Folland: Real Analysis, Chapter 7 Sébastien Picard

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1 Folland: Real Analysis, Chapter 7 Sébastien Picard Problem 7.2 Let µ be a Radon measure on X. a. Let N be the union of all open U X such that µ(u) =. Then N is open and µ(n) =. The complement of N is called the support of µ. b. x supp(µ) iff fdµ > for every f C c (X,[,]) such that f(x) >. (a) Since N is the union of open sets, we know that N is open. Therefore, by inner regularity we have µ(n) = sup{µ(k) : K N, K compact}. For any K N such that K is compact, we see that K is covered by finitely many sets U i X, i =,...,n, such that µ(u i ) =. Therefore µ(k) n µ(u i ) =. i= Since the measure of N is the supremum over all such µ(k), we have µ(n) =. (b) Suppose x supp(µ). Take f C c (X,[,]) such that f(x) = c >. Now, V = f ((c/2,)) is an open set by continuity. Since x V, we have µ(v) >. Hence fdµ fdµ c µ(v) >. 2 V On the other hand, let x X and suppose fdµ > for every f C c (X,[,]) such that f(x) >. Since singletons are compact, {x} is compact. Take any open set U such that x U. By Urysohn s lemma, there exists a function g C c (X,[,]) such that g(x) = and supp(g) U. Then by formula (7.3) in the statement of the Riesz Representation theorem, µ(u) = sup{ fdµ : f C c (X),f U} gdµ >. Problem 7.4 Let X be a LCH space. a. If f C c (X,[, )), then f ([a, )) is a compact G δ set for all a >. b. If K X is a compact G δ set, there exists f C c (X,[,]) such that K = f ({}).

2 c. The σ algebra B X of Baire sets is the σ algebra generated by the compact G δ sets. (a) Since f is continuous, f (a /n, ) is open for all positive integers n. By the equality we see that f ([a, )) is a G δ set. f ([a, )) = f (a n, ), n= To show compactness, we first notice that by continuity f ([a, )) is closed. Since a >, we have f ([a, )) suppf. Since the support of f is compact, and closed subsets of compact sets are compact, we see that f ([a, )) is compact. (b) Write K = U i, i= where U i are open sets. By Proposition 4.3, there exists a precompact open W such that K W U. Define the open sets V n as ( n V n = U i ) W. i= Notice that K = V n and that V V i V i+... From Urysohn s lemma, for each positive integer n, there is an f n C c (X) such that f n, supp(f) V n and f = on K. Define f = k= 2 kf k. The function f(x) is well-defined since the pointwise sum converges absolutely: k= 2 k f k(x) k= 2 k = <. Actually, the sum is absolutely convergent in the uniform norm: k= 2 k f k u k= 2 k =. By Theorem 5., since BC(X) is a Banach space, every absolutely convergent sequence in BC(X) converges, hence f BC(X). The support of each f n is contained in V n W, and hence suppf W. Since W is precompact, f has compact support. Since f, we have f C c (X,[,]). 2

3 We now show K = f ({}). Indeed, if x K, then f n (x) = for all positive integers n, hence f(x) = =. If x / K, then there exist a positive integer i such that x / V 2 k j for all integers j i. Then x / suppf j V j and f j (x) =. Therefore f(x) = i k= 2 kf k <. Hence x / f ({}). (c)theσ algebrab X ofbairesetsisbydefinitiongeneratedbythesets(ref) ([a, )),(Imf) ([a, )), for all a R and all f C c (X). Let A denote the σ algebra generated by compact G δ sets. Let K be a compact G δ set. Then K = f ({}) = f ([, )) for some f C c (X,[,]) by part (b). Then K BX, and therefore A B X by Lemma.. On the other hand, let f C c (X). First, we assume that f is a non-negative real-valued function. In this case, if a >, from part (a) we know that f ([a, )) A, and f ([ a, ) = f ([, )) = X A. Therefore f ([a, )) A for all a R. Next, suppose f C c (X) is real-valued. We can decompose f as f = f χ f f χ f< := f + f. Noticethatf +,f arenon-negativereal-valuedfunctionsinc c (X). Ifa >,wehavef ([a, )) = (f + ) ([a, )) A. Next, notice f ([, )) = (f ) ({}) = ((f ) ((, ])) c, and since (f ) ((, ]) = (f ) ([/n, )), we have f ([, )) A. Lastly, we have f ([ a, )) = (f ) ([,a]) = ( ( c (f ) ((a, )) )c = (f ) ([a+/n, ))). Therefore, f ([a, )) A for all a R. n= If f C c (X) is complex-valued, we decompose f into its real and imaginary components, which are both real-valued functions in C c (X). Then (Ref) ([a, )) A, and (Imf) ([a, )) A, for all a R. It follows that BX A by Lemma.. n= Problem 7. If µ is a Radon measure and f L (µ) is real-valued, for every ǫ > there exists an LSC function g 3

