Vector Integration and Disintegration of Measures. Michael Taylor

Transcription

1 Vector Integration and Disintegration of Measures Michael Taylor 1. Introduction Let and be compact metric spaces, and π : a continuous map of onto. Let µ be a positive, finite Borel measure on. The following result gives the disintegration of µ. Proposition 1.1. Let λ = π µ (positive Borel measure on ). For λ-a.e. y, there is a probability measure ν y on such that supp ν y π 1 (y) and (1.1) µ = ν y dλ(y), in the sense that for each u C(), (1.2) u(z) dµ(z) = u(z) dν y (z) dλ(y). With F (y) = ν y, we will have (1.3) F L w (, λ, M()). Here M() is the Banach space of finite (signed) Borel measures on, and the subscript w means that F is weak -measurable. Part of the proof of Proposiiton 1.1 involves the duality (1.4) L 1 (, λ, C()) = L w (, λ, M()). Here L 1 (, λ, C()) consists of strongly measurable functions. We devote 2 to a discussion of results yielding (1.4). We prove Proposition 1.1 in 3, and give another proof, for the special case = X, π(x, y) = y, in 4. More general results (involving more elaborate arguments) can be found in [Ed] and [B]. 2. Duals of vector L 1 -spaces In this section we prove the following, which implies (1.4). 1

2 2 Proposition 2.1. Let be a compact metric space, λ a finite, positive Borel measure on. Let V be a separable Banach space. Then (2.1) L 1 (, λ, V ) = L w (, λ, V ). Remark. If V is separable and reflexive, then L w (, λ, V ) = L (, λ, V ), but V = C() is almost never reflexive. The proof that the right side of (2.1) is contained in the left side is fairly straightforward. It comes down to showing (2.2) f, g = f(y), g(y) dλ(y) is well definjed for f L 1 (, λ, V ), g L w (, λ, V ), with estimates. That (2.2) is well defined when g is as above and f is a simple function (with values in V ) is clear, and then a simple limiting argument gives the result for general f L 1 (, λ, V ), since it is the L 1 -norm closure of the space of simple functions. For the converse inclusion in (2.1), suppose γ : L 1 (, λ, V ) R is a continuous linear map. Then, for each v V, we have (2.3) γ v : L 1 (, λ) R, γ v (f) = γ(vf). Since L 1 (, λ) = L (, λ), this yields a bounded linear map (2.4) T γ : V L (, λ). Hence the reverse inclusion in (2.1) follows from: Proposition 2.2. With, λ, and V as in Proposition 2.1, given a bounded linear map (2.5) T : V L (, λ), then there exists (2.6) g L w (, λ, V ) such that for each v V (2.7) T v(y) = v, g(y), λ-a.e. y. This result is a consequence of the following.

3 Lemma 2.3. In the setting of Proposition 2.2, given ε > 0, there exists a compact such that λ( \ ) < ε and a continuous linear map Φ : V C() such that for all v V, (2.8) T v = Φv, λ-a.e., sup Φv T v L. Proof. Select V 0 V to be countable, dense, and a vector space over Q. Pick a Q-basis {e k : k N} for V 0. Select a representative T e k in L (, λ) for each T e k L (, λ). Extend by Q-linearity, to obtain T v k L (, λ), where V 0 = {v k : k N}. Then T v k is a representative of T v k for each v k V 0. Use Luzin s theorem to get a compact B k such that λ( \ B k ) 2 k 4 ε and such that T v k is continuous on B k. Let B = k B k, so T v k is continuous on B for each v k V 0. The set of y B such that T v k (y) > T v k L for some v k V 0 has λ-measure 0, so is contained in an open set O of λ-measure < 2 4 ε. Set = B \ O. Then λ( \ ) < ε. We define (2.9) Φ : V 0 C() as Φv k = T v k, for each v k V 0. We have arranged that Φ is linear (over Q) on V 0, and that (2.8) holds, for all v V 0. It follows that there is a unique continuous extension to Φ : V C(), and that this does the job. 3 Proof of Proposition 2.2. Using Lemma 2.3, we can take disjoint compact ν such that λ( \ ν ν ) = 0 and linear maps Φ ν : V C( ν ) such that (2.10) T v ν = Φ ν v, λ-a.e., sup Then define g as in (2.6) (2.7) by Φ ν v T v L. (2.11) v, g(y) = Φ ν v(y), y ν, and, say, g(y) = 0 for y \ ν ν. 3. Proof of Proposition 1.1 We now prepare to disintegrate the measure µ on described in the introduction. To begin, if λ = π µ vanishes on some nonepmty open set, let O be the largest such and replace by \ O (which we now denote ). Next, for each k N, take

