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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
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
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 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 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 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
136
Metric Spaces
[Ch. 7
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:
Sec. 6]
Subspaces
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.
138
Metric Spaces
[Ch. 7
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.
139
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
140
Metric Spaces
[Ch. 7
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.
141
Sec. 7] Baire Category
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
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Fundamental Notions
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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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Bases and Countability
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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.
Sec. 3]
The Separation Axioms and Continuous Real-Valued Functions
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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 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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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
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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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Compact Spaces
[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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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
164
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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
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
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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
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
178
Compact Spaces
[Ch.
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]
179
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)] :
180
Compact Spaces
[Ch.
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
182
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
Banach Spaces
[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
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
186
Banach Spaces
[Ch. 10
for all n > N. Thus for n > N, IIA. — AI! = sup 11(A. — A)x11
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).
188
Banach Spaces
[Ch. 10
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) 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]
Topological Vector Spaces
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]
Topological Vector Spaces
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.]
200
Banach Spaces
[Ch. 10
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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[Ch. 10
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 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
Banach Spaces
[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
Banach Spaces
[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)
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 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
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.
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) _. 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)
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 a a.e. on X — Ba.
Sec. 3]
225
Integration
Proof: Let C = U B a — B o . Then oi 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
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 .
{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
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 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 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 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,
I, then o--finiteness is not required: 30. Theorem: Let F be a bounded linear functional on LP() with 1
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,
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
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)
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,))
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)
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).
294
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 ( 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( 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
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
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[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,
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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
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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
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[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
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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 2( -2)/ (a2 7'
If we replace a by a 7" 2 and S by a2 +
13 112 ,
7'
02)21 . 7'
this becomes
(3 2 > 2(p-2)/p( ap
op‘2123 )
332
Mappings of Measure Spaces
Or ap
[Ch. 15
+ 0 P < 2(2—p)/2(a 2 + 0 2)p/2 .
Since
E2
0
0 there is an n < 1 such that whenever u and y are elements of X with i lull = Ilyil = 1 and llu — vil > a we have Ilu ± vil < 2n. Use the integrated form of Lemma 14 to show that LP is uniformly convex if p > 2 and that we can t take n = (1 — EPP.
Epilogue
In the first part of the book we considered most of the central theory of functions of a real variable. This material falls roughly into three classes: sets of real numbers and their properties; measure and integration; and properties of real valued functions such as continuity and differentiability. The books of Saks [20] and Hobson [12] give an excellent account of this classical theory, particularly of those results which depend strongly on special properties of the real numbers. A very elegant presentation of this classical theory is given by Part I of Riesz-Nagy [18]. The books by Carathéodory [3] and de la Vallée Poussin [24] will give the reader some feeling for the earlier treatments of this subject. One topic from the classical theory of functions of a real variable which has been omitted is the theory of Fourier series, and a good reference here is Zygmund [25]. Another topic we have omitted is the classical theory of functions of several real variables. Although much of this theory is subsumed under the general theory of integration in Part Three, especially that dealing with Fubini's Theorem, there are many properties depending on the special structure of R. These include much of the theory of partial derivatives, change of variables, and integration by parts. Some of this material is covered in Saks [20] and Edwards [6]. In Chapters 7, 8, and 9 we considered various aspects of general topology. For a more detailed treatment of this subject Kelley [14] is an excellent reference, and you will find there further references to the literature. In addition to the topics we have discussed briefly, there is a chapter on uniform spaces which generalize the metric spaces and the topology of linear spaces so that the notions of completeness and uniform continuity are meaningful. For deeper properties of metric spaces and Borel sets, Kuratowski [16] is good. 335
336
Epilogue
In Chapter 10 we discussed briefly some of the geometrical aspects of Banach spaces, and topological linear spaces were touched upon to the extent needed for the weak and weak* topologies in Banach spaces. The theorems (Hahn-Banach, closed graph, uniform boundedness, Alaoglu, Krein-Milman) are some of the most useful general tools at the disposal of the analyst. Useful theorems that we did not mention here are the fixed-point theorems of Schauder and Leray. A thorough catalogue of the "geometrical" properties of such spaces is given in Day [4], which also includes an excellent guide to the literature. The book by Banach [1] is very readable, although it suffers from the fact that general topology was not in existence at the time it was written and the weak topological notions are treated entirely in terms of sequential convergence. A thorough treatment of topological vector spaces can be found in Kelly-Namioka [15] or in Schaefer [21]. A central topic in modern analysis that we have not mentioned here is the theory of linear operators on a Banach or Hilbert space. Discussion of these from various points of view may be found in Dunford-Schwartz [5], Edwards [6], Hille-Phillips [11], Riesz-Nagy [18]. In Part Three we have given a reasonably comprehensive discussion of abstract measure and integration. The central theorems are the convergence theorems, the Radon-Nikodym theorem, the extension theorem for measures on an algebra, product measures and the Fubini-Tonelli theorems, the equivalence of abstract integration with that based on a measure, and the representation of bounded linear functionals on C(X). For a full discussion of measure theory based on o--rings and of measure theory in locally compact spaces the reader should consult Halmos [8], which gives a very readable and thorough treatment of measure theory. For systematic treatment of the Daniell theory of integration in locally compact spaces, the book by Loomis [17] is a good reference. An important branch of measure theory which we have not touched upon is the theory of Haar measure, which is the theory of measure invariant under a group of transformations on the measure space. A nice introduction is given in Banach's appendix to Saks [20], a full treatment in Halmos [8], and a treatment in terms of the Daniell integral in Loomis [17]. Another important aspect of measure theory is ergodic theory, which is the study of the properties of a measure preserving transformation and its iterates. The reader will find readable treatment in Hopf [12] and Halmos [10]. There are a number of applications of real analysis to the theory of functions of a complex variable, and the reader will find a nice treatment of many of them in Rudin [19].
