Recall that if X is a compact metric space, C(X), the space of continuous (real-valued) functions on X, is a Banach space with the norm

Transcription

1 Chapter 13 Radon Measures Recall that if X is a compact metric space, C(X), the space of continuous (real-valued) functions on X, is a Banach space with the norm (13.1) f = sup x X f(x). We want to identify the dual of C(X) with the space of (finite) signed Borel measures on X, also known as the space of Radon measures on X. Before identifying the dual of C(X), we will first identify the set of positive linear functionals on C(X). By definition, a linear functional (13.2) α : C(X) R is positive provided (13.3) f C(X), f 0 = α(f) 0. Clearly, if µ is a (positive) finite Borel measure on X, then (13.4) α(f) = f dµ is a positive linear functional. We will establish the converse, that every positive linear functional on C(X) is of the form (13.4). It is easy to see that every positive linear functional α on C(X) is bounded. In fact, applying α to f(x) a and to b f(x), we see that, when a and b are real numbers (13.5) f C(X), a f b = aα(1) α(f) bα(1), 179

2 Radon Measures so (13.6) α(f) A f, A = α(1). To begin the construction of µ, we construct a set function µ 0 on the collection O of open subsets of X by (13.7) µ 0 (U) = sup {α(f) : f U}, where we say (13.8) f U f C(X), 0 f 1, and supp f U. Here, supp f is the closure of {x : f(x) 0}. Clearly µ 0 is monotone. Then we set, for any E X, (13.9) µ (E) = inf {µ 0 (U) : E U O}. Of course, µ (U) = µ 0 (U) when U is open. Lemma The set function µ is an outer measure. Proof. By Proposition 5.1, it suffices to show that (13.10) U j O, U = j 1 U j = µ 0 (U) j 1 µ 0 (U j ), so that we have an analogue of (5.5). Suppose f U. We need to show that α(f) µ 0 (U j ). Now, since supp f = K is compact, we have K U 1 U l for some finite l. We claim there are g j U j, 1 j l, such that g j = 1 on K. Granted this, we can set f j = fg j. Then f j U j, so α(f j ) µ 0 (U j ). Hence (13.11) α(f) = α(f j ) µ 0 (U j ), as desired. Thus Lemma 13.1 will be proved once we have Lemma If K X is compact, U j are open, and K U 1 U l = V, then there exist g j U j such that g j = 1 on K. Proof. Set U l+1 = X \ K. Then {U j : 1 j l + 1} is an open cover of X. Let {g j : 1 j l + 1} be a partition of unity subordinate to this cover. (See Exercise 9 at the end of this chapter.) Then {g j : 1 j l} has the desired properties. Now that we know µ is an outer measure, we prepare to apply Caratheodory s Theorem.

3 13. Radon Measures 181 Lemma The outer measure µ is a metric outer measure. Proof. Let S j X, and assume (13.12) dist(s 1, S 2 ) 4ε > 0. Take δ > 0. Given U S = S 1 S 2, U open, such that µ 0 (U) µ (S) + δ, set U j = U {x X : dist(x, S j ) < ε}. It follows that (13.13) S j U j, U 1 U 2 U, U 1 U 2 =. Now, whenever f U 1 U 2, we have f = f 1 + f 2 with f j = f Uj U j, which implies µ 0 (U 1 ) + µ 0 (U 2 ) = µ 0 (U 1 U 2 ). Hence (13.14) µ (S 1 ) + µ (S 2 ) µ 0 (U 1 ) + µ 0 (U 2 ) = µ 0 (U 1 U 2 ) µ 0 (U). Thus µ (S) µ (S 1 ) + µ (S 2 ) δ, for all δ > 0, which, together with subadditivity of µ, yields the desired identity µ (S) = µ (S 1 ) + µ (S 2 ). It follows from Proposition 5.8 that every closed set in X is µ -measurable. Hence, by Theorem 5.2, every Borel set in X is µ -measurable, and the restriction of µ to B(X), which we denote µ, is a measure. We make a few useful comments about µ. First, in addition to (13.7), we have (13.15) µ(u) = sup {α(f) : f U}, for U open, where (13.16) f U f C(X), 0 f 1, f = 0 on X \ U. To see that (13.7) and (13.15) coincide, take f U. Then set f j = ξ j (f), where ξ j (s) is defined by 0 for 0 s 1 j, ξ j (s) = 2s 2 j for 1 j s 2 j, s for s 2 j. It follows that f j U and f j f (uniformly); hence α(f j ) α(f). Using the identity (13.15), we can establish the following.