4 and USC function h such that h f g and (g h)dµ < ǫ. First, we decompose f into its positive and real parts: f = f χ f f χ f< := f + f. Since f L (µ), the sets {x : f + > } and {x : f > } are σ-finite by Proposition 2.2. Hence by Proposition 7.4, there exists functions g,g 2 such that g is LSC and g f +, g 2 is USC and g 2 f, and ǫ 4 + f + dµ g dµ, ǫ 4 + f dµ g 2 dµ. We know that g 2 is LSC, so by Proposition 7. we have that g := g g 2 is LSC. Futhermore, g f. Again by Proposition 7.4, there exists functions h,h 2 such that h is USC and h f +, h 2 is LSC and h 2 f, and ǫ 4 + f + dµ h dµ, ǫ 4 + f dµ h 2 dµ. Since h 2 is USC, we have that h := h h 2 is USC. Furthermore, f h. We now compute (g h)dµ = g dµ g 2 dµ h dµ+ f + dµ+ ǫ 4 f dµ+ ǫ 4 h 2 dµ f + dµ+ ǫ 4 + f dµ+ ǫ 4 = ǫ. Problem 7. Suppose that µ is a Radon measure on X such that µ({x}) = for all x X, and A B X satisfies < µ(a) <. Then for any α such that < α < µ(a) there is a Borel set B A such that µ(b) = α. We define 4

5 P = {B A : B = U A, U open,and µ(b) α}. First, we show that P contains an element that is not the empty set. Since A is non-empty, there exists an element x A, and = µ({x}) = inf{µ(v) : {x} V, V open}. Therefore, there exists an open set V containing x such that µ(v) < α. Hence V A P. We can partially order P by inclusion. Let C be a totally ordered non-empty subset of P. We show that C has an upper bound in P. Indeed, take U = B γ C It is clear that U is an upper bound of C. Since the union of open sets is open, we have that U A is open in A. Since µ(u) µ(a) <, we see that U is a σ-finite set, and hence by Proposition 7.5 we have B γ. µ(u) = sup{µ(k) : K U,K compact}. For any K compact such that K U, we see that K is covered by finitely many B γ C. Since C is totally ordered by inclusion, K B γ for some B γ C. Hence µ(k) µ(b γ ) α. By taking the supremum over all such K, we see that µ(u) α. Hence U P. By Zorn s lemma, the set P contains a maximal element B. We claim that µ(b) = α. By contradiction, assume that α µ(b) >. Take x A\B, which exists since the measure of A\B is positive. By outer regularity, there exists an open set V such that x A V and µ(v) < α µ(b). But then B (A V) is open in A and µ(b (A V)) µ(b)+α µ(b) α. Hence B B (A V) and B (A V) P, contradicting the maximality of B. 5