Bibliography
Théorie des Opérations Linéaires (Monografje Matematyczne, Vol. 1), Warsaw, 1932. [2] G. BIRKHOFF and S. MAC LANE, A Survey of Modern Algebra (3rd ed.), New York, Macmillan, 1965. [3] C. CARATHÉODORY, Vorlesungen fiber reelle Funktionen, LeipzigBerlin, 1927. [4] M. M. DAY, Normed Linear Spaces (Ergebnisse der Mathematik und ihrer Grenzgebiete, No. 21), Berlin, Julius Springer, 1958. [5] N. DUNFORD and J. SCHWARTZ, Linear Operators, New York, Interscience, 1958. [6] R. E. EDWARDS, Functional Analysis, Theory and Applications, New York, Holt, Rinehart and Winston, 1965. [7] A. M. GLEASON, Fundamentals of Abstract Analysis, Reading, Mass., Addison-Wesley, 1966. [8] P. R. HALMOS, Measure Theory, New York, Van Nostrand, 1950. [9] P. R. HALMOS, Naive Set Theory, Princeton, Van Nostrand, 1960. [10] P. R. HALMOS, Entropy in Ergodic Theory, Chicago, Univ. of Chicago Press, 1959. [11] E. HILLE and R. PHILLIPS, Functional Analysis and Semi-Groups (American Mathematical Society Colloquium Publications, Vol. 31), Providence, 1957. [12] E. W. HonsoN, The Theory of Functions of a Real Variable and the Theory of Fourier's Series (3rd ed.), Vol. 1, Cambridge, 1927. [13] E. HOPE, Ergodentheorie (Ergebnisse der Mathematik und ihrer Grenzgebiete, No. 5), Berlin, Julius Springer, 1937. [14] J. L. KELLEY, General Topology, New York, Van Nostrand, 1955. [1] S. BANACH,
338
Bibliography
[15] J. L. KELLEY, I. NAMIOKA, et al., Linear Topological Spaces, Princeton, Van Nostrand, 1963. [16] C. KURATOWSKI, Topologie I (Monografje Matematyczne, Vol. 3), Warsaw, 1933. [17] L. H. Loomis, An Introduction to Abstract Harmonic Analysis, New York, Van Nostrand, 1953. [18] F. RIESZ and B. NAGY, Functional Analysis (English ed.), New York, Ungar, 1956. [19] W. RUDIN, Real and Complex Analysis, New York, McGraw-Hill, 1966. [20] S. SAKS, Theory of the Integral (Monografje Matematyczne, Vol. 7), Warsaw, 1937. [21] H. H. SCHAEFER, Topological Vector Spaces, New York, Macmillan, 1966. [22] P. C. SUPPES, Introduction to Logic, Princeton, Van Nostrand, 1957. [23] P. C. SUPPES, Axiomatic Set Theory, Princeton, Van Nostrand, 1960. [24] C. J. DE LA VALLÉE POUSSIN, Intégrales de Lebesgue, Fonctions d'ensemble, Classe de Baire, Paris, 1916. [25] A. ZYGMUND, Trigonometrical Series (Monografje Matematyczne, Vol. 5), Warsaw, 1935.
Index of Symbols
A&B A V B TA A=B A B (x) (3x) 1 N xe A AcB {x e X: P(x)} —
0
{x} {x, y}
-x, Y) [a, b] (a, b) X X Y
A and B A or B not A if A, then B A if and only if B for all x there is an x Q.E.D. set of natural numbers x is an element of A A is a subset of B set of x in X with P(x) empty set singleton unordered pair ordered pair closed interval open interval direct product of X and Y
XX),
direct product
XA g 0f
direct product composition restriction off to A
X
f IA
Page 2 2 2 2 2 2 2 4 6, 33 6 6 6 7 7 7 7 38 38 7, 128, 150, 264
18, 150
339
150 9 9
340
Index of Symbols
(x,) 1
finite sequence infinite sequence set of subsets of P
(P(X)
A nB AuB 21 A AB
n
Ace
A
n {A: A
c e;
R x V y
XA y lim 50-1 9s) 1(1) aft
XA C[0, 1] Rn Sx , a
E ,u 1 y a.e. [i.t] 1,
