Real Analysis, Second Edition

I f(t)g(t)i < f(t)I P + (1 —

and integrating both sides we have (2)

f

fg-1< x f ifr + (1

—

X) f 'gi g =

1.

The inequality is trivial if II f!( = 0 or l!gli= O. Let f and g be any O. Then Eilf li p and elements of L P and Lq with II 0 and g/ligli g both have norm 1. Substituting them in (2) gives

f

1

fg

l=

f

Ifl

II

Ilf11, rglIg

<1

— Ig(t)I, and equality can Equality in (1) can occur only if I f(t)( P hold in (2) only if it holds almost everywhere in (1). Hence equality holds only if (Leg f flIP a.e. I

rP =

The Classical Banach Spaces

114

[Ch. 6

3. Minkowski Inequality: 1ff and g are in LP, then so is f + g and

Ilf + gllp

IlfIlp +

Proof: The cases p = 1 and p = co are straightforward and are left to the reader. Hence we assume 1 < p < co. IgIP), we have f + g in L. Also Since j f + glP 2P(1 fjP

flf +

gP

iv+

+

By the H61der inequality,

fif

gi

' ifi

f if + g

' igl

gi p_i )112,

< <

and

= Ulf+ gr'-'1"

Hof+

+ glIp}",

= since p = q(p — 1). Hence

Ilf+

(11fIl3, + lIglI3,)(11f + g113,)P ig

Ilf + gllp

IlfIlp + lIglIp. 1

or

Problems 5. For 1 < p < co, we denote by IP the space of all sequences (E,Z= 1 P < oc. Show that if (S,) g /P and (ny) e /q with such that y= ; + = 1, then

I timid

I I(G)Ilp • 1 1(ny>II,,

where

(11(t)llp)P

Ê ii

and

11(ny)II. = sup Inv'.

Sec. 3]

Convergence and Completeness

115

This is the Hblder inequality for sequences. 6. Prove the Minkowski inequality for sequences:

li(Ep

n9)11p

5- ii(tv)iip + 11(004.

Here we have 1 < p < cc.

3

Convergence and Completeness

The notion of convergence for a sequence of real numbers generalizes to give us a notion of convergence for sequences in a normed linear space. Definition: A sequence (fi ) in a norrned linear space is said to converge to an element f in the space if, given e > 0, there is an N such that for all n > N we have 11 f — f„It < e. If Jet, converges to f we write f = Lim f7, or fn f.

Another way of formulating the convergence off, to f is by noting —f11 —> O. Convergence in the space L P 1

0, there is an N such that for all n > N and all m > N we have < E. It is easily verified that each convergent sequence is a — Cauchy sequence. Definition: A normed linear space is called complete if every Cauchy sequence in the space converges, that is, if for each Cauchy

116

The Classical Banach Spaces

[Ch 6

sequence (f0 in the space there is an element f in the space such that fn f. A complete normed linear space is called a Banach space. A series (fn ) in a normed linear space is said to be summable to a sum s if s is in the space and the sequence of partial sums of the series converges to s; that is, s —

E

i=i

If this is the case, we write s = absolutely summable if

E

ft

Eft.

0.

The series ( fa) is said to be

fnii < 00 .

n= For a series of real numbers we know that absolute summability implies that the series is summable. While this is not true in general for series of elements in a normed linear space, the following proposition shows that this implication holds if the space is complete.

4. Proposition: A normed linear space X is complete if and only if every absolutely summable series is summable. Proof: Let X be complete and ( fn ) an absolutely summable series of elements of X. Since Ell fnl I = m. < 00, there is, for each

e > 0, an N such that

EfI <

n=1■1

E. Let sn =

E j; be the partial

z--=1

sum of the series (fn). Then for n > m > N we have

Ilsn —

sndl =

E

Ilfi II < E.

Hence the sequence (sa) of partial sums is a Cauchy sequence and must converge to an element s in X, since X is complete. Let (fn) be a Cauchy sequence in X. For each integer k there is an integer nk such that 1 fn — fmII < 2—k for all n and m greater than nk , and we may choose the nk 's so that nk+1 > nk . Then (fr,)17-1 is a subsequence of ( fn ), and if we set g 1 = fni , and g k = fn,— for k > 1 we obtain a series (gk ) whose k' partial sum is f„,. But we have 2 —k+ 1 if k > 1. Thus Eiigkl! E 2—k+1 + 1. Hence the series (g0 is absolutely summable, and so by our hypothesis there is an element f in X to which the partial sums of the series converge. Thus the subsequence (fnk) converges to f.

Sec.

3]

117

Convergence and Completeness

We shall now show thatf = lim f„. Since ( fn ) is a Cauchy sequence, given E > 0, there is an N such that II fn — f.11 < €72 for all n and m larger than N. Since fnk f, there is a K such that for all k > K we have 11 — fil < E/2. Let us take k so large that k > K and nk > N. Then

llf

—

fi!

I fn — fn,11 + Ilfn,

—

Thus for all n > N we have II f„ — fil <

fil E,

+ =

and so fn

f. I

5. Theorem (Riesz-Fischer): The LP spaces are complete. Proof: Since the case p = Go is elementary, it is left to the reader, and we assume 1 < p < Go . By virtue of the preceding proposition we need only show that each absolutely summable series in 1,9 is summable in LP to some element of L P.

Let ( f„) be a sequence in LP with Eiifi1 = M < , and define n=1

functions gn by setting g(x) = E ifk(x)i. From the Minkowski inequality we have

11g.11

k=1

Ilfkll

M.

Hence (g,)9

5,_ mP

.1

For each x, (g„(x)) is an increasing sequence of (extended) real numbers and so must converge to an extended real number g(x). The function g so defined is measurable, and, since gn > 0, we have g9

mP

by the Fatou Lemma. Hence gP is integrable, and g(x) is finite for almost all x.

fk(x) is an absolutely

For each x such that g(x) is finite the series k=

summable series of real numbers and so must be summable to a real number s(x). If we set s(x) = 0 for those x where g(x) =oo , we have defined a function s which is the limit almost everywhere of the partial

118

The Classical Banach Spaces

[Ch. 6

n SUMS sn = E fk. Hence s is measurable. Since fsn (x)i < g(x), we lc= 1 have Is(x)1 < g(x). Consequently, s is in LP and we have isn (x) — s(x)( P < 2' [g(x)]9 .

Since 2Pg9 is integrable and Js(x) — S(x)I P converges to 0 for almost all x, we have

I is

— sI P --* 0

by the Lebesgue convergence theorem. Thus tIsn — sil P ---> 0, whence Ifs„ — sll ---* O. Consequently, the series ( f,.) has in D' the sum s. I

Problems 7. Prove that every convergent sequence is a Cauchy sequence. 8. Let (A) be a sequence of functions in L. Prove that (A) converges to f in LG.' if and only if there is a set E of measure zero such that A converges to f uniformly on E. 9. Prove that Lc° is complete. 10. Prove that P' is complete (1 < p < cc) (see Problem 5). 11. Let C = C[0, 1] be the space of all continuous functions on [0, 1] and define If fll = max I f( x)f. Show that C is a Banach space. 12. We denote by PD the space of all bounded sequences of real numbers and define 11(p)11. — sup la Show that /OE' is a Banach space. 13. Show that the space c of all convergent sequences of real numbers and the space c o of all sequences which converge to zero are Banach spaces (with the P° norm). 14. Let f be a function in L9, 1

0 there is a continuous function yr, and a step function zi, such that 1ff — ioll < e and If f — Ntiff < E. 15. Let (fn ) be a sequence of functions in LP, 1 < p < oo , which converge almost everywhere to a function f in LP. Show that (A) converges to f in L 9 if and only if Ilfill ---* If!!. (For p = 1 this is just Problem 4.14.) 16. Let (f„) be a sequence of functions in LP, 1 < p < oo, which converge almost everywhere to a function f in V', and suppose that there is a constant M such that Ilf„11 < M for all n. Then for each function g in Lq we have f fg = lim f f„g. Is the result true for p = I?

119

Sec. 4] Bounded Linear Functionals on the LP Spaces

17. Let fn, —*f in LP, 1 < p < oo , and let (gr,) be a sequence of measurable functions such that Ign1 < M, all n, and gn --> g a.e. Then gnf, —* g f in LP.

4

Bounded Linear Functionals on the

LP

Spaces

We define a linear functional on a normed linear space X to be a mapping F of the space X into the set of real numbers such that F(af + (3g) = aF(f) + oF(g). We say that the linear functional is bounded if there is a constant M such that IF( f)I __< M • I If I for all f in X. The smallest constant M for which this inequality is true is called the norm of F. Thus VII = sup

1F(f)I 3 Ilfll

as f ranges over all nonzero elements of X. If g is a function in Lq, we can define a bounded linear functional F on LP by setting F(f) = f fg.

The functional F is clearly linear, and the Holder inequality states that IIFII < ligh. In fact, we actually have IIFII = de,• To see this for the case 1 < p < oc, we set' f = I gl q1P sgn g.

Then I f I P = IgI q = fg. Hence f is in LP and II fli p = (lIgligrP. Now F(f) = f fg = f le = (ileq)q = lIglIglIfIlp, and so I F must be at least as great as II e q. We state this result as a proposition, the cases p — 1 and p = Go being left to the reader (see Problem 18). 6. Proposition: Each function g in Lq defines a bounded linear functional F on LP by

F(f) = f fg. We have 11F II = ligil q. 2

For any real number x we define sgn x = 1 if x >0, sgn 0 = 0, and sgn x = —1 if x <0.

120

The Classical Banach Spaces

[Ch. 6

The goal of the present section is to show that for 1 < p < 00 the converse of this proposition holds, that is, that we obtain every bounded linear functional on D' in this manner. We shall find it useful to first establish the following lemma. 7. Lemma: Let g be an integrable function on [0, 1], and suppose that there is a constant M such that

I

1 fg __ m ilf I lp for all bounded measurable functions f. Then g is in D, and lIgII, < M.

Proof Let us first assume 1 < p < oc. We define a sequence of bounded measurable functions by setting g(x) =

Ig(x)

if Ig(x)1 < n

0

if Ig(x)1 > n,

and letting fn = ign riP sgn gn . Now iifnllp = (lIgn110q1 P, and igni q --- fn • gn = f. • g. Hence

(l(gn h)q = ffng < Mdfn li p = M(11g7,110q/P . Since q — q/ p = 1,

Ng. ii, _. Al and

fignr ..... M.

Since ign iq converges to igiq almost everywhere, we have

f 'gi g < lim f Igni g < Mq by the Fatou lemma. Thus g 8 D, and ligilq M. For the case p = 1, let E = {x: fg(x) I > M ± E} , and set f = (sgn g)x E. Then II fill = mE, and

MmE = Miifili ... Thus mE = 0, and ile. < M. I

f fg _>_ (M + f)mE.

Sec. 4]

Bounded Linear Funchonals on the LP Spaces

121

We are now in a position to give the following characterization of the bounded linear functionals on D' for 1 < p < co : 8. Riesz Representation Theorem: Let F be a bounded linear functional on D', 1 < p < cc . Then there is a function g in L 9 such that F(f) = f fg. We have also WI( = lIgl( g . Proof: Let X s be the characteristic function of the interval [0, s]. We begin our investigation of F by observing what it does to X s . For each s the value of F(X 8) is a real number 4,(s), and this defines a function 4) on [0, 1]. Now I maintain that 4) is absolutely continuous. For let {(s„ si.)} be any finite collection of nonoverlapping subintervals of [0, 1] of total length less than b. Then

E 14D(s) — cqs,)i = where f = E (x„, — x„) sgn (4)(s'i ) — ct(s,)). , Since (II f 11) P

< 6, we have E 14,0'.0 —

,)1 = F(f)

11F1111fIlp < IIF(IVP. This shows that the total variation of 4) is less than e over any finite collection of disjoint intervals of total length 6 if we take 6 --- &'/F'. Thus 4) is absolutely continuous. By Theorem 5.13 there is an integrable function g on [0, 1] such that 4,(s) = fos g. Thus F(X 3) = foi g • X,.

122

The Classical Banach Spaces

[Ch. 6

Since every step function on [0, 1 ] is (equal except at a finite number of points to) a suitable linear combination EciX so we must have F(4) = 101 gtk for each step function 4", by the linearity of F and of the integral. Let f be any bounded measurable function on [0, 1]. Then it follows from Proposition 3.22 that there is a sequence (Ip„) of step-functions which converge almost everywhere tof. Since the sequenced f — thil P) is uniformly bounded and tends to zero almost everywhere, the bounded convergence theorem implies that 11 f — 0 nit i, --> 0. Since F is bounded, and IF(f) — F(lin)i = IFU — COI we must have

liFii

Ilf — 0.11p,

F(f) = lim Since gly„ is always less than igl times the uniform bound for the sequence (4/n), we have f fg = lim f gC, by the Lebesgue convergence theorem. Consequently, we must have f fg = F(f) for each bounded measurable function f. Since IF(f)I —‹ liFil I fll P , we have g in Lg and ligl 1 15_ VII by Lemma 7. Thus we have only to show that F(f) = ffg for each f in LP . Let f be an arbitrary function in _U. Then by Problem 14 there is for each e > 0 a step function 0 such that III f — Oli p < e. Since 11, is bounded we have F(0) — 1 Og. Hence 1F(f) — f fg

F(f) — F(0) + fg — f fgl

f( - f)gl < liFil If — oh, + Il gti g iif — In, F( /' — 0)1 +

< (iin + ligh)e•

Sec. 4]

Bounded Linear Functionals on the L 2 Spaces

123

Since E is an arbitrary number, we must have F(f) = f fg. The equality 11F11 --= ligli q follows from Proposition 6. 1 In the problems the reader is asked to carry out a similar representation for the bounded linear functionals on 17), 1 < p < oo , c, and c o . In Theorem 14.8, we give a representation for the bounded linear functionals on C. Unfortunately, the bounded linear functionals on L (and on r) do not admit of a similar representation.

Problems 18. a. Let g be an integrable function on [0, 1]. Show that there is a bounded measurable function f such that 11f II 0 and

f fg = lei 11fII.. '

.

b. Let g be a bounded measurable function. Show that for each e > 0 there is an integrable function f such that

ffg

(11g11. - 011fIli.

[Hint: f may be taken to be a suitable characteristic function.] 19. Find a representation for the bounded linear functionals on 12 , 1 < p
Part Two

0

Abstract Spaces

7 1

Metric Spaces

Introduction

The system of real numbers has two types of properties. The first type consists of the algebraic, dealing with addition, multiplication, etc. The other type consists of properties having to do with the notion of distance between two numbers and with the concept of a limit. These latter properties are called topological or metric, and the object of the present chapter is to study these properties in a general space in which the notion of distance is defined. We make the

following definition: Definition: A metric space (X, p) is a nonempty set X of elements (which we call points) together with a real-valued function p defined on X X X such that for all x, y, and z in X. 0; i. p(x, y) p(x, y) = O f and only if x = y; = P(y, x); P(x, iv. p(x, y) < p(x, z) p(z, y). The function p is called a metric. An obvious example of a metric space is the set R of all numbers with p(x, y) = ix — yl. A second example is the n-dimensional Euclidean space Rn whose points are the n-tuples x = (x 1 , . . • , x.) of real numbers and p(x,y ) = r (x 1 - 1) 2 + • 127

+ (x. - y.) 2 1 112 .

128

Metric Spaces [Ch. 7

For Rn, property (iv) of the metric is merely the statement that the length of a side of a triangle is less than the sum of the lengths of the other two sides. Consequently, (iv) is usually called the triangle inequality. Other examples of metric spaces are the normed linear spaces of the last chapter if we set P(x, Y) = 11x — Y11,

the triangle inequality being equivalent to 11x + yil —‹ 11x11 + VII. It should be emphasized that a metric space is not the set X of its points, since it is in fact the pair (X, p) consisting of the set of its points together with the metric p. For example, the set of all n-tuples of real numbers can also be made into a metric space by use of the metric p* given by P* (x, Y) = Ix). — Yi1 ± • • • + lx n — Ynii,

and this is not the metric space Rn (if n > I). Often we are interested in only one metric for a given set of points, and in such cases we sometimes use the symbol X to denote both the set of points and the metric space (X, p). If we have two metric spaces (X, p) and ( Y, (7), we can form a new metric space called the Cartesian product X X Y whose set of points is the set X X Y = {(x, y): x e X, y e Y} and whose metric T is

given by T((xi, y), (x2, Y2)) = [P(xi, x2) 2 -1- 0-(y „ y 2)1 112

It is readily verified that T has all the properties required of a metric and that Rin X Rn = Rm+n . Any nonempty subset of a metric space is itself a metric space if we restrict the metric to it. For example, the space C of the last chapter is a subspace of L. Problems 1. a. Show that the set of all n-tuples of real numbers becomes a metric space under each of the following metrics:

P* (x, Y) = ix i — Yi 1 + • • • + ixn — Y7,1 max {ixi — Yil, • • • , 1xn — Yni} •

P+(x, Y) =

Sec. 2]

129

Open and Closed Sets

b. For n = 2 and n = 3 describe the sets {x: p(x, y) < 1), {x: p*(x, y) < 11, and Ix: p±(x, y) < 11. 2. By the spheroid (or sphere) centered at x and having radius we mean the set Sz,3 = 1Y: XX, Y) < al. Prove that if 0 < e < — p(x, z), then Sz,, C S. 3. Pseudometrics. A pair (X, p) is called a pseudometric space if p satisfies all the conditions of a metric except that p(x, y) = 0 need not imply x = y. Show that p(x, y) = 0 is an equivalence relation and that if X* is the set of equivalence classes under this relation, then p(x, y) depends only on the equivalence classes of x and y and defines a metric on X*. 2

Open and Closed Sets

We shall find that a number of the properties of sets of real numbers apply immediately to sets in a metric space. Throughout the present section all sets mentioned are subsets of a given metric space (X, p). The following propositions and definitions correspond to those in Section 5 of Chapter 2, and the reader is asked to check that the proofs given there are valid for metric spaces. Definition: A set 0 is called open if for every x e 0, ô > O such that all y with p(x, y) < 3 belong to O. 1. Proposition: The sets X and 0 are open; the intersection of any two open sets is open; and the union of any collection of open sets is open. Definition: A point x e X is called a point of closure of the set E if for every > 0 there is a point y e E such that p(x, y) < a. We use to denote the set of points of closure of E. Clearly E c E. 2. Proposition: If A c B, then A and (A n B) c A n B.

C -E.

Definition: A set F is called closed if

Also (Au B) = -21u7B,

= F.

3. Proposition: The closure of any set E is closed; that is, E =

7.

130

Metric Spaces

[Ch. 7

4. Proposition: The sets 0 and X are closed; the union of any two closed sets is closed; and the intersection of any collection of closed sets is closed. 5. Proposition: The complement of an open set is closed; the complement of a closed set is open. Definition: A metric space X is called separable if it has a subset D which has a countable number of points and which is dense in X, that is, for which .D X. —

Since the set of rational numbers is a countable dense subset of R, we see that R is separable. The following proposition shows that the Lindel6f theorem holds for a metric space if and only if it is separable. 6. Proposition: A metric space X is separable if and only if there is a countable family {0,} of open sets such that for any open set 0 c X, 0= U Oi. oico

Proof: If X is separable, let D be a countable dense set. By the spheroid at x with radius 5 we mean the set Sx,S = {Y: P(X3 Y) < 5 } •

Let {0, } consist of those spheroids S x,6 for which x is in D and 5 is rational. Then f0j is a countable collection of open sets. If 0 is any open set and y c 0, then we want to show that for some 0, we have y c 0, C 0. Since 0 is open, there is a spheroid Sy, 6 such that S y ,6 C 0. By taking 3 even smaller we may assume 5 is rational. Since y is a point of closure of D, there is a point x E D such that p(x, y) < 5/2. Hence Sx0312 C Su o5 C

0.

But Ss ,112 is one of the {0,} , and the "only if" part of the theorem is proved. Suppose, on the other hand, we are given the countable collection {0,}. Let xi be a point of 0„ and let D be the set of all these points x,. We shall now see that D is dense. Let x be any point of X and S

Sec. 3]

Continuous Functions and Homeomorphisms

131

any spheroid centered at x. Then we must show that S contains a point of D. But S is an open set (Problem 6), and so we must have some 0, such that x e 0, c S. Hence x, e S, and we see that x e TD. I

Problems 4. a. Show that C is a closed subset of L (see Chapter 6).

b. Show that the set of all integrable functions which vanish for 0 < t
c. Show that the set of all measurable functions x(t) with f x < 1 is an open subset of L'. 5. Show that E = F for closed sets F. The interior E° of a set E

n

ECF

is the set of those y e E for which there is a (5 > 0 such that p(y, z) < (5 z e E. a. Show that E°= U o. OCE

b. Show that

—CE) = (E)°. 6. a. Show that each spheroid is open. b. Show that the sets {x: p(x, y) < SI are closed.

c. Is the set in (b) always the closure of the spheroid

{x: p(x, y) < (5 } ? 7. Which of the spaces Rn, C, L', L' are separable?

3

Continuous Functions and Homeomorphisms

A function f on a metric space (X, p) into a metric space ( Y, 0" is a rule which associates to each x e Xa unique y e Y. We also call f a mapping of X into Y and shall use the terms function and mapping interchangeably. By a mapping f of X onto Y we mean, as usual, that for any y e Y there is some x e X such that f(x) = y; that is to say, that Y is the range off. Then as in Section 6 of Chapter 2 we have the following definition and propositions :

132

Metric Spaces

[Ch. 7

Definition: The function f is said to be continuous at x if, for every e > 0, there is a 6 > 0 so that if p(x, y) < 6, then cr[f(x), f(y)] < e. The function f is called continuous if it is continuous at each x e X. 7. Proposition: A function f from a metric space X to a metric space Y is continuous if and only if for each open set 0 in Y the set f -1 [0] is an open set in X. 8. Proposition: If f is a continuous mapping from X to Y and if g is a continuous mapping from Y to Z, then the mapping g o f from X to Z is also continuous.

A one-to-one mapping f of X onto Y is called a homeomorphism between X and Yiff is continuous and the mapping f -1 inverse to f is also continuous. The spaces X and Y are said to be homeomorphic if there is a homeomorphism between them. The study of topology is essentially the study of those properties which are unaltered by homeomorphisms, and such properties are called topological. By virtue of Proposition 7 a one-to-one correspondence between X and Y is a homeomovhism if and only if it makes open sets in X correspond to open sets in Y and conversely. Thus the property of being an open subset of a space is a topological one. Since a closed set is the complement of an open set, it follows that being a closed subset of X is a topological property. In fact, every property which can be defined by means of open sets is a topological one, and hence so is the property of being a continuous function: that is, if f is a continuous function on X and h: X —> Y is a homeomorphism between X and Y, then fo 11 -1 is a continuous function on Y. Not all properties in a metric space are preserved under a homeomorphism, however. For example, the distance between two points is usually altered by a homeomorphism. A homeomorphism which leaves distance unchanged, that is, one for which cr[h( x i ), h x 2 ) ] = P(x 1, x2 ) (

for all x 1 and x2 in X, is called an isometry between X and Y. The spaces X and Y are called isometric if there is an isometry between them. From an abstract point of view two isometric metric spaces are exactly the same, an isometry amounting merely to a relabeling of the points. A notion which arises naturally in this connection is that of equivalent metrics: Two metrics p and cr on the same set X

Sec. 4]

Convergence and Completeness

133

of points are called equivalent if the identity mapping of (X, p) onto (X, a) is a homeomorphism. Thus two metrics are equivalent if and only if they define the same open sets, that is, if a set is open with respect to one whenever it is open with respect to the other. Problems 8. Show that the function h on [0, 1) given by h(x) = x/(1 homeomorphism between [0, 1) and [0, 00). 9.

—

x) is a

Let E be a set and x a point in a metric space. Define p(x, E) = inf p(x, y). ycE

a. Show that for a fixed E the functionf given byf(x) -- p(x, E) is continuous. b. Show that {x: p(x, E) = 0} = E. 10. a. Prove that two metrics on a set X are equivalent if and only if given x e X and e > 0 there is a 6 > 0 such that for all y e X p(x, y) G 6 = a(x, y) < e

and a(x, y) < 6

p(x, y) < e.

b. Show that the following set of metrics for the set of n-tuples of real numbers are equivalent: P(x, Y) =

[(xi — Yi) 2 + • • • + (xn — Yn) 2 i ii 2

P* (x, Y) = ixi — Yii + • • • + ixn — Yni p +(x, y) = max {ix i — y 1 1, ... , 1x,, —

c. Find a metric for the set of n-tuples which is not equivalent to these. 11. Show that if p is any metric on a set X, then a = p/(1 + p) is an equivalent metric for X. Prove that (X, a) is a bounded metric space; that is, Œ(x, y) < 1 for all x and y in X. 4

Convergence and Completeness

Just as for the case of real numbers we say that a sequence (xn ) from a metric space X converges to the point x in X (or has x as a limit), if given E > 0, there is an N such that p(x, x n) < e for all

134

Metric Spaces

[Ch.

7

n > N. This definition can be rephrased in geometric terms by saying that .x„) converges to x if every spheroid about x contains all but a

finite number of terms of the sequence. We often write x lim x„ or x„ x to mean that x is the limit of the sequence (x n ). If we have merely the weaker condition that each spheroid about x contains infinitely many terms of the sequence, then we say that x is a cluster point of the sequence (x n ). Thus x is a cluster point of (x n) if given e > 0 and given N there is an n > N such that p(x, x,) < E. Thus if x is the limit of (x„), then x is a cluster point of (x„), but the converse is not necessarily true. A number of properties concerning limits and cluster points are given in the problems. A sequence (xn) from a metric space is called a Cauchy sequence if given e > 0 there is an N such that for all n and m larger than N we have p(x„, x,„) < E. If (x„) converges to some point x, then given > 0 we may choose N so large that p(x, x) < 6/2 for n > N. Hence for all n, m > N we have P(x., x.)

p(x„, x)

p(x„„ x) < e,

and (x 7,) is a Cauchy sequence. The converse statement that every Cauchy sequence converges is not true in an arbitrary metric space (for example, in the space of rational numbers with the usual metric). If a metric space has the property that every Cauchy sequence converges (to some point of the space), we say that the space is complete. The Cauchy criterion for real numbers merely states that the space R of real numbers is complete. Other examples of complete spaces are given by the Banach spaces discussed in Chapter 6. If X is an incomplete metric space, it can always be enlarged to become complete. A precise statement of this fact is given by the following theorem. One proof of this theorem is outlined in Problem 17 and another in Problem 10.16. 9. Theorem: If (X, p) is an incomplete metric space it is possible to find a complete metric space X* in which X is isometrically embedded as a dense subset. If X is contained in an arbitrary complete space Y, then X* is isometric with the closure of X in Y.

Problems 12. Show that a sequence (xn ) in a metric space has x as a cluster point if and only if there is a subsequence (x,,,,) of (xn ) which converges to x.

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13. Show that a sequence (x) in a metric space converges to x if and only if every subsequence of (x,i) has x as a cluster point. Hence (xn) converges to x if every subsequence has in turn a subsequence which converges to X. 14. Let E be a set in a metric space X. If xis a cluster point of a sequence from E, then x e 7e, while, if x e 7E, there is a sequence from E which converges to X. 15. If a Cauchy sequence (x,z ) in a metric space has a cluster point x, then (xn ) converges to x. 16. If X and Y are metric spaces andf a mapping from X to Y, thenf is continuous at x if and only if for each sequence (xn ) in X converging to x we have (f(x„)) converging to f(x) in Y. 17. a. If (x) and (yn) are Cauchy sequences from a metric space X, then p(xn, yn) converges. b. The set of all Cauchy sequences from a metric space X becomes a pseudometric space (cf. Problem 3) if p*( ( 4,), (yn)) = lirn p(xn, yn). c. This pseudometric space becomes (as in Problem 3) a metric space X* when we identify elements for which p* = 0, and X is isometrically embedded in X. d. The metric space (X*, p*) is complete. [Hint: If (x„) is a Cauchy sequence from X we may assume (by taking subsequences) that P(xn, xn+1) < 2'. If ((xnm)--1):=1 is a sequence of such Cauchy sequences which represents a Cauchy sequence in X*, then the sequence is a Cauchy sequence from X which represents the limit of the Cauchy sequences from X*.] e. Use Propositions 10, 11, and 12 of the next two sections to complete the proof of Theorem 9. 18. Prove that the Cartesian product of two complete metric spaces is complete.

5 Uniform Continuity and Uniformity Let f be a mapping from the metric space (X, p) to the metric space ( Y, a). We say that f is uniformly continuous if given e > 0 there is a 8 > 0 such that for all x and x' in X with p(x, x') < 3 we have o-(f(x), f(x')) < e. The function h defined on [0, 1) by h(x) = x1(1 — x) is continuous but not uniformly continuous. Moreover, this function h takes the Cauchy sequence given by xii = 1 — 1/n into the sequence y7, = n — 1 which is not a Cauchy sequence. Thus the image under a continuous function of a Cauchy sequence need

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not be a Cauchy sequence. However, we have the following proposition: 10. Proposition: Let f be a uniformly continuous mapping of the metric space X into the metric space Y. If (x„) is a Cauchy sequence in X, then (f(x, i)) is a Cauchy sequence in Y.

A homeomorphism f between metric spaces X and Y is called a uniform homeomorphism if both f and f -1 are uniformly continuous. It follows from Proposition 10 that the property of being a Cauchy sequence is preserved under uniform homeomorphisms, as is the property of being complete. Properties which are preserved under uniform homeomorphisms are called uniform properties. In addition to the properties of being a Cauchy sequence and of completeness we have also uniform continuity as a uniform property. These three properties are not topological properties, since the function h defined by h(x) = x/(1 — x) is a homeomorphism between the incomplete space [0, 1) and the complete space [0, Do) which takes a Cauchy sequence into a sequence which is not a Cauchy sequence and its inverse carries the uniformly continuous function sin back into a function which is not uniformly continuous on [0, 1). Two metrics p and a for a set X of points are said to be uniformly equivalent if the identity map from (X, p) to (X, a.) is a uniform homeomorphism. Thus a and p are uniformly equivalent if given E > 0 there is a 3 > 0 so that for all x and y we have p(x, y) < 8 a(x, y) < E and a(x, y) < 8 p(x, y) < e. We conclude this section by stating the following useful extension theorem for uniformly continuous mappings. Its proof is outlined in Problem 20. 11. Proposition: Let (X, p) and (Y, a) be metric spaces with Y complete. Let f be a uniformly continuous mapping from a subset E of X into Y. Then there is a unique continuous extension g of f from E to E; that is, there is a unique continuous mapping g from E into Y such that g(x) = f(x) for x e E. Moreover, g is uniformly continuous.

Problems 19. Prove Proposition 10. 20. Prove Proposition 11 by the following steps:

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137

a. If (x„) is a sequence from E which converges to a point x e E, then (f (x„)) converges to a point y e Y (cf. Proposition 10). b. The point y in (a) depends only on x and not on the sequence (x„). Thus if we define y = g(x) we have defined a function g on E which is an extension of f. c. The function g is uniformly continuous on E. d. If h is any continuous function from r to Y which agrees with J on E, then h = g. 21. a. Show that the metrics in Problem 10b are uniformly equivalent. b. Find a metric for the set of n-tuples of real numbers which is equivalent but not uniformly equivalent to the usual metric. c. If (X, p) is any metric space, the metric cr = p/ (1 p) is uniformly equivalent to p. 22. a. Boundedness is a metric but not a uniform property (cf. Problem 21c). b. A metric space X is said to be totally bounded if given e > 0 there are a finite number of spheroids of radius e which cover (that is, whose union is) X. Show that total boundedness is a uniform property. c. Show that total boundedness is not a topological property. [Consider [0, 1) and [0, GO.]

6 Subspaces If (X, p) is a metric space and S is a subset of X, then S becomes a metric space if we restrict p to S, that is to say, if we take as the distance between two points of S their distance as points of X. When we consider S as a metric space with this metric, we call S a subspace of X. For example, the rationals are a subspace of R, and the set {(x, 0) } in R2 is a subspace isometric with R. The space C is a subspace of L. If E is a subset of S, then we may consider the closure of E in S or in X; that is, we may wish to consider the set of all points of X which are points of closure of E or else the set of all points of S which are points of closure of E. These sets are in general different. For example, let X be the space R and S the interval (0, 1). Then if E is the interval (0, 1], the closure of E in R is the interval [0, i], while its closure in S is just (0, IL that is, E is closed relative to S. Thus we see that the closure of a set as well as the properties of a set being closed or open are all relative to the space containing the set. However, we do have the following relations between these notions.

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12. Proposition: Let X be a metric space and S a subspace of it. Then the closure of E relative to S is n S, where E denotes the closure of E in X. A set A c S is closed relative to S f and only if A = SnE with F closed in X. A set A c S is open relative to S if and only if A = Sn 0 with 0 open in X. Proof: If x is a point of closure of E in X, then it is a point of closure of E in S if it belongs to S. Hence the closure in S of E is E n S. If A is closed in S we must have A = S n 7, while on the other hand if F is closed in X, then the closure in S of S n F is

Sn(SnF)cSn(Snl)cSnF, whence S n F is closed relative to S. If A is open relative to S, then S — A is closed relative to S and we haveS—A =SnF, or SnA =Sn(—F) and —Fis open in X. Similarly, if 0 is open, then SnO is the complement in S of S n ( - 0), which is closed in S.1 13. Proposition: Every subspace of a separable metric space is separable. Proof: Let X be a separable metric space and S a subspace. Then by Proposition 6 there is a countable collection {o} of open sets in X such that each open set in X is a union of some subcollection of {02 } . By Proposition 12 the collection {0, nS} is a countable collection of open subsets of S such that every open subset of S is a union of a subcollection of them. Hence S is separable by Proposition 6. I In contrast to the relative properties discussed above, there are some properties which are intrinsic. For example, the property of x being a point of closure of E holds in any subspace of X containing x and E as soon as it holds in one of them. Another such property is that of being complete, since the definition of completeness of a space is given in terms of points in the space. However, the following proposition gives some relations between complete sets and closed sets.

14. Proposition: If a subset A of a metric space X is complete, then it is closed. On the other hand, a closed subset of a complete metric space is itself complete.

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Sec. 7] Baire Category

Problem 23. Prove Proposition 14. [Hint: Problem 14 is helpful.]

7 Baire Category

In this section we shall go more deeply into certain aspects of metric spaces. We begin by defining a set E to be nowhere dense if —E is dense. This is equivalent to stating that r contains no spheroid. For example, the integers are nowhere dense in R, and the Cantor ternary set is nowhere dense in [0, 1]. A set E is said to be of first category (or meager) if it is the union of a countable collection of nowhere dense sets. A set which is not of first category is said to be of second category, and the complement of a set of first category is called residual. We shall show that a complete metric space is of second category when considered as a subset of itself. We begin with the following theorem: 15. Theorem: Let X be a complete metric space and {O n} a O n is not countable collection of dense open subsets of X. Then empty.

n

Proof: Let x1 be a point of 0 1 and Si a spheroid (of radius r i) which is centered at x i and contained in 0 1 . Since 02 is dense, there must be a point x2 in 02 n Si . Since 02 is open, there is a spheroid S2 centered at x2 and contained in 02, and we may take the radius r2 of S2 to be smaller than r 1 and smaller than r — p(x 1 , x2). Then c Si . Proceeding inductively, we obtain a sequence S,i) of spheroids such that 3, c Sn_i and S O n and whose radii (rn ) tend to zero. Let (x,i ) be the sequence of centers of these spheres. Then for n, m > N we have x n e SN and xn, e SN. Hence p(x., xm) < 2rN , and (xii) is a Cauchy sequence, since r„ —> 0. By the completeness of X there is a point x such that xi, x. Since x n e SN+1 for n > N, we have x e EN+1 C SN C ON. Hence x e (1 On . I

16. Corollary (Baire Category Theorem): A complete metric space is not the union of a countable collection of nowhere dense sets. Proof: Let {E„} be a countable collection of nowhere dense sets. Then O = —En is a dense open set, and so there is a point x e 0.. But this means x o U E. I

n

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As an application of the Baire Category Theorem we establish the following theorem, which is known as the uniform boundedness principle: 17. Theorem: Let v be a family of real-valued continuous functions on a complete metric space X, and suppose that for each x c X there is a number Mx such that I f(x)1 < Mx for all f c F. Then there is a nonempty open set 0 c X and a constant M such that I f(x)1 < M for all f c and all x c O.

Proof: For each integer m, let E,„,f — {x: 1 f(x)1 < m}, and set En, = Emj. Since each f is continuous, E,,,,f is closed, and con-

n

sequently En, is closed. For each x e X, there is an m such that I f(x)! < m for all f e 5; that is, there is an m such that x e Ent. Hence X= 0 Em. M=1

Since X is a complete metric space, there is a set En, which is not nowhere dense. Since this En, is a closed set, it must contain some spheroid O. But for every x c O we have j f (x)1 < m for all f e F. I

Problems 24. a. Prove that a closed set F is nowhere dense if and only if it contains no open set. b. Prove that E is nowhere dense if and only if for any nonempty open set 0 there is a spheroid contained in 0 — E. 25. a. Prove that if E is of first category and A c E then A is also of first category. b. Prove that if (En ) is a sequence of sets of first category then En, is also of first category. U 26. If X is a complete metric space and E is a set of first category in E, then —E is dense in X. 27. a. Show that on [0, 1] there is a nowhere dense closed set having Lebesgue measure 1 — 1/n. b. Construct a set of first category on [0, 1] which has measure 1.

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28. A point x in a metric space is called isolated if the set {x} is open. Prove that a complete metric space without isolated points has an uncountable number of points. 29. a. If Ê is dense and F a closed set contained in E, then F is nowhere dense. b. If E and Ê are both dense, then at most one of them is an c. The set of rational numbers in [0, 1] is not a 9,5 . d. Is there a real-valued function on [0, 1] which is continuous on the rationals and discontinuous on the irrationals? 30. Let C be the space of continuous functions on [0, I] and set F,,, {f: (3x0) with If(x) — f(x 0)1 n(x — x o) for all x, xo x < 11. a. Show that F„ is a closed subset of C. b. Show that F„, is nowhere dense. [Hint: Any g e C can be approximated to within e/2 by a polygonal function ço, and ço can be approximated to within E/2 by a polygonal function 1,t whose right-hand derivative is everywhere greater than n in absolute value.] c. Show that the set D of continuous functions which have a finite derivative on the right at at least one point of [0, 1] is a set of first category in C. d. There is a nowhere differentiable continuous function on [0, 1]. 31. Let (f,,) be a sequence of continuous functions from a metric space X to a metric space Y such that for each x in X the sequence (f,i (x)) converges to a pointf(x) in Y. Thenf defines a mapping of X into Y. Show that there is a set E of first category in X such that f is continuous at each point x e X — E. Hence if X is complete, f is continuous at a dense set of . points in X. [Hint: Let = fx: 0- (fa(X), fk(X)) < 1/m for all k Let Fn°,„, be the interior of F„,,„. Set E = U iFn,m tt,m

8 1

Topological Spaces

Fundamental Notions

In Chapter 7 we discussed properties of metric spaces and found that a number of theorems depended only on the properties of open and closed sets. In the present chapter we study spaces in which the notion of an open set is fundamental and other notions are defined in terms of it. Such spaces are called topological spaces and are more general than metric spaces. Perhaps you ask: Why not stick to metric spaces? It is true that metric spaces are simpler, but there are many examples of spaces of functions where certain topological notions have a natural meaning not consistent with the topological concepts derived from any metric that might be put on the space. Important examples are given by the weak topologies in Banach spaces. We proceed with a formal definition. Definition: A topological space (X, 3) is a nonempty set X of points together with a family 5 of subsets (which we shall call open) possessing the following properties: i. X E 3, 0 E 3; ii. 0 1 E 5 and 02 e 3 imply 0 1 n 02 E 3; iii. O , E 5 implies U Oa E 3. The family 5 is called a topology for the set X.

The properties in this definition are all satisfied by open sets in a metric space (X, p), and hence to each metric space (X, p) we can 142

Sec.

11

Fundamental Notions

143

associate a topological space (X, 3), where 3 is the family of open sets in (X, p). A topological space which is associated in this manner to some metric space is called metrizable, and the metric p is said to be a metric for the topological space. From a logical standpoint the distinction between a metric space and its associated topological space is essential, since different metric spaces can give rise to the same topological space. Two such metric spaces are of course equivalent, and we often disregard the distinction between a metric space and its associated topological space in cases where no confusion is likely to arise. In some cases we shall not trouble even to distinguish between a topological space ( X, 3) and the set X of its points, using X to denote both. It is to be remembered, however, that metric and topological spaces are couples, and in many cases it is necessary to express this fact explicitly. Given any set X of points there are always two topologies that can be defined on X. One is the trivial topology in which the only open sets are 0 and X. A second possible topology is the discrete topology: Every subset of X is an open set. In terms of the notion of open set we may define other topological properties, for example: A subset F of X is called closed if F is open. 1. Proposition: The sets 0 and X are closed. The union of any two closed sets is closed. The intersection of any collection of closed sets is closed.

If A is a subset of a topological space (X, 3), we can define a topology 8 for A by taking for s those subsets B of A for which there is a set 0 E 3 such that B = A n O. We call 8 the topology inherited from 3 and call the topological space (A, s) a subspace of (X, 3). This is consistent with our usage concerning metric spaces. A sequence (x,t) in a topological space is said to converge to the point x, or to have the limit x, if given any open set O containing x there is an integer N such that xn E O for all n > N. Similarly, a sequence (xn,) is said to have x as a cluster point if given any open set 0 containing x and any integer N there is an integer n > N such that xn, g O. Thus if (x n) has a subsequence which converges to x, then x is a cluster point of (xn). The converse of this statement is not always true in an arbitrary topological space. In view of Proposition 7.7 we may give a definition of a continuous function on a topological space which agrees with the usual concept

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of continuity in the case in which the topological space is also a metric space. Definition: A mapping f of a topological space (X, 3) into a topological space (Y, 8) is said to be continuous if the inverse image of every open set is open, that is, if 0 e s = f'[0] e 3. It should be noted that if f is a continuous function on a space X, then the restriction fi of f to a subspace A of X is a continuous function on A, for f 1-1 [0] =A n f -1 [0] and must be open for open 0 by the continuity off and the definition of open sets in A. Definition: A homeornorphisrn between two topological spaces is a one-to-one continuous mapping of X onto Y for which f -1 is continuous. The spaces X and Y are said to be homeornorphic if there is a homeomorphism between them. From an abstract point of view two homeomorphic topological spaces are indistinguishable, the homeomorphism amounting to a mere relabeling of the points of one set by the points of a second set. Thus the concept of homeomorphism plays the same role for topological spaces that isometry plays for metric spaces and isomorphism plays for algebraic systems. Suppose that 3 and s are two topologies for the same set X. Then s is said to be stronger than 3 if D 3. Thus 8 is stronger than 3 if and only if the identity mapping of (X, g) into (X, 3) is continuous. The trivial topology for a set X is the weakest possible topology on X, while the discrete topology is the strongest possible topology. If 8 and 3 are two topologies for a set X, then s n 3 is also a topology. In fact, if {3alif is any collection of topologies, then 3a

n

is a topology. Thus if e is any collection of subsets of X, then the intersection of all topologies containing e is a topology containing e. This topology is the weakest topology such that all of the sets of e are open.

2. Proposition: Let X be a nonempty set of points and e any collection of subsets of X. Then there is a weakest topology 3 which contains C.