4 Radon Measures Lemma If K X is compact, then (13.17) µ(k) = inf {α(f) : f C(X), f χ K }. Proof. Denote the right side of (13.17) by µ 1 (K). It suffices to take the inf of α(f) over f K, where (13.18) f K f C(X), 0 f 1, f = 1 on K. Comparing this with (13.16), we see that f K 1 f X \ K, so (13.19) µ 1 (K) = inf {α(1) α(g) : g X \ K} = µ(x) µ(x \ K). On the other hand, since K is µ -measurable, µ(x \ K) + µ(k) = µ(x), so the identity (13.17) is proved. We are now ready to prove Theorem If X is a compact metric space and α is a positive linear functional on C(X), then there exists a unique finite, positive Borel measure µ such that (13.20) α(f) = f dµ for all f C(X). Proof. We have constructed a positive Borel measure µ, which is finite since (13.7) implies µ(x) = α(1). We next show that (13.20) holds. It suffices to check this when f : X [0, 1]. In such a case, take N Z + and define (13.21) ϕ j (x) = min (f(x), jn 1 ), 0 j N, f j (x) = ϕ j (x) ϕ j 1 (x), 1 j N. We have f j = f and (13.22) Hence (13.23) 1 N χ K j f j 1 { N χ K j 1, K j = x X : f(x) j }. N 1 N µ(k j) f j dµ 1 N µ(k j 1).

5 13. Radon Measures 183 We claim that also (13.24) 1 N µ(k j) α(f j ) 1 N µ(k j 1). To see this, first note that if K j 1 U is open, then Nf j U, so Nα(f j ) µ 0 (U). This implies the second inequality of (13.24). On the other hand, Nf j K j, so Lemma 13.4 gives the first inequality of (13.24). (13.25) Summing (13.23) and (13.24), we have 1 N 1 N N µ(k j ) j=1 f dµ 1 N N µ(k j ) α(f) 1 N j=1 N 1 j=0 N 1 j=0 µ(k j ), µ(k j ). Hence (13.26) α(f) f dµ 1 N µ(k 0) 1 N µ(k N ) 1 N µ(x). Letting N, we have (13.20). Only the uniqueness of µ remains to be proved. To see this, let λ be a positive Borel measure on X such that (13.27) α(f) = f dλ for all f C(X). Let K X be compact, and apply this to (13.28) f ν (x) = ( 1 + ν dist(x, K) ) 1 ; fν χ K. By the Monotone Convergence Theorem, we have f ν dµ χ K dµ and fν dλ χ K dλ, so α(f ν ) µ(k) and α(f ν ) λ(k). Hence µ(k) = λ(k) for all compact K. Now, by (5.60), for every positive Borel measure λ on a compact metric space X, we have (13.29) E B(X) = λ(e) = sup {λ(k) : K E, K compact}. This proves uniqueness. Generally, if X is a compact Hausdorff space and λ is a finite (positive) measure on B(X), λ is said to be regular if and only if (13.29) holds. The implication of Exercises of Chapter 5 is that every finite Borel measure is regular when X is a compact metric space. If X is compact but not

6 Radon Measures metrizable, a finite measure on B(X) need not be regular. The generalization of Theorem 13.5 to this case is that, given a positive linear functional α on C(X), there is a unique finite regular Borel measure µ such that (13.20) holds. (Note that if λ is any finite measure on B(X), then (13.27) defines a positive linear functional on C(X), which then gives rise to a regular Borel measure.) For this more general case, the construction of µ is the same as was done above in (13.7) (13.9), but the proof that µ yields a regular measure on B(X) is a little more elaborate than the proof given above for compact metric spaces. Treatments can be found in [Fol] and [Ru]. We want to extend Theorem 13.5 to the case of a general bounded linear functional (13.30) ω : C(X) R. We start with an analogue of the Hahn decomposition. Lemma If ω is a bounded (real) linear functional on C(X), then there are positive linear functionals α ± on C(X) such that (13.31) ω = α + α. Proof. We first define α + on (13.32) C + (X) = {f C(X) : f 0}. For f C + (X), set (13.33) α + (f) = sup {ω(g) : g C + (X), 0 g f}. The hypothesis that ω is bounded implies ω(g) K g K f when 0 g f, so (13.34) 0 α + (f) K f, for f C + (X), where K = ω. Clearly, for c R, (13.35) f C + (X), c > 0 = α + (cf) = cα + (f). Now, suppose f 1, f 2 C + (X). If g C + (X) and 0 g f 1 + f 2, we can write g = g 1 +g 2 with g C + (X) and 0 g j f j. Just take g 1 = min ( g, f 1 ). Hence (13.36) f 1, f 2 C + (X) = α + (f 1 + f 2 ) = α + (f 1 ) + α + (f 2 ).

