1.3. Appendix: A duality argument in Cb(X × Y ) 41 probability measures. Let be a nonnegative linear form on Cb(X × Y ) such that, for all bounded continuous functions ϕ on X and ψ on Y , (1.25) , ϕ + ψ = X ϕ + Y ψ dν. Then is a nonnegative measure, and in particular is an element of Π(μ, ν). Remark 1.26. Let be a nonnegative linear form on Cb(X ×Y ). Thanks to Lemma 1.24, we can write = π + R, where both π and R are nonnegative, and deduce that for any pair (ϕ, ψ) of nonnegative functions in Cb(X) × Cb(Y ), (1.26) X×Y ϕ(x) + ψ(y) dπ(x, y) X ϕ + Y ψ dν. But this alone does not immediately imply that π Π(μ, ν), as shown by Exercise 1.23. The proof of Lemma 1.25 will in fact use in an essential way the identity , 1 = 1. Proof of Lemma 1.24. Let us write = π + R. Since is nonnegative, so is its restriction π to C0(X × Y ). So we just have to prove that π on Cb(X × Y ), in the sense of linear functionals. Let u(x, y) be any continuous bounded nonnegative function, and let χn(x, y) be an increasing sequence of continuous cut-off functions with compact support, 0 χn(x, y) 1, such that for all (x, y) X × Y , χn(x, y) −→ 1 as n ∞. The existence of the sequence (χn) follows from the assumption of σ-compactness and Urysohn’s lemma. Then we can write X×Y u = lim n→∞ X×Y uχn = lim n→∞ , uχn , u , where the first equality is a consequence of the monotone convergence the- orem, and the inequality on the right is a consequence of being nonnega- tive. Proof of Lemma 1.25. As mentioned above, we shall use in a crucial way the identity , 1 = 1, which is a particular case of (1.25). Let (Kn) (resp. (Ln)) be an increasing sequence of compact sets in X (resp. Y ), such that Kn+1 is a neighborhood of Kn and Kn = X (resp. Ln+1 is a neighborhood of Ln and Ln = Y ). We introduce an increasing sequence of cut-off functions ϕn C0(X) with 0 ϕn 1 on X, ϕn = 1 on Kn and ϕn = 0 on Kn+1 c and similarly, an increasing sequence of cut-off functions ψn C0(Y ) with 0 ψn 1 on Y , ψn = 1 on Ln and ψn = 0 on Ln+1. c The existence of these sequences follows from Urysohn’s lemma. In particular, we have (1 ϕn+1)ψn + (1 ψn+1)ϕn 1
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