Setup
I'm revisiting some notes from optimal transport and came across a result I'm not sure how to prove. It is an elementary result that, given a measure space ($\Omega, \mathcal{F}, \mathbb{P}$), a random variable $X: \Omega \rightarrow \Omega'$ with law($X$) = $\mu$ = $\mathbb{P}\circ X^{-1}$, and $f \in L^{1}(\Omega', \mu)$, that
$$
\int\limits_{\Omega} f(X) ~d\mathbb{P} = \int\limits_{\Omega'} f(x) ~d\mu
$$
by using the push forward measure and definition of Lebesgue integration.
Question
When it comes to coupling, how do I show
$$
\mathbb{E}[c(X,Y)] = \int c(X,Y) ~d\mathbb{P} = \int c(x,y) ~d\pi
$$
where $\pi$ is a coupling of $(\mu_{1},\mu_{2})$, i.e. has marginals $\mu_{1}$ and $\mu_{2}$?
Further
This is mainly to understand to following equality
$$
\inf\limits_{\pi\in\Pi(\mu_{1},\mu_{2})} \int c(x,y) ~d\pi(x,y) = \inf\limits_{\substack{X,Y\\law(X)=\mu_{1}\\law(Y)=\mu_{2}}}\mathbb{E}[c(X,Y)]
$$
where $X$ and $Y$ are random variables and $\Pi(\mu_{1},\mu_{2}) = \{\pi : \mathcal{F}_{1} \times \mathcal{F}_{2} \rightarrow \mathbb{R}_{0}^{+} \cup\{\infty\} ~~\vert~ \pi ~\text{couples}~ \mu_{1}~ \text{and}~\mu_{2}\}$.
Best Answer
From “Optimal Transport Old and New”(Villani):
It’s important to notice that the random variables $X$ and $Y$ are defined on the same probability space, and they need not be independent!
In fact for most cost functions $(X,Y)$ will not be an optimal coupling if they are.
Schematically the situation looks like this:
Where we assume without loss of generality that $\Omega = \mathcal{X}\times\mathcal{Y}$.
It might be helpful to consider the following discrete example, where we assume the sigma algebra is the power set and $I=[0,1]$:
Let’s say the cost matrix $C$ is given by $c_{i,j} = \frac{1}{4} - \pi_{i,j}$. Then $\pi$ is an optimal coupling.
Find an example for an optimal coupling $(X,Y)$. What would happen if $X$ and $Y$ were independent?
While this is not a proof, I hope it helps you to understand why the infima are equivalent.