6 Problem 7.2 Some examples of nonreflexivity of C (X): a. If µ M(X), let Φ(µ) = x X µ({x}). This sum is well defined, and Φ M(X). If there exists a nonzero µ M(X) such that µ({x}) = for all x X, then Φ is not in the image of C (X) in M(X) = C (X). b. At the other extreme, let X = N with the discrete topology; then C (X) = l and (l ) = l. (Note: C (N) is usually denoted by c.) (a) First, we show that the sum is well defined. Let A n = {x X : µ({x}) /n}. If card(a n ) is infinite, then letting A n be a countably infinite subset of A n we see that µ(x) µ({x}) (/n)+(/n)+(/n)+ =. x A n Hence card(a n ) is finite for all positive integers n. Now define A µ = {x X : µ({x}) > }. Since A µ = A n, we see that A µ is a countable set. Hence by countable additivity we have Φ(µ) x A µ µ ({x}) = µ (A µ ) µ (X) = µ <. This shows that the sum is well-defined, and furthermore that Φ is a continuous operator on M(X). Linearity of Φ follows from the fact that terms in an absolutely convergent series can be rearranged: Φ(cµ+ν) = (cµ+ν)({x}) = (cµ)({x})+ν({x}) = c µ({x})+ ν({x}). x A µ+ν x A µ+ν x A µ x A ν Hence Φ M(X). Supposethereexistsanonzeroµ M(X)suchthatµ({x}) = forallx X, andthatφ C (X). Recall the Dirac measure δ x defined by δ x (y) = if y x and δ x (x) =. The Dirac measure is a Radon measure, and by definition of Φ, we have Φ(δ x ) =. However, as an element of C (X), we have the action Φ(δ x ) = Φ dδ x = Φ(x). From this we conclude that Φ(x) = for all x X. But by definition of Φ, we have Φ( µ ) = since µ {x} = for all x X. As an element of C (X) we have the action 6

7 Φ( µ ) = d µ = µ (X). This forces us to conclude that µ (X) =, which implies that for all Borel sets E we have µ(e) µ (E) µ (X) =, a contradiction since µ is non-zero. (b) We can identify functions f C (N) with sequences {f n } such that f n as n. The norm f is given by f = max n f n. We define a map Ψ : l C (N) given by This is well defined, since n= Ψ({a n })(f) = a n f n. n= a n f n f a n <. We show Ψ is a bijection by producing an inverse. First, we define { if n = i g i (n) = if n i Then Ψ (µ) = {a n }, where a i = Ψ({a n })(g i ) = n= g i dµ = µ({i}). This is well defined, since a n = µ({n}) n= n= µ ({n}) µ (N) <. n= It is easy to see that Ψ is linear, hence it only remains to show that Ψ is an isometry. We have already seen that Ψ({a n })(f) f {a n } l, hence Ψ({a n }) {a n }. We define the functions h i C (N) as (using the sgn function defined on page 46): Now since h i (n) = { (sgn an ) if n i if n > i Ψ({a n }) = sup Ψ({a n })(f), f C (N), f = 7

8 and h i =, we conclude that i Ψ({a n }) Ψ({a n })(h i ) = a n. Letting i, we see that Ψ({a n }) = {a n } and hence Ψ is an isometric isomorphism. The fact that (l ) l is a direct consequence of Theorem 6.5. n= Problem 7.24 Find examples of sequences {µ n } in M(R) such that a. µ n vaguely, but µ n. b. µ n vaguely, but fdµ n for some bounded measurable function f with compact support. c. µ n and µ n vaguely, but there exists x R such that F n (x) (notation as in Proposition 7.9). (a) Let µ n = δ n, where δ n is the Radon measure defined in the solution to the previous exercise. Then for all f C (R), we have fdµ n = f(n) as n. Hence µ n vaguely. However, µ n = δ n (R) = for all positive integers n. (b) Let µ n = δ /n δ /n. Then for any f C (R), we have fdµ n = f(/n) f( /n) as n. Hence µ n vaguely. However, the function f(x) = (sgnx) χ [ 2,2] is bounded and measurable with compact support, and fdµ n = = 2 for all positive integers n. (c) Let µ n = δ n. Then µ n and for any f C (R), we have fdµ n = f( n) 8

