1.1. General duality 29 From π0 we construct a π ∈ Π(μ, ν) in a natural way, by gluing together π0 with a little bit of π∗: π = π∗[X0 × Y0] π0 + 1(X0×Y0)c π∗ (check that π ∈ Π(μ, ν)!). So I[π] = π∗[X0 × Y0] I0[π0] + (X0×Y0)c c(x, y) dπ∗(x, y) ≤ I0[π0] + 2 c ∞δ = inf I0 + 2 c ∞δ. It follows that inf π∈Π(μ,ν) I[π] ≤ inf I0 + 2 c ∞δ. Now introduce the functional J0(ϕ0,ψ0) = X0 ϕ0 dμ0 + Y0 ψ0 dν0, defined on L1(dμ0)×L1(dν0). By Step 1 of the proof, we know that inf I0 = sup J0, where the supremum runs over all admissible couples (ϕ0,ψ0) ∈ L1(dμ0) × L1(dν0), i.e. those which satisfy ϕ0(x) + ψ0(y) ≤ c(x, y) for almost all x, y. In particular, there exist an admissible couple of functions ϕ0, ψ0 such that J0(ϕ0, ψ0) ≥ sup J0 − δ. Our problem is to construct from (ϕ0, ψ0) a couple (ϕ, ψ) which would be very eﬃcient in the problem of maximization of J(ϕ, ψ). It will be useful to ensure that the inequality ϕ0(x) + ψ0(y) ≤ c(x, y) is valid for all x and y, not just almost all. We can ensure this, provided that we allow ϕ0 and ψ0 to take values in R ∪ {−∞}. Indeed, we can introduce negligible sets Nx and Ny such that the inequality holds true for (x, y) ∈ Nx c × Ny c, and redefine the values of ϕ, ψ to be −∞ on Nx, Ny respectively. In a first step, we will control ϕ0, ψ0 from below at some point in X ×Y . Without loss of generality, we assume that δ ≤ 1. Since J0(0, 0) = 0, we know that sup J0 ≥ 0, and hence J0(ϕ0, ψ0) ≥ −δ ≥ −1. By writing J0(ϕ0, ψ0) = X×Y [ϕ0(x) + ψ0(y)] dπ0(x, y), where π0 is any element of Π0(μ0,ν0), we deduce that there exists (x0,y0) ∈ X0 × Y0 such that (1.13) ϕ0(x0) + ψ0(y0) ≥ −1.

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