I'm looking at this expression: $$ \sup_{f,g}\int f(x)\,d\mu(x) + \int g(y)\,d\nu(y) - \int (f(x)+g(y))\,d\pi(x,y) $$ where $f\in L_1(\mu)$ and $g\in L_1(\nu)$.
I understand why this is $0$ when $\pi\in \Gamma(\mu,\nu)$ (joint coupling of $\mu$ and $\nu$). But I don't see why this is $\infty$ otherwise. Can someone explain in detail how this is unbounded if $\pi$ isn't a coupling?
Suppose that $\mu'$ is the natural projection of $\pi$ on the first coordinate, then without loss of generality, whenever $\pi$ is not a coupling, there is $A$ such that $(\mu-\mu')(A)>0$ (otherwise it is true for $\nu$ and the natural projection of $\pi$ on the second coordinate). Take $f_n(x)=n 1(x\in A)$, and $g_n(x)=0$, then your expression goes to $+\infty$ as $n\to\infty$, also all $f_n$ are in $L_1(\mu)$.