Suppose $(\pi_n)_{n \in \mathbb{N}}$ is a sequence of measures on two Polish spaces $X$ and $Y$, each with marginals $\mu$ and $\nu$. I want to prove that if it exists, the limit of the sequence, denoted $\pi$, has marginals $\mu$ and $\nu$ also. I use the following definition of weak convergence:
A sequence of measures $(\mu_n)$ in $\mathcal{M}$ is said to converge weakly to a measure $\mu \in \mathcal{M}$ if \begin{align*} \lim_{n \to \infty}\int \lvert f \rvert \ d\mu_n = \int \lvert f \rvert \ d\mu \end{align*} for all Lipschitz continuous, bounded functions $f$.
Proof: There exists a sequence $(f_n)$ of Lipschitz continuous bounded functions increasing to $\mathbb{1}_E$ where $E = A \times Y$ for all measurable $A \subset X$. Hence (applying Monotone Convergence theorem):
$\pi(A \times E) = lim_{j \rightarrow \infty}\int f_j \ d\pi = lim_{j \rightarrow \infty} \ lim_{n \rightarrow \infty}\int f_j \ d\pi_n $
I want to be able to exchange the two limits to complete my argument and show that the marginals are equal. However I am not sure how to argue this (or if it is even true).
Any help is much appreciated.