46 1. The Kantorovich Duality u = 1 0 1u≥s ds (check!). In particular, whenever (ϕ, ψ) Φc, one can write (ϕ, ψ) = 1( 0 1ϕ≥s, −1ψ≤−s ) ds. Since ϕ is upper semi-continuous, the set 1ϕ≥s is closed, for all s [0, 1]. So we only have to check that for all s, 1ϕ≥s(x) 1ψ≤−s(y) 1C(x,y). The only nontrivial case is when ϕ(x) s and ψ(y) −s in this situation (x, y) should belong to C. But this case implies ϕ(x)+ψ(y) 0, in particular c(x, y) ϕ(x) + ψ(y) 0. Since c takes values in {0, 1}, we deduce that c(x, y) = 1, which indeed means (x, y) C. The proof is complete.
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