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145

Problems 1. a. Given a set X, can you define a metric on X so that the associated topological space is discrete? Trivial? b. Let X be a space with a trivial topology. Find all continuous mappings of X into R. c. Let X be a space with a discrete topology. Find all continuous mappings of X into R. 2. We say that x is a point of closure of a set E if for every 0 e 3 with x c 0 we have 0 n E 0. Prove that the set F of points of closure of E is the intersection of all closed sets containing E. 3. Prove that a set A C X is open if and only if given x c A there is an open set 0 such that x c 0 C A. 4. Prove that a mapping of X into Y is continuous if and only if the inverse image of every closed set is closed. 5. Show that if f is a continuous mapping of X into Y and g a continuous mapping of Y into Z, then g of is a continuous mapping of X into Z. 6. Prove that the sum and product of two real-valued continuous functions are themselves continuous. 7. a. Let F be a closed subset of a topological space and (x,) a sequence of points from F. Show that if x is a cluster point of (x„) then x e F. b. Show that iff is continuous and x = lim xn then (f(xn)) has the limit f(x). c. Show that if f is continuous and x is a cluster point of (xn ) then f(x) is a cluster point of (f(xn)).

2

Bases and Countability

A collection 43 of open subsets of a topological space X is called a base for the topology 3 of X if for each open set 0 in X and each x c 0 there is a set B e 43 such that x eBc O. A collection 63x of open sets containing a point x is called a base at x if for each open set 0 containing x there is a B e 43z such that x eBc O. Thus a collection 43 of open sets is a base if and only if it contains a base at each point x e X. If X is a metric space, the spheroids form a base, and the spheroids centered at x form a base at x. If 43 is a base for the topology 3, then 0 e 3 if and only if for each X e 0 there is a B e 63 with X eBc O. The "only if" part of

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the statement follows from the definition of a base, while if for each x in 0 we have x eBc 0, then 0 must be the union of those B in 133 with B C 0, and 0 is open since it is a union of open sets. We often find it convenient to specify a topology for a set X by specifying a base 63 of open sets and using the preceding criterion to define the open sets. Conditions on a collection 133 in order that it be a base for some topology are given by the following proposition: 3. Proposition: A collection 63 of subsets of a set X is a base for some topology on X if and only if each x in X is contained in some B, and if x e B1 n B2 then there is a B3 e 63 such that x e B3 C B1 n B2Proof: That these conditions are necessary follows from the definition of a base and the fact that X and B 1 n B2 must be open. Suppose now that 63 satisfies these conditions. If we set 3 {0: (x e 0)(V3 e 63)(x c B c 0)1, then 0 e 3, and the union of sets in 3 will be in 3, and the first condition on 63 implies that X e 3. To show that the intersection of two sets 0 1 and 02 in 3 is itself in 3, let x e 0 1 n 02. Then there are sets Bl and B2 in ea such that x e Bi C 0 1 and x e B2 C 02. Let B3 be a set in 63 with x g B3 C B1 n B2Then x e B3 C 01 n 02, and it follows that 0 1 n 02 is open. 1

A topological space is said to satisfy the first axiom of countability if there is a countable base at each point. Every metric space satisfies the first axiom of countability, since the spheroids centered at x and having rational radii are countable in number and form a base at x. A space is said to satisfy the second axiom of countability if there is a countable base for the topology. Thus Proposition 7.6 states that a metric space satisfies the second axiom of countability if and only if it is separable.

Problems 8. a. Let 63 be a base for the topological space (X, 3). Then x e r if and only if for every B e 63 with x e B, there is ay eBn E. b. Let X satisfy the first axiom of countability. Then x e E if and only if there is a sequence from E which converges to x. c. Let X satisfy the first axiom of countability. Then x is a cluster point of a sequence (xn ) from X if and only if (x7,) has a subsequence which converges to x.

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147

9. A mapping f of a topological space X into a topological space Y is said to be continuous at x if for any open set 0 containing f(x) there is an open set U containing x such that f[U] C O. a. Show that f is continuous if and only if it is continuous at each point of X.

b. Let

a base at x and ey be a base at y = f(x). Then f is x if and only if for each C c e y there is a B y 63, such that continuous at B Cf 1 [C}. 63x be

10. Let 0 be any collection of subsets of X. Let 63 consist of X and all finite intersections of sets in e. Show that 133 is a base for the weakest topology which contains C. 11. Let X be an uncountable set of points, and let 3 consist of the empty set and all subsets of X whose complement is finite. Show that 3 is a topology for X and that the space (X, 3) does not satisfy the first axiom of countability. 12. Let X be the set of real numbers, and let 63 be the set of all intervals of the form [a, b). Show that 63 is the base of a topology 3 for X. (This topology is called the half open interval topology.) Show that (X, 3) satisfies the first but not the second axiom of countability and that the rationals are dense in X. Is (X, 3) metrizable? -

3 The Separation Axioms and Continuous Real-Valued Functions The properties of topological spaces are in general quite different from those of metric spaces, and it is often convenient to suppose that our topological space satisfies some additional conditions which are true in metric spaces. Consider the following set of conditions on a topological space: T1. Given two distinct points x and y, there is an open set which contains y but not x. T2. Given two distinct points x and y, there are disjoint open sets O l and 0 2 such that x c O l and y c 0 2. T3. In addition to T1 , given a closed set F and a point x not in F, there are disjoint open sets O l and 02 such that x c O l and F c 02. T4. In addition to T1 , given two disjoint closed sets F1 and F25 there are disjoint open sets 0 1 and 0 2 such that F1 c 0 1 and F2 C 02.

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These are called separation axioms, and all are satisfied in a metric space. A topological space which satisfies T2 is called a Hausdorff space, one which satisfies T3 is called a regular space, and one which satisfies T4 is called a normal space. The following proposition tells us that the condition T 1 is equivalent to the statement that each set consisting of a single point is closed. With this in mind we see that T4 T3 T2 = T1.

4. Proposition: A topological space X satisfies T 1 if and only if every set consisting of a single point is closed. Proof: If each set {x} is closed, given two distinct points x and y, we may take 0 = —{x} . Then 0 is an open set containing y but not x. Suppose that T 1 holds. Each y c {x} is contained in an open set 0 c {x}. Thus the set — {x} is the union of the open sets contained in it and so must be open. Hence {x } is closed. I

Important consequences of normality are the following propositions, whose proofs are left to the reader (Problems 18 and 19). 5. Urysohn's Lemma: Let A and B be disjoint closed subsets of a normal space X. Then there is a continuous real-valued function f defined on X such that 0
If X is any set of points and a is any collection of real-valued functions on X, there is always a weakest topology on X such that every function in a is continuous, for let C = {E: E = .f-1 [0], f c a and 0 an open subset of RI and apply Proposition 3. This topology is called the weak topology generated (or induced) by a. If the functions in a are all continuous in some topology 3, then the weak topology generated by a is in general weaker than 3. For these topologies to coincide, there must be enough continuous real-valued functions. One condition which guarantees this is the following: Given a closed set F and a point x o F there is a function f c a with f(x) = 1 and f 0 on F. If this condition is satisfied when a is

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the space C(X) of all continuous real-valued functions on X, then we say that X is completely regular, provided that X satisfies T1 . It follows from Urysohn's lemma that each normal space is completely regular, and it is easy to see that complete regularity implies regularity. For this reason the condition defining complete regularity is

sometimes called T31. The following theorem characterizes separable metric spaces. It can be proved using the concepts of the next section (see Problem 30). 7. Urysohn Metrization Theorem: Every normal topological space satisfying the second axiom of countability is metrizable.

Problems 13. a. Show that every metric space is Hausdorff. b. Show that every metric space is normal. [Hint: If F 1 and F2 are disjoint closed sets, let 0 1 = {x: p(x, F1) < P(x, F2)} and 02 = {x: p(x, F 2) < p(x, F1)} 14. Let X consist of the numbers [0, 1] and an element 0', and take as a basis for a topology the sets (a, 13), [0, 0), (a, 1], and {O' } u (0, 0). Show that these are a base for a topology on X which is T1 but not Hausdorff.

15. Let f be a real-valued function on a topological space X. Show that f is continuous if for each real number a the sets {x: f(x) < a} and {x: f(x) > a} are open. Show thatf is continuous if for each real number a the set {x: f(x) > a} is open and the set {x: f(x) > a} is closed. 16. If f and g are continuous real-valued functions on a topological space X, then the functions f + g, fg, f V g, and f A g are continuous. (Here as usual (f V g)(x) max f (x), g(x).) 17. Let (A) be a sequence of continuous functions from a topological space X to a metric space Y. If (fit ) converges uniformly to a function f, then f is continuous. 18. a. Show that a Hausdorff space is normal if and only if given a closed set F and an open set 0 containing F there is an open set U such that F c U and U C O. b. Let F be a closed subset of a normal space contained in an open set O. By repeating the result in (a) indefinitely, show that it is possible to construct a family {Ur} of open sets, one corresponding to each rational in (0, 1) of the form r =p • 2, such that F C UT C 0 and V, C U, for r < s.

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8

c. Let {Ur} be the family constructed in (b) with U i = X. Let f be the real-valued function on X defined by f(x) = inf {r: x e Ur} . Then fis a continuous function, 0 < f < 1, with f m 0 on F and f m 1 on Ô. d. Let X be a Hausdorff space. Prove that Xis normal if and only if for every pair of disjoint closed sets A and B on X there is a continuous real-valued function f on X such that 0 < f < 1, f 0 on A and f -=-- 1 on B. 19. Prove Tietze's Extension Theorem by using the following steps: a. Let h = f/(1 ± Ifl). Then Ihj < 1. b. Let B = {x: h(x) < —}, C = {x: h(x) > il . Then by the Urysohn lemma there is a continuous real-valued function h 1 on X which is — i on B and on C, while Ih i (x)1 < for all x c X. Clearly Ih(x) — hi(x)1 < 1 for all x c A. c. By induction there is a continuous function hn on X such that 2n-1

h(x)1 < 3n for all x e X and Ih(x) —

2n

h(x) I < k, for all x e A.

d. The sequence (ha) is uniformly summable to a continuous function k on X, lki < 1 and k = h on A. Set g = k/(1 — jkl). 20. Let a be a family of real-valued functions on a set X. Show that a base for the weak topology on X generated by a is given by the sets of the form Ix: If(x) — f(y) l < e, for some e > 0, y e X, and some finite set fi, . . . ,f„ of functions in 5} . Show that this topology is Hausdorff if and only if given any pair {x, y} of distinct points in X there is an f e a with f(x) X f( y). 21. Let 5 be a family of real-valued continuous functions on a topological space (X, 3). Show that the weak topology generated by 5 is 3 if for each closed set F and each x 0 F there is an f e 5 with f(x) = 1 and f m. 0 on F. 22. Show that every completely regular space is regular. 23. Prove that every subset of a Hausdorff space is Hausdorff.

4

Product Spaces

and ( Y, 8) are two topological spaces, we define a topology on the product X X Y by taking as a base the collection of all sets of the form 0 1 X 02, where 0 1 e 3 and 02 e S. This is called the product topology for X X Y. If X and Y are metric spaces, then the product topology agrees with the topology given by the product metric. If (XOE, 5OE ) is any indexed family of topological spaces, we If (X, 5)

Sec. 4] Product Spaces

151

define the product topology on X Xa by taking as a base all sets of the « form X Oa where Oa e aa and Oa = Xa except for a finite number « of a. If the X. are all the same space X and indexed by an index set A, we write X" for X X. « If (XŒ, 3a ) is a collection of topological spaces and Y their product, we define for each a a mapping Ira (called a projection) of Y into Xa by letting 71-Œ (x) be the a-th coordinate of x. Each 7ra is continuous, and the product topology in Y is the weakest topology such that each ra is continuous. If A is countable and X is metrizable, then X" is metrizable. Since only the number of elements in A is important in determining X A , we usually write X' (or X N) for a countable product. If we denote the discrete space with two elements by 2, then 2° is homeomorphic to the Cantor set. If we use to to denote not only a countable set but also a countable set with discrete topology, then cow is a topological space which is homeomorphic to the set of irrational numbers. If I = [0, 1], then /A is called a cube. The cube lw is metrizable and is called the Hilbert cube. Let X be any set and a family of functions f on X with 0 < f < 1 and such that for any two distinct points x and y in X there is an f c iI with f(x) f(y). Then I can be identified with a subset of Ix if we let each f correspond to the element whose x-th coordinate is f(x). The mapping of g into /which takes f into f (x) is simply the projection 7rx restricted to the image of g. The topology that g inherits as a subspace of /x is called the topology of pointwise convergence. On the other hand, X can be identified with a subset of P by letting each x correspond to the element whose f-th coordinate is f (x). The topology of X as a subspace of /5 is the weak topology generated by . If X is a topological space and each f in a is continuous, then the mapping of X onto its image in P is continuous, and, if has the property that for each closed subset F c X and each x 0 F there is an f c with f (x) --= 1 and f F---- 0 on F, then X is homeomorphic with its image in Jr

Problems 24,

space.

Prove that the direct product of Hausdorff spaces is a Hausdorff

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25. Prove that the collections taken for bases in defining product topologies satisfy the conditions of Proposition 3. Show that if (X, p) and ( Y, o. ) are two metric spaces then the product topology on X X Y is the same as the topology induced by the product metric. 26. Show that X A is the set of all functions mapping A into X with a base for its topology given by open sets of the form If: f(a i ) e 0 1 , f(a2) e 02, • • • , f(an) E On}, where {a 1 , . . , an } is some finite subset of A and {O 1 , , On} a finite collection of open subsets of X. Prove that a sequence (fn ) converges to f in XA if and only iffn (a) converges to f( a) for each a in A. 27. Show that, if X is metrizable and A is countable, then XA is metrizable. [Hint: X can always be metrized by a bounded metric p. Define a metric a on XA by o- (x, y) = 2 p(x a, ya).]

E

cxeA

28. Show that each projection re, is continuous and that the product topology on X A is the weakest topology such that each Ira is continuous. 29. Show that r is homeomorphic to the Cantor ternary set. 30. a. Show that the correspondence described in the text between a topological space X and its image in F1 is a homeomorphism if g has the property that, given a closed F and x g F, there is an f e g with f[F] = 0 and f(x) = 1. b. Show that, if X is a normal space satisfying the second axiom of countability, then we can find a countable family 5 of continuous functions with the property in (a). c. Prove the Urysohn Metrization Theorem.

5 Connectedness A topological space X is said to be connected if there do not exist two nonempty disjoint open sets 0 1 and 02 such that X = 0 1 u 0 2 . Such a pair of open sets is called a separation of X. Since each set is the complement of the other, they are closed sets as well as open sets. Any pair of disjoint nonempty closed sets whose union is X is a separation for X, since each of these sets must also be open. A space X is connected if and only if the only subsets of X which are both open and closed are the sets 0 and X. A subset E of X is said to be connected when it is a connected space in the topology it inherits from X: Thus E is connected if there do not exist open sets 0 1 and 02 in X, both meeting E such that Ec 0 1 u 02 and E n 0 1 n 0 2 = 0.

153

Sec. 5] Connectedness

8. Proposition: Let f be a continuous mapping of a connected space X onto a topological space Y. Then Y is connected. Proof: Let 0 1 and 0 2 be a separation of Y. Then f -1 [ 0 1 ] and f -1 [02 ] are disjoint open sets in X whose union is X. Since f is onto, neither f [0 1 ] nor f —1 [0 2] is empty, and so this pair is a separation of X. Thus if Y is not connected, Xis not connected, and the proposition follows by contraposition. 1

9. Proposition: Let f be a real-valued continuous function on a connected space X. Let x and y be two points in X and c a real number such that f(x) < c < f(y). Then there is az e X such that f(z) = c. Proof: If f does not assume the value c, then f—'[(— oo, c)] and f -1 [(c, co)] are disjoint open sets whose union is X. They are nonempty, since x belongs to the first and y to the second. Thus X is not connected. 1

10. Proposition: A subset E of R is connected if and only if it is either an interval or a single point. Problems 31. Let A be a connected subset of a topological space, and suppose Then B is connected. 32. a. Let E be a connected subset of R having more than one point. Prove that E is an interval. [If x and y are in E, x < y, then [x, y] C E. Let a = inf E, b = sup E. Then (a, b) CE c [a, M.] b. Prove that an interval in R is connected. [Let I = (a, b) and let 0 be a subset of I which is both open and closed in I. Show that sup {y: (x, y) C 0} = b, and use Problem 31 to take care of nonopen ACBCA.

intervals.] 33. A space X is said to be arcwise connected if given two points x and y in X there is a continuous map f of [0, 1] into X such that f(0) = x and f(I)

Y a. Show that an arcwise connected space is connected. b. In the plane R 2 consider the subspace X

{(x, y): x 0, — 1 < y < I} u {(x, y): y = sin 1/x,0 < x < 1 } .

Show that X is connected but not arcwise connected.

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c. Show that each connected open set G in Rn is arcwise connected. [Let x e G and let H be the set of points in G which can be connected to x by a polygonal arc. Then H is open and closed in G.] 34. Let X be a topological space and {E,2} a collection of connected subsets of X each of which contains a fixed point x. Then U Ea, is connected. 35. Show that the direct product of connected topological spaces is itself connected.

*6 Absolute gs's The purpose of this section is to characterize those topological spaces which are homeomorphic to complete metric spaces: They are the metrizable spaces whose dense homeomorphic images in every Hausdorff space are ga's. We state this property as a theorem. Some suggestions for its proof are given in Problems 37 and 38 and an application is given by Problem 39. This characterization is utilized in Chapter 15 to show that in a sense the complete metric spaces are absolutely measurable. 11. Theorem: Let X be a complete metric space and if, a homeomorphism of X onto a dense subset E of a Hausdorff space F. Then E is a g6. Conversely, if E is a 96 in a complete metric space, then E is homeomorphic to a complete metric space.

Problems

36. Show that if the Hausdorff space F of Theorem II is metrizable, then we do not need to require that E be dense. 37. Let E be a dense subset of a Hausdorff space F and g a homeomorphism of E onto a complete metric space X. Then E is a Ç. (For each positive integer n, let On be the subset consisting of ally c F such that some open set containing y has an image under g whose diameter is less than 1/n. Then On is an open set, and the homeomorphism g can be 00

extended to a continuous mapping h of G = fl O n into X, since X is n-.=1 complete. But g-1 o h must be the identity, whence E = G.)

Sec.

7]

Nets

155

n

38. Let ( Y, a) be a complete metric space, and let E = O„ be a 95. For each n construct a continuous real-valued function ça,, on Y with 0 < ça,, < 1 such that 0„ = {x: ço„(x) > 0}. For x and y in E let P(x, Y) = o(x, Y) ±

E 2-4 1,pn(x) —

X [1,,,n(x) — my) + kp.(x)'.(y)11 — i. Then p is a metric for E which is equivalent to a- and makes E a complete metric space. 39. Let E be a dense go in [0, 1], and suppose that E is also dense. Then E is homeomorphic to the product space co‘'). [Hint: Show that E = On, where 0, 1_ 1 c On, each component of 0„ containing infinitely many components of 0„+1, and each component of On having a diameter in the complete metric for E which tends to zero. Then the natural correspondence between elements of Low and decreasing sequence (4,) of components of On is a homeomorphism onto E.]

n

*7 Nets

By a directed system we mean a set A together with a relation < satisfying the following conditions: i. If a < fi and [3 -< i, then a -< 'Y. ii. If ce, j c A, there is ay eA with a -< Y and fi < -Y. One example of a directed system is the set N of positive integers with -< replaced by Another commonly used directed set is the set of all open sets containing a point x, with 0 1 < 0 2 defined to mean 01 D 02. A net is a mapping of a directed system into a topological space X. If the directed system is the integers, we have a sequence, and nets may be thought of as generalizations of sequences. We usually write xa for the value of the net at a and (.xa for the net itself. A point x e X is said to be the limit of a net (xa if for each open set 0 containing x there is an a o c A such that xa e 0 for all a > a o. A point x is called a cluster point of (xa ) if given 0 containing x and given a e A there is a fi > a such that x0 e 0. For sequences these notions coincide with our earlier notions of limit and cluster point.

12. Proposition: A point x is a point of closure of a set E if and only if it is the limit of a net (xOE ) from E.

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Proof: The "if" part follows directly from the definitions of limit and point of closure. Hence we assume x is a point of closure of E. We take as our directed system A the collection of open sets which contain x and set 0 1 -< 0 2 if 01 D 02. Since x is a point of closure of E, for each 0 c A there is a point xo in 0 n E. Then (x0 ) is a net from E, and it converges to x, since, given 0 containing x, we have x0 e 0 for all 0' > 0. I Problems

40. Prove that X is Hausdorff if and only if every net in X has at most one limit. [To prove the "if" part, let x and y be two points which cannot be separated and let the directed system be the collection of all pairs (A, B) of open sets with x 6 A, y 6 B. Choose x(A,B) to be in A n B and show that both x and y are limits of this net.] 41. Prove that a function f from X to Y is continuous at x if and only if for each net (xa ) which converges to x the net (f(x 0)) converges to f(x). 42. Let X be any set and f a real-valued function on X. Let A be the system consisting of all finite subsets of X, with F < G meaning F c G. For each F A, let yF = E f(x). Prove that the net (yF) has a limit if and xeF only iff(x) = 0 except for x in a countable subset { xn} and Elf xn)j < oo . (

In this case, lirn YF

=

E f(xn).

n

43. Let X X X„, Then a net (xs) in X converges to x if and only if

each coordinate of x(i converges to the corresponding coordinate of x.

9

Compact Spaces

I Basic Properties

Many of the important properties of the interval [0, 1] follow from the Heine-Borel theorem. We introduce a class of topological spaces in which the conclusion of the Heine-Borel theorem is valid and show that many properties of [0, 1] are also true for these spaces. These spaces are called compact spaces. To give an exact definition, we say that a collection cu of open sets in a topological space is an open covering for a set K if K is contained in the union of the sets in cu. A topological space X is said to be compact if every open covering cu. of X has a finite subcovering, that is, if there is a finite collection such that X = U Oz. A subset K of a topot— logical space is called compact if it is compact as a subspace of X. In view of the definition of the topology of a subspace, this is equivalent to saying that a subset K of Xis compact if every covering ctt of K by open sets of X has a finite subcovering. The Heine-Borel theorem states that every closed and bounded subset of real numbers is compact. If cu is an open covering of a space X, then the collection g• of complements of sets in 'it is a collection of closed sets whose intersection is empty, and conversely. Thus a space X is compact if and only if every collection of closed sets with an empty intersection has a finite subcollection whose intersection is empty. A collection of {0 1, 02,

ON} C 91

157

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[Ch. 9

sets in X is said to have the finite intersection property if any finite subcollection of g has a nonempty intersection. Hence we have the

following proposition: 1. Proposition: A topological space X is compact if and only if every collection g of closed sets with the finite intersection property has a nonempty intersection. The notion of compactness is intimately connected with that of being closed, as the following proposition shows. Thus compactness may be viewed as an absolute type of closedness. 2. Proposition: A closed subset of a compact space is compact. A compact subset of a Hausdorff space is closed. Proof: Let X be compact, F a closed subset of X, and cu. an open covering for F. Then cu, u {P} is an open covering for X and so must have a finite subcovering {P, 0 f , . , ON }. Then the sets 0 1 , 02, . • • , ON cover F, and so cu has a finite subcovering. Suppose now that X is a Hausdorff space and K a compact subset of X. We shall show that k is open. Let y c K. Since X is Hausdorff, for each x c K there are disjoint open sets Ox and Nx such that x e Ox and y c N. The sets (0x : x c K) form an open covering of K, and so there is a finite subcollection fO xi , Ox2 , , Ox) which covers K. Let N=

n

Nx.

Then N is an open set containing y and not meeting any of the sets O. Since KcU O, N does not meet K and so is contained in K. Thus k is open and K closed. I 3. Corollary: Every compact set of real numbers is closed and bounded.

Proof: Since R is Hausdorff, a compact subset K of R must be closed. Moreover, the intervals In = (—n, n) form an open covering of K and so a finite number of them must cover K. Hence K must be bounded. I

4. Proposition: The continuous image of a compact set is compact.

Sec. 2]

Countable Compactness and the Bolzano-Weierstrass Property

159

Proof: Let f be a continuous function which maps the compact set K onto a topological space Y. If 'It is an open covering for Y, then the collection of sets f -1 [0] for all 0 e cu is an open covering of K. By the compactness of K, there are a finite number 0 1 , , On of sets of Cu.such that the sets f [Os] cover K. Since f is onto, the sets 0 1 , , 0„ cover Y. I 5. Proposition: A one-to-one continuous mapping of a compact space onto a Hausdorff space is a homeomorphism. Proof: Let X be compact, Y Hausdorff, and f a one-to-one continuous mapping onto Y. In order to show that f is a homeomorphism it is only necessary to show that it carries open sets into open sets or equivalently closed sets into closed sets. But if F is a closed subset of X it is compact by Proposition 2, and so f[F] is compact by Proposition 4 and so must be closed by Proposition 2. I

Problems 1. a. Prove that a compact Hausdorff space is regular. b. Prove that a compact Hausdorff space is normal. 2. Let f be a continuous mapping of the compact space X onto the Hausdorff space Y. Then any mapping g of Y into Z for which g 0 f is continuous must itself be continuous. 3. a. Prove that if (X, 3) is a compact space then (X, 5 1 ) is compact for all 31 weaker than 3. b. Show that if (X, 5) is a Hausdorff space then (X, 3 2) is a Hausdorff space for all 3 2 stronger than 3. c. Show that if (X, 5) is a compact Hausdorff space then any weaker topology is not Hausdorff and any stronger topology is not compact. 2 Countable Compactness and the Bolzano-Weierstrass Property

A weaker notion than compactness is countable compactness: A space X is said to be countably compact if every countable open covering has a finite subcovering. Since every space satisfying the second axiom of countability has the property that every open

160

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[Ch. 9

covering of it has a countable subcovering, it follows that countable compactness is equivalent to compactness in the presence of the second axiom of countability. The proof of Proposition 4 applies in the countably compact case to give the following proposition: 6. Proposition: The continuous image of a countably compact space is countably compact.

A topological space X is said to have the Bolzano-Weierstrass property if every sequence (x„) in X has at least one cluster point, that is, if there is an x e X such that for each open set 0 containing x and for each N there is an n > N with xn c O. 7. Proposition: A topological space has the Bolzano-Weierstrass property if and only if it is countably compact.

Proof: We first observe that X is countably compact if and only

if every countable family of closed sets with the finite intersection property has a nonempty intersection. Suppose now that X has the Bolzano-Weierstrass property and that g• = {Fij is a countable family of closed sets with the finite intersection property. Since the intersection H„ =

n 1,, is empty for no n, we may choose for each

k=1

n an element x n c H. By the Bolzano-Weierstrass property, the sequence (x,a ) has a cluster point x. But x„ e F for all n > i, and so x must belong to F, since F, is closed. Thus x belongs to every F and

so to their intersection. Suppose, on the other hand, that X is countably compact and that (xi) is a sequence from X. Let B„ be the set fx„, x.±1, , 1. Then {} is a countable collection of closed sets with the finite intersection property, and so there is a point x which belongs to B. The point x is a cluster point of the sequence, since given N and any open set 0 containing x we have x c RN, and so there must be an x n e 0 with n > N.

n

A somewhat old-fashioned concept which resembles the BolzanoWeierstrass property is that of sequential compactness. A space X is said to be sequentially compact if every infinite sequence from X has a convergent subsequence. The relationship of sequential compactness to countable compactness is given by the following proposition.

Sec. 2]

Countable Compactness and the Bolzano-Weierstrass Property

161

Problem 6 gives an example of a space which is sequentially compact but not compact, and Problem 27 an example of a compact space which is not sequentially compact. Problem 7 shows that a space may be compact without being either separable or first countable. 8. Proposition: A sequentially compact space is countably compact. A countably compact space satisfying the first countability axiom is sequentially compact. Proof: Sequential compactness implies the Bolzano-Weierstrass

property, which is equivalent to countable compactness. The second part is an immediate consequence of Problem 8.8c. I 9. Proposition: Let f be a continuous real-valued function on a countably compact space X. Then f is bounded and assumes its maximum and minimum.

This proposition can be proved by using Proposition 6 and the fact that every countably compact subset of R is closed and bounded. However, we can give a direct proof which proves more. A real-valued function f on a topological space is called upper semicontinuous if for each real number a the set fx: f(x) < cej is open. If f is continuous, both f and —f are upper semicontinuous. This implies that Proposition 9 is a corollary of the following proposition: 10. Proposition: Let f be an upper semicontinuous real-valued function on a countably compact space X. Then f is bounded (from above) and assumes its maximum.

Proof: The sets O n = {x: f(x) < n} form a countable open covering for X and so there must be a finite subcovering {O,. . , O} . But this implies X c ON. Hence f(x) < N for all x, and f is bounded from above. Let = sup ( f(x): x e X]. Then the sets = {x: f(x)

—

form a countable collection of closed sets with the finite intersection property. Hence there is a y belonging to every F. Then f(y) = 0, and f assumes its maximum at y. I

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[Ch. 9

11. Proposition (Dini): Let (fn) be a sequence of upper semicontinuous real-valued functions on a countably compact space X, and suppose that for each x e X the sequence (f„(x)) decreases monotonically to zero. Then (fn) converges to zero uniformly. Proof: Choose e > 0, and let On = {x: fn (x) < e} . Since fn is upper semicontinuous, 0„ is open. Since fn (x) --> 0 for each x, we have X c U O n . By the countable compactness of X, there are a finite number of open sets {o, . — , 0 } whose union contains X. But this implies that ON = X, and hence fN (x) < E for all x. If n > N, we have 0 < fn (x) < fN (x) < E, and the sequence ( f,.) converges to 0 uniformly. I

Problems 4. a. A real-valued function f is called lower semicontinuous if —fis semicontinuous. Show that a real-valued function on a space X uper is continuous if and only if it is both upper and lower semicontinuous. b. Show that if f and g are upper semicontinuous, so is f + g. e. Let (fn) be a decreasing sequence of upper semicontinuous functions which converge pointwise to a real-valued function f Then f is upper semicontinuous. d. Let (fn) be a decreasing sequence of upper semicontinuous functions on a countably compact space, and suppose that lim fn(x) = f(x) where fis a lower semicontinuous real-valued function. Then f is continuous and (fn ) converges to f uniformly. e. Show that if a sequence (f„) of upper semicontinuous functions converges uniformly to a function f, then f is also upper semicontinuous. 5. Let X be a normal topological space. Then the following are equivalent:

i. X is countably compact. ii. Every continuous real-valued function on X is bounded. iii. Every bounded continuous real-valued function on X assumes its maximum. 6. Let X be the set of ordinals less than the first uncountable ordinal, and let 43 be the collection of sets of the form {x: x < a}, {x: a < x < b } , and {x: a < x} . a. Show that 03 is a base for a topology for X.

Sec. 3]

Compact Metric Spaces

163

b. Show that X is sequentially compact but not compact. [Hint: Use the well ordering of the ordinals.] c. Show that if f is a continuous real-valued function on X, then there is an x 0 such that f( x) = f(x) for all x > xo. [Hint: Show that the set of x for which f(x) < Inn f is countable.] 7. Let Y be the set of ordinals less than or equal to the first uncountable ordinal 52, and let 63 be the collection of sets of the form {x: x < a}, {x: a < x < b } , and {x: a < . a. Show that 63 is a base for a topology for X. b. Show that X is compact but neither separable nor first countable. Compact Metric Spaces

3

In this section we investigate special properties of compactness that are true in metric spaces. We shall show that in a metric space the notions of compactness, countable compactness, and sequential compactness coincide. 12. Lemma: Let X be a countably compact metric space. Then given E > 0 there are a finite number of points x l , . . . , xAr of X such that for each x e X there is an x k with p(x, x k) < E.

Proof: Suppose that no such finite collection of points exists. Then we can choose an infinite sequence (x n ) of points in X so that p(x„, xn,) > e for m n. Since each spheroid of radius e/3 can contain at most one term of this sequence, the Bolzano-Weierstrass property does not hold, and X is not countably compact. I

13. Proposition: A countably compact metric space is separable. Proof: For each positive integer n, let Fn, be a finite set of points such that for each x e X there is a y c Fr, with p(x, y) < 1/n. Then U F

a countable dense subset of X. I

n= 1

14. Corollary: For a metric space the notions of compactness, countable compactness, and sequential compactness are equivalent. Proof: Since a metric space satisfies the first axiom of countability, the notions of countable compactness and sequential compactness

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[Ch. 9

coincide by Proposition 8. Compactness trivially implies countable compactness. If X is countably compact, it must satisfy the second axiom of countability by Proposition 13, and so be compact. I A metric space X is said to be totally bounded if for each e > 0 there is a finite collection of spheroids of radius e which covers X. Every compact space is totally bounded, and every subset of a totally bounded space is totally bounded. The following proposition characterizes compactness in terms of completeness and total boundedness : 15. Proposition: A metric space X is compact if and only if it is both complete and totally bounded.

Proof: If X is compact it is trivially totally bounded. If (x„) is a Cauchy sequence in X, then (x„) must have a cluster point. But a

Cauchy sequence which has a cluster point converges to the cluster point. Thus X is complete. Suppose that X is complete and totally bounded. To show that X is compact, it suffices to show that each infinite sequence (x,i) has a convergent subsequence. Since X is totally bounded, we may cover X by a finite number of spheres of radius 1. Among these spheres there must be a sphere S i which contains infinitely many terms of the sequence (xn). Covering X by a finite number of spheres of radius 1, we can find among them a sphere S2 such that S1 n S2 contains infinitely many terms of the sequence (xn). Continuing, we obtain a sequence (Sk ) of spheres, Sk having radius 1/k, such that S i n • • • n Sk contains infinitely many terms of the sequence. Since there are infinitely many terms of the sequence in Si n • • • n Sk, we may choose nk so that nk > nk _ 1 and x n, c Si n • • • n Sk. Then (x„,) is a subsequence of (x„), and it must be a Cauchy sequence, since p(x„,, x„,) < 2/N for k, I > N. Since X is complete, this subsequence converges. I 16. Proposition: Let f be a continuous mapping of a compact metric space X into a metric space Y. Then f is uniformly continuous.

Proof: Given e > 0 and x e X, there is a Sz > 0 such that P(x, < az implies o- (f(x), f()) < e/2. Let Oz be the spheroid (y: p(x, y) < 115z1. Then {Ox : x e X} is an open covering of X and so

Sec. 4]

165

Products of Compact Spaces

has a finite subcovering {0 x1 , . Let ô = min {(5 . 1 , • • , xj • , Then ô > O. Given two points y and z in X such that p(y, < ô, the point y must belong to some O xo and hence p(Y, xi) < la x,. Consequently, p(z, xi) p(z, y) + p(y, xi) < }ax, < ô . Thus we have 0- ( f (y), f( xi)) < e/2 and 0- ( f(z), f (x i)) < e/ 2. This implies that (f(z), f( y)) < e, showing that f is uniformly continuous on X. I -

,

Problems

8. Let X be a metric space, K a compact subset, and F a closed subset. Then FnK = Ø if and only if p(F, K) > 0, i.e., if there is a .5> 0 such that p(x, y) > for all x c F and all y e K. [Consider the function p(x, F) = inf p(x, y).] yeF

9. Let X be a compact metric space and clt an open covering of X. Then there is a 5 > 0 such that every spheroid of radius is contained in some element of cll.

4 Products of Compact Spaces

In this section we prove the theorem of Tychonoff that a product of compact spaces is compact. It is probably the most important theorem in general topology. Most applications in analysis need only the special case of a product of (closed) intervals, but this special case does not seem to be easier to prove than the general case. We begin with two lemmas concerning the finite intersection property. 17. Lemma: Let a be a collection of subsets of a set X, and suppose that a has the finite intersection property. Then there is a collection 63 D a such that 63 has the finite intersection property and is maximal with respect to this property; that is, no collection properly containing 63 has the finite intersection property.

Proof: Consider the family of all collections containing a and having the finite intersection property. This family is partially ordered by inclusion. By the Hausdorff maximal principle there is a maximal linearly ordered subfamily a. Let 61 be the union of the

166

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[Ch. 9

collections in a. If B 1 , . , B,, are in (33, then each B, belongs to some e, e a. Since a is linearly ordered by inclusion, one of the collections ek contains the others, and so all B, belong to ek. Since B, Ø. Thus 03 has the ek has the finite intersection property, finite intersection property. If 63' D ed and 63' has the finite intersection property, then 63' contains every e in a and so must belong to a by the maximality of a. Thus 63 is a union of collections one of which is 63', and so 63' C 63. This shows that 63 is maximal with respect to the finite intersection property. I

n

18. Lemma: Let 63. be a collection of subsets of X which is maximal with respect to the finite intersection property. Then each intersection of a finite number of sets in 03 is again in 63, and each set which meets each set in 63 is itself in (B. Proof: Let 631 be the collection of all sets which are finite intersections of sets in 63. Then da' is a collection having the finite intersection property and containing 68. Thus 63' = o3 by the maximality of 63. Suppose that a set C meets each element of 63. Since da contains each finite intersection of sets in 63, it follows that da u {C} has the finite intersection property. By maximality 63 U {C} = ea, and SO C 8 63. I 19. Theorem (Tychonoff): Let (Xa) be an indexed family of compact topological spaces. Then the product space X X„ is compact in the product topology. Proof: Let ra be the mapping of X to X,, which assigns to each its a-th coordinate. Then the sets obtained by taking finite intersections of sets of the form 7,7 1 [0 with Oa open in X,, form a base 97, for the topology of X. Let a be any collection of closed subsets of X with the finite intersection property, and let 63 be a collection of (not necessarily closed) sets which contains a and is maximal with respect to the finite intersection property. Let 63„ be the collection of subsets of X,, of Then (Bo, has the finite intersection the form a (B) with B E property, and by the compactness of X,, it is possible to choose a point xa belonging to n [ir a (B)J, that is, a point xa which is a point of 63 x e X

Sec. 4] Products of Compact Spaces

167

closure of each set 7r a [13]. Let x be that point in X whose a-th coordinate is x,. Consider a set S which is of the form 7r,7 1 [0a] for some a and some open set 0a in Xa with xa e Oa. Since xa is a point of closure of 7i-a[B] for each B in 63, the set S must intersect each B in 63. By Lemma 18 we must have S e 63. Each set containing x in the base 97, for the topology of X is a finite intersection of sets of this form and so must be in (33 by Lemma 18. Let F be a closed set in 63. Then F meets each N e 07, with x e N. Consequently, x is a point of closure of F and so is in F. Hence x belongs to each set in a, and a has nonempty intersection. I

Problems 10. Each closed and bounded set in Rn is compact. 11. Prove without using the axiom of choice that if X is compact and I is a closed interval, then X X / is compact. [Hint: Let 91 be an open covering of X X I, and consider the smallest value of t e / such that for each t' < t the set X X [0, /1 can be covered by a finite number of sets in Use the compactness of X to show that X X [0, t] can also be covered by a finite number of sets in cit and that if t < 1, then for some t" > t, X X [0, t"] can be covered by a finite number of sets in cud 12. Prove that the product of a countable number of sequentially compact spaces is sequentially compact. [If (xn) is a sequence in the product, choose a subsequence (x71) whose first coordinate converges, choose a subsequence (4) of this whose second coordinate converges, etc. Then the "diagonal" sequence (4) converges in the product space.] 13. A product Pi of unit intervals is called a (generalized) cube. Prove that every compact Hausdorff space Xis homeomorphic to a closed subset of some cube. [Let be the family of continuous real-valued functions on X with values in [0, 1 . Let Q = X If. Then the mapping g of X into Q ]

fE5

which takes x into the point whosef-th coordinate is f(x) is one-to-one into Q and continuous.] 14. Let Q be a cube, and letf be a continuous real-valued function on Q. Then, given c > 0, there is a continuous real-valued function g on Q such that If — gl < E and g is a function of only a finite number of coordinates. [Hint: Cover the range off by a finite number of intervals of length E and look at the inverse images of these intervals.]

Compact Spaces

168

5

[Ch. 9

Locally Compact Spaces

A topological space X is called locally compact if for each x e X there is an open set 0 containing x such that 0 is compact. Thus X is locally compact if the collection of open sets with compact closures form a base for the topology of X. Every compact space is locally compact, while the Euclidean spaces Rn are examples of spaces which are locally compact but not compact. If X is a locally compact Hausdorff space, we can form a new space X* by adding to X a single point co not in X and taking a set in X* to be open if it is either an open subset of X or the complement of a compact subset in X. Then X* is a compact Hausdorff space, and the identity mapping of X into X* is a homeomorphism of X and X* — {w} . The space X* is called the Alexandroff one-point compactification of X, and w is often referred to as the point at infinity in X*. Some useful properties of compact subsets in a locally compact Hausdorff space are given by the following propositions, whose proofs are left to the reader (see Problems 16, 17, and 18). 20. Proposition: Let K be a compact subset of a locally compact Hausdorff space X. Then there is an open set 0 containing K with0 compact. Given such a set 0, there is a continuous nonnegative function f on X which vanishes outside 0 and is identically 1 on K. If K is also a gs, we may take f < 1 in R. 21. Proposition: Let K be a compact subset of a locally compact Hausdorff space and {0,1 an open covering for K. Then there are a finite number of nonnegative continuous functions f 1 , . . ,f. on X, that each f, vanishing outside a compact set and outside some O is identically 1 on K. f2 + • • • + fn

Problems 15.

a. Prove that the subsets of X* which are either open subsets of X

or the complements of compact subsets of X form a topology for X*, that is, that the intersection of two such sets is such a set and the union of any collection of such sets is such a set. b. Show that the identity mapping from X to the subspace X* — {0.,} is a homeomorphism. c. Show that X* is compact and Hausdorff.

Sec.

169

5] Locally Compact Spaces

16. Let X be a locally compact space and K a compact subset of it. Show that there is an open set 0 D K such that 5 is compact. [Hint: For each point x e K there is an Or containing x with 5x compact. Let 0 be the union of a finite number of these Or which cover K.] 17. Let X be a locally compact Hausdorff space and K a compact set. Then there is a continuous real-valued function on X which is identically 1 on K and for which the set 0 = {x: f(x) 0} has compact closure. [Hint: Use Problem 16 and the Urysohn Lemma.] Prove Proposition 20. 18. Prove Proposition 21. [For each x in K there is a nonnegative continuous function g which is positive at x and vanishes identically outside some Oa . Choose a finite number g l , . . . , gn of such functions so that g = g1 + • • - ± g„,is positive on K. Let h be a continuous function on X which is equal to 1/g on K, and set f, = hg,.] 19. Show that the Alexandroff one-point compactification of R is homeomorphic to the boundary of a sphere in R+ 1 • 20. Show that the one-point compactification of the space X in Problem 6 is the space Y in Problem 7. 21. a. Let 0 be an open subset of a compact Hausdorff space. Then 0 is locally compact. b. Let 0 be an open set in a compact Hausdorff space X. Then the mapping of X to the one-point compactification of 0 which is the identity on 0 and takes each point in X — 0 into co is continuous. 22. A continuous mappingf from a topological space X to a topological space Y is called proper if the inverse image of every compact set is compact. Let X and Y be locally compact Hausdorff spaces, andf a continuous mapping of X into Y. Let X* and Y* be the one-point compactifications of X and Y, and f * the mapping of X* into Y* whose restriction to X is f and which takes the point at infinity in X* into the point at infinity in Y. Then f is proper if and only if f* is continuous. 23. a. Let X be a locally compact space. A subset F of Xis closed if and only if F n K is closed for each closed compact set K. b. The above conclusion is true if X is a Hausdorff space satisfying the first axiom of countability instead of being locally compact. 24. The Moore Manifold. Let X be the set whose elements are the points of the open right half-plane (i.e., {(x, y): x > 0)) and the lines in the plane with nonnegative slope. Write m(1) and b(1) for the slope and y-intercept of the line 1. We define a base for a topology for X by taking the open disks in the half-plane and the sets V, = {l: Im(1) — mol < e, b(l) = b0 )

u {(x,Y): KY — boVx ±

mol < e, x <

f},

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[Ch. 9

a. Show that X is a connected Hausdorff space and that each point of X is contained in an open set which is homeomorphic to an open subset of R 2. (Such a space is called a two-dimensional manifold or surface.) b. Show that X (or any manifold) is locally compact, completely regular, and satisfies the first axiom of countability. c. Show that X has a countable dense subset but does not satisfy the second axiom of countability. d. Show that X is not normal.

