Radon Measures

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) kfk = sup jf(x)j: x2X 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 2 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 Z 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 2 C(X); a ≤ f ≤ b =) aα(1) ≤ α(f) ≤ bα(1); 179 180 13. Radon Measures so (13.6) jα(f)j ≤ Akfk; A = α(1): To begin the construction of µ, we construct a set function µ0 on the collection O of open subsets of X by (13.7) µ0(U) = sup fα(f) : f ≺ Ug; where we say (13.8) f ≺ U () f 2 C(X); 0 ≤ f ≤ 1; and supp f ⊂ U: Here, supp f is the closure of fx : f(x) 6= 0g: Clearly µ0 is monotone. Then we set, for any E ⊂ X; ∗ (13.9) µ (E) = inf fµ0(U) : E ⊂ U 2 Og: ∗ Of course, µ (U) = µ0(U) when U is open. Lemma 13.1. The set function µ∗ is an outer measure. Proof. By Proposition 5.1, it suffices to show that (13.10) Uj 2 O; U = Uj =) µ0(U) ≤ µ0(Uj); j[≥1 Xj≥1 so that we have an analogue of (5.5). Suppose f ≺ U: We need to show that α(f) ≤ µ0(Uj): Now, since supp f = K is compact, we have K ⊂ U1 [ · · · [ U` Pfor some finite `: We claim there are gj ≺ Uj; 1 ≤ j ≤ `; such that gj = 1 on K: Granted this, we can set fj = fgj: Then fj ≺ Uj; so α(fj)P≤ µ0(Uj): Hence (13.11) α(f) = α(fj) ≤ µ0(Uj); X X as desired. Thus Lemma 13.1 will be proved once we have Lemma 13.2. If K ⊂ X is compact, Uj are open, and K ⊂ U1[· · ·[U` = V; then there exist gj ≺ Uj such that gj = 1 on K: P Proof. Set U`+1 = X n K: Then fUj : 1 ≤ j ≤ ` + 1g is an open cover of X: Let fgj : 1 ≤ j ≤ ` + 1g be a partition of unity subordinate to this cover. (See Exercise 9 at the end of this chapter.) Then fgj : 1 ≤ j ≤ `g has the desired properties. Now that we know µ∗ is an outer measure, we prepare to apply Carathe- odory's Theorem. 13. Radon Measures 181 Lemma 13.3. The outer measure µ∗ is a metric outer measure. Proof. Let Sj ⊂ X; and assume (13.12) dist(S1; S2) ≥ 4" > 0: ∗ Take δ > 0: Given U ⊃ S = S1 [ S2; U open, such that µ0(U) ≤ µ (S) + δ; set Uj = U \ fx 2 X : dist(x; Sj) < "g: It follows that (13.13) Sj ⊂ Uj; U1 [ U2 ⊂ U; U1 \ U2 = ;: Now, whenever f ≺ U1 [ U2; we have f = f1 + f2 with fj = f ≺ Uj, Uj which implies µ0(U1) + µ0(U2) = µ0(U1 [ U2): Hence ∗ ∗ (13.14) µ (S1) + µ (S2) ≤ µ0(U1) + µ0(U2) = µ0(U1 [ U2) ≤ µ0(U): ∗ ∗ ∗ Thus µ (S) ≥ µ (S1) + µ (S2) − δ; for all δ > 0; which, together with ∗ ∗ ∗ ∗ subadditivity of µ ; yields the desired identity µ (S) = µ (S1) + µ (S2): 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 re- striction 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 fα(f) : f 4 Ug; for U open, where (13.16) f 4 U () f 2 C(X); 0 ≤ f ≤ 1; f = 0 on X n U: To see that (13.7) and (13.15) coincide, take f 4 U: Then set fj = ξj(f); where ξj(s) is defined by 1 0 for 0 ≤ s ≤ ; j 2 1 2 ξj(s) = 2s − for ≤ s ≤ ; j j j 2 s for s ≥ : j It follows that fj ≺ U and fj % f (uniformly); hence α(fj) % α(f): Using the identity (13.15), we can establish the following. 182 13. Radon Measures Lemma 13.4. If K ⊂ X is compact, then (13.17) µ(K) = inf fα(f) : f 2 C(X); f ≥ χK g: Proof. Denote the right side of (13.17) by µ1(K): It suffices to take the inf of α(f) over f < K; where (13.18) f < K () f 2 C(X); 0 ≤ f ≤ 1; f = 1 on K: Comparing this with (13.16), we see that f < K , 1 − f 4 X n K, so µ1(K) = inf fα(1) − α(g) : g 4 X n Kg (13.19) = µ(X) − µ(X n K): On the other hand, since K is µ∗-measurable, µ(X n K) + µ(K) = µ(X); so the identity (13.17) is proved. We are now ready to prove Theorem 13.5. 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 Z for all f 2 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 2 Z+ and define −1 'j(x) = min (f(x); jN ); 0 ≤ j ≤ N; (13.21) fj(x) = 'j(x) − 'j−1(x); 1 ≤ j ≤ N: We have fj = f and P 1 1 j (13.22) χKj ≤ fj ≤ χKj−1 ; Kj = x 2 X : f(x) ≥ : N N n N o Hence 1 1 (13.23) µ(K ) ≤ f dµ ≤ µ(K 1): N j Z j N j− 13. Radon Measures 183 We claim that also 1 1 (13.24) µ(Kj) ≤ α(fj) ≤ µ(Kj 1): N N − To see this, first note that if Kj−1 ⊂ U is open, then Nfj ≺ U; so Nα(fj) ≤ µ0(U): This implies the second inequality of (13.24). On the other hand, Nfj < Kj; so Lemma 13.4 gives the first inequality of (13.24). Summing (13.23) and (13.24), we have N N−1 1 1 µ(K ) ≤ f dµ ≤ µ(K ); N j Z N j Xj=1 Xj=0 (13.25) N N−1 1 1 µ(Kj) ≤ α(f) ≤ µ(Kj): N N Xj=1 Xj=0 Hence 1 1 1 (13.26) α(f) − f dµ ≤ µ(K0) − µ(KN ) ≤ µ(X): Z N N N Letting N ! 1; 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 Z for all f 2 C(X). Let K ⊂ X be compact, and apply this to −1 (13.28) fν(x) = 1 + ν dist(x; K) ; fν & χK: By the Monotone Convergence Theorem, we have fν dµ & χK dµ and fν dλ & χK dλ, so α(fν ) & µ(K) and α(fν) &R λ(K): HenceR µ(K) = λR(K) for allR compact K. Now, by (5.60), for every positive Borel measure λ on a compact metric space X; we have (13.29) E 2 B(X) =) λ(E) = sup fλ(K) : K ⊂ E; K compactg: 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 10{13 of Chapter 5 is that every finite Borel measure is regular when X is a compact metric space. If X is compact but not 184 13. Radon Measures metrizable, a finite measure on B(X) need not be regular. The generaliza- tion 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) ! : C(X) ! R: We start with an analogue of the Hahn decomposition.

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