Weak limit of coupling is a coupling.

Question

Let $X, Y$ be Polish spaces with probability measure $\mu, \nu$. Let $(\pi_n)$ be a sequence of couplings of ($\mu,\nu)$ that converges weakly to $\pi$. Show that $\pi$ is a coupling of $\mu,\nu$ too.

Answer 1

Define the canonical projections $e_1,e_2$ on $X\times Y$ by $e_1(x,y) = x$ and $e_2(x,y)=y$. By assumption, we have that the pushforwards $e_1(\pi_n)$ and $e_2(\pi_n)$ satisfy $$ e_1(\pi_n) = \mu, \quad e_2(\pi_n) = \nu, \quad n \in \mathbb{N}.$$ Clearly, both $e_1$ and $e_2$ are continuous, so by the continuous mapping theorem, we obtain $$ e_1(\pi) = \lim_{n\to\infty} e_1(\pi_n) = \mu, \quad e_2(\pi) = \lim_{n\to\infty} e_2(\pi_n) = \nu.$$

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Note: here we are using that because $e_1$ is continuous, the map $\pi \mapsto e_1(\pi)$ is continuous from $\mathcal{P}(X\times Y)$ to $\mathcal{P}(X)$, and the same goes for $e_2$.