*6

The Stone-tech Compactification

Let X be a completely regular topological space and 5 the family of continuous real-valued functions f on X with fj < 1. If we let / = [ I, 1], then it follows from Problem 8.30 that X is homeomorphic with a set E c I. Let F = E. Since r is a compact Hausdorff space, the set F is a compact Hausdorff space, and, identifying X with E, we have X a dense open subset of F. The space F is called the Stone-tech compactification of X and is denoted by 0(X). We summarize some of its properties in the following proposition. 22. Proposition: Let X be a completely regular topological space. Then there is a unique compact Hausdorff space 0(X) with the following properties:

i. The space X is a dense open subset of 0(X). ii. Each bounded continuous real-valued function on X extends to a continuous function on 0(X). iii. If X is a dense open subset of a compact Hausdorff space Y, then there is a unique continuous mapping io of 0(X) onto Y such that (c(x) = x for all x c X.

Problems

Prove Proposition 22: a. If f is a bounded continuous real-valued function on X with I fi < 1, then f is the restriction to X of 71-f, and rf is continuous on 0(X). b. Use the fact that Y is a subset of Is, where 9 is the space of continuous g on Y with igi < 1, to show 25.

Sec. 7]

The Stone-Weierstrass Theorem

171

c. Show that 13(X) is unique in the sense that if Z is another space with the same properties, then there is a homeomorphism so of Z with 0(X) such that yo(x) = x for all x c X. 26. Let X and Y be the spaces in Problems 6 and 7. Show that 0(X) = Y. 27. Let N be the set of natural numbers. Discuss 0(N). Show that a sequence from N converges in 13(N) if and only if it converges in N. Hence f3(N) is compact but not sequentially compact.

7 The Stone Weierstrass Theorem -

Let X be a compact Hausdorff space. We denote by C(X) the set of all continuous real-valued functions on X. Since X is normal, it follows from Urysohn's Lemma that there are enough functions in C(X) to separate points; that is, given two distinct points x and y in X, we can find an f in C(X) such that f (x) f (y). The set C(X) is a linear space, since any constant multiple of a continuous real-valued function is continuous and the sum of two continuous functions is continuous. The space C(X) becomes a normed linear space if we define H fH = max 1 f(x)i, and a metric space if we set p(f, g) = H f — gH. As a metric space C(X) is complete. The space C(X) has also a ring structure: The product fg of two functions f and g in C(X) is again in C(X). A linear space A of functions in C(X) is called an algebra if the product of any two elements in A is again in A. Thus A is an algebra if for any two functions f and g in A and any real numbers a and b we have af ± bg in A and fg in A. A family A of functions on X is said to separate points if given distinct points x and y of X there is an f in A such that f(x) f(y). In the present section we study the closed subalgebras of C(X) and prove that if A is a subalgebra of C(X) which separates points, contains the constant functions, and is closed, then A = C(X). The space C(X) also has a lattice structure: If f and g are in C(X), so is the function f A g defined by ( f A g)(x) = min [ f(x), g(x)] and the function f V g defined by (f v g)(x) = max [ f(x), g(x)J. A subset L of C(X) is called a lattice if for every pair of functions f and g in L we also have f v g and f A g in L. It is convenient to investigate subalgebras of C(X) by first investigating lattices of functions. The following proposition can be thought of as a generalization of the Dini theorem:

172

Compact Spaces

[Ch. 9

23. Proposition: Let L be a lattice of continuous real-valued functions on a compact space X, and suppose that the function h defined by h(x) = inf f(x) feL

is continuous. Then given E > 0, there is a g in L such that 0 g(x) — h(x) < E for all x in X. Proof: For each x in X there is a function fx in L such that fz (x) < h(x) €73. Since fx and h are continuous, there is an open set 0 z containing x such that IM.Y) — f(x)1 <

and

ih(Y) — h(x)1 <

for all y c O. HenceL(Y) — h(Y) < E for all y in O. Now the sets 0, cover X, and by compactness there are a finite number of them, say {0,,, , 0,), which cover X. Let g = f 1 A fx, A • • ' A L.. Then g e L, and given y in X we may choose i so that y c O whence ,

g(y) — h(y) < f(y) — h(y) <

E.

I

24. Proposition: Let X be a compact space and L a lattice of continuous real-valued functions on X with the following properties: i. L separates points; that is, if x y, there is an f e L with f(x) f(Y). ii. If f e L, and c is any real number, then cf and c f also belong to L.

Then given any continuous real-valued function h on X and any E > 0, there is a function g e L such that for all x e X 0 < g(x) — h(x) < e.

Before proving the proposition, we first establish two lemmas. 25. Lemma: Let L be a family of real-valued functions on a compact space X which satisfies properties (i) and (ii) of Proposition 24. Then given any two real numbers a and b and any pair x and y of distinct points of X, there is an f e L such that f(x) = a andf(y) = b.

Sec. 7]

The Stone-Weierstrass Theorem

173

Proof: Let g be a function in L such that g(x) g(y). Let a—b bg(x) — ag(y) f = g(x) — g(y) g + g(x) — g(y)

Then f e L, by property (ii), and f (x) = a, f (y) = b. I 26. Lemma: Let L be as in Proposition 24, a and b real numbers with a < b, Fa closed subset of X, and p a point not in F. Then there is a function f in L such that f > a, f(p) = a, and f(x) > b for all x e F. Proof: By Lemma 25 we can choose, for each x e F, a function fx such that fx(p) = a and f(x) = b ± 1. Let O,, — {y: fx (y) > b} . Then the sets 10 x1 cover F, and since F is compact, there are a finite number {O . . . , Ox j which cover F. Let f = fx, V - • • V f. n. Thenf e L, f(p) = a, and f > b on F. If we replace f by f Va, then we also have f> a on X.1 Proof of Proposition 24: Since L is nonempty, it follows from (ii) that the constant functions belong to L. Given g e C(X), let L' = {f:feL and f> g}. Proposition 24 will follow from Proposition 23 if we can show that for each p e X we have g(p) = inff(p), f e L'. Choose a positive real number 77 . Since g is continuous, the set F = Ix: g(x) > g(p) + 77)

is closed. Since X is compact, g is bounded on X, say by M. By Lemma 26 we can find a function f e L such that f > g(p) ± n, f(p) = g(p) + n, and f (x) > M on F. Since g < g(p) + n on P, we have g 0, there is a polynomial P in one variable such that for ails e [ —1, 1] we havelP(s) — Is!! < e. .3

Proof: Let

E can

be the binomial series for (1 — t) 1 / 2 . This

series converges uniformly for t in the interval [0, 1]. Hence, given E > 0, we can choose N so that for all t e [0, 1] we have

174

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[Ch. 9

N

1( 1 — 0 112 — QN(t)I < E, where QN = QN( 1 — S 2 ).

S c [— I, I]. I

E n=0

can. Let P(s)

=

Then P is a polynomial in s, and I151 — P(s)I < E for

28. Theorem (Stone-Weierstrass): Let X be a compact space and A an algebra of continuous real-valued functions on X which separates the points of X and which contains the constant functions. Then given any continuous real-valued function f on X and any E > 0 there is a function g in A such that for all x in X we have Ig(x) — f(x)1 < E. In other words, A is a dense subset of C(X). Proof: Let :4 denote the closure of A considered as a subset of C(X). Thus A consists of those functions on X which are uniform limits of sequences of functions from A. It is easy to verify that A is itself an algebra of continuous real-valued functions on X. The

theorem is equivalent to the statement that A = c(x). This will follow from Proposition 24 if we can show that Al is a lattice. Let f e A, and 11/.11 _- 1. Then given E > 0, I fi — P ( f)11 < E where P is the polynomial given in Lemma 27. Since A is an algebra containing the constants, P( f) E AI, and since :4 is a closed subset of C(X), we have ifl e A. If now f is any function in A, then pi fll has norm 1, and so If1/11f11 and hence also 1 f I belong to A. Thus A contains the absolute value of each function which is in A. But

f y g = (f+ g) + if- gl and

f

A

g= (f+ g) - If - gl.

Thus Al is a lattice and must be C(X) by Proposition 24. I 29. Corollary: Every continuous function on a closed bounded set X in Rn can be uniformly approximated on X by a polynomial (in the coordinates). Proof: The set of all polynomials in the coordinate functions forms an algebra containing the constants. It separates points, since given two distinct points in Rn, one of the coordinate functions takes different values on these points. Hence Theorem 28 applies. I

Sec. 7]

175

The Stone-Weierstrass Theorem

Problems 28. Let f be a continuous periodic real-valued function on R with period 27; that is, f(x -I- 27) = f(x). Show that, given e > 0, there is a N

finite Fourier series

(p,

given by 99(x) ---- ao ±

E

(an, cos nx + br, sin nx),

such that I99(x) — f(x) I < E for all x. [Hint: Note that periodic functions are really functions on the circumference of the unit circle, and that cos mx cos nx = 1.[cos (m ± n)x — cos (m — n)x], etc.] 29. Let A be an algebra of continuous real-valued functions on a compact space X, and assume that A separates the points of X. Then either A = C(X) or there is a point p e X and A = ff: f e C(X), f(p) = 01. 30. Let a be a family of continuous real-valued functions on a compact Hausdorff space X, and suppose that 5 separates the points of X. Then every continuous real-valued function on Xcan be uniformly approximated by a polynomial in a finite number of functions of 5. 31. a. Let X be a topological space and A a set of real-valued continuous functions on X. Define x y if f(x) = f(y) for all f e A. Show that ----- is an equivalence relation. b. Let )7' be the set of equivalence classes of = and 99 the natural map of X into X. Show that for each f e A there is a unique real-valued function J on fi' such that f = Jo (p. c. Let X' have the weak topology generated by these f. Then 99 is

continuous.

d. If X is compact, then so is jt, and the functions f in (b) are continuous.

e. Let X be a compact space and A a closed subalgebra of C(X) containing the constant functions. Then there is a compact Hausdorff space )7 and a mapping 99 of X onto X' such that A is the set of all functions f of the form Jo 99 with J e C(X ). 32. Let X and Y be compact spaces. Then for each continuous realvalued function f on X X Y and each E > 0, we can find continuous realvalued functions g 1 , . . . , gr, on X and h i , . . . , hn on Y such that for each (x, y) e X X Y we have

I f(x, y) — Ê gr(x)MY)I <

E.

i=1

The Weierstrass theorem, which states that a continuous function on a cube in R° can be uniformly approximated by a polynomial, can be proved directly by giving an integral formula for the approximating 33.

176

Compact Spaces

[Ch. 9

polynomials. 1 Show that this special case implies the general StoneWeierstrass Theorem by showing that the functions of norm 1 in the algebra a give a mapping of X into the infinite-dimensional cube X {If :fe a, if M = 1 } . Use the Tietze Extension Theorem and Problem 14 to show that each continuous function on the image of X can be approximated by a polynomial in (a finite number of) the coordinate functions.

The Stone-Weierstrass Theorem gives precise information about approximation by functions in an algebra of continuous functions. A natural question is the nature of functions that can be approximated by a ring of real-valued continuous functions, that is to say, when we no longer postulate the possibility of multiplying by arbitrary real numbers. The next three problems give some results in this direction. 34. Let I be the interval [— 1, 1] in R and f a continuous real-valued function on /such thatf(— 1),f(0), andf(1) are integers andf(1) f( — 1) mod 2. Then given c > 0, there is a polynomial P with integral coefficients such that If(x) — P(x)I < e for all x e I. Hints:

a. Let ço be the polynomial defined by ço(x) = x x(1 — 2x)(1 — x). Then ço is a monotone increasing function whose fixed points are 0, and 1. b. Choose e > O. Then some iterate son of ço is a polynomial with integral coefficients which is monotone increasing on [0, 1] and such that kon (x) —< e for x e [e, 1 — e]. c. Given a number a, 0 < a < 1, and any e > 0, then there is a polynomial 4/ with integral coefficients (and no constant term) such that 0 < 4/(x) < 1 in [0, 1] and 14/(x) — ai < E for all x in [e, 1 — e].

d. Let P be a polynomial with integral coefficients, and suppose that P(-1) = P(0) = P(1) =O. Let 0 be any real number. Then for each E > 0, 13P can be uniformly approximated to within E on [ 1, 1 1 by a polynomial having integral coefficients and no constant term. e. Reduce the statement of the problem to (d) and the StoneWeierstrass Theorem. —

35. a. Let X be a set and R a ring of real-valued functions on X. Let R be the ring of all real-valued functions which can be uniformly approximated by functions in R. If Je R, and sup If(x)1 < 1, then cf c R for each real number c. b. Let X be a compact Hausdorff space and R a ring of continuous real-valued functions on X such that 1 e R, and for each pair of distinct 1

See, for example, R. Courant and D. Hilbert, Springer, Berlin, 1931.

Mathemattsche Phystk,

Bd. 1, pp. 55-57,

Sec. 8]

177

The Ascoli Theorem

points x and y there is a function f in R such that f(x) f(y), and if(z)( < 1 for all z e X. Then every continuous real-valued function on X can be uniformly approximated by functions in R. 36. a. The statement of Problem 34 can be improved slightly. Show, for example, that we may take the interval I to be any closed interval contained in (— Y2). [The polynomial x 2 — 1 has absolute value at most 1 on I. Apply Problem 35b.] b. Show that we cannot take I in Problem 34 to be [ —2, 2]. [Hint: If P is a polynomial with integral coefficients, 1 ir

*8

f

2 p( x ) (4 x 2)112 dx

-2

is an integer.]

The Ascoli Theorem

It is often useful in analysis to know conditions under which a sequence of functions has a subsequence which is convergent in some sense. The following notion plays a central role in such questions: A family of functions from a topological space X to a metric space ( Y, o-) is called equicontinuous at the point x c X if given E > 0 there is an open set 0 containing x such that c[ f (x), f (y)] < E for all y in 0 and all f e a. The family is said to be equicontinuous on X if it is equicontinuous at each point x in X. We are going to show that if (fn ) is a sequence from an equicontinuous family of functions, and if at each point x of X there is a convergent subsequence of ( f„(x)), then (fn ) has a subsequence which converges uniformly on each compact subset of X. We begin with several lemmas. 30. Lemma: Let (fn ) be a sequence of mappings of a countable set D into a topological space Y such that for each x c D the closure of the set {f„(x): 0 < n < 3.0} is sequentially compact. Then there is a subsequence (fnk ) which converges for each x in D.

Proof: Let D = fxkl . By the sequential compactness of the closure of { fjx 1 ): 0 < n < Do} we can pick a subsequence (AO of (fn ) such that (fi (x)) converges. We can now pick a subsequence (f2n ) of ( fi,,) such that (f2 ,(x2)) converges. Continuing in this fashion we obtain a subsequence ( f,) convergent on x,. Consider the "diagonal" sequence (fnn ). We have (fnn )77, a subsequence of ( f,7,), and so ( fn„(x,)) converges. 1

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9

31. Lemma: Let (f„) be an equicontinuous sequence of mappings from a topological space X to a complete metric space Y. If the sequences (f„(x)) converge for each point x of a dense subset D of X, then ( .1,0 converges at each point of X, and the limit function is continuous. Proof: Given x in X and E > 0, we can find an open set 0 containing x such that oi f7,(x), f, i(y)] < e/3 for all y in O. Since D is dense, there must be a point y eDn 0, and since (f.(Y)) converges, it must be a Cauchy sequence, and we may choose N so large that olf,(y), f m (y)] < E/3 for all m, n N. Then offn(x),

Mx

offn(X), fix(Y)] offn(Y), fni(01

)]

offm(X), fm(Y)1

<E

for all m, n > N. Thus (fn (x)) is a Cauchy sequence, and converges by the completeness of Y. Let f(x) = lim f, j (x). To see that f is continuous at x, let e > 0 be given. By equicontinuity there is an open set 0 containing x such that cr[f,z (x), ffl (y)] < e for all n and all y in O. Hence for all y in 0 we have off(x), f(Y)J = lim o[/(x), fn(Y)]

and f is continuous at x. 1 32. Lemma: Let K be a compact topological space and (fn ) an equicontinuous sequence of functions to a metric space Y which converge at each point of K to a function f. Then (fa) converges to f uniformly on K Proof: Choose e > O. By equicontinuity each x in K is contained in an open set 0, such that cr[fn(x),fn(Al < e/3 for ally in 0, and all n. From this it follows that also olf(x), f(Al < e/3 for ally in O. By the compactness of K there is a finite collection {0„ 1 , . . , O z,} of these sets which covers K. Choose N so large that for all n > N we have cr[f„(xi), f(x)] < e/3 for each xi corresponding to this finite collection. Then for any y in K there is an i < k such that J' e Os .. Hence offn(Y), .f(y)]

offn(y), fn(xi)]

off,i(x i), f(x)]

off(y), f(x i)]

<E for n > N. Thus (fa) converges to f uniformly on K. I

Sec. 8]

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The Ascoli Theorem

These three lemmas taken together imply the following theorem: 33. Theorem (Ascoli): Let 5 be an equicontinuous family of functions from a separable space X to a metric space Y. Let (f,) be a sequence in g such that for each x in X the closure of the set {f„(x): 0 < n < 00} is compact. Then there is a subsequence (f,,,) which converges pointwise to a continuous function f, and the convergence is uniform on each compact subset of X. 34. Corollary: Let g be an equicontinuous family of real-valued functions on a separable space X. Then each sequence (fn ) in g which is bounded at each point (of a dense subset) has a subsequence (f„,) which converges pointwise to a continuous function, the convergence being uniform on each compact subset of X. Problems 37. The set of all functions from X to Y is precisely the product of X copies of Y and is often denoted by Yx • The product topology in Yx is called the topology of pointwise convergence. a. Show that a sequence (fa) of mappings of X into Y converges to fin the topology of pointwise convergence if and only if (f,(x)) converges to f(x) for each x in X. b. Let be an equicontinuous family of functions from a topological space X to a metric space Y. Then the closure of in the topology of pointwise convergence is also equicontinuous. 38. Let X be a space which is either locally compact, or Hausdorff and satisfies the first axiom of countability. Let (fn) be a sequence of continuous functions from X to a metric space Y which converge to a function f uniformly on each compact subset K of X. Then fis continuous. [Hint: Use Problem 23.] 39. Let X be a separable, locally compact metric space, and ( Y, a) any metric space. Show that: a. There is a countable collection {On} of open subsets of X such that Dn is compact and X = IvI O. b. The set of functions from X into Y becomes a metric space if we

define o-* (f, g) ----

E 2—ncr,V, g),

where sup , cr[f(x), g(x)] cr (f, g) = o„ i 4- a[f(x), g(x)] :

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9

c. The sequence (fn ) converges to f uniformly on compact subsets of X if and only if o- *(f, fn) -- O. 40. Let X and Y be metric spaces. A sequence (fn ) of mappings from X to Y is said to converge continuously to f at x if for each sequence (xn) such that x = lim xn we have f(x) = limfn(xa). We say (fa) converges continuously to f if it converges continuously at each x in X. a. Show that if each fn is continuous on X and if (A) converges continuously to fat each x e X, then f is continuous. b. Show that (fa ) converges continuously to f if and only if (A) converges to f uniformly on each compact subset of X. 41. A real-valued function f on [0, 1] is said to be Holder continuous of order a if there is a constant C such that If(x) — f(y)I < Clx — yl'. Define Ilfll,„ = max If(x)I + sup 1 f(x) — MI Ix — yla Show that for 0 < a < 1, the set of functions with 1 fil a < 1 is a compact subset of C[0, 1].

Il 0 Banach Spaces 1

Introduction

We are going to study a class of spaces which are endowed with both a topological and an algebraic structure. A set X of elements is called a vector space (or linear space, or linear vector space) over the reals if we have a function + on X X X to X and a function • on R x X to X which satisfy the following conditions: i. x + y — y + X. ii. (x + y) + z = x + (y + z). iii. There is a vector 0 in X such that x + 0 = x for all x in X. iv. X (x + y) = Xx + Xy; X c R, x, y e X. v. (X + ,u)x = Xx + px; X, p c R, x e X. vi. X(,ux) = (Xp)x; X, p e R, x c X. vii. 0 • x = 0, 1 • x = x.

We call + addition and • multiplication by scalars. It should be noted that the element 0 defined in (iii) is unique, for if Ce also has this property, then 0 = 9 ± 0' = 0' + 0 = 0'. The element ( —1)x is called the negative of x and written —x. We have x + (— x) — 1 - x ± (-1)x = (1 — 1)x = 0 • x = O. A nonnegative real-valued function is called a norm if i. 114 = ID <=> x =-- O. ii. Ilx + Yll 114 + Ilyll. iii. ilaxII = la! 11x11. 181

11 11 defined on a vector space

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Banach Spaces

[Ch. 10

A normed vector space becomes a metric space if we define a metric p by p(x, y) = Ilx — yll. When we speak about metric properties in a normed space we are referring to this metric. If a normed vector space is complete in this metric, it is called a Banach space. Examples of Banach spaces were given in Chapter 6. Another example is C(X), the space of all continuous real-valued functions on a compact space X. We restate here Proposition 6.4, and note that the proof given in Chapter 6 is valid in any normed vector space. 1. Proposition: A normed vector space is complete if and only if every absolutely summable sequence is summable.

A nonempty subset S of a vector space X is a subspace or linear manifold if X i x i + X 2x2 belongs to S whenever x 1 and x2 do. If S is also closed as a subset of X, then it is called a closed linear manifold. The intersection of any family of linear manifolds is a linear manifold. Hence, given a set A in X, there is always a smallest linear manifold containing A. We often denote this manifold by {A}. If A is any set in X, we use A + x to denote the set of all elements z of the form z — x + y, y e A. The set A + x is called the translate of A by x. The set XA is the set of all elements of the form Xx with x e A, and A + B is the set of all elements of the form x + y with x in A and y in B.

Problems 1. Show that if x,,, —4 x, then 11xn11 --, 11x11. 2. Two norms 11 11 1 and 11 112 are called equivalent if there is a positive constant K such that K-1 11xli 1 < 11x 112 < K 114 1 . If the norms are equivalent, then the metrics derived from them are uniformly equivalent. Show that the metrics introduced in Problem 7.10b for IV are derived from norms for Rn, and these norms are all equivalent. 3. Show that ± is a continuous function from X X X into X and that • is a continuous function from R X X into X. 4. Show that a nonempty set M is a linear manifold if and only if M ± M --= M and XM ----- M for each X. 5. a. Prove that the intersection of a family of linear manifolds is a manifold.

Sec.

1]

183

Introduction

b. Prove that there is a smallest linear manifold {A} containing a given set A. c. Show that {A} consists of all finite linear combinations of the form X ix • • • -I- Xnx,, with x, e A. 6. a. If M and N are linear manifolds, so is M N, and M N {M U N}.

b. If M is a linear manifold, so is M. 7. Show that the set P of all polynomials on [0, 1] is a linear manifold

in C[0, 1]. Is it closed? Give an example of a closed linear manifold in C[0, 1]. 8. A linear manifold M is said to be finite-dimensional if there are a finite number of elements x1 ,. . . , x,„ such that M ={ x 1 . • • , x,i}. Prove that every finite-dimensional linear manifold in a normed vector space X must be closed. 9. Let S be the spheroid of radius 1 centered at O; that is, S = Ix: lx < 1 1. Prove that S is open and that

= {x: 11x11

1) .

We call S the open unit sphere (or ball) and 3 the unit sphere (or ball). 10. A nonnegative real-valued function 11 11 defined on a vector space X is called a pseudonorm if Ilx -1- yll 114 11y11 and 11ax11 = l ai 11x11. Show that the relation x = y defined by 11x — yll = 0 is an equivalence relation compatible with addition and multiplication by scalars, and that, if x y, then 11x11 = 11Y1I. Let X' be the set of equivalence classes of X under Then X' becomes a normed vector space if we define ax' 01 as the (unique) equivalence class which contains ax -1- 13y for x e x' and y e y' and define 11x'11 = 11x11 for x e x'. The mapping go of X onto X' which takes each element of X into the equivalence class to which it belongs is a homomorphism (called the natural homomorphism) of X onto X'. What is the kernel of go? Illustrate this procedure with the LP spaces on [0, 1]. 11. Let X be a normed linear space (with norm 11 11) and M a linear manifold in X. Show that lxlj i inf 11x — m11 defines a pseudonorm meM

on X. Let X' be the normed linear space derived from X and the pseudonorm 11 11 using the process described in Problem 10. The natural map go of X onto X' has kernel M. Prove that go takes open sets into open sets. The space X' is usually denoted by X/M and called the quotient space of X modulo M. 12. Show that, if X is complete and M is a closed linear manifold of X, then X/M is also complete. [Hint: Use Proposition 1.]

184

2

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[Ch. 10

Linear Operators

A mapping A of a vector space X into a vector space Y is called a linear mapping, a linear operator, or a linear transformation if A(a1X1

+ a2X2) =-- a1AX1 + a2AX2

for all xl , x2 in X and all real a l , a2. If X and Y are normed vector spaces, we call a linear operator A bounded if there is a constant M such that for all x we have liAxli < M lix11. We call the least such M the norm of A and denote it by IA Thus 3 .21jj = sup

iiAxii

xgx JfxJf x*8

Since A(ax) = «Ax, we have also

11.211 -- sup 11A4 = sup II AxII. ilxil< 1 114=i A bounded linear transformation A from X to Y is called an isomorphism between X and Y if there is a bounded linear transformation B from Y to X such that AB is the identity on Y and BA the identity on X. The following proposition relates the notions of boundedness and continuity for linear operators: 2. Proposition: A bounded linear operator is uniformly continuous. If a linear operator is continuous at one point, it is bounded.

Proof: Suppose A is bounded. Then

1}Ax i — AX211 G 1.2111 • 11X1 — X2

E

for all x 1 and x2 in X with 11x 1 — x211 < E/11A11. Thus A is uniformly continuous. Suppose now that A is a linear operator which is continuous at xo. Then there is a 6 > 0 such that liAx — Axd < 1 for all x such that I Ix — xod < 3. For any z in X, set w = nz/lIzji, where 0 < n < 3. Then n—Az = Aw = A(w + x 0 ) — A(x0) and

iizli

±1 iiAzii = ilA(w ± x 0) — A(xo)i

<1, Uz(i

Sec. 2]

185

Linear Operators

since ilw + xo — xoli = *11 = n < 6. Consequently, IlAzil < n -1 114, and A is bounded. 1

3. Proposition: The space 63 of all bounded linear operators from a normed vector space X to a Banach space Y is itself a Banach space. Proof: If A and B are in 63, we define «A ± OB by (aA + f3B)x = aAx + OBx; it is easily seen to be a linear operator. Now

lixAll = sup (IxAxil = Ixi sup IlAxli = 411=1 4—..1

NAN

and

IIA ± AI = sup 1lAx ± Bxli < sup (f(Axil ± IIBXII) 511x11=1

114=i

IA! ± IA.

Thus any linear combination of two bounded linear operators is again a bounded linear operator. If liAli = 0, then !!Ax 1! _<_ I I A II • 1! xl! -= 0, and so Ax = O. Thus iiAl = 0 only for the operator 0, which maps every x into O. Thus (( (( satisfies all the requirements of a norm and we have only to show that if Y is complete so is 63. Let (A n ) be a Cauchy sequence from 53. For each x e X we have 11,4„x — A„,,x11 _.< 1121„ — A mli • lixil, and so (A ux) is a Cauchy sequence in Y and must converge to an element y in Y. Call this element Ax. It follows from the definition of Ax that A(xx) = xAx and A(x i + x2) = Ax i + Ax 2 . To show that the linear operator A is bounded, we observe that, given e > 0, there is an N such that for all ni, n > N we have 10„ — /4„,H < e. Hence liA„ii < AN + e for all n ? N, and so

iiAxii = lim iiAnxii _•_ GAid ± OW. Thus A is bounded. For each x in X we have HArix — Ax 11 = lim 11/1„x — Anixli < lim iii.,,, — A nd( ((.4 m-100

5_ 6114

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[Ch. 10

for all n > N. Thus for n > N, IIA. — AI! = sup 11(A. — A)x11 <

E.

11.11-1.

Thus A n --3 A and aa is complete.

1

Problems

13. Show that if A n —> A and x,„ x, then A nxn, —± Ax. 14. The kernel of an operator A is the set {x: Ax = 0 }. Prove that the kernel of a linear operator is a linear manifold and that the kernel of a continuous operator is closed. 15. a. Let X be a normed linear space and M a closed linear manifold. Then the natural homomorphism ço of X onto X/M has norm 1. b. Let X and Y be normed linear spaces and A a bounded linear operator from X into Y whose kernel is M. Then there is a unique bounded linear operator B from X/M into Y such that A = B 0 47. Moreover, HAI! = IA16. Let X be a metric space, and let Y be the space consisting of those real-valued functions f on X which vanish at a fixed point xo e X and satisfy if(x) — f(y)1 < Mp(x, y) for some M (depending on f). Define lifi 1 = sup i f(x) — AO i . Then Y is a normed linear space. For each P(x, Y) x c X, the functional Fx defined by F(f) = f(x) is a bounded linear functional on Y, and 1lFx — FYI! = p(x, y). Thus X is isometric to a subset of the space Y* of bounded linear operators from Y to R. Since Y* is complete by Proposition 3, the closure of this subset gives a completion of Y, and we have another proof of Theorem 7.9. 3 Linear Functionals and the Hahn-Banach Theorem

A linear functional on a vector space X is a linear operator from X to the space R of real numbers. Thus a linear functional is a realvalued function f on X such that f(ax + Oy) = of (x) + i3f(Y). The first question with which we shall be concerned is that of extending a linear functional from a subspace to the whole space X in such a manner that various properties of the functional are preserved. The principal result in this direction is the following:

Sec. 3]

Linear FunctionsIs and the Hahn-Banach Theorem

187

4. Theorem (Hahn-Banach): Let p be a real-valued function defined on the vector space X satisfying p(x + y) < p(x) + p(y) and p(ax)---- ap(x) for each a > O. Suppose that f is a linear functional defined on a subspace S and that f(s) < p(s) for all s in S. Then there is a linear functional F defined on X such that F(x) < p(x) for all x, and F(s) --- f(s) for all s in S.

Proof: Consider all linear functionals g defined on a subspace of X and satisfying g(x) < p(x) whenever g(x) is defined. This set is partially ordered by setting g 1 -< g2 if g2 is an extension of g l , that is, if the domain of g 1 is contained in the domain of g2 and g 1 --- g2 on the domain of gl . By the Hausdorff Maximal Principle there is a maximal linearly ordered subfamily {ga } which contains the given functional f We define a functional F on the union of the domains of the ga by setting F(x) = g(x) if x is in the domain of ga. Since {g a} is linearly ordered, this definition is independent of the ga chosen. The domain of F is a subspace and F is a linear functional, for if x and y are in the domain of F, then x e domain ga and y E domain go for some a, 13. By the linear ordering of {ga}, we have either ga ..< go or gfl -‹ ga, say the former. Then x and y are in the domain of go, and so xx ± py is in the domain of go and so in the domain of F, and F(xx ± uy) = go (Xx + uy) = Xg o (x) ± ,ugo (y) = XF(x) + ,uF(y). Thus F is an extension of f Moreover, F is a maximal extension. For if G is any extension of F, gc, < F < G implies that G must belong to {ga} by the maximality of {g al. Hence G < F, and so G = F. It remains only to show that F is defined for all x e X. Since F is maximal, this will follow if we can show that each g which is defined on a proper subspace T of X and satisfies g(t) < p(t) has a proper extension h. Let y be an element in X — T. We shall show that g may be extended to the subspace U spanned by T and y, that is, to the subspace consisting of elements of the form Xy + t with t c T. If h is an extension of g, we must have h(Xy ± t) = xh(y) ± h(t) = xh(y) + g(t),

and so h is defined as soon as we specify h(y). For t h t2 e T we have g(ti) + g(t2) = g(ti ± t2) P(t1 + t2)

P(II. - y) + P(t2 ±

Y).

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Hence —p(t i — y) + g(t1 )

p(t2 + y) — g(t 2),

and so

inf [p(t ± y) — g(t)].

sup [—p(t — y) + g(t)] teT

teT

Define h(y) = «, where

a

is a real number such that

sup [—p(t — y) ± g(t)]

inf [p(t ± y) — g(t)].

a

We must now show that

h(Xy

t) = X«

g(t)

p(Xy

t).

If x > 0, then Xa

g(t) = X[a

g(t/ X)]

< X[{p(t/X ± y) — g(t1X)} = Xp(t/ X + y) = p(t

If x =

g(t1X)]

Xy).

ju < O, then —

g(t) =

a

g(t/ A))

< 1.4({p(t/ — y) — g(t/ p,)) =

Thus h(Xy of g. 1

t) p(Xy

g(t1,u))

— Y) = P(t t) for all x, and h is a proper extension

The Hahn-Banach Theorem has a wide range of applications, many of them involving a clever choice of the subadditive function p. Propositions 6 and 7 and Theorem 20 are applications of this sort. The following proposition is a generalization of the Hahn-Banach Theorem which is useful in certain applications (cf. Problems 20 and 21). By an Abelian semigroup of linear operators on a vector space X we mean a collection G of linear operators from X to X such that if A and B are in G, then AB = BA and AB is in G. We also assume that the identity operator belongs to G. 5. Proposition: Let X, S, p, and f be as in Theorem 4, and let G be an Abelian semigroup of linear operators on X such that for every A in G we have p(Ax) < p(x) for all x in X, while for each s in S we have As in S and f(As) = f(s). Then there is an extension F off to a

Sec. 33

189

Linear Functionals and the Hahn-Banach Theorem

linear functional on X such that F(x) < p(x) and F(Ax) = F(x) for all x in X. Proof: Define a function q on X by setting q(x)

. 1 mf - p(A i x + • • • + A ux), n

where the inf is taken over all finite sequences (A 1 , , A n) from G. We clearly have q(x) p(x) and q(ax) = aq(x) for a O. For any x and y in X and any e > 0, we can choose (A1, . . . , An) and (B 1 ,. , Bm), so that 1 - p(A ix + • • • + Anx) < q(x) + n

and 1

PAY + • • • + AnY)
E.

Then q(x + y) <

nm

i=1 1=1

A iB,(x + y))

5_1 p(E B ; (E nm -

n

+

1 nm

p (E A i (E BiY)) J-1

m p (E A ,x) + p (E B.iy)

m

< q(x) + q(y) + 26.

Since e is arbitrary,

q(x + y) < q(x) + q(y).

For s in S, 1 ./(5 ) = -n-

M

1 + • • • + Ans) 5_ - p(A is + • • • + Ans).

n

Hence f(s) < q(s), and we may apply Theorem 4 with p replaced by q to obtain an extension F of f to all X such that F(x) < q(x) < p(x). It remains only to show that F(Ax) = F(x). Now q(x - Ax)

1 - p((x - Ax) + A(x - Ax) + • • • + A n (x - Ax)) n 11 n +1 = - p(x - A X) 5_ -[p(x) + p(-x)].

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Banach Spaces

[Ch. 10

Thus g(x Ax) < O. Since F(x) — F(Ax) = F(x — Ax) g(x — Ax) <

0,

we have F(x) < F(x), and applying this to —x, we get F(x) = F(Ax). I

6. Proposition: Let x be an element in a normed vector space X. Then there is a bounded linear functional f on X such that f(x) =

f I 11x11. Proof: Let S be the subspace consisting of all multiples of x, and define f on S by f(Xx) = X 114, and set p(Y) = Then by the Hahn-Banach Theorem there is an extension of f to be a linear we have functional on X such that f(y) !(iii. Since f(—y) f I 5_ 1. Also f(x) = 11x11 f • 1411. Thus Il fII - 1 and f(x) 11f 1 . 11x11. I 7. Proposition: Let T be a linear subspace of a normed linear space X and y an element of X whose distance to T is at least 6, that is, an element such that Ily — Ill > 5 for all t e T. Then there is a bounded linear functionalf on X with Il fil < 1, f(y) = 6, and such thatf(t) = 0 for all t in T. Proof: Let S be the subspace spanned by T and y, that is, the subspace consisting of all elements of the form ay t with t e T. Define f( ay t) = ab. Then f is a linear functional on S, and, since //all ab, we have f(s) Hsll on S. By the = jal • Hay Hahn-Banach Theorem we may extend f to all of X so that f(x) IX. But this implies 11f II < 1. By the definition of f on S, we have f(t) = 0 for t e T and f(y) = 6.1

The space of bounded linear functionals on a normed space X is called the dual (or conjugate) of X and is denoted by X*. Since R is complete, the dual X* of any normed space Xis a Banach space by Proposition 3. Two normed vector spaces are said to be isometrically isomorphic if there is a one-to-one linear mapping of one of them onto the other which preserves norms. From an abstract point of view, isometrically isomorphic spaces are identical, the isomorphism merely amounting to a renaming of the elements. In Chapter 6 we saw that the dual of LP was (isometrically isomorphic to) Lq for

Sec. 3] Linear Functionals and the Hahn-Banach Theorem

191

1 < p < 00 and that there was a natural representation of the bounded linear functionals on LP by elements of L. We are now in a position to show that a similar representation does not hold for bounded linear functionals on L[0, 1]. We note that C[0, 1] is a closed subspace of L"[O, 1]. Let f be that linear functional on C[0, 1] which assigns to each x in C[0, 1] the value x(0) of x at O. It has norm 1 on C, and so can be extended to a bounded linear functional F on L'[0, 1]. Now there is no y in L 1 [0, 1] such that F(x) = xy dt for all x in C, for let (x n ) be a sequence of continuous functions on [0, 1] which are bounded by 1, have x„(0) = 1, and are such that x(t) 0 for all t O. Then for each y e x„y —> 0, while F(x) = I. If we consider the dual X** of X*, then to each x in X there corresponds an element
192

Banach Spaces

[Ch. 10

F defined on X such that F(s) = f(s) for s in S and 1F(x)1 p(x) for all x in X. Proof: We first note that X can be considered as a real vector space if we simply ignore the possibility of multiplying by complex constants. A mapping F from X to the complex numbers which is linear in the real sense is linear in the complex sense if and only if F(ix) = iF(x) for each x. On S define g and h by taking g(s) to be the real part of f(s) and h(s) the imaginary part. Then g and h are linear in the real sense and f = g + ih. Sincefis linear in the complex sense, g(is) ih(is) = f(is) = if(s) = ig(s) — h(s), and we see that h(s) = —g(is). Since g(s) < if(s)1 p(s), we can extend g to a functional G on X which is linear in the real sense and satisfies G(x) < p(x). Let F(x) = G(x) — iG(ix). Then F(s) = f(s) for s in S. Since F(ix) = G(ix) — iG(—x) = i[G(x) — iG(ix)], we have F linear in the complex sense. For any x, choose co to be a complex number such that = 1 and coF(x) = IF(x)I• Then IF(x)I = coF(x) = F(wx) = G(cox) p(cox) = p(x). I

Problems 17. Show that a linear functionalf on a normed linear space is bounded if its kernel is closed. (The kernel off is {x: f(x) = 0}.) 18. Let T be a linear subspace of a normed linear space X and y a given element of X. Show that inf ilY tii = sup ff(y): 11f11 = teT f(t) = 0 all te T1. 19. Prove Proposition 7 by taking S to be the subspace consisting of multiples of y, f(Xy) = XS, and p(x) = inf Ilx — 11. teT

20. Let r be the space of all bounded sequences. Use Proposition 5 to show that there is a linear functional F on l with the following properties: i. lim

Mtn

lim En . F[(E.)1 n.)] F[(En)] F[(77 71 )].

iii. F[(42 )] = iv. If ?hi = E+ i, then F[()] = The functional F is called a Banach limit and is often denoted by LIM.

Sec. 4]

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193

21. Use Proposition 5 to show that there is a set function g defined for all bounded sets of R with the following properties: i. If A n B = 0, AB. U B) = gA t) = iii. If A C B, MA < AB. iv. If A is Lebesgue measurable, then AA is the Lebesgue measure of A.

[Hint: It is easier to work with integrals than sets.] 22. Show that a Banach space X is reflexive if X* is reflexive. [Hint: If ço[X] is not all of X**, then there is a nonzero functional y e X*** such that y(x) 0 for all x e ço[X].] 23. If S is a linear subspace of a Banach space X, we define the annihilator S ° of S to be the subset S° = ly e X*: y(s) = 0 for all s e SI. If T is a subspace of X*, we define T° = {x e X: t(x) = 0 for all t e a. Show that S° is a closed linear subspace of X. b. Show that S" = c. If S is a closed subspace of X, then S* is isomorphic to X'/S° . d. If S is a closed subspace of a reflexive Banach space X, then S is reflexive. 24. Let X be a vector space and P a subset of X such that x, y e P implies x ± y 6 P and ax e P for a > O. Define a partial order in X by defining x < y to mean y — x e P. A linear functional f on X is said to be positive (with respect to P) if f(x) > 0 for all x e P. Let S be any subspace of X with the property that for each x e X there is an s e S with x < s. Then each positive linear functional on S can be extended to a positive linear functional on X. [Hint: The transfinite part of the proof is the same as that for the Hahn-Banach Theorem. The possibility of extending a functional to a space containing one more element is even simpler than for the Hahn-Banach Theorem.] 25. Let f be a mapping of the unit sphere S = {x: 11,4 < 1} into R such that f(ax + 0y) = af(x) )3f(y) whenever x, y and ax )33/ are in S. Show that fmay be extended to all of X so that it is a linear functional.

4 The Closed Graph Theorem

A mapping from one topological space to another is called an open mapping if the image of each open set is open. Thus a one-to-one continuous open mapping is a homeomorphism. We shall show that a continuous linear transformation of one Banach space onto another

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is always an open mapping, and use this to give criteria for the continuity of a linear transformation. We begin with a lemma. 9. Lemma: Let A be a continuous linear transformation of the Banach space X onto the Banach space Y. Then the image by A of the unit sphere in X contains a sphere about the origin in Y.

Proof: Let S = {x: 11x11 < l/2). Since A is onto and

U

X

k=1

we have Y = U kA(9 1 ). k=1

But Y is a complete metric space, and so is not of first category in itself. Consequently, A(S i) cannot be nowhere dense, and A(S i) contains some sphere, say {Y: 11Y — P il < 71}

Then A(S i) — p contains the sphere {_I': !IA < But A(S 1 ) — p c A(S 1 ) — A(S 1) c 2A(S 1 )

A(S0).

Thus A(S0) contains a sphere about the origin of radius n snd so by the linearity of A, A(S) contains a sphere about the origin of radius n/2n. We now proceed to show that A(S0) contains a sphere of radius n/2 about the origin. Let y be an arbitrary point of Y with yjj < n/2. Since y e A(S i), we can choose .X1 E Si such that A(x1)11 <

Similarly, we may choose x2 e

S2

•

such that

Ify — A(x j ) — A(x2 )11 <

Sec. 4]

The Closed Graph Theorem

195

and we continue so that we choose x r, e Sr, with A(Xk) -E k-1

Since ilXkii < 1/2k,

2n+1

E x k is absolutely convergent and

x=

E Xk

k=i

belongs to S o . Moreover, A(x) = A(ix) =

A(Xk) =

17.