7 13. Radon Measures 185 We claim that α + has an extension to a linear functional on C(X), which would necessarily be positive. In fact, given f C(X), write (13.37) f = f 1 f 2, f j C + (X), α + (f) = α + (f 1 ) α + (f 2 ). A given f C(X) has many such representations as f = f 1 f 2 ; showing that α + (f) is independent of such a representation and defines a linear functional on C(X) is a simple application of (13.36); compare the proof of Proposition 3.7. Note that, if we take f 1 = f + = max(f, 0) and f 2 = f = max( f, 0), we see that (13.38) α + (f) K max ( f +, f ) K f. Finally we set α = α + ω. It remains only to show that f C + (X) α (f) 0, i.e., α + (f) ω(f). But that is immediate from the definition (13.33), so the lemma is proved. We can combine Lemma 13.6 and Theorem 13.5 to prove the following, known as the Riesz Representation Theorem. Theorem If X is a compact metric space and ω is a bounded (real) linear functional on C(X), then there is a unique finite signed measure ρ on B(X) such that (13.39) ω(f) = for all f C(X). Furthermore, X f dρ, (13.40) ρ = ρ (X) = ω, so there is an isometric isomorphism (13.41) C(X) M(X). Here, M(X) denotes the linear space of finite signed measures on B(X), with norm given by the first identity in (13.40). This is also known as the space of finite Radon measures on X. For the proof, write ω = α + α, as in (13.31), take finite positive measures µ ± on B(X) so that α ± (f) = f dµ ±, and set ρ = µ + µ. Thus we have the identity (13.39).

8 Radon Measures We need to prove (13.40). Let ρ = ρ + ρ be the Hahn decomposition of ρ, so ρ + ρ and (13.42) ω(f) = f dρ + f dρ, for all f C(X). Consequently (13.43) ω(f) f ρ + (X) + f ρ (X) = ρ f, so we have ω ρ. To prove the reverse inequality, let δ > 0. Suppose ρ ± are supported on disjoint Borel sets X ±. Let K ± be compact sets in X ± such that ρ ± (K ± ) ρ ± (X) δ. We have K + K =, so say dist(k +, K ) = ε > 0. Let U ± = {x : dist(x, K ± ) < ε/4}, so U + U =. Using a simple variant of (13.28), we can construct ϕ ± ν U ± such that ϕ ± ν χ K ±. Hence, as ν, (13.44) ω ( ϕ + ν ϕ ) ν ρ + (K + ) + ρ (K ) ρ 2δ, while ϕ + ν ϕ ν = 1. This proves the reverse inequality, ρ ω, and establishes (13.40). This inequality also establishes uniqueness, so Theorem 13.7 is proved. Though it is not needed to prove Theorem 13.7, we mention that µ + µ is the Hahn decomposition of ρ (provided α + is given by (13.33)). The proof is an exercise. Applying Alaoglu s Theorem (Proposition 9.2 and Corollary 9.3) to V = C(X), we have the following compactness result: Proposition If X is a compact metric space, the closed unit ball in M(X) is a compact metrizable space, with the weak topology. By the definition of weak topology given in Chapter 9, we see that a sequence µ ν in M(X) converges weak to µ if and only if (13.45) f dµ ν f dµ, for each f C(X). This topology on M(X) is also called the vague topology; one says µ ν µ vaguely provided (13.45) holds for all f C(X). There is an important subset of M(X), the set of probability measures. An element µ of M(X) is called a probability measure if and only if µ is positive and µ(x) = 1. We use the notation (13.46) Prob(X) = {µ M(X) : µ 0, µ(x) = 1}. It is easy to see that (13.47) Prob(X) = {µ M(X) : µ 1, µ(x) = 1}. Hence Prob(X) is a subset of the unit ball of M(X) which is closed in the weak topology. This has the following useful consequence.

