42 1. The Kantorovich Duality on the whole of X × Y , for all n and (1 − ϕn+1)ψn + (1 − ψn+1)ϕn −−−→ n→∞ 0, pointwise on X × Y . Let us introduce the quantity (1.27) An = , (1 − ϕn+1)ψn + (1 − ψn+1)ϕn . On one hand, it is equal to X×Y (1 − ϕn+1)ψn + (1 − ψn+1)ϕn dπ(x, y) + R, (1 − ϕn+1)ψn + (1 − ψn+1)ϕn ≤ X×Y (1 − ϕn+1)ψn + (1 − ψn+1)ϕn dπ(x, y) + R, 1 , which converges as n → ∞ to R, 1 = , 1 − X×Y dπ = 1 − X×Y dπ. Thus, (1.28) lim sup n→∞ An ≤ 1 − π[X × Y ]. On the other hand, (1.27) can be rewritten as , ϕn + , ψn − , ϕn+1ψn − , ϕnψn+1 . By (1.25) and the fact that R is supported at infinity, this is also X ϕn dμ + Y ψn dν − X×Y ( ϕn+1ψn + ϕnψn+1 ) dπ, which converges as n → ∞ to 1 + 1 − 2 X×Y dπ. Thus, (1.29) lim n→∞ An = 2 − 2 X×Y dπ. From (1.28) and (1.29), it follows that π[X × Y ] ≥ 1. In particular, R, 1 = , 1 − π, 1 ≤ 0, which implies that R, 1 = 0, so that in fact R = 0. This concludes the proof. With Lemma 1.25 in hand, the proof of Proposition 1.22 will be straight- forward.
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