Thus y e A(S 0) and so

1

Y:

<

C A(S 0 ). I

10. Proposition: A continuous linear transformation A of a Banach space X onto a Banach space Y is an open mapping. Thus in particular, if A is one-to-one, it is an isomorphism.

Proof: Let 0 be an arbitrary open set of X and y any point of A[0]. Then there is some x e 0 such that y = A(x). Since 0 is open there is a sphere S containing x and contained in O. But Lemma 9 states that A[S x] must contain a sphere about the origin or that A[S] must contain a sphere about y. Thus y is contained in a sphere which is contained in A[0], and so A[0] is open. I —

11. Proposition: Let X be a linear vector space which is complete in each of the norms and Hi Ili, and suppose there is a constant C such that 11.x11 < lx111 for all x e X. Then the norms are equivalent. That is to say, there is a second constant C' such that

11141 .5_ c'llxII for all x e X.

Proof: The identity map of (X, Ili Ili) onto (X,II II) is a one-to-one continuous linear transformation and so must be an isomorphism by Proposition 10. Therefore the inverse mapping must be bounded. I

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12. Closed Graph Theorem: Let A be a linear transformation on a Banach space X to a Banach space Y. Suppose that A has the property that, whenever (x) is a sequence in X which converges to some point x and (Ax„) converges in Y to a point y, then y = Ax. Then A is continuous.

Proof: Define a new norm in X by

111x111 = 114 + 111x11. Then X is complete in the norm I I H. For if iiixp xqiii —> 0, then i lx p — x,11 —+ 0 and IlAx, — Ax,11 —> O. Hence by the completeness of X and Y we have x c X and y c Y such that Ix — x11 0 and Ax, —> O. By the hypothesis of our theorem y = Ax. Hence 0, and X is complete with respect toll' HI. But now llIx — xl Proposition 11 applies, and there is a C' such that

11xII

IlAx11

ClIx11.

Thus

IlAx 1

c llx 11,

and A is bounded. 1 The graph of a mapping of X into Y is just the set of all (x, Ax) in X X Y. The hypothesis of Theorem 12 merely states that the graph of A is closed. Another consequence of the theory of category is the following proposition, which is known as the principle of uniform boundedness.

13. Proposition: Let X be a Banach space and

a

a family of bounded linear operators from X to a norrned space Y. Suppose that for each x in X there is a constant Ms such that liTx11 < Mr for all T in a. Then the operators in a are uniformly bounded; that is, there is a constant M such that ITII < M for all T in a.

Proof: For each T the function f defined by f(x) = IITxII is a real-valued continuous function on X. Since the family of these functions is bounded at each x in X and X is complete, there is by Theorem 7.17 an open set 0 in X on which these functions are uniformly bounded. Thus there is a constant M' such that lTxlI< M'

Sec. 5]

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197

for all x e O. Let y be a point in O. Since 0 is open, there is a sphere S ---- {x: 11x — yli < 6} of some radius 6 centered at y and contained in O. If IA _._. 5, then Tz — T(y ± z) — Ty with y + z in S c O. Hence 117'4 _< ilT(y ± z)jj + IITyll < M' ± M. ConM' ± My sequently VII __< for all T in ff". I

a

Problems 26. Let (Ty ) be a sequence of continuous linear operators on a Banach space X to a normed vector space Y, and suppose that for each x in X the sequence (7,0c) converges to a value Tx. Then T is a bounded linear operator. 27. Let A be a bounded linear transformation from a Banach space X to a Banach space Y, and let M be the kernel and S the range of A. Then S is isomorphic to X/M if and only if S is closed. 28. Let S be a linear subspace of C[0, 1] which is closed as a subspace of L 2 [0, 1]. a. Show that S is a closed subspace of C[0, 1]. b. Show that there is a constant M such that for all f c S we have ilf112 -• 11f11. and 11f11. _._ milf112. c. Show that for each y c [0, 1] there is a function Icy in L 2 such that for each f e S we have f(y) = f k(x)f(x) dx. The space S can be shown to be finite dimensional (See Problem 41 or 55). 29. a. Give an example of a discontinuous operator A from a normed linear space X to a Banach space Y such that A has a closed graph. b. Give an example of a discontinuous operator A from a Banach space X to a normed linear space Y such that A has a closed graph.

*5 Topological Vector Spaces

Just as the notion of a metric space generalizes to that of a topological space, so the notion of a normed linear space generalizes to that of a topological vector space: A linear vector space X with a topology 5 on it is called a topological vector space if addition is a continuous function from X X X into X and multiplication by scalars is a continuous function from R X X into X. It follows from

Banach Spaces

198

[Ch. 10

the continuity of addition that translation by an element x is a homeomorphism and that the translate x + 0 of an open set 0 is open. Any topology with this property on a vector space is said to be translation invariant. If 3 is a translation invariant topology on X and 63 a base for 3 at 8, then the sets of the form x + U, U e 63, form a base for 3 at x. Thus it suffices to give a base at 0 in order to determine a translation invariant topology. A base at 0 is often called a local base. The following proposition gives conditions on a set 63 which guarantee that it will be a base for a topology for a topological vector space and states that we can always find a base satisfying these conditions. 14. Proposition: Let X be a topological vector space. Then we can find a base 63 at 0 which satisfies the following: i. ii. iii. iv. v.

If If If If If

U, V e OE, then there is a WeOE with Wc Un V. U e OE and x e U, there is a V e 63 such that x + V c U. U e OE, there is aVeOE such that V + V c U. U e (33 and x e X, there is an a e R such that x e a U. U e (33 and 0 < lai < 1, then aU c U and aU e 63.

Conversely, given a collection (33 of subsets containing 0 and satisfying the above conditions, there is a topology for X making X a topological vector space and having 63 as a base at O. The topology will be Hausdorff if and only if

vi.n {U e 63} = {0}. The proof of the proposition is left to the reader. We note that if X is a normed linear space we may take 63 to be the set of spheres about 0, and the proposition gives us a base for the general case which has many of the properties possessed by the collection of spheres. In a topological vector space we can compare the neighborhoods at one point with those at another by translation. Thus it is possible to speak of uniform properties: A mapping f from a topological vector space X into a topological vector space Y is said to be uniformly continuous if for any open set O containing the origin in Y there is an open set U containing the origin in X such that for each x e X we have f[x + U] c f (x) + O. It is readily seen that a linear transformation from X to Y is uniformly continuous if it is con-

Sec. 5]

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199

tinuous at one point. A linear mapping ço of X onto Y is called a (topological) isomorphism if cp and io-1 are both continuous. From an abstract point of view isomorphic spaces are the same. The following theorem tells us that the only topology on a finite-dimensional vector space which makes it into a topological vector space is the usual one. 15. Proposition (Tychonoff): Let X be a finite-dimensional topological vector space. Then X is topologically isomorphic to Rn for some n.

Some suggestions for the proof of this proposition are given in Problem 33. Useful corollaries are given in Problems 34, 35, and 36. Problems 30. Prove Proposition 14: a. A collection CB of subsets containing 0 is a base at 0 for a translation invariant topology if and only if (i) and (ii) hold. b. Addition is continuous from X X X to X if and only if (iii). c. If multiplication by scalars is continuous (at (0, 0)) from R X X to X, then (iv). d. If X is a topological vector space, then the family 63 of all open sets U which contain 0 and such that aU C U for all a with l al < 1 is a local base for the topology and satisfies (v). [If 0 is any open set containing 0, the continuity of multiplication implies that there is an open set V containing 0 and an e > 0 such that XV c 0 for all I XI < e. Then U = U XV is open, Be Uc 0, and OEU c U for each a with l al < 1.] ixl< e. If da satisfies the conditions of the proposition, then it generates a topology in which multiplication by scalars is continuous from R X X to X. [Show that (iv) and (v) imply continuity at (0, x) and (a, 0), and use (iii).] f. If X is T1 , then (vi) holds. If (vi) and (iii) hold, then X is Hausdorff. 31. a. Show that a linear transformation from one topological vector space to another is uniformly continuous if it is continuous at one point. b. Show that a linear functional f on X is continuous if there is an open set 0 such that f[0] R. [Hint: You can take 0 to satisfy property (v) of Proposition 14.]

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32. Let X be a topological vector space and M a closed linear subspace. Let so be the natural homomorphism of X onto X/M, and define a topology on X/M by taking 0 to be open if and only if so-1 [0] is open in X. Then this topology makes X/M a topological vector space and so a continuous open map. When we speak of X/M as a topological vector space, we always mean X/M with this topology. 33. Prove Proposition 15: a. If X is n-dimensional, there is a continuous one-to-one linear map so of Ir onto X. b. Let S and B be the subsets of RI' defined by S = {y: 1111 = 1 1, B = fy: 113,11 < 11. Then go[S] is closed and X — ço[S1 open. e. There is an open set U of X containing 0 such that aU C U for each lal < land U C X — go[S]. d. The set U in (c) is contained in q,[B], and so so-1 is continuous. 34. Show that a finite-dimensional subspace M of a topological vector space X is closed. [Hint: If x y M, let N be the finite-dimensional subspace spanned by x and M. Then N has the usual topology, and so xis not a point of closure of M.] 35. Let A be a linear mapping of a finite-dimensional topological vector space X into a topological vector space Y. Then A is continuous. [Hint: The range of A is finite-dimensional and hence has the usual topology.] 36. A linear mapping A from a topological vector space X to a finitedimensional topological space Y is continuous if its kernel M is closed. [Hint: A = B 0 ço, where so is the natural mapping of X on X/M and B is continuous by Problem 35.] 37. Prove that every locally compact topological vector space X is finite-dimensional. [Hint: Let V be a neighborhood of 0 with T7 compact and aVC V for each a, lal G 1. If we cover T7 by a finite number of translates x i + V, . . . , xr, + V, then x l, . . . , xt, are a basis for X.]

*6

Weak Topologies

If X is any vector space and a a collection of linear functionals on X, we define the weak topology generated by a to be the weakest topology such that each f in 5 is continuous (cf. Problem 8.20). This topology is easily seen to be translation invariant, and a base at 0 for this topology is given by the sets {x: 1 fi(x)I < e, i = 1, . . . , n} ,

Sec. 6]

Weak Topologies

201

where e > 0 and {fi , . . JO is a finite subset of if. Since the family of all such sets satisfy the conditions of Proposition 14, this topology makes X a topological vector space. A sequence (or net) (xn ) converges to x in this topology if and only if f(x) f(x) for each f e T. If X is a normed vector space and the functionals in F are all continuous (that is, if a c X*), then the weak topology generated by if is weaker (has fewer open sets) than the norm topology of X. We usually call the metric topology generated by the norm the strong topology of X and the weak topology on X generated by X* the weak topology of X. Thus we speak of strongly closed and strongly open sets when referring to the strong topology and weakly open and weakly closed sets for the weak topology. Every weakly closed set is strongly closed but not conversely. Every strongly convergent sequence (or net) is weakly convergent. While not every strongly closed set is weakly closed, we do have the following proposition, a generalization of which is given by Corollary 23. 16. Proposition: A linear manifold M is weakly closed if and only if it is strongly closed. Proof: Since every weakly closed set is strongly closed, we have only to show that if M is strongly closed it is also weakly closed. Suppose M is strongly closed and x is a point not on M. We must show that x is not a point of closure of M in the weak topology. Since x is not a point of closure of M in the strong (metric) topology we have inf lix — sil > 6 > O. Hence by Proposition 7 there is a seM

continuous linear functional f which vanishes on M and does not vanish at x. Now {y: f(y) 0} is an open set in the weak topology which contains x but does not meet M. Hence x is not a weak point of closure of M. I If we apply the notion of weak topology to the dual X* of a normed space X we see that the weak topology of X* is the weakest topology for X* such that all of the functionals in X** are continuous. The weak topology for X* turns out to be less useful than the weak topology for X* generated by X (or more precisely, by yo[X] where so is the natural embedding of X into X**). This topology is called the weak* topology for X* and is even weaker than the weak topology.

202

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Thus a weak* closed subset of X* is weakly closed, and weak convergence implies weak* convergence. A base at 0 for the weak* topology is given by sets of the form If: f f WI < e, i = 1, . . . , nl where {x1 , . . . , x } is a finite subset of X. If X is reflexive, then the weak and weak* topologies of X* coincide. Some of the importance of the weak* topology stems from the following theorem: 17. Theorem (Alaoglu): The unit sphere S* = { f: IIIf It _. 11 of X* is compact in the weak* topology. Proof: If f c S*, then I f(x)I <_ 11x11, and so f(x) e [ - 11x11, 114. Let Iz = [ 11x11, 114]. Then each f c S* corresponds to a point in P = XL, since the latter is by definition the set of all functions ze.x f on X such that f(x) e I. Thus we may think of S* as a subset of P, and the definition of the topology for P shows that the topology S* inherits as a subspace of P is the weak* topology of S*. Since P is compact by the Tychonoff theorem, S* will be compact if it is a closed subset of P. Let f be a point of closure of S* in P. Then f is a mapping of X into R. Since Ig(x)1 < 11x11 for g e S*, and evaluation at x is a continuous function on P, we have I f(x)1 _ < 114. Let x, Y, and z be three points of X such that z --= ax + 0y. For each e > 0 the set N = Ig e P: Ig(x) — f(x)I < e, rg(y) — f(y)I < e, and ig(z) — f(z)I < el is an open subset of P containing f. Since f is a point of closure of S*, we can find a g in S* n N. Since this g is linear (being in S*) we have g(z) = ag(x) + ,3g(y). Hence it follows that I f(z) — a f(x) — 0f(y)1 < e(1 + la! ± I01). This inequality holding for each positive e, we have f(z) = af(x) ± g(y), and f is linear on X. Thus f is in S*, and so S* is closed. I -

Problems 38. a. Show that if x„ -- x weakly, then (11xn ii) is bounded. b. Let (x„) be a sequence in P, 1 < p < oc, and let xn = (tnt,n):=1. Show that (xn) converges weakly to x = (E„,,) if and only if (ilx,ill) is bounded and for each m we have t,,,,n ---- Em. c. Let ()en) be a sequence in LP[O, 1], 1 < p < oc. Show that (xn) converges weakly to x if (11x.„11) is bounded and (xn) converges to x in measure (cf. Problem 6.16).

Sec. 7] Convexity

203

d. In P', 1 < p < oo, let xr, be that sequence whose nth term is one and whose remaining terms are zero. Then (xn ) does not converge in the strong topology, but x„, —+ 0 in the weak topology. e. Let (x„) be as in (d), and define Yn m = nx„,. Then the set F = m > n) is strongly closed. [Hint: The distance between any two points of F is at least 1. Hence F contains no nonconstant sequences which converge in the strong topology.] f. The set F in (e) has 6 as a weak closure point. However, there is no sequence (z,„) from F which converges weakly to zero. 39. a. Let S be a bounded subset of a normed space X. Let 5 be a set of functionals in X* and let 5.0 be a dense subset of ff• (dense in the sense of the norm topology on X*). Then g• and ff•c, may generate different weak topologies for X, but these topologies are the same on S; that is, S inherits the same topology from each. b. Let S* be the unit sphere in the dual X* of a separable Banach space X. Then the weak* topology on S* is metrizable. (Caution: This does not mean the weak* topology on X* is metrizable!) 40. Show that every weakly compact set is bounded in the norm topology. 41. Let S be the linear subspace of C[0, 1] given in Problem 28. a. Show that if fi, —* f weakly in L 2, then fn(y) f(y) for each y c [0, 1]. b. If fin weakly in L 2, then is bounded, and hence f7, ---> f strongly in L 2 by the Lebesgue convergence theorem. c. The space S is a locally compact subspace of L 2 and hence finitedimensional. ,

unto

*7

Convexity

A subset K of a vector space X is said to be convex if, whenever it contains x and y, it also contains Xx + (1 — N)y for 0 < X < 1. The set {z: z = Xx + (1 — X)y for 0 < X < 11 is called the line segment joining x and y. The points x and y are its endpoints, and a z for which 0 < X < 1 is called an interior point of the segment. Thus a set K is convex if whenever it contains x and y it contains the line segment joining x and y. Every linear manifold is convex, and the unit ball in a normed space is convex. The following lemma gives some basic properties of convex sets. Further properties are given in

204

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[Ch. 10

Problems 42, 43, and 44. The proof of the lemma is straightforward and is omitted. 18. Lemma: If K1 and K2 are convex sets, so also are the sets K 1 n K2, XKI, and Ki + K2.

A point xo is said to be an internal point of a set K if the intersection with K of each line through xo contains an open interval about xo. Thus xo is internal to K if, given x e X, there is an E > 0 such that xo Xx e K for all X with IX! < E. Let K be a convex set which contains 0 as an internal point. Then we define the support function p of K (with respect to o) by p(x) = inf {x: X -1 x e K, X > . We have the following properties for this support function: 19. Lemma: If K is a convex set containing 0 as an internal point, then the support function p has the following properties: j.

p(Xx) = Xp(x) for X _> O. P(4 + P(Y). P(x {x: p(x) < 1} c KC {x: p(x) < 1}

Proof: The first and third properties follow immediately from the definition of p. To prove the second, suppose that x —ix and if ly belong to K Then

(X + 12) -1 (x + Y) = X(X

A) -1 (X -1x)

m(X

A) -1 ( 2-1Y)

belongs to K, since K is convex. Thus p(x + y) < X + A, and taking infima over all admissible X and A we obtain p(x + y) < p(x)

IAA I

Two convex sets K1 and K2 are said to be separated by a linear functional f if there is a real number a such that f(x) < a on K1 and f (x) > a on K2. 20. Theorem: Let K 1 and K2 be two disjoint convex sets in a vector space X, and suppose that one of them has an internal point. Then there is a nonzero linear functional f which separates K1 and K2.

Proof: Let x1 be an internal point of K1. Then K1 — K2 is convex, and the point xo = x1 — x2 is an internal point of K1 — K2 for any

Sec. 7] Convexity

205

x2 in K2. Let K K 1 — K2 — x0. Then K is a convex set containing O as an internal point. Since K1 and K2 are disjoint, 19 e K1 — K2, and so —xo e K. Let p be the support function of K (with respect to 6). Then p(— x 0) > 1. Let S be the one-dimensional subspace of X consisting of all multiples of x0. On S define f by f(ax 0) = —a. Then f(s) p(s) and p satisfies the condition of the Hahn-Banach Theorem by Lemma 19. Thus we may extend f to a linear functional defined on all of X so that f(x) < p(x) for all x. Thus if x e K, we have f(x) < 1. Let x e K1 and y e K2. Then x — y — xo e K, and we have

f(x) — AY)

—

f(x0) = f(x — y — xo) < I.

Since f(x 0) = —1, we have f(x) < f(y). This being true for each x e K1 and each y in K2, we have sup f (x) < inf f(y). Thus f xeKi

yeK2

separates K1 and K2, andfis a nonzero functional sincef(x0) = —1.1 A topological vector space is called locally convex if we can find a base for the topology consisting of convex sets. The following proposition gives a convenient criterion for ensuring that a given topology 5 for a vector space X will make X into a locally convex topological vector space. Suggestions for its proof are given in Problem 46.

21. Proposition: Let oz, be a family of convex sets in a vector space X. Then the following conditions are sufficient for the translates of sets in t to form a base for a topology which makes X into a locally convex topological vector space: i. If N e at, each point of N is internal. ii. If N 1 and N2 are in at, there is an N3 in at with N3 C N1 n N2. If N is in at, then for each a with 0 < lai < 1 we have aN c 9t. Moreover, in each locally convex topological vector space there is a base it at 0 which satisfies these conditions.

It follows from this proposition that the weak topology on a vector space X which is generated by a family of linear functionals makes X into a locally convex topological vector space. Also a normed vector space is a locally convex topological vector space.

206

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[Ch. 10

22. Proposition: Let X be a locally convex topological vector space and F a closed convex subset. Let xo be a point of X not in F. Then there is a continuous linear functional f on X such that

f(x0) < inf AX). xe

Proof: By translating by —x 0, we reduce the proposition to the case that x o = O. Since o is not a point of closure of F, there is a convex open set N which contains 0 but does not meet F. Let 0 — N n N). Then 0 is an open convex set containing 0 and disjoint from F, and —0 = O. Since o is an internal point of 0 (see Problem 44a), there is by Theorem 20 a nonzero linear functional f such that sup f(x) < inf f(y) = a. Thus f(x) < a for x in 0, and since (—

xe0

ye?

x e 0 implies —x e 0, we have — f (x) < a on O, whence I f(x)1 < a on O. Thus for each e > 0, we have I f(x)i < e on the set 0' = (€a -1 )0. But 0' is an open set containing 0, and so f is continuous at O. Since f is linear and continuous at 0, it is continuous everywhere. It remains only to show that a > 0. Sincef is a nonzero functional, there is an x such that f(x) > 0. Since 0 is an internal point of 0, we can choose X > 0 so that Xx is in O. Then

0 < Xf(x) = f(Xx) <

a.

I

23. Corollary: Let K be a convex set in a locally convex topological space. Then K is strongly closed if and only if it is weakly closed. 24. Corollary: Let x and y be distinct points of a locally convex topological vector space X. Then there is a continuous linear functional f such that f(x) f(y). Let K be a convex subset of a vector space X. A point x in K is called an extreme point if it is not an interior point of any line segment lying in K. Thus x is extreme if whenever x = Xy ± (1 — X)z with 0 < X < 1, we have y K or z K. We are going to show that every compact convex set has extreme points, but we first consider some preliminary notions. A subset S of a convex set K is called a supporting set of K if it is closed and convex and has the property that if an interior point of a line segment in K belongs to S then the entire line segment belongs to S. Thus an extreme point is a supporting set consisting of exactly one point.

Sec. 7]

207

Convexity

25. Lemma: Let f be a continuous linear functional on a closed convex set K. Then the set S of points where f assumes its maximum on K is a supporting set of K. Proof: The set S is convex, for if f(x) = m and f(y) = m, then f(Xx + (1 — X)y) = m. If the line segment joining x to y is in K and f assumes its maximum m at the point z = Xx + (1 — X)y, then m = Xf(x) + (1 — X) f(y), and since f(x) and f(y) are not greater than in, we must have f(x) =[(y) = m. But this implies that f(Ax + (1 — p.)y) = m, and so the entire line segment joining x to y is in S. I The intersection of all convex sets containing a set E is a convex set which contains E and which is contained in every convex set containing E. This set is called the convex hull of E. The intersection of all closed convex sets containing E is a closed convex set which contains E and which is contained in every closed convex set containing E. This set is called the closed convex hull of E. 26. Theorem (Krein Milman): Let K be a compact convex set in a locally convex topological vector space X. Then K is the closed convex hull of its extreme points. -

Proof (Kelley): We assume that K is not empty. It follows from the definition of supporting sets that the intersection of a collection of supporting sets of K is a supporting set of K and that if S is a supporting set of K and T a supporting set of S then T is a supporting set of K. Given any nonempty supporting set S of K, the family of all nonempty supporting sets of K is partially ordered by inclusion, and by the Hausdorff maximal principle there is a maximal linearly ordered family 8 of nonempty supporting sets with S in 8. Since K is compact, the intersection T of all members of s is nonempty and hence is itself a nonempty supporting set of K. It is, moreover, a minimal nonempty supporting set, for if T properly contained a supporting set the family s would not be maximal. Thus any supporting set contains a minimal nonempty supporting set. Now a minimal nonempty supporting set can contain only one point. For if a supporting set S contains the distinct points x and y, there is a continu-

208

Banach Spaces

[Ch. 10

ous linear functional f such that f(x) > f(y). Then the subset of S where f attains its maximum is by Lemma 25 a supporting subset of S and hence of K. Since K is compact, it is a nonempty supporting subset of K which does not contain y. If a supporting set consists of just one point, that point must be extreme. Hence we have shown that every nonempty supporting set contains an extreme point. Since the subset of K where a linear functional assumes its maximum is a nonernpty supporting set, we conclude that the maximum of a continuous linear functional on K is equal to its maximum on the set E of extreme points of K. Let C be the closed convex hull of the extreme points of K, and suppose x 0 C. By Proposition 22 there is a continuous linear functionalf such that f(x) > max f(y) = max f(y). yee

Thus x

yeK

K, and we have K c C. Hence K = C. I

Problems 42. Let A be a linear operator from the vector space X to the vector space Y. Then the image of each convex set (or linear manifold) in X is a convex set (or linear manifold) in Y and the inverse image of a convex set (or linear manifold) in Y is a convex set (or linear manifold) in X. Give an example to show that a nonconvex set may have a convex image. 43. Show that the closure of a convex set K in a topological vector space is convex. 44. a. Show that each interior point of a convex subset of a topological vector space is internal. [Hint: Use the continuity of multiplication.] b. Show that in R" each internal point of a convex set is an interior point. c. Give an example of a set in the plane which has an internal point that is not an interior point. 45. Let K be a convex set containing 6, and suppose that x is an internal point of K. Then for some X > 0 the set x XK is contained in K. [Hint: Choose X > 0 so that (1 — X) lx is in K.] 46. Prove Proposition 21. [Hint: If N is convex, then IN + N C N. Use Proposition 14 and its proof.] 47. Strongest locally convex topology. Let X be a vector space and 63 the collection of all convex sets V containing 0 such that for each x e X

Sec. 7]

209

Convexity

there is an a > 0 with ax e V. Then IT. is a local base for a locally convex topology on X, and this topology is stronger than any other locally convex topology on X. 48. a. In LP[O, 1], 1 < p < oo , every x with IX = 1 is an extreme point of the unit sphere S = 114 < 11.

b. In L'[0, lithe extreme points of the unit sphere are those x such that lx(i)1 = 1 a.e. c. The unit sphere in L 1 [0, 1] has no extreme points. d. L 1 [0, 1] is not the dual of any normed linear space. e. What are the extreme points of the unit sphere in P? f. What are the extreme points of the unit sphere in C(X), X a compact Hausdorff space? Show that C[0, 1] is not the dual of any normed linear space. 49. Let X be the vector space of all measurable real-valued functions on [0, 1] with addition and multiplication by scalars defined in the usual way. i x(t)1 Define a(x) = dt. fl 1 ± l x(t)I

a. We have o-(x ± y) < r(x) a(x — y), p is a metric for X.

cr(y). Hence if we define p(x, y) =

b. In this metric xn x if and only if xn --> x in measure. c. X is a complete metric space (cf. Problem 4.25). d. Addition is a continuous mapping of X X X into X. e. Multiplication is a continuous mapping of R X X into X. [Since X is a metric space, it suffices to prove that if xn X then x and X„ Xnxn Xx. This follows from the Bounded Convergence Theorem for convergence in measure.] f. Show that the set of step functions is dense in X. g. There is no nonzero continuous linear functional on X. [Show that there is an n such that f(x) = 0 whenever x is the characteristic function of an interval of length less than 1/n. Hence f(x) = 0 for all step functions x.] h. The space X is a topological vector space which is not locally convex. i. Let s be the space of all sequences of real numbers, and define

2- 161 Prove the analogues of (a), (c), (d), and (e). What 1 + Itv i is the most general continuous linear functional on s? .

210

Banach

Spaces

[Ch.

10

8 Hilbert Space

By a Hilbert space we mean a Banach space H in which there is defined a function (x, y) on H X H to R with the following properties': i. (ceiXi a2X29 Y) = al(X1, Y) (x, y) = (y, x). (x, x) = 11x11 2 .

ct2(X29 Y).

We call (x, y) the inner product of x and y. Two examples are immediate: One is the space Rn with

The other is the space L2 with (x, y) =

f x(t)y(t) dt.

Since lxii > 0 with equality only for x = 0, we have 0

lix — XYII 2 = (x — XY, x — XY) = (x, x) — 2X(x, y)

X 2(y, y).

If X > 0, we have 2(x, y)

Setting X =

XIIY112.

we obtain (x, .Y)

11 ,4 • VII,

and we see that equality can only occur when y = 0 or x = Xy for some x > O. This inequality is variously known as the Schwarz, Cauchy-Schwarz, or Cauchy-Buniakowsky-Schwarz inequality. A consequence of this inequality is that the linear functional f defined and from this it follows that by f(x) = (x, y) is bounded by (x, y) is a continuous function from H X H to R. We say that two elements x and y of H are orthogonal if (x, y) = O. We write x 1 y to mean x and y are orthogonal. A set 8 in H is 1

have defined here a real Hilbert space. In analysis it is generally more convenient to deal with a complex Hilbert space, that is, one in which the scalars are complex numbers, the inner product is complex-valued, and (ii)is replaced by(ii'): (x,y) =

We

Sec. 8]

211

Hilbert Space

called an orthogonal system if any two different elements ço and 1,t, of s are orthogonal, that is, (ço, 1,0 = 0. An orthogonal system s is called orthonormal if 11(1011 = 1 for each ço in S. Any two elements of an orthonormal system are at distance -0 from each other. Hence if H is separable, every orthonormal system in H must be countable. Henceforth we shall deal only with separable Hilbert spaces. Thus each orthonormal system may be expressed as a sequence ((pp ), which may be finite or infinite We define the Fourier coefficients (with respect to ((p p )) of an element x in H to be ap = (x, ço). For any n we have 2 n

2

0<

X—

E

ario9

X11 2 — 2

=

E

av (x,

v=1

v=1

= 114 2 —

n

(I:0+ v=1 A=1

E ce.

v=1

Thus

E 442 ^ 11x11

v=1

and, since

n

was arbitrary, we have Bessel's inequality:

E

p=1

On the other hand, let

(ap )

be any sequence of real numbers with

E ap2 < oo . Then the sequence Zn =

is a Cauchy sequence, since for m > z„, — Zn =

n

E

av
n+1

and we have

Ilzm — znII 2 = n+1 E a92, 2

If there are only a finite number N of elements in (), we adopt the convention

that

p=i

means

E

v=1

for N < < 00.

212

Banach Spaces

[Ch. 10

which must tend to zero since Ea v2 converges. By the completeness of H there is an element y such that y = lim z n, and we write oo

y =

E aviop.

v=1.

Since the inner product is continuous, we have (y, cpp) = lim (z 7i, iop) = ay.

We have thus shown that for any x there is a y of the form y = E apiop which has the same Fourier coefficients as x. v 1 —

When does this y equal x? If we look at y — x, we see that all its Fourier coefficients are zero. Hence y = x if the orthonormal system (cp„) has the property that, if (z, (pr) = 0 for all y, then z = O. An orthonormal system with this property is called complete (or total). A complete orthonormal system is clearly a maximal one, while, if (4op ) is a maximal orthonormal system, it must be complete. For, if (z, io„) = 0 all y and z o 0, then zazil can be added to (40p). The Hausdorff Maximal Principle implies the existence of a maximal orthonormal system. We have thus established the following proposition: 27. Proposition: In a separable Hilbert space every orthonormal system is countable and there is a complete orthonormal system. If (cp„) is any complete orthonormal system and x any element of H, we have x

= E apio„ v=- 1

where a„ = (x,‘,0,,). Moreover, PO =

E a.

In a separable Hilbert space there are two alternatives: Either every complete orthonormal system has an infinite number of elements or else there is a complete orthonormal system with a finite number N of elements. In the latter case such a system is a basis (in the vectorspace sense) of H by Proposition 27. Hence H is a finite-dimensional vector space, and any system of N + 1 elements is linearly dependent. Consequently, every orthonormal system can have at most N elements in it. From this it follows that every complete orthonormal

Sec. 8]

213

Hilbert Space

system must have N elements. We have thus proved that in a separable Hilbert space H every complete orthonormal system has the same number of elements. We call this number the dimension of H. (Thus if H has a countably infinite complete orthonormal system we say dim H = N o.) An isomorphism (P of a Hilbert space H onto a Hilbert space H' is a linear mapping of H onto H' such that ((f,x, (Py) = (x, y). Thus an isomorphism between Hilbert spaces is an isometry. Each ndimensional Hilbert space is isomorphic to Rn, since the mapping defined by 4)(x) = (a1 ,. , an), where a, = (x, ço„) is an isomorphism. Similarly, every N o-dimensional Hilbert space is isomorphic to 12 . Since L 2 [0, r] is separable and {cos vt} is an infinite orthogonal system, we see that the dimension of L 2 is No, and so L 2 is isomorphic to /2

28. Proposition: Let f be a bounded linear functional on the Hilbert space H. Then there is a y eH such that f(x) = (x, y) for all x. Moreover, IfM = Proof: (We consider only the case of separable H. For nonseparable H see Problem 52.) Let (pi,) be a complete orthonormal system for H, and set bp = f (cpp). Then for each n we have

E bi! =

v=i

f(

v=1

11f11 •v—iE bAc'v

bvçov)

11f11

bv2r 2.

11f11 2, and so E b,2, 11f11 2 < Go. Hence there is an element y = E /Jo,. We have VII 11f11. v= 1 x, and so Let x be any element of H. Then E av ,p,

Thus E b,2,

p=

f(x) = lim f (E a v(p) = V= I

By the Schwarz inequality 11 f II

E ab. 1

ii.Y11. I

214

Banach Spaces

[Ch. 10

Problems 50. Show that the inner product is continuous; that is, if xi, --> x and Yn --* y, then (xn, yr,) --)• (x, Y). 51. a. Show that {cos vi, sin pi} is (when suitably normalized) a complete orthonormal system for L 2 [0, 27r] (cf. Problems 6.14 and 9.28). b. Every function in L 2 [0, 27r] is the limit in mean (of order 2) of its Fourier series (cf. Section 6.3). 52. a. Show that for each x in a nonseparable Hilbert space there are

only a countable number of nonzero Fourier coefficients (with respect to a fixed orthonormal system). b. Show that Proposition 27 is still valid in a nonseparable Hilbert space except that every complete orthonormal system is uncountable. c. Show that Proposition 28 is still valid in a nonseparable Hilbert space. d. Show that if His any infinite-dimensional Hilbert space, then the number n of elements in a complete orthonormal system in H is the smallest cardinal n such that there is a dense subset of H with n elements. Hence every complete orthonormal system in H has the same number of elements. We call this number the dimension of H. e. Show that two Hilbert spaces are isomorphic if and only if they have the same dimension. f. Show that there is a Hilbert space of each dimension. 53. Let P be a subset of H. By the orthogonal complement P1 of P we mean the set {y: y I. x all x a P}. a. Show that P1 is always a closed linear manifold. b. Show that P±± is the smallest closed linear manifold containing P. c. Let M be a closed linear manifold. Then each x e H can be written uniquely in the form x = y + z with y e M and z a MI. Moreover, 11x11 2 = 111'11 2 + 1iz11 2 . 54. Let (x,,) be a bounded sequence of elements in a separable Hilbert space. Then (x.) contains a subsequence which converges weakly. 55. Let S be a subspace of L 2 [0, 1], and suppose that there is a constant K such that 1f(x)1 < Klifil for all x 6 [0, 1]. Then the dimension of S is at most K2 . [Hint: If (f 1 , . . . ,f,,) is any finite orthonormal sequence in S,

then t r. 1fi(x)1 2 _< K2.] t=1

Part Three

0

General Measure

and Integration Theory

11 1

Measure and Integration

Measure Spaces

The purpose of the present chapter is to abstract the most important properties of Lebesgue measure and Lebesgue integration. We shall do this by giving certain axioms which Lebesgue measure satisfies and base our integration theory on these axioms. As a consequence our theory will be valid for every system satisfying the given axioms. We begin by recalling that a c-algebra 63 is a family of subsets of a given set X which contains 0 and is closed with respect to complements and with respect to countable unions. By a set function we mean a function which assigns an extended real number to certain sets. With this in mind we make the following definitions: Definition: By a measurable space we mean a couple (X, 63 ) consisting of a set X and a cr-algebra 63 of subsets of X. A subset A of X is called measurable (or measurable with respect to 63) if A e Definition: By a measure u on a measurable space (X, d3) we mean a nonnegative set function defined for all sets of 1s3 and satisfying A(0) = 0 and 12, (ID E)z — J=1

,uE

for any sequence E, of disjoint measurable sets. By a measure space (X, 6-3 , )1) we mean a measurable space (X, G3) together with a measure )1 defined on (B. 217

218

Measure and Integration

[Ch. 11

This second property of /A is often referred to by saying that A is countably additive. We also have that A is "finitely additive"; that is,

for disjoint sets Ei belonging to CA, since we may set Ei = Ø for i > N.' One example of a measure space is (R, OR, m), where R is the set of real numbers, Mt the Lebesgue measurable sets of real numbers, and m Lebesgue measure. Another measure space results if we replace R by the interval [0, 1] and Mt by the measurable subsets of [0, 1]. A third example is (R, CA, m), where (id is the class of Borel sets and m is again Lebesgue measure. Another example is the counting measure (see Problem 3.4). A slightly bizarre example is the following. Let X be any uncountable set, (id the family of those subsets which are either countable or the complement of a countable set. Then (B is a o--algebra and we can define a measure on it by setting AA = 0 for each countable set and AB = 1 for each set whose com-

plement is countable. Two further properties of measures are given by the following propositions: 1. Proposition: If A c 63, B c

(id,

and A C B, then

ALA <

Proof: Since B = A U [B

A]

is a disjoint union, we have ,uB = AA 2. Proposition: If Ei e

A) > A. I

,u(B

(id, mEi < ck

Ei) =

and Ei D Ei+1 , then

lim

A (z1 =--

A set function A defined on an algebra of sets and satisfying A(23) = 0 and A(A U B) = 1.1A /213 for disjoint sets A and B in the algebra is called a finitely additive measure. Since our definition (and the usual usage) requires a measure to be countably additive, it follows that a finitely additive measure is not in general a measure, although every measure is, of course, a finitely additive measure.

Sec. 1]

Measure Spaces

Proof: Set E =

219

n E. Then

i=1

E l = E u 1.1 (E1 i= 1

and this is a disjoint union. Hence = pE

E

E1-3-1).

i=1

Since E, — E, +1 U (E,

is a disjoint union, we have y(E, E, +1) = yE, — Hence y E,=

yE+ E (yEi — n-1

==

li

E 7,-00 i=i

cuEi — yEi+i)

= pE pE1 — lim yEn

,

whence the proposition follows. I

3. Proposition: If Ei r 63, then < E yEi.

Ei) In-1

Proof: Let Gn = E. are disjoint. Hence

U yGn

. Then G„ C E,, and the sets G. <

ttEn,

while pG.

tt(U E,) = n=1

E /2E,,. I n=1

220

Measure and Integration

[Ch. 11

A measure is called finite if pt(X) < Go. It is called or finite if there is a sequence (Xn) of sets in 63 such that -

X

= U Xn n=i

and AX„ < 00. By virtue of Proposition 1.2 we may always take (Xn ) to be a disjoint sequence of sets. Lebesgue measure on [0, 1] is an example of a finite measure, while Lebesgue measure on (— cr, , cr..) is an example of a a--finite measure. The counting measure on an uncountable set is a measure which is not a-finite. Other examples are given in Problem 46. A set E is said to be of finite measure if E e 03 and AE < c4. A set E is said to be of cr finite measure if E is the union of a countable collection of measurable sets of finite measure. Any measurable set contained in a set of o--finite measure is itself of a-finite measure, and the union of a countable collection of sets of a--finite measure is again of a--finite measure. If is a--finite, then every measurable set is of a--finite measure. Roughly speaking, nearly all the familiar properties of Lebesgue measure and Lebesgue integration hold for arbitrary a.-finite measures, and many treatments of abstract measure theory limit themselves to a--finite measures. However, many parts of the general theory do not require the assumption of 0-finiteness, and it seems undesirable to have a development that is unnecessarily restrictive. A notion weaker than that of a--finiteness is semifiniteness. A measure pt is said to be semifinite if each measurable set of infinite measure contains measurable sets of arbitrary large finite measure. Thus every a--finite measure is semifinite, while the measure which assigns 0 to countable subsets of an uncountable set X and coc to the uncountable sets is not semifinite. Measures which are not semifinite tend to be somewhat bizarre, and there is a temptation to restrict oneself to the consideration of semifinite measures. Unfortunately, many of the natural operations on measures, such as the extension procedures of the next chapter, take us out of the class of semifinite measures. A measure space (X, 63, pt) is said to be complete if 03 contains all subsets of sets of measure zero, that is, if B e 63, cB = 0, and A c B imply A e 83. Thus Lebesgue measure is complete, while Lebesgue measure restricted to the a--algebra of Borel sets is not complete. The following proposition, whose proof is left to the reader (Problem -

221

Sec. 1] Measure Spaces

shows that each measure space can be completed by the addition of subsets of sets of measure zero. The measure space (X, a o , /.Lo) given in the proposition is called the completion of (X, 63, /.4).

2),

4. Proposition: If (X, 63, /2) is a measure space, then we can find a complete measure space (X, 63 0, A o) such that i. (33 c (S3 0 . E c (33= E = 110E. E e (33 0 <=>E = Au BwhereB eG3andAcC,C e,uC = O.

is a measure space, we say that a subset E of X is locally measurable if En B e ci3 for each B e ci3 with AB < go. The collection e of all locally measurable sets is a a--algebra containing The measure is called saturated if every locally measurable set is measurable, i.e., is in (B. Every a--finite measure is saturated. A measure can always be extended to a saturated measure, but, unlike the process of completion, the process of saturation is not uniquely determined: There is no difficulty with the a--algebra, for we simply take the a--algebra e of all locally measurable sets, but the measure can often be extended to e in different ways. Some details are given in Problem 8, and we shall return to this topic in the next chapter. If (X, 63,

A)

Problems 1. Let {A n}

be a countable collection of measurable sets. Then

C1

(

=

k

U

n—,00

Ah)

k=1

2. Let {(X«, A.) } be a collection of measure spaces, and suppose that the sets {X a} are disjoint. Then we can form a new measure space (called their union) by letting X = U Xa, G3 {B: (0)[B n Xa e

and defining

kt(B) = E,ua(B

n

a. Show that 63 is a a--algebra.

is a measure. c. Show that y is a--finite if and only if all but a countable number of

b. Show that A the

are zero and the remainder are o--finite. 3. a. Show that ,u(Ei A ED = 0 implies AE I = AE2 provided that Ei and E2 e (B. 0 b. Show that if y is complete, then E l e 63 and ,u(E i A E2) imply E2 e (B.

222

Measure and Integration

[Ch. 11

4. Let (X, 03, p) be a measure space and Y c 03. Let 03y consist of those sets of 03 which are contained in Y. Set p.yE = pE if E c (B y . Then ( Y, 03y, gy) is a measure space. The measure gy is called the restriction of p to Y. 5. Let (X, 03) be a measurable space. a. If p. and u are measures defined on 03, then the set function X defined on OB by XE = aE ± vE is also a measure. We denote X by ,u ± v. b. If p, and u are measures on 03 and p. > u, then there is a measure X on 03 such that p = v + X. c. If 7/ is o--finite, the measure X in (b) is unique. d. Show that in general the measure X need not be unique but that there is always a smallest such X. 6. a. Show that each o--finite measure is semifinite. b. Show that each measure p is the sum pi ± p2 of a semifinite measure p i and a measure p 2 which assumes only the values 0 and GO The measure p i is always unique, and there is a smallest p 2 . 7. Prove Proposition 4. [First show that the family 63 0 defined by (iii) is a o--algebra. If E e 630, show that AA is the same for all sets A c 03 such that E= A uB with B a subset of a set of measure zero. Use this fact to define p o and show I.L 0 is a measure.] 8. a. Show that each o--finite measure is saturated. b. Show that the collection of locally measurable sets is a o--algebra. c. Let (X, 133,0 be a measure space and e the o--algebra of locally measurable sets. For E e e set rcE = I.LE if E e 63 and gE = cc if E Y 63. Show that (X, e, FL) is a saturated measure space. d. If p is semifinite and E e e, set pE — sup {AB: B c 03, B C E} . Show that (X, 6, p) is a saturated measure space and that A is an extension of A. e. Use Problem 5d to show that, even if pc is not semifinite, there is a smallest extension p of A to 0, and that (X, 6, 1.0 is saturated. f. Give an example to show that 7.4- and p._ may be different. 9. o--Rings and o--Algebras. Some authors prefer to define a measure on a o--ring 01, that is, a collection of subsets of X such that if A and B are in CA so is A — B, and if (An ) is a sequence from Oi then U A n is in 01. Thus a cr-ring a is a o--algebra if and only if X e R. Some of the relations between o--rings and o--algebras are given below: a. Let a be a o--ring which is not a o--algebra, let 03 be the smallest o--algebra containing GI, and set Cir = {E: E-- ' e 61 } . Then 63 = GI U 61' and .

a n a' = Ø.

