onto the range of T.

Chapter 10

Sobolev Spaces

We now define spaces H1,P(JR n ), known as Sobolev spaces. For U to belong to H1,P(JR n ), we require that U E LP(JRn ) and that U have weak derivatives of first order in LP(JRn): (10.1) where (10.1) means (10.2) If U E COO (JR n ), we see that OjU = oU/OXj, by integrating by parts, using (7.67). We define a norm on H1,P(JR n ) by

(10.3)

IluIIH1,P = IlullLP + I: IlojullLP. j

We claim that H1,P(JR n ) is complete, hence a Banach space. Indeed, let (u v ) be a Cauchy sequence in H1,P(JR n ). Then (u v ) is Cauchy in LP(JRn ); hence it has an LP- norm limit U E LP(JRn ). Also, for each j, OjU v = fjv is Cauchy in LP(JRn ), so there is a limit fj E LP(JRn), and it is easily verified from (10.2) that OjU = fj. We can consider convolutions and products of elements of H1,P(JR n ) with elements of COO (JR n ) and readily obtain identities

(10.4)

-

129

130

10.

Sobolev Spaces

and estimates (10.5)

II'P * ulIHl,P

11'P11£1lluIIHl,p,

:::::;

II1};uIIHl,P :::::; 111};llu lluIIHl,P Xl

+ L 118j 1};IILoo IluIILP,

oo

for 'P, 1}; E C (lR n ), u E H 1 ,P(lRn ). For example, the first identity in (10.4) is equivalent to

-11 ;~ 11 =

(x) 'P(x - y)u(y) dy dx

1};(x)

;~ (x -

y) u(y) dy dx,

V 1}; E C

oo (lR n ),

an identity that can be established by using Fubini's Theorem (to first do the x-integral) and integration by parts, via (7.67). If p < 00, u E H 1 ,P(lRn ), and ('Pj) is an approximate identity of the form (7.64), with 'Pj E C (lR n ), then we can show that

oo

(10.6)

'Pj

*U

---t

using (10.4) and (7.65). Given

E

U

in H1,p- norm,

> 0, we can take j such that

Then we can pick 1}; E C (lRn) such that 111};('Pj * u) - 'Pj * ullHl,p < E. Of course, 'Pj * u is smooth, so 1};(rpj * u) E COO (lRn). We have established

oo

Proposition 10.1. Forp E [1, (0), the space

Coo (IRn) is dense in H

1 ,P(lRn ).

Sobolev spaces are very useful in analysis, particularly in the study of partial differential equations. We will establish just a few results here, some of which will be useful in Chapter 11. More material can be found in [EG], [Fol], [T1], and [Yo]. The following result is known as a Sobolev Imbedding Theorem. Proposition 10.2. If p > n or if p

= n = 1, then

(10.7) For now we concentrate on the case p E (n, (0). Since dense in H 1 ,P(lRn ), it suffices to establish the estimate (10.8)

COO (IRn)

is then

10.

Sobolev Spaces

131

In turn, it suffices to establish (10.9) o

To get this, it suffices to show that, for a given