1.1. General duality 31 = π∗[X0 × Y0] X0×Y0 ϕ0(x) + ψ0(y) dπ∗0(x, y) + (X0×Y0)c ϕ0(x) + ψ0(y) dπ∗(x, y) ≥ (1 − 2δ) X0 ϕ0 dμ0 + Y0 ψ0 dν0 − (2 c ∞ + 1) π∗[(X0 × Y0)c] ≥ (1 − 2δ) J0(ϕ0, ψ0) − 2(2 c ∞ + 1)δ ≥ (1 − 2δ) J0(ϕ0, ψ0) − 2(2 c ∞ + 1)δ ≥ (1 − 2δ)(inf I0 − δ) − 2(2 c ∞ + 1)δ ≥ (1 − 2δ) ( inf I − (2 c ∞ + 1)δ ) − 2(2 c ∞ + 1)δ. Since δ is arbitarily small, we conclude that sup J(ϕ, ψ) ≥ inf I, which was our goal. Note that the functions ϕ0, ψ0 are continuous (and even uniformly con- tinuous) on the whole of X, Y since c is uniformly continuous. Therefore, it does not matter whether we take the supremum of J over Φc ∩ L1 or over Φc ∩ Cb. By the way, this also shows that ϕ0 and ψ0 are measurable, a property which otherwise would not be obvious to establish. 3. We finally turn to the general case. Write c = sup cn, where cn is a nondecreasing sequence of nonnegative, uniformly continuous cost functions. Upon replacing cn by inf(cn,n), one can assume that each cn is bounded. Now let In be defined on Π(μ, ν) by In[π] = X×Y cn dπ. From Step 2 we know that (1.14) inf π∈Π(μ,ν) In[π] = sup (ϕ,ψ)∈Φcn J(ϕ, ψ). We will conclude the argument by showing that (1.15) inf π∈Π(μ,ν) I[π] = sup n inf π∈Π(μ,ν) In[π] and that, for each n, (1.16) sup (ϕ,ψ)∈Φcn J(ϕ, ψ) ≤ sup (ϕ,ψ)∈Φc J(ϕ, ψ). Indeed, the combination of (1.14), (1.15) and (1.16) will imply that inf π∈Π(μ,ν) I[π] ≤ sup (ϕ,ψ)∈Φc J(ϕ, ψ), and we already know that the reverse inequality is always true.

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