Sec.

2]

Measurable Functions

223

b. If A is a measure on 61, define A on 63 by TtE = I.LE if E e ca and 61'. Then g is a measure on 63. c. Define A on 63 by AE = ,LLE if E e 61 and

AE = oo if E e

AE =

sup {AA: A C E, A e 01}

61'. Then A is also a measure on B. _ d. More generally, for any nonnegative extended real number 0 define ,up on CB by ,upE = AE if E e 61 and AgE = AE + 13 if E e 61'. Then /213 is a measure on 63. e. If v is any measure on 63 which agrees with A on Ca, then v = ,uo for some 3> O.

if E e

2

Measurable Functions

The concept of a measurable function on an abstract measurable space is almost identical with that for functions of a real variable. Consequently, those propositions and theorems whose proofs are essentially the same as those in Section 5 of Chapter 3 are stated without proof. The reader should verify that the proofs in Chapter 3 can be carried out in the abstract case. Throughout this section we assume that a fixed measurable space (X, 63) is given. 5. Proposition: Let f be an extended real-valued function defined on X. Then the following statements are equivalent:

i. {x: f(x) < a} e 63 for each a. 11. {X: f(x) _.<. a} e 63 for each a. ill. {X: f(x) > al e 63 for each a. iv. {x: f (x) a} e 63 for each a. (See Proposition 3.18.) Definition: The extended real-valued function f defined on X is called measurable (or measurable with respect to (33) if any one of the statements of Proposition 5 holds. 6. Theorem: If c is a constant and the functions f and g are measurable, then so are the functions f + c, cf,f + g, f • g, and f v g. Moreover, if (f7i ) is a sequence of measurable functions, then sup f,„ inf, f. , lim f„, and lim fn are all measurable. (See Theorems 3.19 and 3.20.)

224

Measure and Integration

[Ch. 11

By a simple function we mean as before a finite linear combination ,p(x) = E c zx E,(x) =1

of characteristic functions of measurable sets E. 7. Proposition: Let f be a nonnegative measurable function. Then there is a sequence ((,o7,) of simple functions with ço„±i > (ion such that f = lim ço„ at each point of X. If f is defined on a a--finite measure space, then we may choose the functions (p, so that each vanishes outside a set of finite measure. 8. Proposition: If bt is a complete measure and f is a measurable function, then f = g a.e. implies g is measurable.

(See Proposition 3.21.) The sets {x: f(x) < a} are sometimes called ordinate sets for f. They increase with a. The following two lemmas will be useful later. They state that, given a countable collection [Ba} of measurable sets which increase with a, we can find a measurable function f which nearly has these for ordinate sets in the sense that {x: f(x) <

c Ba c {x: f(x)

.

9. Lemma: Suppose that to each a in a countable set D of real numbers there is assigned a set Ba e 63 such that B a c Bo for a < #. Then there is a measurable extended real-valued function f on X such that f < a on B a and f > a on X — B a. Proof: For each x c X, let f(x) = inf [a: x c Ba}, where as usual inf 0 = c/o . If x c Ba, f(x) < a. If x Ba, x Bo for each # < a and so f (x) > a. Thus we have only to show that f is measurable. But for any real a, {x: f(x) < a} = U B. For if f(x) < a, then x c Bg (3
for some 13 < a by the definition of f. On the other hand, if x c Bp for some 13 < a, then f(x) < # < a. I 10. Lemma: Suppose that for each a in a countable set D of real numbers there is assigned a set Ba c 63 such that A(B, — Bo) = 0 for a G #. Then there is a measurable function f such that f < a a.e. on B a and f > a a.e. on X — Ba.

Sec. 3]

225

Integration

Proof: Let C = U B a — B o . Then oi<0

/I

C = 0. Let K, = B a U C.

If a G 0, we have BL B'o (B a — Bo) C = Ø. Thus by Lemma 9 there is a measurable f such that f < a on K and f > a on X BÇ . Thus f < a on Ba and f > a on X B a except for X e C. I Problems 10. Prove Proposition 7. [For each pair (n, k) of integers let 22n

En

{x: k 2— n < f(x) G (k

1)21, and set (pn,= 2—n

E0 kx.E.

11. Prove Proposition 8, and show that it is false if the word 'complete' is omitted. 12. Let (X, 03, ,u) be a measure space and (X, (B o, ,u 0) its completion. Then a function f is measurable with respect to (g o if and only if there is a function g measurable with respect to 03 such thatf -= g almost everywhere in the sense that there is a set E e 63 with AE = 0 and f = g on X — E. g(x)} to be in e3. Note that this does not require the set {x: f(x) 13. Show that the requirement that D be countable in Lemmas 9 and 10 is unnecessary. [Replace an uncountable D by a countable subset E which is dense in D.] 14. Show that if D is dense in R, then the function f in Lemma 9 is unique and the function f in Lemma 10 is unique to within equivalence; that is, any other function satisfying the conclusion of the lemma is equal to f almost everywhere. 15. Let D be the rationals and f the function in Lemma 9. Express the sets {x: f(x) G a}, {x: f(x) < {x: f(x) = in terms of the sets {B}.

3

Integration

Many definitions and proofs of Chapter 4 depend on only those properties of Lebesgue measure which are also true for an arbitrary measure in an abstract measure space and carry over to this case. If E is a measurable set and cp a nonnegative simple function, we define

E 1-1

n E),

226

Measure and Integration

[Ch. 11

where

E

iO(X) =

cix E,(x).

1

It is easily seen that the value of this integral is independent of the representation of io which we use. If a and b are positive numbers and io and tp nonnegative simple functions, then f aio btp = a f b ftP.

In Chapter 4 we first defined the integral for bounded measurable functions and then defined the integral of a nonnegative measurable function f to be the supremum of fg dj as g ranged over all bounded measurable functions which vanished outside a set of finite measure. Unfortunately, this definition is not quite suitable in the case of general measures, for, if (B =- {X, 0 } and AO = 0, AX = 00 , then we certainly want f i d = CO But the only measurable function g which vanishes outside a set of finite measure is g 0, and hence sup fg d = O. To avoid this difficulty, we define the integral of a nonnegative measurable function directly in terms of integrals of nonnegative simple functions. .

Definition: Let f be a nonnegative extended real-valued measurable Then f f cl A is the supremum function on the measure space (X, 63, of the integrals f çe di.4 as go ranges over all simple functions with 0 <
It follows immediately from this definition that f < g implies ff< fg and that f cf = c ff. In other respects, however, this definition is somewhat awkward to apply, since we are taking a supremum over a large collection of simple functions, and it is not apparent from the definition that f (f + g) = f f + fg. Consequently, we begin our treatment of the integral by first establishing the convergence theorems. These then enable us to determine the value of f f by taking the limit of f ion for any increasing sequence of simple functions which converge to f. We begin with Fatou's Lemma: 11. Theorem (Fatou's Lemma):

be a sequence of nonnegative measurable functions which converge almost everywhere on a set E to a function f Then Let

Sec. 3]

227

Integration

Proof: Without loss of generality we may assume that f,,(x) —f (x) for each X F E. From the definition of ff it suffices to show that, if io < lim fn. is any nonnegative simple function with (to f, then

L

If= o, then there is a measurable set A c E with i.LA = oo such that cp > a > 0 on A. Set A n = {x e E: fk (x) > a for all k > n} . Then (A n) is an increasing sequence of measurable sets = 00 . whose union contains A, since cp < lim f„,. Thus Jim Since f a,uA n, we have lirn f f. = 00 = f 0 B If JE cp < oo , then there is a measurable set A c E with 12,4 < 00 such that cp vanishes identically on E — A. Let M be the maximum of cp, let e be a given positive number, and set

J

E

.

--= Ix e E: fft(x) > (1 — e)io(x) for all k

Then (A n) is an increasing sequence of sets whose union contains A, and so (A — A n ) is a decreasing sequence of sets whose intersection is empty. By Proposition 2 lim /.4(A — A n) = 0, and so we can find an n such that ,u(A A k) < E for all k > n. Thus for k > n

f

E fk

k fk

(1 - E >(1-

fA k

E )

JE

-

LA:,0

IL, — fE so + m] • Hence -

E

UE (P M

Since E is arbitrary, lim

.I.

E`P'

12. Monotone Convergence Theorem: Let (fn ) be a sequence of nonnegative measurable functions which converge almost everywhere to a function f and suppose that fn < f for all n. Then f = lim f fn.

228

Measure and Integration

[Ch. 11

Proof: Sincef, < f, we have ff, < ff. Thus by Fatou's Lemma we have
We are now able to establish some of the standard properties of the integral. 13. Proposition: If f and g are nonnegative measurable functions and a and b nonnegative constants, then faf+bg= aff+big.

We have

If

0

with equality only if f = 0 a.e. Proof: To prove the first statement, let (io n) and (1,1,„) be increasing sequences of simple functions which converge to f and g. Then (a(pn, -I- bon ) is an increasing sequence of simple functions which converge to af bg. By the Monotone Convergence Theorem

f af bg = lim f &lion Ern (a Icp„ b f On ) = a

ff+ b f g.

Clearly f f O. If ff = 0, let A„ = {x: f(x) > 1/n} . Then f > X, and so I.LA„ = f XA. = O. Since the set where f> 0 is the union of the sets A„, it has measure zero. 1 Taking this proposition together with the Monotone Convergence Theorem gives us the following corollary: 14. Corollary: Let (fn ) be a sequence of nonnegative measurable functions. Then

f

j

f n=1

=

71=1

j

A nonnegative function f is called integrable (over a measurable set E with respect to j.i) if it is measurable and

f <

Sec. 3]

229

Integration

An arbitrary function f is said to be integrable if both f+ and integrable. In this case we define

JE f =

r are

JEfJE

Some of the properties of the integral are contained in the following proposition, whose proof is left to the reader. 15. Proposition: lf f and g are integrable functions and E is a measurable set, then

J (cif + c2g) = E

ii. if' !hi <

iii. if f

cif f+ c2f

g;

if!

and h is measurable then h is integrable; g a.e., then if g.

16. Lebesgue Convergence Theorem: Let g be integrable over E, and suppose that (fa) is a sequence of measurable functions such that on E

1.f.(x)i

g(x)

and such that almost everywhere on E f(x) f(x).

Then

fE f = lim fE Proof: Apply Fatou's Lemma to the sequences g + f„ and g — f„. 1

Problems 16. Prove Proposition 15. 17. Suppose that 1.1, is not complete, but that we define a bounded function f to be integrable over a set E of finite measure if sup f çdp = inf f to9f

E

E

for all simple functions 40 and %,G. Show thatf is integrable if and only if it is measurable with respect to the completion of A.

230

Measure and Integration

[Ch. 11

18. Let f be an integrable function on the measure space (X, G3, ,u). Show that given e > 0, there is a .5 > 0 such that for each measurable set E with ,uE < we have

fE f l < 19. Let 0'0 be a sequence of measurable functions. We say that converges in measure to f if given e > 0, there is an N such that Afx: if(x) — f(x)I >

<e

for n > N. We say that (fn ) is a Cauchy sequence in measure if given 0, there is an N such that

E>

Afx: if(x) — f rn(x)I >

4<e

for in, n > N. Show that a necessary and sufficient condition that a sequence (fn ) of measurable functions converge in measure to some function f is that it be a Cauchy sequence in measure. If (fn) converges tof in measure, then there is a subsequence (fnp ) which converges to f almost everywhere, and hence almost everywhere convergence may be replaced by convergence in measure in the convergence theorems. 20. Show that if f is integrable then the set {x: f(x) 0) is of a-finite measure. 21. a. Let (X, 03, i.L) be a measure space and g a nonnegative measurable function on X. Set pE f g du. Show that y is a measure on 63. b. Let f be a nonnegative measurable function on X. Then f f dy--= f fg d,u. [Hint: First establish this for the case that g is simple and then use the Monotone Convergence Theorem.] 22. A function f on a measure space (X, 03, ,u) is called locally measurable if the restriction off to each E in 03 with uE < 00 is measurable, that is, if f xE is measurable. a. Show that f is locally measurable if and only if it is measurable with respect to the a-algebra of locally measurable sets. b. If we define integration for nonnegative locally measurable functions f by taking ff to be the supremum of f go as ço ranges over all simple functions less than f, then ff = ff du, where u is the extension of /2 given in Problem 8d.

Sec. 4]

*4

231

General Convergence Theorems

General Convergence Theorems

In the preceding section we discussed the behavior of the integrals of a convergent sequence of functions. These were all integrals with respect to a fixed measure ,u. We can generalize by allowing the measure to vary also. Let (X, 63) be a measurable space and (un) a sequence of set functions defined on 63. We say that converges setwise to the set function AL if for each E e 63 we have 1.4E = lim With this notion we have the following two propositions, which generalize Fatou's Lemma and the Lebesgue Convergence Theorem.

17. Proposition: Let (X, 63) be a measurable space, (pa) a sequence of measures which converge setwise to a measure u, and ( fn ) a sequence of nonnegative measurable functions which converge pointwise to the function f. Then

ffd iu <11m f Proof: Setwise convergence of An, to

f

g

dp, = lim

implies that dti,„

for any simple function so. From the definition of ff du it suffices to prove that Fp tip < Jim ff„ 4, for any simple function so < f. Suppose fyo dp, < ci. Then cp vanishes outside a set E of finite measure. Let E be a positive number, and set E

(1 — E)(p(x) for all k > .

{x: fk (x)

Then (En) is an increasing sequence of sets whose union contains E, and so (E E„) is a decreasing sequence of measurable sets whose intersection is empty. Thus by Proposition 2 there is an m such that ,u(E Em) < E. Since ,u(E En) = lirn p k(E Em), we may choose n > m so that pk (E Em) < E for k > n. Since E Ek C E we have uk (E Ek) < E for k > n. Thus

f fk diak

(1 — e) Ek cp dp.k

L k fk dpk

(1 ?_.

e) f — e) fE

dAk

f E-E k S° dAk Me,

232

Measure and Integration [Ch. 11

where M is the maximum of so. Thus lim f fk d,uk _> fs so d,u — e[M + f so d]•

Since E was arbitrary,

fs so d,u 5_ lim f fk The case when

f so gu =

(Auk.

c/o is similarly handled. I

18. Proposition: Let (X, 63) be a measurable space and (du n) a sequence of measures on 63 which converge setwise to a measure du. Let (fn) and (gn) be two sequences of measurable functions which converge pointwise to f and g. Suppose I fnl < _ g n and that

lim f g n d,u.„, -= fg di < c. 0.

Then

Ern ff n 47, --= f f dp. Proof:

Apply Proposition 17 to the sequences gn + fn and

gn — fn. I Problems

23. Let (X, 63) be a measurable space and (du n) a sequence of measures on 63 such that for each E c 63, du n+ IE > E. Let ,uE = firn ME. Then A is a measure on 63. 24. Give an example of a decreasing sequence (A n ) of measures on a measurable space such that the set function ,u defined by AE = lim du nE is not a measure. 25. Let (X, 63) be a measurable space and (An ) a sequence of measures on 63 which converge setwise to a set function ,u. If du X < wo, then /.4 is a measure. 5 Signed Measures

In this section we consider some of the possibilities which may arise if a measure is allowed to take on both positive and negative values.

233

Sec. 5] Signed Measures

We first note that if A i and I/2 are two measures defined on the same measurable space (X, 63), then we may define a new measure /23 on (X, 63) by setting 123 (E)

= c l /2 i (E)

c2 > O.

c2/22(E)

What happens if we try to define a measure by

vE

= A.t i

E

— A2

E?

The first thing that may occur is that P is not always nonnegative, and this leads to the consideration of signed measures which we shall define later. A more serious difficulty comes from the fact that p is not defined when 1/ 1 E = 1/ 2 E = cr.,. For this reason we should have either /.4. 1 or 12 2 finite. With these considerations in mind we make the following definition: Definition: By a signed measure on the measurable space (X, 03) we mean an extended real-valued set function 11 defined for the sets of 63 and satisfying the following conditions: assumes at most one of the values ii. v(0) = O. j. y

P U

2=1

+ cc , —

Go.

E PE, for any sequence E, of disjoint measurable

2=1

sets, the equality taken to mean that the series on the right converges absolutely if P(U is finite and that it properly diverges otherwise. Thus a measure is a special case of a signed measure, but a signed measure is not in general a measure. We say that a set A is a positive set with respect to a signed measure y if A is measurable and for every measurable subset E of A we have vE > O. Every measurable subset of a positive set is again positive, and if we take the restriction of 11 to a positive set we obtain a measure. Similarly, a set B is called a negative set if it is measurable and every measurable subset of it has nonpositive p measure. A set which is both positive and negative with respect to i is called a null set. A measurable set is a null set if and only if every measurable subset of it has u measure zero. The reader should carefully note the distinction between a null set and a set of measure zero: while every null set must have measure zero, a set of measure zero may well be a union of two sets whose measures

234

Measure and Integration

[Ch. 11

are not zero but are negatives of each other. Similarly, a positive set is not to be confused with a set which merely has positive measure. We have the following lemmas concerning positive sets. Similar statements hold, of course, for negative sets. 19. Lemma: Every measurable subset of a positive set is itself positive. The union of a countable collection of positive sets is positive. Proof: The first statement is trivially true by the definition of a positive set. To prove the second statement, let A be the union of a sequence (A n ) of positive sets. If E is any measurable subset of A, set En = E n An n A-in • • • n A 1 .

Then En is a measurable subset of A n and so vEn > O. Since the E we have En are disjoint and E = H .... —n) vE -=

E

vEn > O.

n=1

Thus A is a positive set. I 20. Lemma: Let E be a measurable set such that 0 < vE < Go. Then there is a positive set A contained in E with vA > O. Proof: Either E itself is a positive set or it contains sets of negative measure. In the latter case let n i be the smallest positive integer such that there is a measurable set Ei c E with 1 vE i < — — ni

Proceeding inductively, let nk be the smallest positive integer for which there is a measurable set Ek such that k-1 ] Ek C

E —[t.i E, i=1

and

If we set

1 vEk < — • nk . A=E—UEk, k=1

Sec. 5]

235

Signed Measures

then E=

ALTO

k=1

Since this is a disjoint union, we have OD

vE =

PEk k=1

with the series on the right absolutely convergent, since

1 Thus E nk

vE

is finite.

converges, and we have nk —* 00. Since pEk < 0 and

we must have PA > O. To show that A is a positive set, let > 0 be given. Since nk —* 00, we may choose k so large that (nk — < E. Since vE > 0,

[k

Ac E— U j=1

A can contain no measurable sets with measure less than — (n k 1) — ', which is greater than —e. Thus A contains no measurable sets of measure less than — e. Since E is an arbitrary positive number, it follows that A can contain no sets of negative measure and so must be a positive set. I

21. Proposition (Hahn Decomposition Theorem): Let 11 be a signed measure on the measurable space (X, 03). Then there is a positive set A and a negative set B such that X = Au B and A nB = 0. Proof: Without loss of generality we may assume that - IL 00 is the infinite value omitted by v. Let x be the supremum of v A over all sets A which are positive with respect to v. Since the empty set is positive„X > O. Let (A t ) be a sequence of positive sets such that X = lim vA„ and set

A= ÛA,. z=.1

By Lemma 18 the set A is itself a positive set, and so X > A — A, C A and so p(A A,) > O. Thus vA = vA, v(A A,) >

Hence

vA >

X, and so vA

=

X, and X <

vA. But

236

Measure and Integration [Ch. 11

Let B = —A, and suppose E is a positive subset of B. Then E and A are disjoint and E u A is a positive set. Hence X>

v(E u A) = vE vA = vE +

X,

whence vE = 0, since 0 < X < oo. Thus B contains no positive subsets of positive measure and hence no subsets of positive measure by Lemma 20. Consequently, B is a negative set. I A decomposition of X into two disjoint sets A and B such that A is positive for y and B negative is called a Hahn decomposition for v. Proposition 21 states the existence of a Hahn decomposition for each signed measure. Unfortunately, a Hahn decomposition is not unique. If {A, B} is a Hahn decomposition for y, then we may define two measures v+ and y — with v = v+ — v — by setting v ±(E) = v(E n A) and v — (E) = —v(E n B). Two measures v 1 and v2 on (X, 63) are said to be mutually singular (in symbols v 1 _L v 2) if there are disjoint measurable sets A and B with X= A uB such that v i (A) = v2(B) = 0. Thus the measures v+ and v — defined above are mutually singular. We have thus etablished the existence part of the following proposition. The uniqueness part is left to the reader.

22. Proposition: Let v be a signed measure on the measurable space (X, go. Then there are two mutually singular measures VI - and v on (X, 63) such that v = v+ — v — . Moreover, there is only one such pair of mutually singular measures. The decomposition of y given by the proposition is called the Jordan decomposition of v. The measures v+ and v — are called the positive and negative parts (or variations) of v. Since y assumes at most one of the values + co and — oo, either v+ or v — must be finite. If they are both finite, we call y a finite signed measure. The measure tyl defined by 114E) = v +E v — E

is called the absolute value or total variation of v. A set E is positive for y if v —E = 0. It is a null set if 'PRE) = 0.

Sec.

5]

237

Signed Measures

Problems 26. a. Give an example to show that the Hahn decomposition need not be unique. b. Show that the Hahn decomposition is unique except for null sets. 27. Show that there is only one pair of mutually singular measures y+ and v — such that v = v+ — v —. [Hint: Show that any such pair determines a Hahn decomposition and apply the results of Problem 26b.] 28. Show that if E is any measurable set, then —v — E < vE < v+E and

IPEI

IPRE).

29. Show that if v 1 and y2 are any two finite signed measures, then so is av i ± Ov 2 , where a and 13 are real numbers. Show that

laP I = lal

IPI

and

Iiii + v 21

Iv 1 l + Iv 2 1,

where v < /.4 means vE < 1.4E for all measurable sets E. 30. We define integration with respect to a signed measure y by defining

1 f dv = f f dv+ — f f dv — . If If I < M,

JE

f dv '. MM(E).

Moreover, there is a measurable function f with I fi < 1 such that

JE"

dv = M(E).

31. a. Let ir. and v be finite signed measures. Show that there is a signed measure A A I/ which is smaller than 1.1 and v but larger than any other signed measure which is smaller than A and v. [Hint: m. A y = 102 + y lil — PD.] b. Show that there is a measure A V v which is larger than A and v but smaller than any other measure which is larger than A and v. Also A V v + /J. A I/ = A 4- v. c. If /I and v are positive measures, then they are mutually singular if and only if A A y = O. —

238

Measure and Integration

[Ch. 11

6 The Radon-Nikodym Theorem

Let (X, (33) be a fixed measurable space. If A and v are two measures defined on (X, (33), we say that A and v are mutually singular (and write p. 1 v) if there are disjoint sets A and B in 63 such that X = A u B and PA = AB = O. Despite the fact that the notion of singularity is symmetric in v and 12, we sometimes say that v is singular with respect to p.. The notion antithetical to singularity is absolute continuity. A measure P is said to be absolutely continuous with respect to the measure p. if v A = 0 for each set A for which AA = O. We use the symbolism v << su for v absolutely continuous with respect to p. In the case of signed measures A and v, we say v << ii. if Ivl << 1A1 and v 1 A if 1 vi 1 kul. Whenever we are dealing with more than one measure on a measurable space (X, (33), the term 'almost everywhere' becomes ambiguous, and we must specify almost everywhere with respect to p or almost everywhere with respect to v, etc. These are abbreviated a.e. [A] and a.e. [v]. If v << p and a property holds a.e. [A], then it holds a.e. [v]. Let p be a measure and f a nonnegative measurable function on X. For E in 63, set vE = JE f dy.

Then P is a set function defined on 63, and it follows from Corollary 14 that P is countably additive and hence a measure. The measure v will be finite if and only iff is integrable. Since the integral over a set of p.-measure zero is zero, we have v absolutely continuous with respect to A. The next theorem shows that, subject to o--finiteness restrictions, every absolutely continuous measure v is obtained in this fashion. 23. Theorem (Radon-Nikodym): Let (X, 63, ii,) be a g finite measure space, and let v be a measure defined on 63 which is absolutely continuous with respect to p. Then there is a nonnegative measurable function f such that for each set E in 63 we have -

vE — f B f di.4.

The function fis unique in the sense that i f g is any measurable function with this property then g = f a.e. [A].

Sec. 6]

239

The Radon-Nikodym Theorem

Proof: The extension from the finite to the o-finite case is not difficult and is left to the reader. Thus we shall assume that 1.‘ is finite. Then y — aA is a signed measure for each rational number a. Let (A a, Ba) be a Hahn decomposition for y — ail, and take A o = X, B o = 0. = Ba n A o. Thus (y ap)(Ba < 0 and Now Ba (y — 01.0(B,, — B o) 0. If 3 > a, these imply 4.13„ = 0, and so by Lemma 10 there is a measurable function f such that for each rational a we have f> a a.e. on A a and f a a.e. on Ba. Since Bo = 0, we may take f to be nonnegative. Let E be an arbitrary set in 0A, and set Ek =

E n (Bk +1 B k ), N

Then E = E. u U

Ek,

E. = E U Bk

N

and this is a disjoint union. Thus

k=0

PE

Since

Ek C Ek±i

n

Ak,

= PE. +

k=0

VEk•

k k+ 1 on we have — f N — N

— yE k < f f < N Rk

Ek

and so

k + 1 r,

k+ 1 k Since — p.Ek < vEk < N ihEk, we have N 1 vEk — — yEk < f f N F

vEk

1

— PEk. N

On E. we have f = oo a.e. If ihEc, > 0, we must have vEc, = oo, since (y — a/. E. is positive for each a. If jiE,0 = 0, we must have vE. = 0, since P A. In either case )

vE,, =f f E c., Adding this equality and our previous inequalities gives PE

—

N

1 f f d < vE + — E.

240

Measure and Integration

[Ch. 11

Since ALE is finite and N arbitrary, we must have vE = f f d,u. I The function f given by Theorem 23 is called the Radon-Nikodym

c . derivative of v with respect to A. It is sometimes denoted by [l diA

24. Proposition (Lebesgue Decomposition): Let (X, 63, /1) be a o--finite measure space and I, a o--finite measure defined on 63. Then we can find a measure v 0 which is singular with respect to and a measure v which is absolutely continuous with respect to ,1.1, such that = v o ± v 1 . The measures v 0 and v are unique. Proof: Since j and r are o--finite measures, so is the measure v. Since both A and v are absolutely continuous with X= A respect to x, the Radon-Nikodym Theorem asserts the existence of nonnegative measurable functions f and g such that for each E e

= fE f dX,

vE = fE g dX.

Let A f(x) > 01 and B = {x: f(x) = 0} . Then X is the disjoint union of A and B, while AB = O. If we define v o by p oE = v(E n B),

we have v o(A) = 0 and so v o i A. Let v i (E) = v(E n A) = fEn.4 g dX '

Then v = ro and we have only to show that v 1 << A. Let E be a set of A-measure zero. Then O = AE =

JE f dX,

and f= O a.e. [X] on E. Since f> 0 on A n E, we must have X(A n E) = O. Hence v(A n E) = 0, and so v i (E) = v(A n E) = O. This establishes the proposition except for the uniqueness, which is left to the reader. I

Sec. 6]

241

The Radon-Nikodym Theorem

Problems 32. a. Show that the Radon-Nikodym Theorem for a finite measure y implies the theorem for a a-finite measure y. [Hint: Decompose X into a countable union of sets X, of finite y-measure and apply the RadonNikodym Theorem to each X, to obtain f. Show f to have the required properties.] b. Show the uniqueness of the function f in the Radon-Nikodym Theorem. 33.

Radon-Wilcodym derivatives. Show that the Radon-Nikodym deriv-

ative [— du] has the following properties: a. If v << and f is a nonnegative measurable function, then dv f f dv = f f[]di.t. rd(vi + P2)1

b.L

rdvil

Ldp.] '

f dv21

L-dAd •

c. If v «Li «X, then

d. If v «ii and y

r du]

rdvirdui

[fix]

LdAd [dxf

v,

then

[di)] c-/Tc

-

[dy did

34. a. Show that if v is a signed measure such that v I p and v y, then I) = O. b. Show that if v 1 and v 2 are singular with respect to 11, then so is c ul c2I)2. c. Show that if 1i 1 and v 2 are absolutely continuous with respect to p. SO is CiVi +

C2 1) 2-

d. Prove the uniqueness assertion in the Lesbesgue decomposition. 35. Extend the Radon-Nikodym Theorem to the case of signed measures. 36. Complex measures. A set function v which assigns a complex number vE to each E in a a-algebra 63 is called a complex measure if

242

Measure and Integration [Ch. 11

v,(2S = 0 and for each countable disjoint union U E, of sets in 63 we have E1 )

=E

vE,

with absolute convergence on the right. a. Show that each complex measure may be expressed as v = Mi — 1.4 2 ± 44 3 — 1g4 where g 12 29 1.1 31 and /24 are finite measures. b. Show that for each complex measure v there is a measure A and a complex-valued measurable function (p with j(pi = 1 such that for each set E in 63, vE = [Hint: Apply the Radon-Nikodym Theorem to the measures gi with respect to the measure pi + /2 2 /23 /24 .] c. Show that the measure g in (b) is unique and that cp is uniquely determined to within sets of p measure zero. d. The measure A in (b) is called the total variation or absolute value of v and is sometimes denoted by iv!. Show that the results of Problem 30 hold for complex measures. e. If v is a complex measure with M(X) = 1 and v(X) = 1, then v is a positive real measure. 37. Alternate proof of the Radon-Nikodym Theorem. We can give a proof of the Radon-Nikodym Theorem which is independent of the Hahn decomposition theorem by using Proposition 10.28, which states that for each bounded linear functional F on a Hilbert space H there is a g in H such that F(f) = (f, g) for all ! in H. This proof is due to von Neumann. The details are outlined below: a. Let g and v be finite measures on a measurable space (X, 03) and set X = g + v. Define F(f) ff du. Then F is a bounded linear functional on L 2(X). b. The function g e L 2(X) such that F(f) = (f, g) has the property that 0 < g < 1, and g(E) = E g dX

v(E) = fE (1 — g) dX. c. If v

g, then X << g, and g = 0 only on a set of g-measure

zero. In this case X(E) = E [Hint: Consider Problem 21.]

g

Sec. 7]

The V' Spaces

d. If v

243

/2, then (1 — g)g-1 is integrable with respect to A and v(E) = (1 — g)g -1

JE

38. Use the following example to show that the hypothesis in the Radon-Nikodym Theorem that A is if-finite cannot be omitted. Let X --= [0, I], Oa the class of Lebesgue measurable subsets of [0, 1], and take 7, to be Lebesgue measure and A to be the counting measure on 63. Then i■ is finite and absolutely continuous with respect to A, but there is no function f such that vE = fE f du for all E c 03. At what point does the proof of Theorem 23 break down for this example?

39. Decomposable measures. Let (X, 63, u) be a measure space. A collection {Xa} of disjoint measurable subsets of X is called a decomposition for is if tiXa < oo for each a and if piE — 0 for every measurable set E such that ,u(E n Xa) = 0 for all a. A measure ti is called decomposable if it has a decomposition. a. If {,Va} is a decomposition for A, and if E is any measurable set, n E), where E is meant in the sense of Problem 2.20. b. If {X„} is a decomposition for a complete measure A, then f is if and only if the restriction off to Xa is measurable locaymesurb for each a. Iff is a nonnegative locally measurable function of X, then

then pE =

Jfdis= E a

Xa

f

c. Let i; be absolutely continuous with respect to A, and suppose that there is a collection {Xa} which is a decomposition for both A and v. Then there is a nonnegative, locally measurable real-valued function f such that for every measurable set E we have

vE = d. The conclusion of (c) is still valid if, instead of assuming {lIa} to be a decomposition for we merely assume that if E e 63 and v(E n Xa) = 0 for all a then pE = 0.

7 The LP Spaces

If (X, 03, pt) is a measure space, we denote by LP(A) the space of all measurable functions on X for which f f1P cl/.4 < 00, considering two functions in LP to be equivalent if they are equal almost everywhere. As in Chapter 6 we define L'(1) to be the space of bounded

244

Measure and Integration

[Ch. 11

measurable functions. For 1 < p < co we set

Ilfllp = {f If I' dil } '', and for p = 00 we set

If

= ess sup f i.

Note that the space L' (y) depends on the choice of p, to determine the norm and the classes of equivalent functions but that this only requires knowing what the sets of measure zero are. The Holder and Minkowski inequalities and the Riesz-Fischer Theorem follow just as in Chapter 6, and we summarize them in the

following theorem: 25. Theorem: For 1
_< oo

e Lq(,(2),

with !

and if f c LP(m), g

the spaces LP 04 are Banach spaces, 1 - = 1, then fg c D(/.4) and

flfgldm The following proposition, whose proof is left to the reader, is a version of Littlewood's second principle:

26. Proposition: Let f e LP(A), 1 < p < 00. Then, given e > 0, there is a simple function („o vanishing outside a set of finite measure such that {jf — cp{1, < c.

It follows from the Wilder inequality that each g e D defines a linear functional F on LP by setting F(f) = f fg

and it is not difficult to show that IFII = Ile. Our goal in the remainder of this section is to establish the converse (Riesz Representation Theorem), which states that each linear functional on LP(/.1.) for 1 < p < co is of this form and, if p, is a-finite, each linear functional on L1 (p.) is of this form. We begin by establishing a lemma.

Sec. 7] The

LP

Spaces

245

27. Lemma: Let (X, 08, /1) be a finite measure space and g an integrable function such that for some constant M,

f gcp dy, for all simple functions (p. Then g

E

L.

Proof: Assume p > 1, and let (V.,i ) be a sequence of nonnegative simple functions which increase to le. Set = (11.,)9 sgn g.

Then çon is a simple function, and

=fr. d„} P.

4.112,=

1+1 Since q;„g- > koni 10.1 9 = /nr q = On, we have

f g 11/1 11çCn 11P

.41

1

itn C1/4

1 1 Since 1 — — = — , P

If

ifrn chilq

or

1 V/i,

<

Mq,

and by the Monotone Convergence Theorem

J. I g

d

W.

The case p = 1 is left to the reader. I

We shall also use the following lemma, whose proof is left to the reader.

246

Measure and Integration

[Ch. 11

28. Lemma: Let (En ) be a sequence of disjoint measurable sets, and for each n let fn be a function in LP(1

side E. Set f = Ê fn . Then f e LP if and only n=1

this case f =

EL,

if E iimP < ..

In

in LP; that is, n

f

—

E

—> 0

fi P

and

Il flIP = n=i i IlfnlIP. 29. Riesz Representation Theorem: Let F be a bounded linear functional on LP(g) with 1

1 1 there is a unique element g in V, where - + - = 1, such that q P F(f) = f fg dg. We have also 114 = lig. Proof: Let us first consider the case of a finite measure ,u. Then every bounded measurable function is in LP 0.4). Define a set function y on the measurable sets by setting

vE = F(X E ). If E is the union of a sequence (En ) of disjoint measurable sets, let )-- an...4. x Then by Lemma 28 and a, = sgn F(X4), and set f = _a the boundedness of F we have

i 11)En] = F(f) < co n=1

and

E

n=1

PE. = F(X E ) = vE.

Thus y is a signed measure, and so by the Radon-Nikodym Theorem there is a measurable function g such that for each measurable set E we have vE = f g dy. Since y is always finite, g is integrable. E

Sec. 7]

The LP Spaces

247

If ye, is a simple function, the linearity of F and of the integral imply that F(ç) = f sog dtt. Since the left side is bounded by VII we have g c D by Lemma 27. Let G be the bounded linear functional defined on LP by G(f) = f fg Then G — F is a bounded linear functional which vanishes on the subspace of simple functions. By Proposition 26 the simple functions are dense in LP, and so G — F = O. Thus for all f c LP F(f) =- f fg and it is readily verified that In = uGh = The function g must determine a unique element of Lg, for if g1 and g2 determine the same functional F, then g l — g 2 must give the zero functional, and so Ilgi - g2 11, = O. Thus g 1 = g2 a.e. To extend the theorem to the a-finite case, let (X„) be an increasing sequence of measurable sets of finite measure whose union is X. The theorem for finite measure spaces implies that for each n there is a function g„ in Lq which vanishes outside X„ such that F(f) =f fg„ for all f e LP which vanish outside Xn . Moreover, g,, < ilf11. Since any function g„ with this property is uniquely determined on X„ except for changes on sets of measure zero and since gn+1 also has this property, we may assume g„ +i = g„ on Xn . For x e Xn set g(x) = g„(x). Then g is a well-defined measurable function and ign i increases pointwise to g Thus by the Monotone Convergence Theorem d,u =-- limf g,. r d

and g e L.

248

Measure and Integration

[Ch. 11

If f e LP, let fn = f on Xn, and f. = 0 otherwise. Then f. --> f pointwise and in L. Since j fgl is integrable and I f,igl < I fgl, the Lebesgue Convergence Theorem gives ffgd = 1imff n gdi.4 = lim f fi gn = lim F(f„,) = F(f). I If p = I, the requirement that be a-finite is necessary. Some extensions and counterexamples are given in Problems 45 and 46. If p > I, then o--finiteness is not required: 30. Theorem: Let F be a bounded linear functional on LP() with 1

F(f) = ffgd. We have VII Proof: It follows from the previous theorem that if E is any measurable set of o--finite measure then there is a unique gE e Lq, vanishing outside E, such that F(f) = fgEfdil for each f e L P which vanishes outside E. The uniqueness of gE implies that if A c E then gA = gE a.e. on A. For each E of o--finite measure set X(E) = f Ig E lq di.t. Then for A c E, we have x(A) < x (E) < Let be a sequence of sets of o--finite measure such that x(E) tends to the maximum value m of X. Then H = U En is a set of a-finite measure, and by the monotonicity of X we have X(H) = m. Let g be defined to be gH on H and 0 elsewhere. Then g e L. If E is any set of a-finite measure containing H, then gE = gH a.e. on H. But f ig = x(E) < x(H) = f ig'j ig , and so gE = 0 a.e. on E — H. Thus gE = g almost everywhere on E.

Sec. 7]

249

The LP Spaces

If f e D', then the set N = f(x) 0} measure, and so is the set E = N u H. Thus F(f) = f fgE d

is a set of a-finite

= f fg d.

The equality of ilFil and lie, follows as in the previous theorem. I

Problems 40. Prove Proposition 26. 41. Prove Lemma 27 for the case p = 1. 42. Prove Lemma 28. 43. For g e V, let F be the linear functional on Ly defined by

F(f) = f fg

Show that VII = 44. a. Let ,u be the counting measure on a countable set X. Show that (p.) = 12) . b. Discuss the spaces P(X) = Cu.), where ,u is the counting measure on a not necessarily countable set X. 45. Let (X, 03, 1.0 be a decomposable measure space (cf. Problem 39). a. Show that for each bounded linear functional F on L 1 (2) there is a bounded locally measurable function g such that F(f) = ffg b. Show that the dual space to L 1 (j2) is the space L(M), where pt is the saturation of j given in Problem 8e. 46. Let A and B be uncountable sets with different numbers of elements, x = a} is called a vertical and let X = A X B. A set of the form {(x, line and a set of the form {(x, y): y = b} a horizontal line. Let 61 be the collection of subsets E of X such that for every horizontal or vertical line L either E n L or E n L is countable. Then 63 is a a-algebra. Let I.LE be the number of horizontal and vertical lines L for which E n L is countable and vE the number of horizontal lines with E n L countable. Then p, and y are measures on 03, and we can define a bounded linear functional F on L 1 (t) by setting F(f) = ff dv. There is no locally measurable function g such that F(f) = ffg

12

Measure and Outer Measure

In this chapter we first consider some of the ways in which a measure can be defined on a a-algebra. In the case of Lebesgue measure we defined measure for open sets and used this to define outer measure, from which we obtain the notion of measurable set and Lebesgue measure. Such a procedure is feasible in general. In the first section we discuss the process of deriving a measure from an outer measure, and in the second section we derive an outer measure from a measure which is defined only on an algebra of sets. The remainder of the chapter is devoted to some applications of this process.

1

Outer Measure and Measurability

By an outer measure ti.* we mean an extended real-valued set function defined on all subsets of a space X and having the following properties: i. i-L * 0

= O.

ii. A C B = 1.1*A < ju*B. co

C.

iii. E CU E, 12*E < E ,*Ei. z=i

The second property is called monotonicity and the third countable subadditivity. In view of (i) finite subadditivity follows from (iii). 250

Sec. 1]

251

Outer Measure and Measurability

Because of (ii), property (iii) can be replaced by

u

E=

E„ E, disjoint

teE <

-1

2--

The outer measure y* is called finite if 12*X < 00. By analogy with the case of Lebesgue measure we define a set E to be measurable with respect to A* if for every set A we have I.L*(A) = ,e (A n E)

,*(A n

Since /.4* is subadditive, it is only necessary to show that

e(A) ,4*( 4 n E) A*(A

n

for every A in order to prove that E is measurable. This inequality is trivially true when 1.1*A = 00, and so we need only establish it for sets A with eA finite. 1. Theorem: The class icB of A*-measurable sets is a a--algebra. If P. is p.* restricted to GI, then Z is a complete measure on CB.

Proof: Trivially, the empty set is measurable. The symmetry of the definition of measurability in E and E shows that P - . is measurable whenever E is. Let E l and E2 be measurable sets. From the measurability of E2, 12 * ( A)

11* (A

n E2)

n E2)

and A* 0) = 1.1*(A

n

E2)

+ Al*(A n E 2 n E1) + ih*(A n E1 n E 2)

by the measurability of E l . Since ' A n [E 1 u

= A n [

u [A n E1 n E2],

we have A* (A

n

[E1

u E2]) < I.L*(A n E2) + /.1*(A n E 2 n E1)

by subadditivity, and so 1.4*A > 412*(A n [E1 u E2]) +

n E1 n E2).

This means that E 1 u E2 is measurable, since —(E 1 u E2) = E1 n

E 2.

252

Measure and Outer Measure

[Ch. 12

Thus the union of two measurable sets is measurable, and by induction the union of any finite number of measurable sets is measurable, showing that 63 is an algebra of sets. H E where EZ is a disjoint sequence of Assume that E measurable sets, and set Gn

= t-i U Et-

Then G„ is measurable, and A * (A) = 1.4* (A

n

bt*(A

since E c O - n . Now Gn n En = the measurability of En we have /I

n

?_

n

+ t.t*(A n

t.t*(A n

En and G„ nE„ = G„_1, and

= t.t*(A n E„)

A*(A

by

n Gn _ 1).

By induction /.4 * (A

n Gn ) = t tt*(A n Ei), t=i

and so tt*(A)

A*(A n

i.t*(A

n E1)

tt*(A n E),

tt*(A n E) -

since AnEc

U

t=i

(AnE1).

Thus E is measurable. Since the union of any sequence of sets in an algebra can be replaced by a disjoint union of sets in the algebra, it follows that 63 is a a-algebra. We next demonstrate the finite additivity of A. Let El and E2 be disjoint measurable sets. Then the measurability of E2 implies that A(E 1 u E2) = tt*(Ei u E2)

u il*E2 + 1-4* ([-E1

n E2)

Finite additivity follows by induction.

