Given a coupling $\pi(\mu,\nu)$, show that $E_\mu f- E_\nu f= E_\pi [f(X) - f(Y)]$ In the lecture notes by for High-Dimensional Probability by Handel, the following is affirmed:
Let $\mu$ and $\nu$ be probability measures, then
$$\mathcal C(\mu,\nu) = \{ \text{Law} (X,Y) : X\sim \mu, Y\sim \nu \}
$$
Therefore, any
$\pi \in \mathcal C (\mu,\nu)$ is called a coupling of $\mu$ and $\nu$.
Hence, the author claims that
$$E_\mu f- E_\nu f= E_\pi [f(X) - f(Y)]$$
my question is how to prove the claim above. At first it seemed easy, but I’m getting confused on how to prove this rigorously.
I imagine that there is some kind of abuse of notation, since, for example, $E_\mu f = \int_\mathbb R f(w) d\mu$ and $E_\pi f(X) = \int_{\mathbb R^2} f(X((w,z)))d\pi$. But since $X:\Omega \rightarrow \mathbb R$, then $X((w,z))$ is ill defined.
 A: Suppose that $X$ and $Y$ take values in $\mathcal{X}$ and $\mathcal{Y}$, respectively. Since $\mu$ and $\nu$ are the marginal measures corresponding to $\pi$ (a prob. distribution on $\mathcal{X}\times\mathcal{Y}$),
\begin{align}
\mathsf{E}_\pi f(X)&=\int_{\mathcal{X}\times\mathcal{Y}} f(x)\pi(d(x,y))=\int_\mathcal{X} f(x)\int_\mathcal{Y} \pi(d(x,y)) \\
&=\int_\mathcal{X}f(x)\mu(dx)=\mathsf{E}_\mu f(X).
\end{align}
Similarly, $\mathsf{E}_\pi f(Y)=\mathsf{E}_\nu f(Y)$.
A: Note that $E_{\pi} [f(X)-f(Y)] = \int f(x) -f(y)d\pi(x,y) = \int f(x)d\pi(x,y) -\int f(y)d\pi(x,y)$ and $$\int f(x)d\pi(x,y) = \int f\circ P(x,y)d\pi(x,y) = \int f(z) d(P_*\pi)(z) = \int f(z) d\mu(z) $$
where $P$ is the mapping $(x,y)\mapsto x$ and $P_*\pi$ denotes the pushforward of $\pi$ by $P$.
A: The other answers have given formal proofs, so let me add:
One intuition for a coupling is that if you only observe $X$ it will look exactly like a sample from $\mu$, and similarly for $Y$. That is, a sample of $\pi_1( ( X,Y))$ ($\pi_1$ is projection on to first factor) sampled from $\pi$ follows the same distribution as a sample $X$ under $\mu$, and similarly for $Y$. This is more than an intuition, it's precisely the definition.
So $\mathbb{E}_{\pi}(f(X)) = \mathbb{E}_{\mu}(f)$ precisely because the distribution of $X$ under $\pi$ is $\mu$.
There's no abuse of notation. Generally in probability theory one is flexible with the measure space behind a a random variable, but in this case $X$ and $Y$ are both defined on the same measure space, with measure $\pi$, so that the distribution of these random variables is $\mu, \nu$ respectively. The coupling refers to the choice of the joint distribution of $(X,Y)$.
It may help to think through some examples here. For instance, can you describe all of the possible distributions of a coupling of two independent coin flips?