9 as n. Hence µ n vaguely. However, F n () = µ n ((,]) = for all positive integers n. Problem 7.27 Let C k ([,]) be as in Exercise 9 in 5.. If I C k ([,]), there exist µ M([,]) and constants c,...,c k, all unique, such that I(f) = k f (k) dµ+ c j f (j) (). (The functionals f f (j) () could not be replaced by any set of k functionals that separate points in the space of polynomials of degree < k.) Take any f C k ([,]). By Taylor s Theorem, we can Taylor expand about and obtain f(x) = k n= f (n) () x x n + n! (x t) k f (k) (t)dt. (k )! Let I C k ([,]). The bounded linear functional I is defined on a dense subset of C[,], hence can be uniquely extended to a bounded linear functional on all of C[,]. By the Riesz Representation Theorem, there exists a Radon measure ν M([,]) such that I(f) = fdν. Then we have I(f) = k n= f (n) () x n dν + n! k = c n f (n) ()+ n= k = c n f (n) ()+ n= k = c n f (n) ()+ n= k = c n f (n) ()+ n= k = c n f (n) ()+ n= where c n = xn /n! dν and g(t) = t x f (k) (t) (x t) k f (k) (t) dt dν(x) (k )! χ [,x] (t) (x t)k f (k) (t) dt dν(x) (k )! χ [,x] (t) (x t)k f (k) (t) dν(x) dt (k )! χ [t,] (x) (x t)k f (k) (t) dν(x) dt (k )! f (k) (t)g(t)dt. t (x t) k (k )! dν(x) dt (x t) k (k )! dν(x). In the preceding computation, the order of integrals was switched by Fubini s Theorem. From the discussion on page 223, we know that dµ = g(t)dt, where dt is Lebesgue measure, defines a Radon measure. 9

10 We must now show uniqueness. For any integer n < k, we substitute the function x n into the formula for I to obtain I(x n ) = n! c n. Therefore, the constants c,...,c k areuniquely determined. If there exists another µ M([,]) with the desired property, then since the c,...,c k are unique, we see that f (k) dµ = f (k) d µ, for all f C k ([,]). To show the two measures are equal, it suffices to show that µ([,c]) = µ([,c]) for all c [,]. If we let f (k) =, we see that µ([,]) = µ([,]). Let c [,). For all positive integers n such that c+/n, we can define functions f n C k ([,]) such that f n (k) = on [,c], f n (k) = on [c +/n,], and f n (k) is linear on [c,c+/n]. Pointwise, we have f n (k) χ [,c]. Then by the Dominated Convergence Theorem, we see that µ([,c]) = c dµ = lim f (k) n dµ = lim f (k) c n d µ = d µ = µ([,c]). Problem 7.3 Let µ and ν be Radon measures on X and Y, not necessarily σ finite. If f is a nonnegative LSC function on X Y, then x f x dν and y f y dµ are Borel measurable and f d(µˆ ν) = f dµdν = f dνdµ. Warning: this solution seems sketchy to me. I don t fell great about it. Suggestions are welcome. We define the functions Φ f : X C and Ψ f : Y C as Φ f (x) = f x dν, Ψ f (y) = f y dµ. By Lemma 7.24, if f C c (X Y), then Φ f and Ψ f are continuous. If f is a nonnegative LSC function on X Y, then so are f x and f y, hence by Corollary 7.3 we have Φ f (x) = f x dν { } = sup g dν : g C c (Y), g f x { } = sup g x dν : g C c (X Y), g f = sup{φ g (x) : g C c (X Y), g f}

11 Since the supremum of a family of measurable functions is measurable, we see that Φ f (x) is a measurable function. The same argument shows that Ψ f (x) is a measurable function. The equality of the supremums of the two different families in the previous argument is justified as follows. If g C c (X Y) and g f, it is immediate that g x C c (Y) and g x f x. On the other hand, if g C c (Y) and g f x, by Urysohn s lemma we can extend g to a function g C c (X Y) with g f, such that g x = g. Next, we notice that by the Fubini-Tonelli theorem for Radon products, if g C c (X Y), then g d(µˆ ν) = g dµdν = g dνdµ since the support of g has finite measure, hence the integrals can be taken over a set of finite measure. By applying Corollary 7.3, if f is nonnegative and LSC, then { } f d(µˆ ν) = sup g d(µˆ ν) : g C c (X Y), g f { } = sup g dµdν : g C c (X Y), g f = f dµdν The same argument shows f d(µˆ ν) = f dνdµ.