9 13. Radon Measures 187 Corollary If X is a compact metric space, the set Prob(X) of probability measures on X is compact (and metrizable) in the weak topology. It is often useful to consider Borel measures on a locally compact space Y, i.e., a Hausdorff space with the property that any y Y has a compact neighborhood. In such a case, there is a Banach space C (Y ), the space of continuous functions on Y that vanish at infinity. We say a continuous function u : Y R belongs to C (Y ) if, for any δ > 0, there exists a compact K Y such that u(y) < δ for y Y \ K. We use the sup norm on C (Y ). We can construct the one point compactification Ŷ = Y { }, declaring a set U Ŷ to be open if either U Y is open or U and Ŷ \ U = K is a compact subset of Y. It readily follows that Ŷ is a compact Hausdorff space. Also, C (Y ) is naturally isomorphic to a closed linear subspace of C(Ŷ ) : (13.48) C (Y ) {u C(Ŷ ) : u( ) = 0}. Furthermore, given any f C(Ŷ ), if we set a = f( ), then f = g + a, g( ) = 0, so (13.49) C(Ŷ ) C (Y ) R, and, for the duals, we have (13.50) C(Ŷ ) C (Y ) R. In case Y has the additional properties of being metrizable and σ- compact, the one-point compactification Ŷ is also metrizable; cf. Exercise 15. Hence we can appeal to Theorem 13.7 to identify C(Ŷ ) with M(Ŷ ), the space of finite signed measures on B(Ŷ ). (Even without these additional hypotheses, C(Ŷ ) can be identified with the space of signed finite regular Borel measures on Ŷ, though one must go to another source, such as [Fol] or [Ru], for a proof.) In the decomposition (13.50), we see that the last factor on the right consists of multiples of δ, the measure defined by δ (S) = 1 if S, 0 otherwise. Consequently we have the identification (13.51) C (Y ) {µ M(Ŷ ) : µ({ }) = 0} M(Y ), where M(Y ) is the space of finite signed measures on B(Y ).

10 Radon Measures Exercises In Exercises 1 3, X is a compact metric space and F : X X is a continuous map. As in (7.32), we set (13.52) F µ(s) = µ F 1(S) = µ ( F 1 (S) ), for a Borel measure µ on X. We set F n = F F (n factors) and similarly define F n µ. 1. Define T : C(X) C(X) by T u(x) = u(f (x)). Show that, under the identification (13.41), the adjoint T : C(X) C(X) is given by (13.52). 2. Assume µ is a probability measure on X. Set (13.53) µ j = F j µ, ν n = 1 n + 1 Show that ν n Prob(X) and n µ j. j=0 F ν n = ν n + 1 ( ) µn+1 µ 0. n Suppose ν nj ν in the weak topology on Prob(X), as j. Show that (13.54) F ν = ν. We say ν is an invariant measure for F. Hint. Given u C(X), one has u F dν nj = u df ν nj = u dν nj + 1 ( u dµ nj +1 n j + 1 u dµ 0 ). 4. Let f L (R). Show that the following are equivalent: (a) h 1 (τ h f f) is bounded in L 1 (R), for h (0, 1]. (b) 1 f = µ for some finite signed measure µ on R. (c) f(x) = x dµ a.e. on R.

11 13. Radon Measures Let f 0 (x) denote the right side of (c) above, defined as in Exercise 1 of Chapter 11. Show that f 0 is the unique right continuous function on R equal to f a.e. 6. Let f : R R be bounded and right-continuous. Show that f has the properties (a) (c) of Exercise 4 if and only if there exists C < such that, for any finite set of real numbers x 0 < x 1 < < x l, (13.55) l f(x j ) f(x j 1 ) C. j=1 Hint. To prove one implication, given ν Z +, set f ν (x) = f(ν) if x ν, f ν (x) = f( ν) if x ν, and f ν (x) = f( ν + 2 ν j) if ν + 2 ν j x < ν + 2 ν (j + 1), 0 j < 2ν2 ν. Show that f ν f and { 1 f ν : ν Z + } is a bounded set of measures on R +. Consider weak limits. One says that f has bounded variation on R if this property holds, and one writes f BV(R). 7. Let f L (R). Show that f is equal a.e. to an element of BV(R) if and only if you can write f = g 1 g 2 a.e., with g j L (R) monotone. In particular, if g L (R) is monotone, then 1 g is a positive measure. Reconsider Exercise 1 of Chapter 5, on Lebesgue-Stieltjes measures, in this light. Hint. Given f BV(R), apply the Hahn decomposition to the signed measure µ arising in Exerise Let f : R R be bounded and continuous. One says f is absolutely continuous provided that, for every ε > 0, there is a δ > 0 with the property that, for any finite collection of disjoint intervals, (a 1, b 1 ),..., (a N, b N ), (13.56) (bj a j ) < δ = f(b j ) f(a j ) < ε. Show that the following are equivalent: (a) f is absolutely continuous. (b) 1 f = g L 1 (R). (c) f(x) = x g(y) dy, g L1 (R). Hint. If (a) holds, first show that f has bounded variation and use Exercises 5 6 to get 1 f = µ. Then show µ is absolutely continuous with respect to Lebesgue measure. Compare Exercises 7 9 in Chapter