11*([Ei

u

n E2)

Sec. 2]

253

The Extension Theorem

If E is the disjoint union of the measurable sets {Ei} , then rt(E)

(

=

i=1

ti=1

and so T4(E)

i=1

rt(E1).

But A(E)

i=1

FL(Ei)

by the subadditivity of a*. Hence g is countably additive and thus a measure since it is nonnegative and AO = A*0 = O. I

Problems 1. Prove the completeness of g. 2. Show that g is saturated. 3. Assume that (Ei) is a sequence of disjoint measurable sets and E = U E. Then for any set A we have ii*(A n E) =

ii*(A n Ei).

2 The Extension Theorem By a measure on an algebra we mean a nonnegative extended realvalued set function defined on an algebra et of sets such that: i. A(0) = 0; ii. if (A i ) is a disjoint sequence of sets in et whose union is also in a, then

(Û

A

=

Thus a measure on an algebra et is a measure if and only if et is a a-algebra. The purpose of this section is to show that, if we start with a measure on an algebra et of sets, we may extend it to be a measure defined on a a-algebra 63 containing a. We shall do this by using

254

Measure and Outer Measure

[Ch. 12

the measure on the algebra to construct an outer measure ,u* and show that the measure r4 induced by u* is the desired extension of u. The process by which we construct u* from 1.2 is analogous to that by which we constructed Lebesgue outer measure from the lengths of intervals: We define = inf

E juA,

(1)

where (A,) ranges over all sequences from a such that E C U A. 2= 1 ,u*. Wefirstablhom cnerig 2. Lemma: If A e a and if (41,) is any sequence of sets in a such that A C U A

,

then ,uA <

E

Br, --- A

A„

Proof: Set n

n 2,i_i n • • • n 71 1.

Then Br, e a and Bfl, C An . But A is the disjoint union of the sequence (BO, and so by countable additivity cc

=

E

03,„<

n=1

E n =1

3. Corollary: If A e a ,u*A = A. ,

4. Lemma: The set function ,u* is an outer measure. Proof: Since A * is clearly a monotone nonnegative set function defined for all sets and p. * 325 = CI, we have only to show that it is

countably subadditive. Let E C U E,. If u*Ei = 00 for any i, we have

E <E,u*E,= 00. If not, given E > 0, there is for each

i a sequence (A„);°_, of sets in a such that E, c U A„ and ,=1

E R A„

< ,u*E,± E/2'.

=1

Then teE

< E RA„ < E /2*Ei ± 2)

2=1

E.

Sec. 2]

255

The Extension Theorem

Since E was an arbitrary positive number,

and

i.t*

is

I

subadditive.

5. Lemma: If A c a then A is measurable with respect to

e.

Proof: Let E be an arbitrary set of finite outer measure and e a positive number. Then there is a sequence (A i) from a such that E c U A, and E IA i < ,u* E + e.

By the

additivity

of 1.1 on a we have 4,4 2) = il(Az

n A) + ,u(A, n M.

Hence co

CO

,a*E + € >

E t (A , n A)

-1-

E

A(Ai n ,i)

> ,.,*(E n A) -1- u*(E n M,

since EnA C U (A, n A)

and EnA- CU (A i n M.

Since e was an arbitrary positive number, ,u*E > ,21, ( E n A) -1- A*(E n A),

and A is measurable.

1

The outer measure ii* which we have defined is called the outer measure induced by IA. For a given algebra a of sets we use a, to denote those sets which are countable unions of sets of a and use a,s to denote those sets which are countable intersections of sets in a,.

256

Measure and Outer Measure

[Ch. 12

6. Proposition: Let ,u be a measure on an algebra a, ,u* the outer measure induced by IA, and E any set. Then for e > 0, there is a set A e a, with E C A and

,u*A < *E +E There is also a set B e a,3 with E c B and ,*E = Proof: By the definition of such that E c U A, and

E

A*

there is a sequence (A i ) from a

A. < pc* E

E.

s =1

Set A = U A. Then pc*A < Ept*A i EgA i. To prove the second statement, we note that for each positive integer n there is a set A, in a, such that E C A„ and ,u*A. < ,u*E ,•;. Let B = fl A1. Then B ct,s and E c B. Since B C A., ,u* B < ,u* A „ < ,u* E Since n is arbitrary, *B < ,u* E. But E c B, and so ,u*B > A *E by monotonicity. Hence bz * B = 12*E.1 If we apply this proposition in the case that E is a measurable set of finite measure, we see that E must be the difference of a set B in a,3 and a set of measure zero. This gives us the structure of the measurable sets of finite measure, and the next proposition extends this to the a-finite case. It can be considered a generalization of the first principle of Littlewood. 7. Proposition: Let be a a-finite measure on an algebra a, and let A * be the outer measure generated by pt. A set E is ,u* measurable if and only if E is the proper difference A — B of a set A in a,3 and a set B with A*B = O. Each set B with 1.4*B = 0 is contained in a set C in cto.6 with ,u*C = O.

Proof: The "if" part of the proposition follows from the fact that each set in a,3 must be measurable, since the measurable sets form a a-algebra, while each set of A *measure zero must be measurable, since Ti is complete. To prove the "only if" part of the proposition, let IX11 be a countable disjoint collection of sets in a with 12X, finite and X = U X. If E is measurable, then E is the disjoint union of the

Sec. 21

257

The Extension Theorem

measurable sets E, = X, n E. By Proposition 6 we can find for each positive integer n, a set A„„ in a, such that E, C A„, and 1 n2z Set

= U An,

Then E C A n, and A„ E C U

[An,

—

E7]. Hence

t=1

g.(A„ E)

u(A ni E,) 7=

E Since An e

a,, the set A

=

fl

n=1

1

1

n2'

n

A,, is in a, and for each n

A — E c A n — E.

Hence (A,

<1

Since this holds for each positive integer n, we must have ,u(A E) = O. I

We summarize the results of this section in the following theorem. 04,

8. Theorem (Carathéodory): Let y be a measure on an algebra and A * the outer measure induced by m. Then the restriction rt of

to the y*-measurable sets is an extension of A to a o--algebra containing a. If u is finite (or cr-finite) so is u. If y is-finite, then Ti is the only measure on the smallest a-algebra containing a which is an extension of A. 1.1 *

Proof: The fact that g is an extension of y from a to be a measure on a a-algebra containing a follows directly from Corollary 3, Lemma 5, and Theorem 1, and it is readily verified that r.k is finite or o--finite whenever 12 is.

258

Measure and Outer Measure

[Ch. 12

To show the unicity of z when I.4 is o--finite, we let CB be the smallest o--algebra containing a and ki some measure on G3 which agrees with ,u on a. Since each set in a, can be expressed as a disjoint countable union of sets in a, the measure 1.1: must agree with 1- 7 on a,. Let B be any set in CB with finite outer measure. Then by Proposition 6 there is an A in a, such that B C A and i.t.*A < ,u*B +

E.

Since B c A,

jo < ra = ,.,*A < ,u*B +

E.

Since E is an arbitrary positive number, we have

03 < 12* B for each B Since the class of sets measurable with respect to ,L* is a o--algebra containing a, each B in eii must be measurable. If B is measurable and A is in a, with B c A and 1.4*A < I.4*B + E, then

,u* A = ju*B + , L * (A ,-.., B), and so

B) <

E,

if 1.4*B < oo. Hence A* B < ,u* A = AA

- _. A B + IAA — B)

< AT +

E.

Since E is arbitrary, we have and so

i.t*B = it- B. If 1.4 is a o--finite measure, let { X,} be a countable disjoint collection of sets in a with X — U X., and ,uX, finite. If B is any set in CB, then

B= U (X, n B)

Sec. 2]

259

The Extension Theorem

and this is a countable disjoint union of sets in 63, and so we have JIB =

E p(X, n B)

pB =

E Tu- (X, n B).

and

Since e(X, n B) <

co,

we have

Ta(X, n B) = p(X, n B). I

This extension procedure not only extends p, to a measure on the smallest o--algebra 63 containing a, but also completes and saturates the measure. If p. is o--finite, the extension to 63 is already saturated, and the extension to the e-measurable sets is merely the completion of A. If p is not a-finite, then the extension to e-measurable sets also saturates i. It should be observed that in this case the extension of p. to (33 need not be unique (Problem 4), although any extension JI must agree with for each set B of 63 for which 1.7B < 00, and we always have JIB < A* B. We shall return to the question of extension and unicity in Sections 5 and 6.

It is often convenient to start with a set function on a collection e of sets having less structure than an algebra of sets. We say that a collection e of subsets of X is a semialgebra of sets if the intersection of any two sets in e is again in e and the complement of any set in e is a finite disjoint union of sets in e. If e is any semialgebra of sets, then the collection a consisting of the empty set and all finite disjoint unions of sets in e is an algebra of sets which is called the algebra generated by e. If p is a set function defined on e, it is natural to attempt to define a finitely additive set function on a by setting =

E

whenever A is the disjoint union of the sets Ei in e. Since a set A in a may possibly be represented in several ways as a disjoint union of sets in e, we must be certain that such a procedure leads to a unique value for IA. The following proposition gives conditions under which this procedure can be carried out and will give a measure on the algebra a.

260

Measure and Outer Measure

[Ch. 12

9. Proposition: Let e be a semialgebra of sets and u a nonnegative set function defined on e with bk = 0 (if 0 E e). Then bt has a unique extension to a measure on the algebra a generated by e if the following conditions are satisfied: i. If a set C in e is the union of a finite disjoint collection {C il of sets in e, then AC = ii. If a set C in e is the union of a countable disjoint collection {C,} of sets in e, then 1.4C < Problems 4. Let X be the set of rational numbers and a the algebra of finite unions of intervals of the form (a, b] with i.t(a, b] = oo and /.425 = O. The extension of p. to the smallest a-algebra containing a is not unique. 5. Prove Proposition 9 by showing: a. Condition (i) implies that, if A is the union of each of two finite disjoint collections {C,) and {D,} of sets in e, then EpC, = A DJ. [Hint: pC, = Ace n Din

E/

b. Condition (ii) implies that p, is countably additive on a [for finite additivity and monotonicity already imply the reverse inequality]. 6. Let a be a collection of sets which is closed under finite unions and finite intersections; an algebra of sets, for example. a. Show that a, is closed under countable unions and finite intersections. b. Show that each set in a„,5 is the intersection of a decreasing sequence of sets in a,. 7. Let g be a finite measure on an algebra a, and m* the induced outer measure. Show that a set E is measurable if and only if for each e > 0 there is a set A c as A C E, such that p*(E — A) < e. 8. If we start with an outer measure ,u* on X and form the induced measure /-.7, on the p*-measurable sets, we can use 72 to induce an outer measure m,+. a. Show that for each set E we have p+E > p*E. b. For a given set E we have p+E = p*E if and only if there is a p*-measurable set A D E with p*A = c. An outer measure which satisfies the criterion in (b) for each set E is called a regular outer measure. Show that every outer measure induced by a measure on an algebra is regular. d. Let X be a set consisting of two points. Construct an outer measure on X which is not regular. ,

Sec. 3]

261.

The Lebesgue-Stieltjes Integral

*3 The Lebesgue-Stieltjes Integral

Let X be the set of real numbers and 63 the class of all Borel sets. A measure y defined on 6a and finite for bounded sets is called a Borel measure (on the real line). To each finite Borel measure we associate a function F by setting

F(x) = m( — 00 , x].

The function F is called the cumulative distribution function of j and is real-valued and monotone increasing. We have p,(a, b] = F(b) — F(a). 1 — ], Proposition 11.2

Since (a, b] is the intersection of the sets (a, b implies that

(a, b] = hm

1

(a, b

and so F(b) = Ern F

1

= F(b+).

Thus a cumulative distribution function is continuous on the right. Similarly, A{b} = lim n

co

(I) — , bl n

lim F(b) F (I) 7Z

1

— n

= F(b) — F(b—). Hence F is continuous at b if and only if the set {b} consisting of b -n], we have alone has measure zero. Since 0 =

n

lim F(n) = 0, n

—co

and hence lim F(x) = 0, —

because of the monotonicity of F. We summarize these properties in the following lemma:

262

Measure and Outer Measure

[Ch. 12

10. Lemma: If A is a finite Borel measure on the real line, then its cumulative distribution function F is a monotone increasing bounded function which is continuous on the right. Moreover, lim F(x) = O. 00

Suppose that we begin with a monotone increasing function F which is continuous on the right. Then we shall show that there is a unique Borel measure A such that (2)

p.(a, b] = F(b) — F(a)

for all intervals of the form (a, b], where we define F(00) = lirn F(x) 2 -000

and F(— Go) = lim F(x). We begin with the following lemma, whose

proof is left to the reader (Problem 9): 11. Lemma: Let F be a monotone increasing function continuous on the right. If (a, b]

C

U

(a„ b0], then

0=1

F(b) — F(a) <

F(b0) — F(a,). 1=-1

If we let e be the semialgebra consisting of all intervals of the form (a, b] and set 14(a, b] = F(b) — F(a), then A is easily seen to satisfy condition (i) of Proposition 9, and since Lemma 11 is precisely the second condition, we see that A admits a unique extension to a measure on the algebra generated by e. By Theorem 8 this A can be extended to a a--algebra containing e. Since the class 63 of Borel sets is the smallest a--algebra containing e, we have an extension of A to a Borel measure. The measure A is a--finite, since X is the union of the intervals (n, n 1] and each has finite measure. Thus the extension of A to 03 is unique, and we have the following proposition: 12. Proposition: Let F be a monotone increasing function which is continuous on the right. Then there is a unique Borel measure A such that for all a and b we have p.(a, b] = F(b) — F(a).

13. Corollary: Each bounded monotone function which is continuous on the right is the cumulative distribution function of a unique finite Borel measure provided F(— oo) = O.

Sec. 3]

The Lebesgue-Stieltjes Integral

263

If ça is a nonnegative Borel measurable function and F is a monotone increasing function which is continuous on the right, we define the Lebesgue-Stieltjes integral of ça with respect to F to be

f cp dF = f cp dp, where A is the Borel measure having F as its cumulative distribution function. If cp is both positive and negative, we say that it is integrable with respect to F if it is integrable with respect to A. If F is any monotone increasing function, then there is a unique function F* which is monotone increasing, continuous on the right, and agrees with F wherever F is continuous on the right (Problem 10), and we define the Lebesgue-Stieltjes integral of cp with respect to F by f (p dF = f cp dF*. b

If F is a monotone function, continuous on the right, then f (p dF L agrees with the Riemann-Stieltjes integral whenever the latter is defined. The Lebesgue-Stieltjes integral is only defined when F is monotone (or more generally of bounded variation as in Problem 11c), while the Riemann-Stieltjes integral can exist when F is not of bounded variation, say when F is continuous and (p is of bounded variation.

Problems 9. Prove Lemma 11. [Choose e > 0. By the continuity on the right of F, choose ?i t > 0 so that F(b, + ni) < F(b) I € 2—i , and choose 3 > 0 so that F(a + (5) < F(a) I e. Then the open intervals (a„ b, + ni) cover the closed interval [a ± (5, b], and the proof proceeds like that of Proposition 3.1. A little extra care must be taken when (a, b] is infinite.] 10. Let F be a monotone increasing function, and define - -

- -

F*(x) = Jim F(y). Then F* F*is a monotone increasing function which is continuous on the right and agrees with F wherever F is continuous on the right. We have (F*)* = F*, and if F and G are monotone increasing functions which agree wherever they are both continuous, then F* = G*.

Measure and Outer Measure

264

[Ch. 12

11. a. Show that each bounded function F of bounded variation gives rise to a finite signed Borel measure y such that v(a, b] = F(b+) — F(a+). b. Compare Theorem 5.4 with the Jordan decomposition of v. c. Extend the definition of the Lebesgue-Stieltjes integral fclo dF to functions F of bounded variation. d. Show that if kpj < M and if the total variation of F is T, then if/PI _< MT.

12. a. Let F be the cumulative distribution function of the Borel measure y, and assume that F is continuous. Then for any Borel set E contained in the range of F, we have mE = v[F-1 (E)], with m Lebesgue measure. [Hint: This is true for intervals, and the uniqueness part of Theorem 8 can be used to derive its truth in general.] b. Generalize to the case of discontinuous cumulative distribution functions. 13. Let F be a continuous increasing function on [a, b] with F(a) = c, F(b) = d, and let ço be a nonnegative Borel measurable function on [c, d]. d

Then f cp(F(x)) dF(x) a

f ,,o(y) dy. [Hint: Use Problem 12a to take

care of the case when (F is a characteristic function and generalize first to simple ço and then to general ço.]

14. a. Show that a measure ih is absolutely continuous with respect to Lebesgue measure if and only if its cumulative distribution function is absolutely continuous.

b. If u is absolutely continuous with respect to Lebesgue measure, then its Radon-Nikodym derivative is the derivative of its cumulative distribution function. c. If F is absolutely continuous, then

f f dF = f fF' dx.

4

Product Measures

Let (X, a, ,u) and ( Y, 63, v) be two complete measure spaces, and consider the direct product X X Y of X and Y. If A c X and B c Y, we call A X Ba rectangle. If A e a and B e 63, we call A XBa

Sec. 4]

265

Product Measures

measurable rectangle. The collection semialgebra, since

61,

of measurable rectangles is a

(A x B) n (C X D) = (A n C) X (B n D) and —(A

x

B)

(Al

B)u (A X b) u (21 X h).

If A X B is a measurable rectangle, we set X(A X B) = ,uA • vB.

14. Lemma: Let {(A, X .81) } be a countable disjoint collection of measurable rectangles whose union is a measurable rectangle A x B. Then X(A X B) = x(Az X B 1).

E

Proof: Fix a point x e A. Then for each y e B, the point (x, y) belongs to exactly one rectangle A, X B,. Thus B is the disjoint union of those B, such that x is in the corresponding A,. Hence

E vB,- X

1 (x) = vB • X A (x),

since I/ is countably additive. Thus by the corollary of the Monotone Convergence Theorem (11.14), we have

L

fB 1 x 41 d = f u(B) • XA d,u

or

E vB, • ALA,

= vB • ,uA. I

The lemma implies that X satisfies the conditions of Proposition 9 and hence has a unique extension to a measure on the algebra at' consisting of all finite disjoint unions of sets in act. Theorem 8 allows us to extend X to be a complete measure on a a--algebra 8 containing at. This extended measure is called the product measure of y and y and is denoted by X v. If and if are finite (or a--finite), so is X y. If X and Y are the real line and and y are both Lebesgue measure, then X y is called two-dimensional Lebesgue measure for the plane. The purpose of the next few lemmas is to describe the structure of the sets which are measurable with respect to the product measure

266

Measure and Outer Measure (Ch. 12

X v. If E is any subset of X X Y and x a point of X, we define the x cross-section Ex by = (y: (x, y) EE},

and similarly for the y cross-section for y in Y. The characteristic function of Ex is related to that of E by X E.(y) = X E(x, y).

We also have collection {Ea}.

(f) x

= — (Ex) and (U Ea)x = U (Ea)x for any

15. Lemma: Let x be a point of X and E a set in 61,8. Then Ex is a measurable subset of Y. Proof: The lemma is trivially true if E is in the class 61 of mea-

surable rectangles. We next show it to be true for E = U E„

where each

E,

E

in

131,.

Let

is a measurable rectangle. Then

z=1

xE(y) = xE(x, y) = sup x E,(x, y)

= sup X ( E0 (y).

Since each E, is a measurable rectangle, X (E,4 (y) is a measurable function of y, and so XE z must also be measurable, whence Ex is measurable. Suppose now that E = E, with E, e CR.,. Then

n

z=i

xE(y) = XE(x, Y)

= inf x Ei (x, y) = inf X(E i)(Y),

and we see that

XE.

is measurable. Thus Ex is measurable for any

E e (Bo a. I

16. Lemma: Let E be a set in function g defined by

61,73

with

g(x) = vEx

/.4

X v(E) < Go. Then the

Sec. 4)

267

Product Measures

is a measurable function of x and fgdj = /.‘ X y(E).

Proof: The lemma is trivially true if E is a measurable rectangle. We first note that any set in 61,7 is a disjoint union of measurable rectangles. Let (E,) be a disjoint sequence of measurable rectangles, and let E = U E,. Set gi (x) = vr(E0x].

Then each g, is a nonnegative measurable function, and g=

Thus g is measurable, and by the corollary of the Monotone Convergence Theorem (11.14), we have

f

f g, d/.4

g

=EAXP(E,)

= pi X v(E).

Consequently, the lemma holds for E e Let E be a set of finite measure in (R,s . Then there is a sequence (E,) of sets in di, such that E,+1 C E, and E = E,. It follows from Proposition 6 that we may take 1.4 X v(E 1) < oo. Let gt (x) = Then g(x) = lim g,(x), and so g is measurable. Since

n

f gi d1 =

X v(E 1 ) < 00,

we have g 1 (x) < oo for almost all x. For an x with g 1 (x) < oo, we have ((E,)x ) a decreasing sequence of measurable sets of finite measure whose intersection is E. Thus by Proposition 11.2 we have g(x) = y(Ex ) = lim v[(E0x ] = lim

Hence g, --> g a.e.,

268

Measure and Outer Measure

[Ch.

12

and so g is measurable. Since 0 < g, < gli the Lebesgue Convergence Theorem implies that Jgdp lim

diu

= lim 1.t X v(E 2) = IA X v(E),

the last equality following from Proposition 11.2. 1 17. Lemma: Let E be a set for which almost all x we have v(E z) = O.

1.1,

X v(E) =

O. Then for

Proof: By Proposition 6 there is a set F in 61,8 such that E c F and ,u X v(F) = O. It follows from Lemma 16 that for almost all x we have v(F) = O. But Ex c Fx, and so vEr = 0 for almost all x since v is complete.1 18. Proposition: Let E be a measurable subset of X X Y such that X v(E) is finite. Then for almost all x the set Ez is a measurable subset of Y. The function g defined by /.4

g(x) =v(Ex) is a measurable function defined for almost all x and

f g diu = 1.4 X v(E). Proof: By Proposition 6 there is a set F in Gi,a such that E c F ,u X v(F) = A X v(E). Let G = F — E. Since E and F are measurable, so is G, and

and

A X 70) = A

X v(E)

A

X v(G).

Since A X v(E) is finite and equal to j2 X v(F), we have ,u X v(G) = O. Thus by Lemma 17 we have vGz = 0 for almost all x. Hence g(x) = vEz = vF, a.e.,

and so g is a measurable function by Lemma 16. Again by Lemma 16 X v(F)

fgd = ti

X v(E). 1

Sec.

41

269

Product Measures

The following two theorems enable us to interchange the order of integration and to calculate integrals with respect to product measures by iteration. 19. Theorem (Fubini): Let (X, a, /2) and (Y, CB, y) be two complete measure spaces andf an integrable function on X X Y. Then i. for almost all x the function jez defined by f(y) = f(x, y) is an integrable function on Y; i'. for almost all y the function fy defined by f(x) -= f(x, y) is an integrable function on X; f(x, y) dv(y) is an integrable function on X;

fy

fx f(x, y) d(x) u.

is an integrable function on Y;

fx [ fy f dv]di.1 = fxxy f d(i.1 X y) = f y [fx f diddy.

Proof: Because of the symmetry between x and y it suffices to prove (i), (ii), and the first half of (iii). If the conclusion of the theorem holds for each of two functions, it also holds for their difference, and hence it is sufficient to consider the case when f is nonnegative. Proposition 18 asserts that the theorem is true if f is the characteristic function of a measurable set of finite measure, and hence the theorem must be true iff is a simple function which vanishes outside a set of finite measure. Proposition 11.7 asserts that each nonnegative integrable function f is the limit of an increasing sequence ((ion ) of nonnegative simple functions, and, since each (p i, is integrable and simple, it must vanish outside a set of finite measure. Thus fx is the limit of the increasing sequence ((gon),) and is measurable. By the Monotone Convergence Theorem fy f(x, y) dv(y) = lim fy cp,i (x, y) dv(y),

and so this integral is a measurable function of x. Again by the Monotone Convergence Theorem fx [fi, f di)] d = lim fx [

cp„ dv]d/2

.f.xy(Pn Cl( bi X y)

= xxy f d(u X y). I

270

Measure and Outer Measure

[Ch. 12

In order to apply the Fubini Theorem, one must first verify that f is integrable with respect to A X y; that is, one must show that f is a measurable function on X X Y and that 11 f I d().t X y) < 00. The measurability of f on X X Y is sometimes difficult to establish, but in many cases we can establish it by topological considerations (cf. Problem 17). In the case when p, and y are a--finite, the integrability of f can be determined by iterated integration using the following theorem: 20. Theorem (Tonal»: Let (X, a, p,) and (Y, 03, y) be two a--finite measure spaces, and let f be a nonnegative measurable function on X X Y. Then

fx defined by fx (y) = f(x, y) is a measurable function on Y. i'. for almost all y the function L defined by f(x) = f(x, y) is a measurable function on X. f(x, y) cly(y) is a measurable function on X. i. for almost all x the function

fy f

X

f(x, y) dp,(x) is a measurable function on Y.

[ f f did d p, f f= X Y

XXY

f C102 X P)

f [f f dA]dy. Y X

Proof: For a nonnegative measurable function f the only point in the proof of Theorem 19 where the integrability of f was used was to infer the existence of an increasing sequence ((p„) of simple functions each vanishing outside a set of finite measure such that f = fim yon . But if p. and y are a--finite, then so is y X 1), and any nonnegative measurable function on X X Y can be so approximated by Proposition 11.7. I

If a and 03 are a--algebras on X and Y, then the smallest a--algebra containing the measurable rectangles is denoted by a X 03. Thus the product measure is defined on a a--algebra containing a x 63, and since p. X r is obtained by the Carathéodory extension process, it is both complete and saturated. If y and P are both a--finite, then the product measure on a X 03 is already saturated and the measurable sets for y X 7) are those which differ from sets in a x o3 by sets of measure zero. Many authors prefer to define product measure to be the restriction of 11 X r to a X 63. The advantage of taking y X 1, to be complete,

271

Sec. 4] Product Measures

as we have done here, is that this does what we want it to for Lebesgue measure: The product of n-dimensional Lebesgue measure with m-dimensional Lebesgue measure is (n + m)-dimensional Lebesgue measure. Since our hypotheses for the Fubini and Tonelli theorems require only measurability with respect to the complete product measure, they are weaker than requiring measurability with respect to a X (33. The price for using these weaker hypotheses is the necessity of including the "almost all" phrases in the conclusion of the theorems. This has to be expected, since changing f arbitrarily for x in a set of measure zero doesn't change the measurability or integrability of f, but fs can be arbitrary for those x. If however, f is measurable with respect to a X (fa, then f,, is measurable for each X. We have also used the completeness of I.,. to show that if (x, y) dv(y) is measurable, for if p, were not complete we could only conclude that this was a function which differed on a subset of a set of measure zero from a measurable function. If, however, f is measurable with respect to a X 63, then it turns out that ff(x, y) dv(y) is measurable with respect to a even if 14 is not complete (provided f is integrable), but the proof of this is surprisingly intricate. For a proof see Halmos [8], p. 143. The examples in the problems show that we cannot omit the hypothesis of the integrability of f from the Fubini Theorem or the hypotheses of o--finiteness and nonnegativity from the Tonelli Theorem. Problem 22 shows the essential role played by the measurability of f in these theorems: If we omit this assumption, even for bounded functions and finite measures, we may have the iterated integrals f[f f 4] dp, and f[f f 4] dv well defined but unequal.

Problems 15. Let X — Y be the set of positive integers, a OE — 6,(X), and let v = g be the measure defined by setting p,(E) equal to the number of points in E if E is finite and 00 if E is an infinite set. (This measure is called the counting measure.) State the Fubini and Tonelli Theorems explicitly for this case. 16. Let (X, a, g) be any 0-finite measure space and Y the set of positive integers with v the counting measure (Problem 15). Then Theorem 20 and Corollary 11.14 state the same conclusion. However, Corollary 11.14 is ----

Measure and Outer Measure

272

[Ch. 12

valid even if y is not a-finite, and hence the Tonelli Theorem is true without a-finiteness if ( Y, 63, y) is this special measure space. 17. Let X = Y = [0, It and let A = y be Lebesgue measure. Show that each open set in X X Y is measurable, and hence each Borel set in X X Y is measurable. 18. Let h and g be integrable functions on X and Y, and definef(x, y) = h(x)g(y). Then f is integrable on X X Y and fxxy fd(y. X v) = f h dy f g dy. x Y

We do not need to assume that y and v are a-finite.) 19. Show that Tonelli's Theorem is still true if, instead of assuming ia and v to be o--finite, we merely assume that {(x, y): f(x, y) 0)- is a set of a-finite measure. 20. The following example shows that we cannot remove the hypothesis that f be nonnegative from the Tonelli Theorem or that f be integrable from the Fubini Theorem. Let X = Y be the positive integers and A = v be the counting measure. Let (Note:

{ 2 — 2 —x f(x, y) = — 2 + 2—z 0

if x = y if x = y + 1 otherwise.

21. The following example shows that we cannot remove the hypothesis that f be integrable from the Fubini Theorem or that A and y are a-finite from the Tonelli Theorem: Let X = Y be the interval [0, 1], with a = (B the class of Borel sets. Let ,u. be Lebesgue measure and y the counting measure. Then the diagonal A = {(x, y) e X X Y: x = yl is measurable (is an 61,3, in fact), but its characteristic function fails to satisfy any of the equalities in condition (iii) of the Fubini and Tonelli Theorems. 22. The following example shows that the hypothesis that f be measurable with respect to the product measure cannot be omitted from the Fubini and Tonelli Theorems even if we assume the measurability of fy and f, and the integrability of ff(x, y) d(y) and ff(x, y) di.t(x). Let X = Y = the set of ordinals less than or equal to the first uncountable ordinal St. Let a = (B be the a-algebra consisting of all countable sets and their complements. Define A = y by letting yE = 0 if E countable, yE = 1 otherwise. Define a subset S of X X Y by S = {(x, y): x < y} . Then Sz and Sy are measurable for each x and y, but iff is the characteristic function of S we have

f[ff(x,

y) di.t(x)]dv(y)

f [ f f(x, y) dv(y)]d,u(x).

Sec. 4] Product Measures

273

If we assume the continuum hypothesis, that is, that X can be put in one-to-one correspondence with [0, 1], then we can take f to be a function on the unit square such that f, and fy are bounded and measurable for each x and y but such that the conclusion of the Fubini and Tonelli Theorems do not hold. 23. Show that if (X, a, y) and ( Y, (B , v) are two g-finite measure spaces, then M X I/ is the only measure on a x 6,3 which assigns the value MA vB to each measurable rectangle A X B. Show that a measure on a x with this property need not be unique, if we do not have a-finiteness. 24. a. Show that if Ega X then Ex cΠfor each x. b. If f is measurable with respect to a x ea, then I:, is measurable with respect to OE for each x. 25. Let X = Y = R and let i.c=v= Lebesgue measure. Then ith X v is two-dimensional Lebesgue measure on X X Y = R 2. We often write dx dy for d(y X v). a. For each measurable subset E of R, let u(E) = {(x,

x — y e E}.

Show that a(E) is a measurable subset of R 2 . [Hint: Consider first the cases when E open, E a 96, E of measure zero, and E measurable.] b. If f is a measurable function on R, the function F defined by y) F(x, = f(x — y) is a measurable function on R 2 . c. If f and g are integrable functions on R, then for almost all x the function ço given by ço(y) = f(x — y)g(y) is integrable. If we denote its integral by h(x), then h is integrable and f

f fi

f igi.

26. Let f and g be functions in L 1 (—x, co), and define f* g to be the function h defined by h(y) = ff(y — x)g(x) dx. a. Show that f* g = g * b. Show that (f* g) * h = f* (g * h). c. For fe L 1 , define f by f(s) = fetstf(t) dt. Then f is a bounded complex function and f*g

f

27. Let f be a nonnegative integrable function on (— Go, Go), and let m 2 be two-dimensional Lebesgue measure on R 2 . Then

m2 {(x, y):

y

f(x)} = m 2 {(x, y): O
Measure and Outer Measure (Ch. 12

274

Let

. Then

9(t) = m (x: f(x)

is a decreasing function and

I: 9(0 di = f(x) dx.

*5

Inner Measure

Let A be a measure on an algebra a and A * the induced outer measure. Then A*.E may be thought of as the largest possible measure for E compatible with A. We can also define an inner measure ,u* which assigns to a given set E the smallest measure compatible with A: Definition: Let A be a measure on an algebra a and A * the induced outer measure. We define the inner measure A* induced by A by setting A*E = sup AA — p.* (A — E),

where the supremum is taken over all sets A e a for which 1,1,* (A — , < 00.

Inner measure was important historically because the measurability of a set was originally characterized using both inner and outer measure. A set E of finite outer measure was defined to be measurable if 12*E = pc* E, and the measurability of sets of infinite outer measure was defined in terms of intersections with sets of finite measure. Even in the case of a finite measure A this procedure is more cumbersome than the elegant approach due to Carathéodory, which we have followed in this chapter. Apart from this historical importance, inner measure is useful for the extension of A from a to an algebra containing a and a given set E (which need not be measurable) and for determining the freedom we have in extending A to a or-algebra containing a. The purpose of this section is to do this and to develop the basic properties of inner measure. 21. Lemma: We have u*E < ,u*E. If E c a, ,u*E = Proof: Since MA < ,u*(A n E)

itt*(A

n E),

ME.

Sec. 5]

275

Inner Measure

we have MA — y*(A n < y*(A n

E) < th*E.

Consequently, y *E < y* E. If E e a, take A = E in the definition of /2*. Then y*E >

ME =

y*E. 1 22. Lemma: If E c F, then 1.4E < Proof: If y*(A

< oc, then y*(A F) < oc, and so F) ?_ MA — y*(A E).

/LW > MA — y* (A

Taking suprema over

A e

a with 1.t*(A E) <

co,

we have

y*F > p*E. I One of the difficulties of using the definition of inner measure is that we must take the supremum of MA — y* (A E) over all A with y*(A E) < Go. The next lemma shows that this expression is monotone in A and enables us to calculate y *E more easily. 23. Lemma: Let A and B be two sets in a with A*(A E) < oc and A*(B E) < oc. If A c B, we have ILA — p,*(A E) MB — ti*(B E). If also E c A, we have equality, and hence p*E — AA — p,*(A E). Proof: Since B E C (B A) U (A E), we have 1.t*(B E) < y(B A) + y*(A E). If E C A, this union is a disjoint one and the measurability of B A gives us equality. Since MB = MA + y(B A), we have MB — y*(B E)

MA — y*(A E)

with equality if E C A.1 24. Corollary: If A g a, then

n E)

m*(1 n

Proof: If 1.t*(A n E) = 00, then AA = oc and there is nothing to prove. Otherwise, set F= AnE. Then A—F= AnB, and I.4*F =MA — I.1*(A n L'), since F C A and y*(A F) < 00.1

276

Measure and Outer Measure

[Ch. 12

The following lemma expresses the additivity of 1.4* on decompositions given by disjoint sequences of sets in a. The corresponding result for outer measure is true even if the sets A, are only assumed to be measurable (see Problems 3 and 32e).

25. Lemma: Let (A,) be a disjoint sequence of sets In a. Then ,u* (E n

z=i

A1)

=

z=i

*(E n Co

oo

Proof: Replacing E by E n U A„ we may suppose that E C U A,. 0=

1

For any B

6

a with „,*(B

uB — u *(BE) =

1

E) < oo, we have

1-1

u(A, n B) — /2 * (A, n B n f)

z—i

E 1.1*(A, n E).

z=i

Taking the supremum over all such B gives i.k *E

E y*(A,

n E).

Let B, be any set in a with e(B, (E n Az)) < oo.

Then — /.1 * (Bi (E n A,)) = PL*(B2 n A, n E) Pt(B)

/11 * (.13 n É),

where B, = B, n A,. Since the sets B,' are disjoint and measurable,

E

n

B, —

1.1 * ( B,

(E n A,)) = Ly i Yi] — < 12*E.

Taking suprema over all choices of B, c a with (E n A,)) <

ri n

Sec.

5]

277

Inner Measure

we have (E n A,) < y* E. Thus m*(E n Ai) < ,u*E. I

26. Theorem: Let 12 be a measure on an algebra a of subsets of X and E any subset of X. If 63 is the algebra generated by a and E and if ri is any extension of 1 2 to 63, then A * E < 1.1E < A*E.

Moreover, there are extensions z and A of At to 63 (and hence also to the a-algebra generated by 63) such that pE = ki*E and tzE = 1.4E. Proof: If (A i) is any disjoint sequence of sets from a with E c U A„ then E = U(A2. n E), and so

=E

12(A,

n E)

E AzA,.

Thus ,OEE < /2*E. If A is any set in a with ,u *(A — E) < oo, then /2(A

<

12*(A E), and <

Hence u*E < The sets B in ■33 are the sets of the form B = (A n E)u (A' n f) with A and A' in a, since the collection of all sets of this form is an algebra contained in 63 and containing a and E. For each B e define ri and A by

i.t*(B n E)

,u* (B

n L')

and 1213 = /2*(B n E)

ti*(B n te ).

Then and A are monotone, nonnegative functions defined on 63, and it follows from Corollary 24 that /IA = 2A = AA for A e a. Thus

278

Measure and Outer Measure

[Ch. 12

the theorem will be proved if we show that 17 and ,u are countably additive on 03. Let B, = (A, n E) U (A', n f) be a disjoint sequence of sets whose union B is also in 63. Then B = (A n E) U (A' n f), where A u A, and A' = U A. If i j, B. n B,= 0, and hence A, n A, n E = Bz n B, n E = 0 and A: n A; nE = B, n B, n E = 0. Let C„ — A n nA 1 n - n and G = A n A'1 n • • • n Then C, and Ç are in a,, and C, n C, = 0, Ç n C,' = 0 for i j. Also C,i nE= A n n A- i n•••nfin_inE=AnnE, since we have A„ n A n E = (A,, n A, n E) u (A. n A, n E) = A„ n E. Similarly, Cn1" A:i nE, and so B„ = (CnnE)u (C:,nE). Now B n E = n È). Since C„ and C are disjoint U (G. n E) and Bnt =U sequences of sets in a, we have ,

A*(B n E) =

E 1 *(( E) = E

ti*(B n

E

n E)

and =

n

=

E 1.1*(B,„ n L')

by Problem 3, and i.t*(B n E) =

E 12* (Cn, n E) = E A *(B„ n E)

and p.*(B n t) =

E 1.4*(q, n E)

E

n

t).

Thus /7,B = LuB„ and /2/3 = EAB„, showing that rt and ,u are countably additive and hence measures on 01. 1 Proposition 6 states that every set with finite outer measure is contained in an a,s with the same outer measure. The following proposition is the analogue for inner measure. 27. Proposition: Let E be a set with p. *E < 00. Then there is a set H e as cr such that H C E and /all =

be a set in a with ,u*(A„ E) < 00 and iA,, — By Proposition 6 there is a G,, e av3 such E) > ,u*E — that G„ D A„ E and AG„ = A*(A„ E). Let H„ = A — Gn. Then H„ e as„ and H„ c E. Moreover, glin = PA. — G,, > /2*E. Thus the proposition is proved if we take H = U 11„. 1 Proof: Let A n

,,

Sec. 5]

279

Inner Measure

28. Proposition: Suppose A*E < oo. Then E is measurable if and only if 12*E = m*E. Proof: Suppose m *E = m*E < GO. Then Propositions 6 and 27 give us measurable sets G and H with GcEcH and Till= TiG. Thus E differs from a measurable set by a set of measure zero and hence is measurable. Suppose that E is a measurable set of finite outer measure. Then E is contained in the union of a disjoint sequence (A n ) of sets of a of finite measure. By Proposition 25 and the measurability of E we have E= E m*(E n A n) = E M An — A*(A 7, n L') n E) =E = E m*E. I 29. Theorem: Let E and F be disjoint sets. Then

i-t*E m*F < ii*(E u

< p* E p*F < p.*(E u

< p*E p*F.

Proof: If m *E or i.i* F is infinite, the first inequality follows from the monotonicity of y*. If m *E and i.i*F are both finite, let G and H be measurable sets with G C E and H C F such that AG = y *E and H = 1./*F. Then G U H is a measurable set of finite outer measure contained in E u F. Thus

m*(E u F) ii*(G u H) = 7.7(G u H) =

rtH

proving the first inequality. If ti*F = Go, the second inequality is trivial. If p*F < cc, let A be any set in a with m*(A (E u F)) < cc. Since A — E C [A — (E u F)] U F, we have ,u*(A

E) < /.1* (A — (E

U

F))

M* F.

Thus 1.1*(A E) < oc, and MA — ,u*(A (E U F))

MA

—

m*E m*F.

E) u*F

280

Measure anti Outer Measure

[Ch. 12

Taking the supremum over A, we get y*(E U

< y*E y*F.

To prove the third inequality, we choose a measurable set G C E with AG = 1.4E. Then the measurability of G implies that pi*E ± 4u* F = TAG ± g*F = 11,*(G u < y* (E U F).

The last inequality is just the subadditivity of outer measure. I

Problems 28. a. If /IX < 00, then p*E = /1,X — b. If a is a o--algebra, then inf {gA: E C A, A c a} and y*E = sup IgA: A c E, A e

.

c. If y is Lebesgue measure on R, then 1.4*E = sup tuF: F C E, F closed). 29. Let E be any set. A measurable set G C E is called a measurable kernel for E if y*(E G) = O. A measurable set H D E is called a measurable cover for E if y*(H E) = O. a. Show that any two measurable kernels (or covers) for E differ by a null set. b. Show that if E is a set of o--finite outer measure, then E has a measurable kernel and a measurable cover. 30. Let g be a measure on an algebra a and E a set with y*E < o. If j3 is any real number with g* E < o < 1.4*E, then there is an extension A of y to a o--algebra containing a and E such that AE = 31. a. If p is a semifinite measure on an algebra a and if G3 is the smallest cr-algebra containing a, then y* restricted to 63 is a semifinite measure and is the smallest extension of y to Gi. b. There may be more than one semifinite extension to G3 (for example, let X = N U {co} and a the algebra generated by the finite subsets of N with y the counting measure). 32. Let a be the algebra of finite unions of half-open intervals of R

Sec. 6]

Extension by Sets of Measure Zero

281

tl 0 = 0 and pA = oc for A O. The class (B of Borel sets is the smallest a-algebra containing a. a. Show that p*E = Go if E Ø. b. Show that p*E = 0 if E contains no interval and that p*E = 00 if E contains an interval. c. The restriction of p * to (53 is not a measure. Hence there is no smallest extension of p, to OB. d. The counting measure on 63 is an extension of p to (B. e. Show that Lemma 25 is no longer true if we replace "sets in a" by "measurable sets". 33. If p* is not a regular outer measure (that is, if it does not come from a measure on an algebra), then we do not get a reasonable theory of inner measure by setting ,u*E = 1.4* X — ti*(E). Let X — la, b, cl, and set p*X = 2, p,*0 = 0, and i.t* E = I if E is not X or 0. a. Calculate ,u*. b. What are the measurable subsets of X? c. Show that there is a nonmeasurable set E with p*E = p*E. d. Show that the first and third inequalities of Theorem 29 fail. 34. Let X — R 2 and a the algebra consisting of all disjoint unions of vertical intervals of the form I = gx, y): a < y < bl. Let pA be the sum of the lengths of the intervals of which A is composed. Then p, is a measure on a. Let E = {(x, y): y = 0}. a. Show that p*E = oc, i.t*E = O. b. Show that every subset of E is an as. c. Assuming that there is no finite nonatomic measure defined on (5)(R), show that each extension of p to a measure on a a-algebra must assign to E the value 0 or co.

and let

*6

Extension by Sets of Measure Zero

The results of Section 2 allow us to extend a measure A on an algebra a to a g-algebra containing a and those of Section 5 provide for the extension from a to a a-algebra containing a and one additional set. It is sometimes useful to be able to extend to a cr-algebra containing a and some collection arc of subsets of X and to extend in such a way that each of the sets in arc has measure zero. A necessary condition for this to be possible is that, whenever we have a set A c a such that A c Me art, then /221 = O. This condition is not in general sufficient, since a countable union of sets in mt may contain

282

Measure arid Outer Measure

an A with positive measure, but, if we assume that countable unions, then the condition is sufficient.

