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Math 641 Lecture #26 ¶6.18,6.19 Complex Measures and Integration the Dual of C0(X)

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Math 641 Lecture #26 ¶6.18,6.19 Complex Measures and Integration the Dual of C0(X) Math 641 Lecture #26 ¶6.18,6.19 Complex Measures and Integration Definition (6.18). Let µ be a complex measure on a σ-algebra M in X. There is (by Theorem 6.12) a meaurable function h such that |h| = 1 and dµ = hd|µ| [actually, h is a L1(|µ|) function that depends on µ]. The integral of a measurable function f on X with respect to µ is Z Z f dµ = fh d|µ|. X X A special case of this is Z Z Z χE dµ = χEh d|µ| = h d|µ| = µ(E), X X E Z which justifies setting µ(E) = dµ for all E ∈ M. E The Dual of C0(X) Definition. Let M(X) be the collection of all regular complex Borel measures on a LCH space X, and equip M(X) with the total variation norm, kµk = |µ|(X). [RECALL: a regular complex Borel measure is a complex measure µ on BX such that |µ| is regular.] Proposition. M(X) is a Banach space. Proof. Homework Problem Ch.6 #3. Problem. Each µ ∈ M(X) defines a linear functional Iµ : C0(X) → C by Z Iµ(f) = f dµ, X where (for the unique measurable h correpsonding to µ) Z Z Z |Iµ(f)| = fdµ = fh d|µ| ≤ |fh| d|µ| X X X Z ≤ kfku 1 d|µ| ≤ kfku |µ|(X). X ∗ Hence kIµk ≤ kµk < ∞, so that Iµ ∈ C0(X) . ∗ Does C0(X) = {Iµ : µ ∈ M(X)}? ∗ Define a map I : M(X) → C0(X) by I : µ → Iµ. Proposition. The map I is linear. Proof. For c ∈ C and µ, λ ∈ M(X) and f = χE (E ∈ M), Z Z Icµ(f) = χE d(cµ) = (cµ)(E) = cµ(E) = c χE dµ = cIµ(f), X X and Z Iµ+λ(f) = χE d(µ + λ) = (µ + λ)(E) = µ(E) + λ(E) X Z Z = χE dµ + χE dλ = Iµ(f) + Iλ(f). X X By linearity of integration, this extends to all simple Borel functions f, and hence to all bounded Borel functions, which contains C0(X). ∗ Riesz Representation Theorem for C0(X) (6.19). If X is LCH, then I : M(X) → ∗ C0(X) is a isometric isomorphism. ∗ Proof. Let Φ ∈ C0(X) . (Injectiveness) Suppose there are µ1, µ2 ∈ M(X) such that Iµ1 = Φ = Iµ2 . 1 For µ = µ2 − µ1 ∈ M(X)) there is a (unique) Borel function h ∈ L (|µ|) such that Z Z Z 0 = Iµ2 (f) − Iµ1 (f) = Iµ2−µ1 (f) = f d(µ2 − µ1) = f dµ = fh d|µ| X X X for all f ∈ C0(X). Hence for any sequence {fn} ⊂ C0(X), Z Z Z |µ|(X) = d|µ| = |h|2d|µ| = hh¯ d|µ| − 0 X X X Z Z ¯ = hh d|µ| − fnh d|µ| X X Z Z ¯ ¯ ¯ = (h − fn)h d|µ| ≤ |h − fn| d|µ| = kh − fnk1. X X 1 Since Cc(X) is dense in L (|µ|) (Theorem 3.14), there is a sequence {fn} ⊂ Cc(X) such ¯ that kh − fnk1 → 0. The inclusion Cc(X) ⊂ C0(X) implies that |µ|(X) = 0; hence µ = 0. (Surjectiveness and Isometricness) Assume that kΦk = sup{|Φ(f)| : kfku = 1, f ∈ C0(X)} = 1 [rescale Φ when kΦk > 0; choose µ = 0 when kΦk = 0]. Suppose there is a positive linear functional Λ on Cc(X) such that |Φ(f)| ≤ Λ(|f|) ≤ kfku for all f ∈ Cc(X). [The existence of Λ will be shown in the next Lecture.] Then by the Riesz Representation Theorem for Positive Linear Functionals on Cc(X) (Theorem 2.14), there is a unique positive Borel measure λ such that Z Λ(f) = f dλ X where λ is outer regular on all BX and inner regular on every open E ⊂ X or any E ∈ M with λ(X) < ∞ [and λ(K) < ∞ for all compact K ⊂ X; i.e. λ is a Radon measure]. Since λ(V ) = sup{Λf : 0 ≤ f ≤ 1, f|V C = 0, f ∈ Cc(X)} for V open (this is the definition of λ in Theorem 2.14), and since X is open, λ(X) = sup{Λf : 0 ≤ f ≤ 1, f ∈ Cc(X)} ≤ sup{kfku : 0 ≤ f ≤ 1, f ∈ Cc(X)} = 1. Thus λ(E) ≤ λ(X) = 1 for all E ∈ BX , and so λ is regular by Corollary A in Lecture #11. 1 On the subspace Cc(X) of L (λ), the functional Φ has norm at most 1: Z |Φ(f)| ≤ Λ(|f|) = |f| dλ = kfk1. X By the Hahn-Banach Theorem, there is a norm-preserving extension of Φ to L1(λ); call it Φ. By Theorem 6.16 (the Representation Theorem for Lp(µ)∗ when 1 ≤ p < ∞), there is a ∞ g ∈ L (λ) such that kgk∞ = kΦk = 1 and Z Φ(f) = fg dλ for all f ∈ L1(λ). X Since each side of this is continuous (with respect to the uniform norm) on Cc(X), and since Cc(X) is dense in C0(X) (in the uniform norm, Theorem 3.17), it follows that R Φ(f) = X fg dλ for all f ∈ C0(X). R R Since X |fg|dλ ≥ | X fg dλ| = |Φ(f)|, it follows that Z Z |g| dλ = sup |fg| dλ : f ∈ C0(X), kfku ≤ 1 ≥ kΦk = 1. X X This together with λ(X) ≤ 1 and |g| ≤ kgk∞ = 1 a.e.[λ] implies that λ(X) = 1 and |g| = 1 a.e.[λ]. The complex Borel measure µ defined by dµ = gdλ satisfies d|µ| = |g|d|λ| = dλ (Theorem R 6.13) and Φ(f) = X f dµ for all f ∈ C0(X); also µ is regular (since λ is regular), and kµk = |µ|(X) = λ(X) = 1 = kΦk = kIµk. Thus I is an surjective linear isometry, and hence an isometric isomorphism..
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