4 4 a finite cover {Uα} k of by balls of radius 2 k, and let {ψα} k be a continuous partition of unity on subordinate to this cover. We have (3.1) ψα k dλ = Cα k > 0, α. Set (3.2) µ k α = 1 Cα k (ψα k π)µ, so each µ k α is a probability measure on, and (3.3) supp µ k α π 1 (U k α). Given y, define ν k y, a probability measure on, by (3.4) ν k y = α ψ k α(y) µ k α. Note that (3.5) supp ν k y Now consider {α:y U k α } π 1 (U k α). (3.6) F k L w (, λ, M()), F k(y) = ν k y. We have {F k : k N} bounded in this L -space, which by Proposition 2.1 is the dual of L 1 (, λ, C()), as in (1.4). Consequently, by Alaoglu s theorem, there exists a limit point (3.7) F L w (, λ, M()), and a subsequence (which we continue to denote F k ), converging weak to F. In other words, for each v L 1 (, λ, C()), we have (3.8) v, F k = v(y, z) dν k y (z) dλ(y), converging to v, F. If we apply this to v(y, z) = u(z), u C(), we get (3.9) u, F k = α u(z)ψα(y) k 1 Cα k ψα(π(z)) k dµ(z) dλ(y).

5 5 Noting that (3.10) 1 Cα k ψ k α(y) dλ(y) = 1, we obtain u, F k = α u(z)ψ k α(π(z)) dµ(z) (3.11) = u(z) dµ(z). Taking k, we replace F k by F in the left side of (3.11), and obtain (3.12) u(z) dν y (z) dλ(y) = u(z) dµ(z), with (3.13) ν y = F (y) M(). This is the desired result stated in Proposition Alternative proof in the product case Here we give a slightly different proof of Proposition in case (4.1) = X, π(x, y) = y, X also being a compact metric space. We are given a measure µ on as before, λ = π µ, and we seek a family ν y of measures on X (defined for λ-a.e. y ) such that for each u C(X ), (4.2) u(x, y) dµ(x, y) = u(x, y) dν y (x) dλ(y). X X In this approach, we start with the estimate (4.3) X u(x)v(y) dµ(x, y) ( ) sup u v L 1 (,λ),

6 6 and deduce that µ defines a continuous linear map (4.4) T µ : C(X) L 1 (, λ) = L (, λ). Again we use Lemma 2.3 to produce disjoint open sets ν such that λ( \ ν ν ) = 0 and continuous linear maps Φ ν : C(X) C( ν ) such that (4.5) T µ v ν = Φ ν v, λ-a.e., sup Φ ν v T µ v L. Then, for each y ν, define ν y M(X) by (4.6) u(x) dν y (x) = Φ ν u(y), u C(X). X We have u(x)v(y) dν y (x) dλ(y) = T µ u(y)v(y) dλ(y) (4.7) X = u(x)v(y) dµ(x, y), X for each u C(X), v C( ), and this in turn yields (4.2). Remark. This approach was used in [T], Chapter 13, Corollary The use of Lemma 2.3 here fills in the details of that argument. References [B] N. Bourbaki, Intégration, Chap. 6, Elements de Mathématiques, Hermann et Cie, Paris, [Ed] R. Edwards, Functional Analysis, Dover, New ork, [T] M. Taylor, Partial Differential Equations, Vols. 1 3, Springer-Verlag, New ork, 1996.