OTC

[Ch. 12

is closed under

30. Proposition: Let /.4 be a measure on a or-algebra a of subsets of X, and let arc be a collection of subsets of X which is closed under countable unions and which has the property that for each A e a with A c Me grc we have AA = O. Then there is an extension A of A to the smallest a-algebra G3 containing a and 5rt such that AM = 0for each M c OTC. Proof: Since the collection of sets which are subsets of a set in

OTC

satisfies the same hypothesis as 5E, we may assume that each subset of a set in art is itself in art. With this assumption the collection 133 = { B: B = A A M, A c a, M c art} is a o--algebra containing a and 01Z, and since each o--algebra containing a and art contains 03, 63 is the smallest a--algebra containing a and arc. If B = A 1 A M1 = A 2 A M2, then A 1 A A 2 = M1 A M2, and so il(A i A A 2) = O. Thus ilA 1 -= mA 2, and, if we define igB to be AA 1, then g is well defined on G3 and is an extension of A. It remains only to show that rt is countably additive. Let B = U B,, B, n B, = Ø. If B, = A, A M„ then A, A A, e arc. Setting A n' = An n 21 n • • • n Ân_i, we have .44; n A; = 0, and A,„ P A n' e arc. Thus B, = A,' P M f , and B = A P M, where A = U A,' and M cU M,'. Thus FLB = MA = Ei.zA: = EAB,. I We observe that the condition that MA = 0 for each A E a with A c M simply states that A *M = O. Thus the proposition states that we can extend the domain of A to include any collection art of sets of inner measure zero provided that art is closed under countable unions. Note that on the or-algebra generated by a and arc we have ri. = M. Thus this proposition gives a generalization of the process of completion which extends the domain of a measure by adding sets of outer measure zero.

Problem 35. Let a be a a-algebra on X and art a collection of subsets of X which is closed under countable unions and which has the property that each subset of a set in grt is in Nt. Show that the collection ff3 = {B:B= APM,AcaandMeart}

is a a-algebra.

283

Sec. 7] Carathéodory Outer Measure

36. Show that Proposition 30 need not be true if

a is only an algebra

of sets.

*7 Carathéodory Outer Measure

Let X be a set of points and r a set of real-valued functions on X. It is often of interest to know conditions under which an outer measure I.4* will have the property that every function in r will be measurable, and it is the purpose of the present section to prove the sufficiency of one criterion for this. Two sets are said to be separated by the function it) if there are numbers a and b with a > b such that ço is greater than a on one and less than b on the other. An outer measure /2 * is called a Carathéodory outer measure (with respect to r) if it satisfies the following axiom: iv. If A and B are two sets which are separated by some function in r, then 11*(A u B) = 1.4*A + i.t*B. 31. Proposition: If i.t* is a Carathéodory outer measure, then every function in r is 1.4*-measurab1e. Proof: Given the real number a and the function ic e r, we must show that the set E = {x: ço(x) > a} is 1.4*-measurable or, equivalently, that given any set A, i.t*(A n E) + 1.4*(il n E).

i.t*A

Since this inequality is trivial if 1.1*./1 = Do, we suppose l.t*A < oo. We begin by setting B = E n A, C = É n A, and B„ = Defining Rn =

B

Ix:

(x e

1 B) & (o(x) > a + -n-)} •

Rn_i, we have B = B,, U[ U Rd• k.---n+1

Now on B„_ 2 we have so > a -I- 11(n 2), while on Rn we have 9 < a + 1/(n — 1). Thus so separates R, and Bn_2 and hence —

284

Measure and Outer Measure

[Ch. 12

k-1

separates

R2k

and U R 2i, since this latter set is contained in

R2k-2.

J=1

Consequently,

k —1 A * [U R 2 1]= -

R 2k + A* [

=E

U R2 11 1=1

y*R2i,

j-=1

by induction. Since U R25

i=1

CB c

A,

we have

E

/./ *

R2j

/1 *fi

j=1

and so the series

E

eR2 1

converges. Similarly, the series

E

eR2j +1

j=-1

converges, and therefore also the series E x A * Rk•

k=1

From this it follows that, given e > 0, we can choose n so large that

E teR,, <

E.

k=n

Then by the subadditivity of A* ,

E

1.4*B < y*Bn

k=n+1 < A* Rn

e

or

ex>

/A

*B — e.

1.1* -Rk

Sec. 7] Carathéodory Outer Measure

285

Now At*A

1.4*(Bn U C) = *B

m*C

since (to separates B„ and C. Consequently, i.L*A

> I.L*B + 1.4*C

—

e.

Since e is an arbitrary positive quantity, At*A

> 1.4*B + m*C. 1

32. Proposition: Let (X, p) be a metric space, and let i.t* be an outer measure on X with the property that I.4*(A u B) = 1.4.*A + i.t*B whenever p(A, B) > O. Then every closed set (and hence every Borel set) is measurable with respect to 41.4*.

Problem 37. Prove Proposition 32. [Let r be the set of functions (p of the form = p(x, E). Show that pt* satisfies (iv) with respect to r, and note that for a closed set F we have F = {x: p(x, F) < 0}]

9(x)

13 1

The Daniell Integral

Introduction

It is sometimes convenient to introduce integration directly without using the concept of measure. This happens when we have an elementary integral I defined on some class L of 'elementary functions' and want to enlarge the class L and extend I so that it has all the properties of the abstract Lebesgue integral, including the convergence theorems. If we take, for example, our class L to consist of the continuous real-valued functions on (— c co) each of which vanishes outside a finite interval and take I to be the Riemann integral, we might expect to obtain the class of Lebesgue integrable functions and the Lebesgue integral by some sort of extension process. Such a process has been carried out by Daniell for this case and generalized by Stone, who also clarified the structure of the extended integral. It is the purpose of the present chapter to describe this extension procedure and to show its connection with measure theory. Let L be a family of real-valued functions on some set X, and suppose that L is a vector lattice, that is, that whenever the functions f and g are in L so also are the functions of ± fig, f V g, and f A g. g) v 0 g, Since fAg=f+g (f (fvg)andfVg we see that a vector space L of functions is a vector lattice if for each h in L we have h V 0 in L. Thus a vector space of functions is a vector lattice if it is closed under the operationf —>f+ f V O. Since i fi (—f)+, each vector lattice contains the absolute value of each function in the lattice. Conversely, if L is a vector space such that ,

—

—

286

—

Sec. 1] I ntroduction

287

jf1 is in L for each f in L, then L is a vector lattice, for f+ = + I fl). A linear functional I on L is said to be positive' if I(ç) > 0 for each nonnegative function ya in L. If I is positive and io < 1,t, then k,o) < 104 A positive linear functional I on L is called a Daniell functional or a Daniell integral if the following condition is satisfied: D. If (io n ) is a sequence of functions in L which decrease to zero at each point, then lim 1(ic„) = O.

This condition is clearly equivalent to each of the following conditions: D'. If (cp„) is an increasing sequence of functions in L, and if ye, is a function in L such that 2 ç < lim (p then l(p) < lim D". If (u„) is a sequence of nonnegative functions in L, and if io is a function in L such that io < Eun, then 1((p)

One example of a Daniell integral is obtained by taking L to be the set of continuous functions on (— 00, 00) each vanishing outside a finite interval and taking I to be the Riemann integral. Another is obtained from a measure g on an algebra a of subsets of X by letting L be the class of simple functions and I the natural integration with respect to /.4 . A third example is given by taking L to be the class of all functions integrable with respect to a measure A. This last example does not quite satisfy the definition we have given, since integrable functions may be extended real-valued. We want to extend our definition to cover this case but must be careful about the fact that f + g is not defined at points where it is of the form 00 — 00 or — 00 + 00 . Thus we shall say that L is a vector lattice of extended real-valued functions on X provided that, whenever f and g are in L, the functions f V g, f A g, and of are in L as well as each function h such that h(x) = f (x) + g(x) whenever the right side is defined. Thus we require all possible choices for the sum f + g to be in L, and we shall write h f g whenever h is any possible choice. By a linear functional on L we mean a mapping I of L into R such that 1(af) = al(f) and 1(h) = l(f) + 1(g) whenever h = f + g. This Properly speaking, we should call / "nonnegative," but the use of "positive" in this connection seems to be standard. 2 Here, of course, lim means pointwise limit. 1

288

The Daniell Integral

[Ch. 13

implies that if h l and h 2 are two functions with h l = f + g and h 2 = f+ g then 1(h 1) = 1(h2). A positive linear functional I on L is called a Daniell integral if it satisfies condition D. In general the limits of functions in L will not be in L, and the procedure of Daniell and Stone, which we are going to develop in this chapter, consists in extending I to a class L 1 containing L and closed under suitable limiting operations. In many ways this extension procedure parallels that of Carathéodory for measures, and in Section 3 we show that under mild restrictions this extended integral is in fact integration with respect to a suitable measure.

Problems 1. Show that the conditions D, D', and D" are equivalent. 2. Let g be a measure on an algebra a, L the class of simple functions on a, and I integration with respect to pt. Show that L is a vector lattice and that I is a Daniell integral. 3. Let L be the family of continuous real-valued functions on (— co, co) each of which vanishes outside a finite interval, and let I be Riemann integration on L. Show that lis a Daniell integral. [Hint: The Dini Theorem is useful.] 4. Let L be a vector lattice of extended real-valued functions on X and I a positive linear functional on L, as defined in this section. Let f c L, and let E = {x: f(x) = co}. Show that any extended real-valued function h which vanishes outside E is in L and I(h) 0.

2 The Extension Theorem We begin by considering the class L. consisting of all those extended real-valued functions on X each of which is a limit of a monotone increasing sequence of functions in L. It is clear that if f Ag where a and are and g are in L,„ then so is the function of nonnegative constants. If ((p,i ) is an increasing sequence of functions from L, then (/(yo n)) must be an increasing sequence of real numbers and so must have a limit (which may be co). It is tempting to try to define lim /((p n) as the value of I for the function f which is the pointwise limit of the sequence (cp„. In order to do this we need to know that this value depends only on the function f and not on the

Sec. 2]

The Extension Theorem

289

choice of the increasing sequence (ion) whose limit is f. That this is so is guaranteed by the following lemma: 1. Lemma: If (v.) and (1,G,.) are increasing sequences from L and if lim ço. < lirn ihn, then lim Ayo n) < lim /01/70Proof: For fixed n, we have („on < lim çon < lim /((p„) lim /(1,t,„) by (D'). Hence lim /((p„) lim /(V,„). I

and so

Thus we can extend the functional I to be an extended real-valued functional on L„ with the property that I(f) < I(g) for f < g and I(af + (3g) = aI(f) 0I(g) for positive constants a and 0 and

functions f and g in L. It is clear that L„ is a lattice, for if (p. Tf and ;/,,, g, then (ion A %t. if A g and (p„ V 11,„ f V g. 2. Lemma: A nonnegative function f belongs to L„ if and only if there is a sequence (Ik v ) of nonnegative functions in L such that f =

E ik v . In this case I(f) = E Y= 1

Y=1

Proof: The "if" part is trivial. On the other hand, let f be nonnegative and (p„ 1f with (p„ e L. By replacing (,0„ by (pi, V 0, we may assume that each (p ,, is nonnegative. Set = = (p. — CO

for n > 1. Then f =

E v=1

and

I(f) = Ern /WO v—i tp)

lirn

E

E

P

=1

3. Lemma: Let (L) be a sequence of nonnegative functions in 00

L. Then the function f =

CO

E

n=1

f,, is in L. and I(f) =

E AL).

n=1

Proof: For each n there is a sequence (1,/,) of nonnegative func-

tions in L such that f. =

E

Hence f =

E v,n

Since the

290

The Daniell Integral

[Ch. 13

set of pairs of integers is countable, f is the sum of a sequence of nonnegative functions in L and so must be in L. Also 1(f) =

E op

E

n=1

I(fn).

For an arbitrary function f on X we define the upper integral 7(f) by setting I(f) = inf I(g), f

ge L„

where we adopt the convention that the infimum of the empty set is + Go . We define the lower integral I by setting l(f) = —7(—f). Elementary properties of these upper and lower integrals are given by the following lemmas. The properties in Lemma 4 follow directly from the definition of I. 4. Lemma: Let h = f + g. Then 7(h) < 7(f) + 7(g), provided the right-hand side is defined. If c > 0, 7(cf) = ci(f). if f < g, then 7(f) < 7(g) and 1(f) < I(g). 5. Lemma: We have I(f).

I (f) <

7(f). tf f e Lu, then l(f) = i(f)=

Proof: We have 0 = 1(0) = I(f f)

7(—f). Hence

l(f) = To prove the second statement, we note that if f e L u then I(f) < I(f) by the definition of I. If g c L u and f < g, then I(f) 1(g) and so I(f) < 7(f), whence RD = l(f). If (p e L, then e L C L, and so 1 ( — (P) = 1( — ça) = 1(0 , and hence = 1(4 But each f e L u is the limit of an increasing sequence çr,„) of functions in L. Since f > ço„ we have I(f) > Aq),) = IGO. Hence l(f) > Jim I(go) = I(f).

Since I(f) < I(f) = I(f), we have AD = l(f). I 6. Lemma: Let (f) be a sequence of nonnegative functions, and 00

let f =

00

E f„. Then I(f) < E i(f). v=1

Sec. 2]

291

The Extension Theorem

Proof: If r(fp) = co for some v, we are done. If not, given E > 0, there is a function g, e L u such thatf, < g, and 1(g,) < (f„) + E 2—'. Since each g, is nonnegative, Lemma 3 implies that the function g = Egy is in L u and that I(g) = EI(g Er(f) -I- E. Since g f, we have )

Ï(f)

E

+

p—i

and the lemma follows since E was an arbitrary positive number. I We shall call a function f on X integrable with respect to I (or Iintegrable) if i(f) = I(f) and this value is finite. We denote the class of functions integrable with respect to I by L 1 . For f in L 1 we shall write I(f) for i(f). Thus we have an extension of our original I to all of L 1 . Properties of L 1 and this extended I are given by the following proposition: 7. Proposition: The set L 1 is a vector lattice of functions containing L, and I is a positive linear functional on L 1 , which extends the functional I on L. Proof: If f is in L 1 , so is cf, since I(cf) = cI(f) = c1(f) = I(cf) for c > 0, and I(cf) = ci(f) = cI(f) 1(cf) for c O. Iff and g are in L 1 , then Î(f+ g) _< l(f) + 1(g)

and —1(f + g) = I(—f — g) < —I(f) — I(g),

that is .1.(f + g)

I(f)

I(g).

Thus + g) = I(f + g) = I(f) I(g),

and so f + g is in L 1 . Consequently, L 1 is a linear space, and I is a linear functional on L 1 . Lemma 5 implies that L 1 D L and that our definition of i on L 1 gives us a positive linear functional which agrees with our original I on LI. To prove that L 1 is a lattice, it suffices to show that iff e LI, then f+ e Li . Ley g L 1 . Then for each E > O there are functions g and h in

292

The Daniell Integral

[Ch. 13

L u such that —h < f < g, while I(g) < I(f) E < co and I(h) < —I(f) E < 09. Since g = (g V 0) (g A 0) and (g A 0) e we have I(g A 0) > oo and I(g V 0) < I(g) — I(g A 0) < oo . Thus the function g i = g V 0 is in L. and AO < oo. Let h i = h AO. Then h 1 E L . , and h 1 < f + < g i . Since g > —h, gi + h i < g -I- h. Consequently, I(g 1 ) I(h 1) < I(g) I(h) < 2E. Since —1(h 1 ) < .1(f+) < 1 - (f+) < 1(g1), we have I (f +) — l(f+) < 2E, and so I(f+) = l(f+), since E was arbitrary. Since 0 < I(f+) < I(gi ) < we have f-+ c L I . Thus L 1 is a lattice. 1 —

The following proposition is the analogue for L 1 of the monotone convergence theorem. It also shows that L 1 and I satisfy condition D' and hence D. 8. Proposition: Let (fa ) be an increasing sequence of functions in L i , and let f = lim fn . Then f c L 1 if and only if Ern I(fn) < oo. In this case I(f) = lim I(f). Proof: Since f > fn , i(f) I(f n). Thus if lim I(fn) = oo, then I(f)= Go, and f L i . Then g > 0, and g = Suppose lim 1(fn) < oo. Set g = f E (f;i+i — fn). Hence by Lemma 6, n=1 I(g) < E i(f„±, — fn ) n=1

= E 1(f.+1) — I(fn) n =1 = lim l(fr,) —

Thus 1 (f) = 1(g)

I(fi)

Since fn < f, we have Rf)

liM kfn).

AA), and so

I(f) ?: Ern I(f).

Thus f(f) = I(f) = Em I(fn). I

Sec. 23

293

The Extension Theorem

9. Corollary: The functional 1 is a Daniell integral on the vector lattice Li.

The following two propositions are the analogues for the integral 1 of Fatou's lemma and the Lebesgue convergence theorem. 10. Proposition: Let (f„) be a sequence of nonnegative functions in L 1 . Then the function inff, is in L i , and the function Jim f, is in Li if lim I(f,) < Go. In this case

I(lim f,) < lim l(f„). Proof: Let gn = fi A f2 A • • • A fn. Then (gn ) is a sequence of nonnegative functions in L I, which decrease to g = inff,. Thus —g n t — g, and since I(—g n) < 0, we must have g e L 1 by Proposition 8.

To prove the rest of the proposition, let h8 = inf f„. Then (10 is a p>8

sequence of nonnegative functions in L 1 which increases to limfp. Since h8 < f, for n < lim 1(178) < hm Afp) < oc. Hence 11m f, e L 1 and /(lim f„) < lim l(fp) by Proposition 8. I 11. Proposition: Let (fn) be a sequence of functions in L 1 and suppose that there is a function g in L 1 such that for all n we have Ifid g. Then if f = limf n, we have

Af) = larn AL). Proof: The functions fn + g are nonnegative, and I(f„ + g) < 21(g). Hence by Proposition 10 we have f + g in L 1 and

1(f+ g)

lim 1(f8 + g) = I(g) + lim I(fn).

Hence < Jim 1(f8).

Since the functions g — fn are also nonnegative, we have 1(g

—

f) < lim I(g — = 1(g)

—

111m 1(fn).

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The Daniell Integral

[Ch. 13

Hence lirn l(f„) < l(f), and so Jim l(f„) exists and is equal to l(f).

Problems 5. A function f belongs to L u if and only if f = g ± io with g e Lu, g > 0, and ç e L.

6. Prove Lemma 4. 7. Show that the ambiguous case in Lemma 4 does not cause difficulties in Lemma 5.

3

Uniqueness

In the present section we show that the extension to L 1 of a Daniell integral 1 on L is unique. We begin by proving a proposition of some interest in its own right which describes the structure of functions in L 1 . It is the analogue for I of Proposition 12.7. Let us denote by L u/ the class of those functions on X which are the limit of a decreasing sequence f„) of functions in Lu with 1(f 1) < oo and lim l(f„) > oo. It follows from Proposition 8 applied to (-1;,) that Lu / C L 1 . Iff is any function on X such that 7(f) is finite, then given n we can find h„ E L,. such that

f < h„ and

1(10 < ir(f)

•

Setting g. --= h 1 A h2 A • • • A h„, we have f < g,. < h„, and so (g.) is a decreasing sequence of functions in L u with 7(f) < Ag.) < 7(f) ± Hence the function g = lim gu is in L.1, while f < g and 7(f) = 1(g). We have thus established the following lemma: 12. Lemma: If f is any function on X with 7(f) finite, then there is a g c L ui such that f g and 7(f) = I(g).

A function f on X is called a null function iff e L 1 and /(1 f l) = O. I(f) = O. If f is a null function and f, 0 < 7( gl) Hence g e L 1 , and g is a null function. 13. Proposition: A function f on X is in L 1 if and only if f is the difference g — h of a function g in Lul and a nonnegative null function

Sec. 4]

295

Measurability and Measure

h. A function h is a null function if and only if there is a null function k in L u1 such that ihl < k. Proof: Iff = g h, then f is the difference of two functions in L 1 so must itself be in L 1 . If Ih < k with k null, then h is a nulland function. If f is in L 1 , then Lemma 12 asserts the existence of g e L ia such that f < g and 1( f) = 1(g). Hence h = g f is a nonnegative function and 1(h) = 0, making h a null function. If h is a null function, then by Lemma 12 there is a function k e L. 1 with 1,/i < k and 1(k) = 1(1111) = O. 1 —

—

14. Proposition: Let I be a Daniell integral on a vector lattice L of functions on X and let J be a Daniell integral on a vector lattice A D L. If 1(f) = .1(f) for all f 6 L, then A 1 D L i and I(f) = J(f)

for all f e L i .

Proof: By applying Proposition 8 twice we see that L ia c A 1 and that 1(f) = f(f) forf e L ia. Hence by the second part of Proposition 13 each function which is null with respect to I must also be null with respect to J. By the first part of Proposition 13 every function f in L 1 must be in A 1 , and l(f) = J(f). I 4

Measurability and Measure

We say that a nonnegative function f on X is measurable (with respect toi) if g A f is in L 1 for each g in Li. 15. Lemma: If f and g are nonnegative measurable functions, so are f A g and f v g. If (fn ) is a sequence of nonnegative measurable functions which converge pointwise to a function f, then fis measurable. Proof: If f and g are nonnegative measurable functions and h is in L 1 , then h A (f A g) = (h A f) A (h A g) and h A (f V g) — (h A f) V (h A g). Hence the measurability of f A g and f V g follows from the fact that L 1 is a lattice. If (fn ) is a sequence of nonnegative measurable functions converging to f and g a function in L l , then (f„ A g) is a sequence of functions in L 1 converging to f A g. Since If, A gl < Igl, we have f A g in L 1 by Proposition 11. 1 16. Lemma: A nonnegative function f on X is measurable with respect to I if (p A f is in L 1 for each (p in L.

296

The Daniell Integral

[Ch.

13

Proof: If cp e L, then cif, A f is in L 1 , and Proposition 8 implies that g Afe L1 for g e Lu and I(g) < oc . It follows from Proposition il that g Afe L 1 for all g L.1. If h is any function in L I , Proposition 13 tells us that h = g — k, where g e L u l and k is a nonnegative null function. Since 0 < g Af— h A f < k, the function h A f differs from the integrable function g A f by a null function. Thus h A f is integrable, proving that f is measurable. I We say that a set A in X is measurable with respect to I if its characteristic function XA is measurable. We say that A is integrable if its characteristic function XA is integrable. Note that a measurable subset of an integrable set is itself integrable. 17. Lemma: If A and B are measurable sets, so are the sets A U B, A n B, and A — B. If (A n ) is a sequence of measurable sets, then the sets A„ and U A„ are measurable. If the function 1 is measurable, then the class a of measurable sets is a o--algebra.

n

Proof: The measurability of A n B and A u B follows from the fact that XAnE = XA A XE and XALJE -= XA V XE• If g is in L 1 , we have g A XA-B = g A XA — g A XAns + g A 0, and the measurability of A — B follows from that of A and B. If A = U A n , then X A = liM (X A i V • • • V XA„)

and the measurability of A follows from Lemma 15. A similar arguA. If 1 is a measurable function, then the set ment holds for X is a measurable set, and the complement of a measurable set is measurable. I

n

18. Lemma: If 1 is a measurable function and f a nonnegative integrable function, then for each real number a the set E = fx:f(x) >

is measurable. Proof: If a is negative, E = X and is measurable, since XE = 1 and 1 is measurable. Hence we assume a > O. If a = 0, set g = f, while if a > 0, let g = (01-1f) — [(or lf) A 1]. Since g is the difference of two functions in L I , g is in L 1 . In either case we have

Sec. 4]

Measurability and Measure

297

g(x) > 0 for x e E, and g(x) = 0 for x e É. Let cp„ = 1 A (ng). Then (p. e L i , and (p. XE. Hence XE is measurable, and so E is

measurable. a 19. Lemma: Let the function 1 be measurable, and define a set function kt, on the class a of measurable sets by = .1(XE) if x E is integrable, and AE

= c

otherwise. Then

1.1

is a measure.

Proof: We have AO = 1(0) = O. If A and B are integrable sets with A C B, we have XA < XB, and so AA < AB. Thus /2 is monotone for integrable sets and consequently for measurable sets. Let (E,) be a disjoint sequence of measurable sets, and let E = U E,. If one of the E, is not integrable, then E is not integrable,

and /.4.E =

=

E /2E2 .

If each E, is integrable, E will be integrable if and only if EAE, < by Proposition 8, since XE = EX E, In either case AE = Ei2E„ and the measure /1, is countably additive. I This measure has the property that the integrable sets are precisely the measurable sets of finite measure. The following theorem tells us that the Daniell integral I on L I is equivalent to the integral with respect to this measure 20. Theorem (Stone): Let L be a vector lattice of functions on X with the property that iff e L then 1 Afe L, and let I be a Daniell integral on L. Then there is a a-algebra a of subsets of X and a measure u on a such that each function f on X is integrable with respect to I if and only if it is integrable with respect to A. Moreover, l(f) = f f

Proof: Let a be the class of sets which are measurable with respect to I. It follows from Lemma 16 that 1 is measurable. Lemma 17 then asserts that a is a a-algebra, and Lemma 18 asserts that each nonnegative 1-integrable function is measurable with respect to a. Since

298

The Daniell Integral

[Ch. 13

each 1-integrable function is the difference of two nonnegative /-integrable functions, every /-integrable function must be measurable with respect to a. Let ki be the measure given in Lemma 19, and let f be a nonnegative function which is integrable with respect to I. For each pair (k, n) of positive integers let Ek,n = {X: f(x) >

Then

Ek, n

,

is measurable, and since XEk,„ -= XEk.,. A (k-1 27f),

we have X Ek.. e L i , and ii(Ek , n) < 00. Set 22

= 2—n

.

E

XEk n •

k=1

Then ion e L 1 and (p. I f Hence 1(f) = lim 1(w,a). But /(çon) = 2—n

E I(XE k „)

k-1 22n

= 2—n = fion

k=1

dp.

Since f f dp = lim

fp,

dp

by the Monotone Convergence Theorem, we have l(f) = f f dp,

and f is integrable with respect to kt. Since an arbitrary f which is 1-integrable is the difference of two nonnegative /-integrable functions, it follows that such an f must also be integrable with respect to and that l(f) = f f dp.

Sec. 4]

Measurability and Measure

299

Iff is a nonnegative function on X which is integrable with respect to /.4, we construct Ek ot and ion as before. Since ff dj2 < co, each Ek , n has finite measure, and so XEk,. and hence ça,, belong to L I . Since yon If and lim /(ion) = ffdit < co, we have f e L 1 by Proposition 8. Thus each f which is integrable with respect to is also integrable with respect to L I 21. Proposition: Let L be a vector lattice of functions on a set X, and suppose that 1 e L. Let 63 be the smallest o--algebra of subsets of X such that each function in L is measurable with respect to (B. Then for each Daniell integral I there is a unique measure A on (B such that for every f c L

1(f) = ffdA. Proof: The existence of u is a special case of Theorem 20 and we

have only to prove the uniqueness of A on (B. Let a be the a-algebra of measurable sets given by Lemma 17. Lemma 18 asserts that each f in L is measurable with respect to a, and so we must have 63 c a. Since 1 e L, the functions XB are in L 1 for each B in a and hence for each B in B. The uniqueness of A on 133 will be established if we can show that A(B) = /( X B ) for each B in (B. If we let A be the set of functions on X which are measurable with respect to (B and integrable with respect to A and set J(f) = f f

for f 6 A, then Proposition 14 implies that ./(f) = 1(f) for feL l n A. But if B e (B, then XB E Li n A, and so = J(XB)

= I(XB).

Thus

/A

is uniquely determined on

133

by I. I

We can still establish the uniqueness of the measure kt in this proposition if instead of assuming 1 e L, we make the weaker assumption that there is an everywhere positive function in L 1 (Problem 10). Without some such assumption the measure need not be unique on CB (Problem 11).

300

The Daniell Integral

[Ch. 13

Problems

8. Let u be a measure on an algebra a of sets, and let L be the family consisting of those functions which are finite linear combinations of characteristic functions of sets in a with finite measure, and let I be integration with respect to A. Discuss the extension of I to L I , and compare this process with the Carathéodory extension for A. 9. Prove directly from the definition that, if fi and f2 are two nonnegative measurable functions, then fi f2 is measurable. 10. Prove that the conclusion of Proposition 21 still holds when the hypothesis that 1 e L is replaced by the hypothesis that there is an everywhere positive function e in L 1 . [Hint: X = U {x: e(x) > 1/n} , and the proof given can be modified to show that A is unique on sets of 63 which are contained in {x: e(x) > 1/n}.] 11. Let X = (— co, cc) U Icol, and let L consist of all functions on X which are Lebesgue integrable on (— oc, oc) and vanish at co. Then L is a vector lattice, and the smallest a-algebra with respect to which each function in L is measurable is the family 133 consisting of all sets B such that B n (-00, oc) is Lebesgue measurable. Let / be defined on L by I(f) (x) dx. Then I is a Daniell integral on L. What is the measure y constructed in Theorem 20? Show that there is another measure v defined on 03 such that for each f in L, I(f) = f f 12. a. Define a measure y on the /-measurable sets by yE = /(x E) if E integrable, and /(xE) = sup II(xA): A C E, A integrable) otherwise. Show that p. is a measure and is the smallest measure such that I(f) = ff dp. b. Show that for this measure At(X) = jjIj = sup {I(f): f e L, f< 11. c. Show that if I III! G oc, then this measure is the unique measure such that I(f) = ff and AL(X) II/11-

14

Measure

and Topology

We are often concerned with measures on a set X which is also a topological space, and it is natural to consider conditions on the measure so that it is connected with the topological structure. There seem to be two classes of topological spaces for which it is possible to carry out a reasonable theory. One is the class of locally compact Hausdorff spaces, and the present chapter develops the theory for this class. The other is the class of complete metric spaces, and some of the significance of this class for measure theory is discussed in Chapter 15. 1 Baire Sets and Borel Sets

Let X be a locally compact Hausdorff space. From the point of view of integration theory the most useful family of functions on X is the family Cc(X) consisting of all continuous real-valued functions which vanish outside a compact subset of X. If f is a real-valued function, we define the support of f to be the closure of the set Ix: f(x) 01. Thus Cc(X) is the class of all continuous real-valued functions on X which have compact support. The class of Baire sets is defined to be the smallest a-algebra 33 of subsets of X such that each function in Ce(X) is measurable with respect to .33. Thus 33 is the o--algebra generated by the sets Ix: f(x) > al with Je Ce(X). If a > 0, these sets are compact Vs. It follows from Proposi301

302

Measure and Topology

[Ch. 14

tion 9.20 that each compact 9 3 is a Baire set. Consequently, 63 is the a--algebra generated by the compact Vs. If X is any topological space, the smallest a--algebra containing the closed sets is called the class of Borel sets. Thus, if X is locally compact, every Baire set is a Bord set. The converse is true when X is a locally compact metric space, but there are compact spaces where the class of Bord sets is larger than the class of Baire sets (Problem 5). The reader should be warned that not everyone uses the same definitions for Baire and Borel sets. When X is compact (or o-compact), everyone agrees that the Baire sets are the smallest a--algebra such that each f in C(X) is measurable and that the Borel sets are the smallest a--algebra containing the closed sets, but definitions differ when X is allowed to be locally compact. Some authors like to take the Baire sets to be the smallest a--ring containing the compact 9 3 's ; others the smallest a--algebra such that each f in C(X) is measurable. Some authors prefer to take the class of Borel sets to be the smallest a--algebra (or a--ring) containing the compact sets. The definitions we have made here seem to me the most useful. A set is called a--compact if it is the union of a countable collection of compact sets. A set which is contained in a compact set is called bounded, and one which is contained in a a--compact set is called a--bounded. The following lemma is useful.

1. Lemma: If B is a Baire set, then either B or 13 is o--bounded. Proof: The collection of sets B such that either B or 13 is a--bounded is a a--algebra. Since it contains all compact sets, it must contain all Baire sets. I A measure p defined on the a--algebra of Baire sets is called a Baire measure if it is finite for each compact Baire set. A measure defined on the Bord sets is called a Borel measure if it is finite for each compact set. In the next two sections we shall show that each positive linear functional on Cc (X) is given by integrating with respect to a suitable Baire measure and that for compact spaces X each bounded linear functional on C(X) is given by a finite signed Baire measure. In Section 4 we show that each Baire measure can be extended to a Bord measure.

Sec. 1] Baire Sets and Borel Sets

303

Problems

1. Let X be a separable locally compact metric space. Show that the class of Baire sets is the same as the class of Borel sets. 2. Show that the collection of compact 93 's is closed with respect to finite unions and intersections. 3. a. Show that each compact set in a locally compact space is contained in an open a--compact set 0 with O compact. b. Show that each a--bounded set is contained in an open a--compact

set. 4. Let X be a locally compact Hausdorff space, and let Co(X) be the space of all uniform limits of functions in Ce(X). a. Show that a continuous real-valued function f on X belongs to Co(X) if and only if for each a > 0 the set {x: I f (x)1 > a} is compact. b. Let X* be the one-point compactification of X. Then Co(X) consists precisely of the restrictions to X of those functions in C(X*) which vanish at co. c. If B is a Baire set in X*, then B n Xis a Baire set in X. 5. Let X be an uncountable set with the discrete topology.

a. What are Ce (X) and Co(X)? b. What are the Baire sets in X? c. Let X* be the one-point compactification of X. What is C(X*)? d. What are the Baire subsets of X*? Show that X* has a compact subset which is not a Baire set. e. Show that there is a Baire measure /2 on X such that i.t(X) = I and ff dm, — 0 for each f in Co(X). 6. Let X and Y be two locally compact Hausdorff spaces. a. Show that each f e CAX X Y) is the limit of sums of the form n

E so,(x)I,L i(y). t=i

[Hint: The Stone-Weierstrass Theorem is useful.]

b. If a and 63 denote the Baire subsets of X and Y, respectively, then the class of Baire subsets of X X Y is a x 63. c. Let X and Y each be the Alexandroff one-point compactification of a discrete uncountable set. Then the algebra of Borel subsets of X X Y is larger than the a--algebra which is the product of the Borel sets of X and the Borel sets of Y.

304

2

Measure and Topology

[Ch. 14

Positive Linear Functionals and Baire Measures

In the present section we show that each positive linear functional on Cc(X) is given by integration with respect to a Baire measure and that this measure is essentially unique. We begin with two lemmas about compact 96 's and Baire measures. 2. Lemma: If K is a compact 95, there is a sequence 4,j ) of functions in Cc (X) such that yon, 1 XK. Proof: By Proposition 9.20 there is a nonnegative function ço e Cc (X) with („0 1 on K and 0 < (p < 1 on K. Let ço,„ = ion. I 3. Lemma: Let p i and /22 be Baire measures on a compact Hausdorff space X. If yiK = /.1 2K whenever K is a compact 96 , then /h i = /2 2. Proof: The semialgebra e generated by the compact 96 's is the collection of all sets of the form C = K 1 n k2 , where K1 and K2 are compact 96 's with K2 C K 1. Since ,u,C = u1K1 — ,u,K2, we see that and /22 agree on e. Propositions 12.9 and 12.8 imply that their extensions to the smallest a-algebra containing e are identical, since /2 1 X = ,u 2X < 00. 1 4. Corollary: Let u i and ,u2 be Baire measures on a locally compact Hausdorff space such that ,u1K = ,u2 K whenever K is a compact 96 . Then /2 1 E = I.L2E for each a-bounded Baire set E. This corollary shows that so far as the a-bounded sets are concerned the value of a Baire measure is determined by its values on compact 96's. It is possible, unfortunately, for two Baire measures to agree on the cr-bounded Baire sets and yet differ on sets which are not a-bounded. I (See Problem 10.) 5. Theorem: Let X be a locally compact Hausdorff space and I a positive linear functional on Cc (X). Then there is a Baire measure /.1 such that for each f in Ce(X) we have l(f) = 1 f di.t. If X is compact, the measure pt is unique. In general /.1 is uniquely determined on the a-bounded Baire sets. 1 This is one of the principal reasons many authors prefer to do measure on if-rings rather than if-algebras, since the smallest if-ring containing the compact 93's contains only if-bounded sets.

Sec. 2]

305

Positive Linear Functionals and Baire Measures

Proof: The space Ce (X) is a vector lattice, and I will be a Daniell integral if it satisfies the Daniell condition D. Let (<1 ) be a sequence of functions in Ce (X) which decrease to zero and let K be the support of (p i . Let 4/ be a nonnegative function in C(X) which is positive on K. Then for a given E > 0 the sets F = {x: Elp(x)} are a decreasing family of closed subsets of K whose intersection is empty. Thus for some N we have FN = 0, and cp„ < el,b for n > N. Thus I((p„) < €411/) for n > N, and I(cp„) —> 0, since e was arbitrary. The existence of a Baire measure so that I(f) = ff d,a for each f e Cc (X) now follows from the Stone Theorem (13.20). To see the uniqueness of ,u, we note that, if K is a compact go, then XK is the limit of a decreasing sequence 47,) of functions in Cc (X). Since p is integrable, the Lebesgue Convergence Theorem asserts that AK = lim f yo n dp = urn i(
Thus ,aK is uniquely determined by I. By Corollary 4 the measure p is uniquely determined on the o--bounded Baire sets and hence on all Baire sets if X is compact. I

The following proposition is the analogue for Baire measures of Proposition 3.15. 6. Proposition: Let p be a &tire measure on a locally compact Hausdorff space X. Then given a o--bounded &tire set E, ,uE = inf i,u0: E C 0, 0 a a-compact open set) and pE = sup {AK: K c E, K a compact 961.

Proof: To prove the first statement, we note that, since E is a-bounded, it is contained in a 0-compact open set O. If pE = Go, then AO = o and there is nothing to prove. Hence we shall assume ,uE < If E is a compact ga , then there is a function (10 c Cc (X) which is 1 on E and 0 < (10 < 1 on E. Let 0„ = {x: yo(x) > 1 — 1/n}. on. Then each On is a 0-compact open set, On D 07,+1, and E =

n

306

Measure and Topofogy [Ch. 14

Since D i is compact, A01 < 00 and A E = lim AO„. Thus for some On we have AO, < ALE + E. Let E = E l n E2 where E l and E2 are compact 96's with E2 C El, and let U be a o--compact open set with ri compact such that E 1 c U and AU < AE1 + e. Set 0 = U n E 2 . Then 0 is the intersection of two 5„'s and so is an 56 . Since 0 is contained in the compact set U, 0 must be o--compact. Since 0 • E = U El, we have = /2(U E1 ) = /Ai — /2E1 e. Thus /10 < AE + e, and the first statement in the proposition holds for sets E in the semialgebra e generated by the compact 91's. The algebra a generated by the compact 9 6's is the collection of all finite unions of sets from e, and a, consists of all countable unions of sets from e. By Proposition 12.6 each set E in the o--algebra generated by a (that is, each Baire set E) with AE < 00 is contained in < AE + - . Let A be the disjoint union of the 2 sequence (C,) from e, and choose o--compact open sets 0, such that an A c a, with

AA

C, c 0, and AO, < AC, +

.Then 0 = U 0, is a o--compact 2'+'

open set, A c0, and A(0

A)

< ;. Thus 0

E and AO < AE + e.

This proves the first statement of the proposition. If E is a o--bounded Baire set, then E C U K, with K, compact. Since each compact set is contained in a compact 98, we may assume each K,

is a Compact 93

and hence a Baire set. Thus En = E n LI K, z= 1

is a bounded Baire set, and E is the union of the increasing sequence (E„) of bounded Baire sets. Hence ALE = lim AtEn , and so ALE — sup {ALB: B C E, B a bounded Baire set}. Let B be a bounded Baire set and C a compact 93 containing B. Since C B is a bounded Baire set, we can find a o--compact open set 0 containing C B such that /AO < AL(C ± E. Let K = C n Ô. Then K C B, and K is a 96 since it is the intersection of two 93 ' s, and compact since it is a closed subset of a compact set. Since Cc Ku 0, we have ALC <

ALK < At(C B) ,uK +

E.

307

Sec. 2] Positive Linear Functionals and Baire Measures

Hence AB < IIK ± e and AB = sup

WC: K C B, K a compact 951.

I

A Baire measure ,u is called regular if for each Baire set E we have ALE = sup ,uK as K ranges over the compact 9 6's contained in E. Thus Proposition 6 states that each Baire measure on a o--compact space is regular. Some further properties of regular Baire measures are given in Problem 10. We have used throughout this section the fact that if a compact set is a 96 it is a Baire set. The converse is also true: If a compact set is a Baire set, it is a 96 . Since we do not need this here, we refer the interested reader to Halmos [8], p. 221, for a proof.

Problems 7. Let 3C be a collection of subsets of X with the property that the union of any two sets in ac is in x and the intersection of any two sets in 3C is in N. Show that the algebra generated by 3C consists of the sets which are finite disjoint unions of sets of the form K 1 n k2 where K 1 and K2 are in 3C and K2 C K1. 8. Show that each compact set in a locally compact Hausdorff space is contained in a compact 96 . 9. An alternate proof of Theorem 5 can be given using the results of Section 123 instead of those from Chapter 13: a. Define an outer measure ,u* on X by setting u*E = inf EA,p,) for all sequences (45,,,) from C(X) with E(p,, > 1 on E. Show that A* is a Carathéodory outer measure with respect to Cc (X) and that hence each Baire set is A* measurable. b. Let ,u, be ,u* restricted to the Baire sets. If f is a nonnegative function in Ce(X) and f > 1 on a Baire set E, then u(E) < l(f). Iff < 1 and the support off is contained in a compact Baire set K, then AK > l(f). c. Show that for any Baire measure with the properties in (b) and any f c C c(X) we have ,

l(f) = f f d,u.

308

Measure and Topology

[Ch. 14

10. a. Show that a finite Baire measure on a locally compact Hausdorff space X is regular if 1.cX = sup 1K: K a compact w. [Problem 11.9 is useful.] b. If bi is a regular Baire measure on X, then for each Baire set E we have iE = inf {110: 0 D E, 0 an open Baire set} . c. Show that the measure i of Theorem 5 can be chosen so that it is regular, and that it is then unique. [See Problem 13.12.] d. Let X be an uncountable set with the discrete topology. Construct a nonregular Baire measure on X. e. Let ,u be a Baire measure on X. Then there is a unique regular Baire measure on X which agrees with ,u on the a-bounded Baire sets. 11. Let kt be a Baire measure on a locally compact space X. Let U be the union of all open Baire sets 0 with 1.4,0 = O. The complement F = of U is a closed set called the support (or carrier) of 12. a. If 0 is an open Baire set with 0 n F 0, then AO > O. b. If K is a compact Baire set with K n F = 0, then AK = O. [Each point of K is contained in an open set of measure zero. Thus by compactness K is contained in an open set of measure zero.] c. If E is a a-bounded Baire set with E n F = 0, then ,I.LE = O. d. If f C c(X) and f > 0, then f f = 0 if and only iff 0 on F. [Hint: The set ri x:f(x) > 0; is a o--bounded Baire set.] e. Give an example to show that F need not be a Baire set. f. It follows from (c) that if X is compact (or o--compact), then = 0 for each Baire set with E n F = Ø. Construct an example to show that this need not be true if X is not o--compact (see Problem 5e).