12 Radon Measures 10 and also Exercise 1 in Chapter Let {U 1,..., U l+1 } be an open cover of a compact metric space X. Show that there exist open sets V j, j = 1,..., l + 1, covering X, such that V j U j. Let h j (x) = dist(x, X \ V j ). Show that h j C(X), supp H j V j U j, and that h = l+1 j=1 h j C(X) is > 0 on X. Deduce that g j (x) = h j (x)/h(x) form a partition of unity of X, subordinate to the cover {U 1,..., U l+1 }. Hint. Set H j (x) = dist(x, X \ U j ). Show that H = H j > 0 on X, H C(X), hence H a > 0. Set V j = {x U j : H j (x) > a/(l+2)}. 10. The countable infinite product Z = j 1 {0, 1} is compact, with the product topology, and metrizable (cf. Appendix A). Let A C(Z) consist of continuous functions depending on only finitely many variables, so an element f A has the form f(x) = f(x 1,..., x k ), for some k Z +. For such an f, set ϕ(f) = 2 k f(x 1,..., x k ). x j {0,1},1 j k Show that ϕ has a unique extension to a positive linear functional α on C(Z). Show that, for f C(Z), α(f) = Z f dµ, where µ is the product measure on Z, discussed in Exercises 7 10 of Chapter Generalize Exercise 10 to other countable products of compact metric spaces, carrying positive Radon measures of total mass one. 12. Let R = I 1 I n be a compact cell in R n, with I ν = [a ν, b ν ]. Consider the positive linear functional on C(R) given by f I(f), where I(f) is the Riemann integral of f, discussed in Chapter 1 (for n = 1) and in the second exercise set at the end of Chapter 7 (for n > 1). Show that the measure on R produced by Theorem 13.5 coincides with Lebesgue measure (on the Borel subsets of R), as constructed in Chapter 2 (for n = 1) and in Chapter 7 (for n > 1). 13. Let X be a compact Hausdorff space. Show that if C(X) is separable, then X is metrizable. Hint. Define ϕ : X C(X) by ϕ(y)(f) = f(y) and use Corollary 9.3.

13 13. Radon Measures Let Y be a locally compact Hausdorff space. Assume Y is σ-compact, i.e., there is a countable family of compact K j Y such that Y = j K j. Show that there is a sequence f k C (Y ) such that each f k has compact support and for each y Y, some f k (y) 0. (Also arrange 0 f k 1.) Note that then f = k 1 2 k f k C (Y ) is > 0 at each point of Y and that U k = {y Y : f(y) > 1/k} satisfies U k open, U k compact, U k Y. 15. Let Y be a locally compact metrizable space. Assume Y is σ-compact. Show that C (Y ) is separable. Deduce that the one-point compactification Ŷ is metrizable. Hint. With U k Y as in Exercise 14, show that V k = {f C (Y ) : supp f U k } is separable and k V k is dense in C (Y ). In Exercises 16 17, take C(X) to be the space of complex-valued continuous functions on the compact metric space X. Let ρ be a complex Borel measure on X, of the form ρ = ν 1 + iν 2 (as in (8.19)), where ν j are finite signed measures, with associated positive measures ν j as in (8.10). Set ν = ν 1 + ν 2 and apply the Radon-Nikodym theorem to ν j << ν to obtain ρ = f ν. 16. Show that { } ρ = sup ρ(s k ) : S k B(X) disjoint k 0 defines a norm on the set M C (X) of complex Borel measures on X. Show that ρ = f d ν. 17. Show that the dual C(X) of C(X) is isomorphic to M C (X), with norm given in Exercise 16.