3

Bounded Linear Functionals on C(X)

Let X be a compact Hausdorff space and C(X) the space of realvalued continuous functions on X. In Section 2 we described the positive linear functionals on C(X), and in this section we consider the bounded linear functionals on C(X). We first note that if F is a positive linear functional on C(X) and if I fl < 1, then IF(f)! 5_ F(! fl) < F(1).

Hence it follows that

I I FII = F (l ).

Sec. 3]

Bounded Linear Functionals on C(X)

309

The next proposition shows that every bounded linear functional on C(X) is the difference of two positive ones. Since the proof makes no use of particular properties of C(X) other than that it is a vector lattice of bounded functions containing 1, we state the proposition in this generality. If L is a vector lattice of bounded real-valued functions on a set X, L becomes a normed linear space when we define Ilf = sup If(x)I. A linear functional is bounded if there is an M such that and we define as usual IlF11 = sup F(f). 11f11<1

7. Proposition: Let L be a vector lattice of bounded functions on a set X, and suppose 1 e L. Then for each bounded linear functional F on L, there are two positive linear functionals F+ and F_ such that F = F+ — F_ and 11F11 = F+(1) F41). Proof: For each nonnegative f in L define

F+(f) = sup F(). 0
Then F+(f) 0, and F+(f) F(f). Moreover, F+(cf) = cF+(f) for c > O. Let f and g be two nonnegative functions in L. If 0 < < f and 0 < < g, then 0 < < f ± g, and so F ±( f + g)

F

F(i).

Taking suprema over all such ço and 1,t, we obtain F+(f ± g) ?_ F+(f) + F(g).

On the other hand, if 0 < < f + g, then 0 < A f) < g, and so

0 < — (itt

F

F(' A f) + FOP F+(f) + F(g).

—

[1P A

A)

Taking the supremum over all such 1,t, we get F±(f + g)

F±(f) F(g),

A f < f and

Measure and Topology

310

[Ch. 14

and so F+(f + g) = F+(f) + F(g). Let f be an arbitrary function in L, and let M and N be two nonnegative constants such that f + M and f + N are nonnegative. Then F+(f + M + N) = F+(f + M) + F+(N) F-4-(f Hence F+(f + M) — F ±(M) = F+(f + N) — F(N). Thus the value of F+(f M) — F +(M)is independent of the choice of M, and we define F+(f) to be this value. The functional F+ is now defined on all of L, and we have easily F+(f + g) = F+(f) F+(g) and F+(cf) = cF+(f) for c O. Since F+(—f) + F+(f) --F+(0) = 0, we have F+(—f) = —F +(f), and F+ is a linear functional on L. Since 0 < F+(f) and F(f) < F+(f) for f > 0, both F+ and the linear functional F_ = F+ — F are positive linear functionals, and F = F+ — F_. We always have 11F F41) F_(1). To !IF +11 establish the inequality in the opposite direction, let io be any function in L such that 0 < ç < 1. Then I2 — 11 < 1, and

IIF ?_ F(Ip — 1) = 2F(p) — F(1). Taking the supremum over all such ç

,

we have

IlF11 > 2F+(1) — F(1) = F+(1) F_(1). Hence IlF11 = F+(l)

F_(1). I

8. Riesz Representation Theorem: Let X be a compact Hausdorff space and C(X) the space of continuous real-valued functions on X. Then to each bounded linear functional F on C(X) there corresponds a unique finite signed Baire measure p on X such that

F(f) = f Jed!) for each f in C(X). Moreover, VII = Iv (X).

Sec. 3]

Bounded Linear Functionals on C(X)

311

Proof: Let F = F± — F_ as in Proposition 7. Then by Theorem 5 there are finite Baire measures I.L i and /2 2 such that

F±(f) = f f dil l

and F_(f) = f f (111 2.

If we set y = A i —

142,

then y is a finite signed Baire measure, and F(f) = f f dp.

Now

IF(f)i ._ f If divi l If!! Iv1(x). Hence 11F11 __ 1v1(X). But PI(X)

121(X) + /2 2(X) = F± (1) + F_(l) = lift

Thus VI = y (X). To show the uniqueness of y, we note that if y 1 and y 2 were both finite signed Baire measures such that f f di', = F(f)

for i = 1,2 and f e C(X), then X = y 1 — y 2 would be a finite signed Baire measure such that

f f dX = 0 for allf e C(X). Let X = X+ — X — be the Jordan decomposition of X. Then integration with respect to X+ gives the same positive linear functional on C(X) as that given by X —, and so by Theorem 5 we must have X+ = X. Hence X = 0, and 1, 1 = y 2 . 1 9. Corollary: Let X be a compact Hausdorff space. Then the dual of C(X) is (isometrically isomorphic to) the space of all finite signed Baire measures on X with norm defined by MPH = iPi(X)-

The fact that the space of finite signed Baire measures on X is the dual of C(X) enables us to conclude a number of things about this

312

Measure and Topology [Ch. 14

space. For example, it follows from Proposition 10.3 that the space of Baire measures is complete, and it follows from Theorem 10.17 that the set of Baire measures with !H (X) < 1 is compact in the weak* topology. Some consequences of this are explored in the problems.

Problems 12. Let L and F be as in Proposition 7. Show that if G and H are two positive linear functionals on L such that F = G — H and G(1) + H(1) < 11E11, then G = F+and H = F. [Hint: Use the definition of F+ to show that G — F+ is a positive linear functional.] 13. Let X be a compact Hausdorff space if = {fa } a family of continuous real-valued functions on X and {C6,} a corresponding family of constants . Suppose that for each finite set {fai , . . . ,A.} there is a signed Baire measure y with IH(X) < 1 such that

f fdv = ca,. Then there is a finite signed Baire measure y with I p1(X) < 1 such that for every fa, fa dp = cc,.

f

14. a. Let X be a compact Hausdorff space and g, fi, • • • „A continuous real-valued functions on X. Suppose that there is a signed Baire measure y on X with 10(X) < 1 such that for each i we have f f, di, = ci. Then there is a signed Baire measure y on X with lyi(X) < 1 such that and

f fi dm = ci fgd,u15_ fg dX

for any signed Baire measure X with I XI(X) < 1 and such that f ft dX ----- c i. b. Suppose that there is a Baire measure y on X with y(X) = 1 and 5f2 dp = c2. Then there is a Baire measure AL on X with y(X) — 1 and 5 f, dy = ci which minimizes fg di.t among all Baire measures which satisfy these conditions. c. Let G, F1 ,. . . , Fn be continuous functions on In--Euclidean m-dimensional space), and let fl , . . . , f,,, be continuous functions on X. Show that if there is a Baire measure y with y(X) ---- 1 such that Fi (f f i dp, . . . , f fni di') --- c„

Sec. 4]

313

The Borel Extension of a Measure

then there is a Baire measure on X which minimizes

G (f dv,

,f

under these restrictions. 15. Let B be the Banach space of signed Baire measures on a compact Hausdorff space X. What are the extreme points of the unit sphere of B?

16. Alternate proof of the Stone - Weierstrass Theorem. We can use the techniques of this section, together with results of Chapter 10, to give a proof of the Stone-Weierstrass Theorem which does not depend on Lemma 9.27. This proof is due to deBranges. Let a be an algebra of real-valued continuous functions on a compact space X which separates points and contains the constants. Let al be the set of signed Baire measures on X such that < 1 and if = 0 for all f e a. Use the Hahn-Banach Theorem and Corollary 9 to show that if al contains only the zero measure, then -.6, = C(X). b. Use the Krem-Milman Theorem and the compactness of the unit ball in C*(X) to show that if the zero measure is the only extreme point of al, then a-L contains only the zero measure. c. Let p. be an extreme point of al. If f e 0 < f < 1, the measures p. and /12 given by dp. j = f cli.z and 4.2 = (I — f) dbl. are in al, bi,H, and /I + 11 2 = pt. Since p. is an extreme point, with lig i d = cp. for some constant c. d. Then f— c = 0 on the support of m. (see Problem lid). e. Since f separates points, the support of p. can contain at most one point. Since f 1 dp, = 0, the support of p. is empty, and p. is the zero measure.

*4

The Borel Extension of a Measure

Previously we have considered only Baire measures on a locally compact Hausdorff space, but it is sometimes convenient to extend a given Baire measure to be a Borel measure. In this section we show that, if the Baire measure is regular, then this can be done in such a way that the analogue of Proposition 6 holds and so that the Borel measure really lives on the Baire sets in the sense that each Borel set of finite measure differs from a Baire set by a set of measure zero. We begin with a lemma.

314

Measure and Topology

[Ch. 14

10. Lemma: Let K be a compact g3, and suppose K c U O i with

O. open. Then K = LI Ki where the K, are compact 96's and K, c O. 1 =1

Proof: By Proposition 9.21 there are continuous nonnegative functions fi with support in 0, such that fi • ± = 1 on K. Let Ci = {x: f(x) > -71,} . Then Ci is a compact 96 contained in Oi and K c tj G. Set K, = Ci n K I

11. Theorem: Let 12, be a regular Baire measure on a locally compact Hausdorff space X. Then there is a unique Borel measure g on X which extends js and for which: i. For each open set 0, /70 = sup {AK: K c 0, K a compact g6

}

For each Borel set E, = inf {170: E c 0, 0 an open set} .

The measure rt also satisfies: iii. For each Borel set E with ).-LE < 00, = sup {17K: K c E, K compact}.

iv. For each Borel set E with i7E < 00, there is a Baire set H and a Borel set N with riN = O and E = H P N. Proof: Such a measure Ti is clearly unique, since (i) specifies its value on open sets and that together with (ii) determines its value for all Borel sets. Define a set function /1* on open sets by setting

p,*0 = sup {ilK: K c 0, K a compact ç}.

Since it is regular, it follows that, if 0 is an open Baire set, then i.z*0 = O. Let 0 = U 0i, and let K be a compact gs contained in j=1 O. Then K = U Ki with Ki c Oi by Lemma 10, and so < EtzK i < Ep*Oi. Taking the supremum over K, we have A*0 < Etz*Oi, and is countably subadditive on open sets. CO

Sec. 4)

The Borel Extension of a Measure

315

Let aa be the collection of subsets M of X with the property that for each e > 0 there is an open set 0 such that Mc0 and 12*0 < e. Any subset of a set in 01Z is again in gc, and the countable subadditivity of tt* implies that the union of a countable collection of sets in on is in NE. If E is a Baire set belonging to NE, then = sup K: K c E, K a compact 9,3}. But for each K C E and each E > 0 there is an open set 0 D K with 12*0 < E. Thus 12K < E by the definition of kt * O. Since E was arbitrary, AK = 0. Consequently, = 0, and the hypotheses of Proposition 12.30 are satisfied. Thus there is an extension g of A to a o--algebra 03 containing art and the Baire sets such that 7.-LM = 0 for M c Each B c GI is of the form H A M with H a Baire set, M c and 72B = AH. Since M is contained in an open set of arbitrarily small measure, it follows from Problem 10 b. that if 13 < co, then there is an open set 0 with B C 0 and 12*0 < Let 0 be an open set with 11*c0 < oc. If K is a compact 93 contained in 0, then 72*(0 = tz* 0 — 12 K. Let WO a sequence of compact 93's in 0 such that 1.4*0 = lim 12Kn . Then H = UKK7, isa n Baire set contained in 0, and, since 0 —Hc0 and H e (317,. Thus we have 0 I2*0, and 0 c 03. Let a be the a-algebra consisting of all sets which are locally measurable with respect to 72, that is, of all sets E such that EnBe@ for each Be@ with gB < co. Since each subset of 03 of finite measure is contained in an open set of finite measure and each open set of finite measure is in it follows that E is in a if and only if the intersection of E with each open set of finite measure is in 63. Consequently, a contains all open sets and hence all Borel sets. Extend g to a by setting TX = co if E e a 63. If we restrict g to the Borel sets, we have a Borel measure which is an extension of A and satisfies properties (i) and (iv) of the theorem. If E is a Borel set not in 03, then every open set 0 containing E has = c,0 and TX = co. Thus (ii) holds for such E. Let E be a set in og. Then E = H A M, where H is a Baire set and M c Since H is contained in an open Baire set 0 1 with 110 1 < E and M in an open set 0 2 with /1 02 < E, we have Ec 0 1 u 0 2 with 2e. This proves (ii). 401 u 0 2) < pE To prove (iii), let TLE < co. Then there is an open set 0 D E { A

—

with 72(0 —

< . By (i) there is a compact set C C 0 with

Measure and Topology

316 E

[Ch. 14 E

C) < — and by (ii) an open set U D (0 — E) with A. 0 < 2 2

Set K = C n U. Then K c E, and 17K > TiC — > TO — E > 2

Problem 17. a. Let f be a Borel measurable function on a locally compact Hausdorff space X and 1.4. a (y-finite Borel measure on X satisfying conditions (i) and (ii) of Theorem 11. Then there is a Baire measurable function g such that f = g a.e. [A. b. What if i.t. is not (7-finite? c. If f > 0, can we always choose g with 0 < g < f?

15 1

Mappings of Measure Spaces

Point Mappings and Set Mappings

Let X and Y be any two spaces and cp a mapping of X into Y. Associated with cp are several mappings of objects associated with Y into corresponding objects associated with X. For example, the set mapping 43 defined by cb(E) = yo-1 [E] is a mapping of the subsets of Y into the subsets of X. This mapping preserves unions, intersections, and complements. It is called the set mapping induced by or adjoint to go. We refer to (if) as a point mapping. If (X, a) and ( Y, 03) are measurable spaces, the point mapping yc, of X into Y is called measurable if ço — 1 [E] c a for each E c 63. Thus cp is measurable if 4, maps 03 into a. If f is a real-valued function on (X, a), then f is a mapping of X into R, and f is measurable with respect to a if and only if f is a measurable mapping of (X, a) into (R, 133) where ild is the a-algebra of Borel sets. A real-valued function of a real variable is Lebesgue measurable if it is a measurable mapping of (R, Mt) into (R, 63), where fflz is the class of Lebesgue measurable sets. It is Borel measurable iff it is a measurable mapping of (R, ea) into (R, (O. Also associated with ço is the mapping (do* of the space of realvalued functions on Y into the space of real-valued functions on X defined by v*(f) = fo ço. The mapping ço* is often called the adjoint of (to, and it preserves sums, products, maxima, etc. If ço is measurable, then ço* takes measurable functions into measurable functions. 317

318

Mappings of Measure Spaces

[Ch. 15

Let a be an algebra of subsets of X and 03 an algebra of subsets of Y. Then a mapping cP of 03 into a such that ci.( Y) = X, cD(É) = —43(E), and c1, (A U B) = cD(A) U cl.(B) is called a (lattice) homomorphism. If a and 03 are a-algebras and cl) has the property that cf. (

E1) =J cP(E1), then cP is called a cr-homomorphism. Every

set mapping induced by a point mapping of X into Y is a a--homomorphism, but we can have cr-homomorphisms which are not induced by any point mapping (Problem 2). Let (X, a) and ( Y, 03) be measurable spaces and 4) a 0--homomorphism of 03 into a. Then induces a mapping ci,* of measures on (X, a) into measures on ( Y, 03) if we define ci, */.1 by (4)* )(E) = y(t(E)). The following proposition may be thought of as a changeof-variable formula. 1. Proposition: Let (p be a measurable point mapping of the measure space (X, a, I.4) into the measurable space (Y, 03). Let 4' be the induced set mapping of 03 into a. Then for each nonnegative measurable function f on Y we have

fy f g = Proof: The proposition is clearly true if f is a characteristic function. From this it follows for f a simple function. Since f f is the supremum of the integrals of all nonnegative simple functions less than f, the proposition follows. I

Problems 1. Let 03 be a family of subsets of Y which contains Y, 0, and each set {y} consisting of a single element. Then each mapping 4) of 03 into the subsets of X which preserves finite intersections and arbitrary unions and for which 4)(Y) = X and 4)(0) = 0 is induced by a point mapping. [The sets 4'({y}) are disjoint and their union is X. Let ço(x) = y for x in Ey .] 2. Let X = Y = [0, 1], and let a be the collection of all subsets of [0, I] which are either countable or the complements of countable sets. Then a is a a-algebra. Let 63 = { Y , Ø}. For E r a, define 43(E) = 0 if E is countable, 4,(E) = Y if E is the complement of a countable set. Then 4) is a a-homomorphism and it is not induced by any point mapping of Y into X. 3. Show that the adjoint mapping ço* can be extended to map the

Sec. 2] Measure Algebras

319

extended real-valued functions on Y into the space of such functions on X. Generalize Proposition 1 to this case. 4. Let (X, a), ( Y, Ca), and (Z, e) be measure spaces and yo: X—* Y and V.: Y —f Z measurable mappings. Show that ti. . io is a measurable mapping of (X, a) into (Z, e). What has this to do with the fact that, if f is a Lebesgue measurable and g a Borel measurable real-valued function, then g of is Lebesgue measurable, but f 0 g need not be? 5. Prove that cD*1.4 is a measure. 6. a. Let so be a measurable point mapping of (X, a, 12) into ( Y, 63, y) and cl) the induced set mapping of 63 into a. Suppose VIA is absolutely continuous with respect to v and that y is a finite (or a-finite) measure. dtz

Define [— to be the Radon-Nikodym derivative of VA with respect dv to v. Then for each nonnegative measurable function f on Y we have (f o io)dtt = f

y

clv

b. Let f be a nonnegative measurable function on [0, 1] and g a monotone absolutely continuous function on [0, 1] with g(0) = 0, g(1) = 1. Then 1 fol f[g(t)]g'(t) dt = f f(t) di. 7. a. Let (X, a) and (Y, Co3) be measurable spaces and 4 a cr-homomorphism from 63 to a. Show that there is a unique linear mapping T4, of the measurable real-valued functions on Y into the measurable realvalued functions on X, which takes nonnegative functions into nonnegative functions, such that for characteristic functions xE we have T,p(xE) = XcE)•

b. Let z be a measure on (X, a) and f a nonnegative measurable function on Y. Then

Ix TD(f) diu = fy f de (A). 2

Measure Algebras

A Boolean algebra is a set of elements on which there are defined two binary operations V and A and a unary operation which satisfy the following rules: i. A VA = A, . A A A = A,

320

Mappings of Measure Spaces

[Ch. 15

ii.AVB=BV A, W.AAB=BAA, iii. (A v B) vC=Av (B V C), iii' . (A A B) A C=A A (B A C), iv. A v (B A C) = (A V B) A (A VC), v (A A C), iv' . A A (B v C) = (A A v. (A A = A' v B', vi. (A')' = A, vii. 30 such that A A 0 = 0 and A v 0 = A, viii. A' A A = 0. One example of a Boolean algebra is an algebra a of subsets of some set X with V, A, ' interpreted to mean u, n, —. We shall sometimes call V, A, 'union', 'intersection', and 'complementation'. A Boolean algebra a becomes partially ordered if we define A < B to mean A A B = A. Then 0 is the smallest element, while X = 0' is the largest element. Moreover, A v B is the smallest element of a which is larger than both A and B. A Boolean algebra a is called a Boolean o--algebra if for each sequence (A n ) of elements of a there is a smallest element B such

that A n < B for all n. This element B is denoted by V A. In a n=1

Boolean a-algebra the element C — (V c.' An' )' is the largest element n= 1

such that C < A n for all n. We write C =

A A.

One example of a

n=1

Boolean a--algebra is given by a a-algebra of subsets of a set X. Another is the following: Let (X, a, kt) be a measure space and 9z, the family of sets of measure zero. Two sets of a are said to be equivalent modulo at if their symmetric difference is in N. Finite or countable unions and intersections of equivalent sets are equivalent, and so the classes of equivalent sets form a Boolean a-algebra. We denote it by a/9z. In fact, the only properties we need of 91, are (i) if A E 9Z, and B e a with B c A, then B e gt, and (ii) if A n e 9i, then U A r,

91. A subset of a Boolean a-algebra a with these properties

n=1

is called a e-ideal, and we may define the Boolean a-algebra a/9z of equivalence classes of a mod N. By a measure algebra, we mean a Boolean cr-algebra a together with a nonnegative real-valued function g defined on a such that

Sec.

2]

321

Measure Algebras

p(A) = 0 if and only if A = 0 and 1.4 (V Az) =

E

/Lit

if

A A A, = 0 for i j. We call ,u a measure on a. If (X, 63, 11) is a finite measure space and oz is the collection of sets of measure zero, then we obtain a measure algebra if we consider kt on the Boolean cr-algebra a = ca/a, that is, if we fail to distinguish between sets of 153 which differ by a set of measure zero. In a Boolean algebra we define the symmetric difference A A B of two elements by A A B = (A A B') V (A' A B). A measure algebra becomes a metric space if we define p(A, B) = I.L(A A B). A', This metric space is always complete, and the mappings A (A, B) ---> A v B, and (A, B) A A B are continuous. A measure algebra is called separable if it is separable as a metric space. A mapping (D of a measure algebra (a, /1) into a measure algebra y) is called an isomorphism into if cD(A') = [43(A )]' , «A 1 y A 2) = «A 1 ) V «A 2) and I.L(A) = y((D(A)). The mapping (D is called an isomorphism if it is onto. If the measure algebras are considered as metric spaces, an isomorphism is an isometry which preserves complements and finite unions. It follows that an isomorphism also preserves countable unions and intersections (Problem 10). An element A 0 in a measure algebra a is called an atom if B < A can occur only for B = A and B = 0. Let ar6 be the class of measurable subsets of [0, 1], oz the class of subsets of measure zero, and m Lebesgue measure. Then (91 6/91, m) is a separable measure algebra without atoms. The following theorem asserts that, apart from isomorphism, it is the only such measure algebra. 2. Theorem (Carathéodory): Let (a, ,u) be a separable measure algebra with p(X) =- 1. Then there is an isomorphism ct. of (a, /1) into the measure algebra (fflz/91, m) induced by Lebesgue measure m on [0, 1]. The isomorphism 49 is onto jf a has no atoms.

Proof: Since (a, A) is separable, there is a sequence (A n ) of elements which are dense in a. Let an be the Boolean algebra obtained by taking all unions of intersections of the sets A 1 , . . • , A.

and their complements, and let a. = U an- Then a. is again a Boolean algebra. For, given any A and B in a„, they belong to an and am, respectively. But if m < n, we have am c an, and so A', A y B, and A A B are all contained in a c a... 71=1

322

Mappings of Measure Spaces

[Ch. 15

We shall define by induction a mapping I. of a. into the algebra g of all finite unions of half open subintervals of [0, 1). The algebra al consists of the four sets 0, A1, Ai, X, and we have ,uA l = uX = 1. Let «A 1) = [0, A 1), = 1),cl)(0) = 0, 1.(X) = [0, 1). Then 4) preserves unions, intersections, complements, and measure. Suppose now that l) has been defined on an _ i so that it

maps an _ i onto the algebra generated by the half open intervals [0, x 1), [x 1 , x2), • • • , [xk, 1) and that 4) is measure preserving and preserves unions and complements. We wish to extend the mapping 4) to an . Let Bo, • • • , Bk be the sets in an_i which are mapped onto the intervals [0, x 1), . . , [xk , 1). Then an consists of all finite unions of the sets Bo, . . • , Bk, and an consists of all finite unions of A n A B o, ,A n A Bk, A n' A B0, . . . ,A, A Bk. For those intersections which are 0, set 4) equal to Ø. For those which are not 0, let 4)(An A B3) = [x3, xi + 1.1 (An A B3 )) and 43(A n' A B = [x; m(A„ A B1), xi+1). Since u(An A B) + 12(A'n A B,) = 1.1(13;) = x 11 — x j, we see that these are properly defined intervals, 4. is measure preserving, and «A n A Bi) U «A A B1) = x1+1) = .t(B1). From this it follows that we can extend I. to all of an so that it preserves unions, complements, and measures. Thus we have defined by induction the mapping I. from a,, to urc/gt so that it is measure preserving. Hence it is an isometry. Since a. is dense in a and the metric space arz/gt is complete, we can extend l) to be an isometry from a into art/a. To see that l) preserves complements, let E be any element in a and choose A e a. so that ,u(E A A) < E. Then A' e a., and ,u(E' A A') = ,u(E A A) < e. Since 43 is an isometry and 43(A') = we have m(4)(E') A 43(A)) < e and m(4)(E) A 43(A)) = m(4)(E) A 4)(A)) < 6. Hence m(4)(E') A 44E)) < 2€ for all e > 0. Thus 43(E') = —43(E) in the algebra arz/91. A similar argument shows 43(E V F) = 4)(E) u 4)(F). Since 4) is an isometry, it is one-to-one into. To identify the range of 4), let E be the set of endpoints of those intervals which were used in defining the mapping 43 on the algebras an . Then maps a.

onto the algebra of finite unions of half-open intervals with endpoints in E. Suppose that r is not all of [0, 1], and let I be one of the open is composed, that is, an interval conintervals of which [0, 1) tained in —1-7 whose endpoints lie in E . Since the endpoints of / lie in .E, I is a limit of intervals in (DN.], and so lies in 1.[a]. Let —

Sec. 2]

323

Measure Algebras

A e a be such that (A) = I, and let B be an element of a with B < A. Then ci (B) c I. Since B can be approximated by elements in ao,„ ck(B) can be approximated by elements in (14a,,,]. But these latter are sets which either contain I or do not meet I. Hence 4)(B) = I or (D(B) = Ø, and we have B = A or B = Ø, since ci is one-to-one. Consequently, A is an atom. We have thus proved that if a has no atoms, then E' = [0, 1]. But if E = [0, 1], then every half-open interval is in [a], and hence ci)[a] contains every Borel set. Since every measurable subset of [0, 1] is the union of a Borel set and a set of measure zero, we have

in this case

cD[a] = TC/91. I

This theorem states that, if (X, 61, ,u) is a separable measure space without atoms for which A(X) = 1, then the corresponding measure algebra is isomorphic to the measure algebra induced by Lebesgue measure on [0, 1 1 . It does not assert the existence of a point mapping between [0, 1] and X, or even of a set mapping of 61 into the measurable sets of [0, 1], but only of a correspondence of sets of modulo null sets with measurable sets modulo sets of measure zero. Thus if we have a set B c 61, we do not assign to it a particular measurable set but only an equivalence class of measurable sets in [0, 1]. In the next sections we shall develop criteria for asserting the existence of point mappings which induce given mappings of measure algebras.

Problems 8. Prove that in a Boolean a-algebra we have B 9. Let

A (V

A„) =

(B

A

A n).

a be a Boolean a-algebra and at a Boolean a-ideal. Show that then A' B' c D-C, and that if (A n A B„) EN. then

if A ABe 01,

(V AO A (V Bn) 10. a. Let (a, ,u) be a measure algebra and (A n ) a sequence of elements

such that A n A A, = 0 for n X

m.

Then V A = lirn V A. (Here n= 1

k—,00 n=

lim means limit in the metric space defined by 1.t.)

Let (a, 1.4) be a measure algebra and oe, elements in a. Then V A = lim V A. k—*co n=1 n=1 b.

(A n )

any sequence of

324

Mappings of Measure Spaces

[Ch. 15

c. Show that if c is an isomorphism of a measure algebra (a, A) CO oo into a measure algebra (CB, y), then cD ( ci An) = V 49(11 . )

11. Prove that a measure algebra is complete as a metric space. [If (A s ) is a Cauchy sequence we may assume that u(A n A A m) < 2—N for n, m > N. Then if B, = 11 V A„ we have 1.4(A, A Bn) < 2—n+ 1 . Now v=n

A

n=1

Br, = lim 13, = lim An.]

12. Show that in a measure algebra the operations ', A, and V are continuous. 13. Show that a measure algebra (as we have defined it with /L(X) < co) can have only a countable number of atoms. Hence any complete separable measure algebra is isomorphic either to an interval (with Lebesgue measure), to a measure space consisting of a countable number of atoms (discrete measure space), or to a measure space which is the union of the preceding two. 14. Discuss measure algebras in which we allow /J. to be an extended real-valued function. 3

Borel Equivalences

If (X, a) and ( Y, 03) are measurable spaces, we may ask for conditions under which they are equivalent in the sense that there is a one-to-one mapping yo of X onto Y such that yo and cp -1 are measurable, that is, such that (p[A] 6 03 for each A c a and yo —l [B] c a for each B c G. In the present section we shall always assume that X and Y are metric spaces and that a and 03. are the algebras of Borel sets. A one-to-one mapping of X onto Y which takes Borel sets into Borel sets and whose inverse takes Borel sets into Borel sets will be called a Borel equivalence. Thus any homeomorphism is a Borel equivalence, but we may have Borel equivalences which are not homeomorphisms. If X is Borel equivalent to Y and Y Borel equivalent to Z, then X is Borel equivalent to Z. We begin by establishing some lemmas: 3. Lemma: Let X be a metric space and E a Borel subset of X. Then the Borel subsets of E are those Borel subsets of X which are contained in E. Proof: Since E is a Borel subset of X, 0 n E is a Borel subset of X for each open subset 0 of X. Hence the Borel subsets of X which

Sec. 3] Borel Equivalences

325

are contained in E form a a-algebra which contains the (relatively) open subsets of E and so must contain every Borel subset of E. On the other hand, the collection of those subsets of X each of which is the union of a Borel subset of E and a Borel subset of X — E is a a-algebra which contains the open subsets of X. Hence it must contain all Borel subsets of X, and we see that a Borel subset of X which is contained in E must be a Borel subset of E.I 4. Lemma: Let X and Y be metric spaces, and let X0 and Yo be countably infinite subsets of X and Y. Then each Borel equivalence between X — X 0 and Y — Yo can be extended to a Borel equivalence between X and Y. Proof: Since X0 and Y0 are countably infinite, we can extend the Borel equivalence between X — X 0 and Y — Yo to be a one-to-one correspondence between X and Y. Since any countable subset of a metric space is an Fo., the union and difference of a Borel set and a countable set are again Borel sets. Thus a subset of X [or Y] is a Borel set if and only if its intersection with X — X0 [or Y — Yo] is a Borel set. Hence the extended correspondence between X and Y is a Borel equivalence. I

If X is a topological space and A any set, we use X A to denote the product space X X, where each X, = X. The set X A is the set of all aeA

mappings of A into X. If X is metric and A is countable, then X A is metrizable (cf. Problem 8.27). We shall use w to denote the set of integers. The following lemma is an immediate consequence of the fact that co X co is countable. 5. Lemma: If X is a topological space, Xc° and (XI" are homeomorphic.

One of the simplest topological spaces is the space whose elements are 0 and 1 and whose topology is discrete. We often denote this space by 2. The product space r is a compact metric space. 6. Proposition: The unit interval [0, 1] is Borel equivalent to 2".

Proof: Let X 0 be that subset of 24' consisting of those elements which have only a finite number of coordinates equal to 0 and of

326

Mappings of Measure Spaces [Ch. 15

those which have only a finite number of coordinates equal to 1. Then X 0 is countably infinite. On 26) — X0 define yo as that mapping

E E,2 -1, where E, is the ith coordinate of x. 2= Then ye, maps 2 — X0 in a one-to-one manner onto those elements of [0, 1] which are not of the form p 2 —n with p and n integral, and cp is readily seen to be a homeomorphism between r — X 0 and these elements of [0, 1]. Since the set of elements of [0, 1] of the form p 2 -n is countably infinite, the proposition follows from Lemma 4. I which sends x into

7. Proposition: Let I be the interval [0, 1 ]. Then Iw and I are Borel equivalent.

Proof: By Lemma 5 the spaces 2w and (r)''' are homeomorphic. Thus by Proposition 6 we have I Borel equivalent to 2w, which is Borel equivalent to (216', which in turn is Borel equivalent to I

8. Theorem: Each complete separable metric space is Borel equivalent to a Borel subset of [0, 1]. Proof: Since I = [0, 1] and I" are Borel equivalent, it suffices to show that each complete separable metric space (X, p) is homeomorphic to a Borel subset of P. We may assume without loss of generality that p is bounded by one. Let (r„) be a dense sequence of points, and let f be that mapping of X into P which assigns to x the point whose ith coordinate is p(x, r„). Then f is a one-to-one mapping of X onto a subset E of P, for, if x y, there is an r, such that p(x, r,) < p(y, r2). Since X and ./w are metric, the mapping f will be continuous if x„ x implies f(x) —f(x). But if x,„ x, then p(x„, r,) p(x, r,) for each i, and so each coordinate of f(x) converges to the corresponding coordinate of f(x). Thus f(x) f (x). (cf. Problem 8.26). The mapping f —1 will be a continuous mapping of E onto X if f(x) f(x) implies xn x. But if f(x) f (x), p(x, r,) for each i. Given E > 0, choose r, so that then p(x„, r,) p(x, r,) < E/3 and choose N so that Ip(x„, r) — p(x, r i)i < e/3 for n > N. Then, for n > N, we have p(x„, r) < 2E/3, and so < E. Consequently, f is a homeomorphism of X onto E. P(xn, It remains only to show that E is a Borel set. But E is dense in F = E. By Theorem 8.11 the completeness of X implies that E is a

Sec.

3]

Borel Equivalences

327

gs relative to F and hence a Borel set relative to F. Thus E is a Borel subset of P. I Actually, a somewhat stronger statement is true: Each complete separable metric space is Borel equivalent to [0, 1] (see Kuratowski [16], p. 227). We shall only make use of the weaker statement given in Theorem 8. 9. Theorem: Let X be a complete separable metric space and A a Borel measure on X such that AX = 1 and such that ,u{x} = 0 for each set fx; consisting of a single point. Then there is a Borel set X 0 c X with ,uX 0 = 0, and a subset Yo of [0, 1] of Lebesgue measure zero, such that there is a Borel equivalence 1,t of X - X 0 with [0, 1] - Y o having the property that p.(1,t -1 [A]) = mA, where m is Lebesgue measure on [0, 1].

Proof: Let (p be a Borel equivalence of X with a Borel subset E of [0, 1], and define the Borel measure v on [0, 1] by vA Let f be the cumulative distribution function of v, that is, f(x) = v[0, x]. Then f is a monotone nondecreasing function on [0, 1] which is continuous on the right. Since f(x) - lim f(t) = v(xl = t--u-14p -i [x]} = 0, we see that f is continuous. Since f(0) = 0 and f(1) = v[0, 1] = AX = 1, f is a continuous mapping of [0, 1] onto [0, 1]. For each y e [0, 1], the set f -l [fyl ] is a closed set, and since f is nondecreasing it is either a closed interval or a single point. Let M be the set of y such that f -1 [{y1 ] is a nondegenerate intèrval. Since these intervals are disjoint, M is countable. Now mE = v(f -l [E]) by Problem 12.12, and so the set N = f-l [M] has v-measure zero. Now f is a homeomorphism of [0, 1] - N onto [0, 1] - M. Since N is a Borel set, the set X 0 = cp -1 [N] is a Borel set, and AX0 = vN - O. Thus the mapping ly = fo (p, is a Borel equivalence of X - X 0 with 1,G[X - X 0]. Since (p is a Borel equivalence of X onto a Borel subset of [0, 1], ço[X - X0] is a Borel subset of [0, 1] contained in [0, 1] - N and hence a Borel subset of [0, 1] - N. Thus X0] = f [ço[X - X 0]] is a Borel subset of [0, 1] - M and hence of [0, 1]. Thus 11, is a Borel equivalence of X - X 0 with a Borel subset of [0, 1]. Since mA = ,u( 1 [A]), we have m(Iti[X - X0]) - 1. ( X - X0) = 1, and we see that the set Y o = [0, 1] - %G[X - X0] is a Borel subset of Lebesgue measure zero. I

Mappings of Measure Spaces

328

[Ch. 15

Problems 15. Let X be a complete separable metric space and y a Borel measure on X. Let E be a Borel subset of X with the property that if A is a Borel subset of E, then AA = 0 or i.t(E — A) = O. Then there is a point x contained in E such that ,u(E {X}) = 0. [Use Theorem 9.] 16. Let y be a Borel measure on a space X. Then A=A0-1-Ai where = 0 for each x c X and ,u 0E =

E

.

xeE

4 Set Mappings and Point Mappings on Complete Metric Spaces

We say that (X, a, 9-c) is a measurable space with null sets if a is a a--algebra of subsets of X and 91 is a a--ideal in a. Each measurable point mapping go of X into [0, 1] induces a cr-homomorphism 43 of the Borel subsets of [0, 1] into the a-algebra a/91 by letting 1.(A) be the equivalence class containing cp -1 [A]. The following theorem states that, conversely, every a-homomorphism of the Borel subsets of [0, I] into a/m is induced in this way by a point mapping go of X into [0, 1]. It is essentially Lemma 11.10 in another guise. 10. Theorem (Sikorski): Let (X, a, 91) be a measurable space with null sets, and let 43 be a a—homomorphism of the family 133 of Borel sets in [0, 1] into the a--algebra a/91 with .1,([0, 1]) = X. Then there is a measurable mapping cp of X into [0, 1] such that for each B c (B we have yo -1 [B] in the equivalence class 4,(B). If p is any other point mapping with this property, then ‘1. = ia except on a subset in N.

Proof: For each rational number a in [0, 1] let A, be a set in the equivalence class ,t.([0, a]). We may take A = X. If a < 0, then I.([0, a]) < 4)([0, 0]), and so the set Ea o = A, — A o is in 9Z. Let E = U Eafi , where a and 13 run through the rationals. Since this ,

a
is a countable union and at is a a-ideal, we have E c N. Set B, A a u E. Then Ba is in the equivalence class .t.([0, a]), and Ba C B o for a < 3. For each x e X, let go(x) = inf (a: x e B al-. Since B 1 = X, ça(x) is defined for each x, and 0 < go(x) < 1. Thus ça is a mapping of X into [0, 1]. For each t we have {x: ça(x) < t} = U B a. a<1

Sec.

4]

Set Mappings and Point Mappings on Complete Metric Spaces

329

Hence ço is measurable, and ço -1 [[0, a]] = Ba. If we let NI, be the u-homomorphism of 63 into a/oz induced by cp, then NI/ agrees with ci) on the closed intervals with rational endpoints. Since and ci) are u-homomorphisms, the family of sets on which they agree is a if-algebra and so must be the family 63 of Borel sets. If o is any other mapping of X into [0, 1] which also induces the u-homomorphism cic then for each pair a and 3 of rational numbers the set {x: ç(x) < a < < ti.(x)j = a]] — IA] and Thus we see that the set tx: ç(x) < o(x)1 = so must be in (10(x) < a < < V/(x)} is in M. Also ix: (p(x) > 0(x); c oz, U a,

and so (x: go(x)

o(x); c

M. I

11. Theorem: Let (X, a, M.) be a measurable space with null sets, Y a complete separable metric space, and cla a Œ-homomorphism of the Borel sets 63 of Y into a/oz with 4)(Y) = X. Then there is a set X0 e T. and a point mapping cp of X — X 0 into Y such that for each B e 63 we have (p —i [B] in the equivalence class cla(B).

Proof: By Theorem 8 there is a Borel equivalence X of Y with a Borel subset E of [0, I]. For each Borel subset B of [0, 1] let if(B) = ci)(X -1 [B n E]). Then there is a mapping o of X into [0, 1] which induces Nir. Let X0 = 0 -1 [É]. Since O -1 [É] = 43(0), we have X0 e Dz. On X — X 0, the mapping X -1 . o is defined and induces the u-homomorphism F. I A u-homomorphism trip from a Boolean if-algebra a to a Boolean if-algebra 63 is called a if-isomorphism if there is a u-homomorphism NIf from 63 to a such that . 43 and 43 . 4, are the identity mappings on a and 63. The following theorem is a generalization of a theorem of von Neumann, who assumed the unnecessary hypothesis that be measure preserving with respect to suitable measures on a and B. 12. Theorem: Let X and Y be complete separable metric spaces, a and cB their Borel sets, and let T-c and oz be o--ideals in a and (B. If cl) is a if-isomorphism of On onto (B/gt then there are sets X 0 e and Yo e at and a one-to-one mapping cp of Y — Yo onto X — X 0 and cp — ' are measurable and 4)(A) = p 1 [A] modulo M. suchtayo

Proof: By Theorem 11 there is a set Y 1 e 91 and a mapping cp of Y — Y 1 into X such that 4)(A) = cp 1 [A] modulo oz. Similarly,

Mappings of Measure Spaces

330

[Ch. 15

there is a set X1 e OR and a mapping ;I/ of X X 1 into Y such that *(B) = irt [B]. Since 430 4/ is the identity on 03/M, the mapping 0 so can differ from the identity mapping on Y — Y1 only on a subset Y2 e DI. Let Yo = Y1 u Y2. Then so is one-to-one on Y — Yo, and so—a [A ] — 43(A) modulo Dz, so[B] — *(B) modulo Mt Thus so[ Y — Yo] differs from X by a set X0 e NC. I —

Unfortunately, the exceptional sets X0 and Yo in the preceding theorem may be unavoidable (see Problem 17), but they may be eliminated in special cases. One is given in Problem 18, another by the following theorem. 13. Theorem: Let X be a complete separable metric space, ea the family of Borel sets of X, and 91 a a-ideal of 63. If 4) is any a-isomorphism of 03/1t onto itself, then there is a one-to-one mapping ie, of X onto itself such that ie, and ie, are Borel measurable and 4)(A) = —so 1 LA] modulo N.

By Theorem 12 there are sets X0 and Yo in Dz. and a one-toone Borel measurable mapping ‘G of X Yo onto X Xo. Let Zo = X0 u Yo, Z1 = ;G[Z o] u 1G — 'Vol Zn+i = ‘qZ.] U 0-1 [Zn]Since ;G and 1,G -1 take sets in 9t into sets in at, we have Z„ e 9z. Set Z = U Zr,. Then Z e Di, 44Z] C Z, and V i [Z] C Z. Thus 1,1' and are one-to-one mappings of X Z into X Z, and since they are inverses of each other they must map X Z onto X Z. Define so by letting so(x) = ;G(x) if x eX—Z and so(x) = x if x e Z. Then so is a Borel equivalence of X onto itself which agrees with L. except on the set Z e Di. 1 Proof:

—

—

—

—

—

—

Problems 17. a. Show that the exceptional set X0 in Theorem 11 may be unavoidable. [Let Y be a countable space with the discrete topology, X the unit interval with 9-1. the family of all sets which contain no rational numbers.] b. Show that the exceptional sets X0 and Yo in Theorem 12 may be unavoidable. 18. Show that we can dispense with the exceptional sets X0 and Yo in Theorem 12 Won and 9-t are o--ideals of (P(X) and aci( Y) and each contains a set with as many elements as there are real numbers.

Sec. 5]

The Isometries of LP

331

19. Let X, Y, a, ea, 51 t, at be as in Theorem 12, I. a a-homomorphism of a/ort into oalgt, and let so be the mapping given by Theorem 11. a. The mapping so is one-to-one except on a set Yo e 9l if and only if there is a cr-homomorphism IF of 33/97, into a/mt with 4)o x1f the identity on 63/91. b. The mapping so is onto all of Xexcept for a set in 511 if and only if cNA) = 0 implies A = O.

5 The lsometries of

LP

We illustrate the use of the theorems in the preceding sections by deriving a characterization of the isometries of LP[O, 1] onto itself, that is, of those linear mappings U of /2[0, 1] into itself such that Uf =11. We begin by establishing two inequalities, the first concerning real (or complex) numbers and the second elements in LP. 14. Lemma: Let and n be real numbers. Then if 2 < p < o ,

IE

ni P +

2 (l+

It

while if 0