Assume that we have a metric space $(A,\rho)$ and a sequence of probabilistic Borel measures $\mu_{n}$ on $A$ that converges weakly to the probabilistic Borel measure $\mu$. Assume also that one is given a sequence $A_{j}$ of closed subsets of $A$. Let $\limsup_{n\to\infty} A_{n}=\bigcap_{j=1}^{\infty}\bigcup_{i=j}^{\infty} A_{i}$ be the usual set-theoretic upper limit of the sequence $A_{j}$. An easy combination of Portmanteau theorem and Fatou lemma gives one that $$\limsup_{m\to\infty}\limsup_{n\to\infty}\mu_{n}(A_m)\leq \mu(\limsup_{m\to\infty} A_{m}).$$

I am interested under what additional assumptions on the measures $\mu_n$ one can get the stronger statement:

$$\limsup_{n\to\infty}\mu_{n}(A_n)\leq \mu(\limsup_{n\to\infty} A_{n})?$$

Note that this is true if $\mu_{n}$ converges strongly to $\mu$ and in this case the assumption that $A_{n}$ are closed is unnecessary. The inequality also trivially holds if we assume that $A_{n}$ is a nested family of closed sets, that is $A_{n+1}\subseteq A_{n}$. On the other hand the inequality does not hold for an arbitrary weakly convergent sequence of probability measures as the following trivial counterexample shows:

$$A=\mathbb R, \rho(x,y)=|x-y|, \mu_{n}=\delta_{\frac{1}{n}}, \mu=\delta_{0}, A_{n}=\{\frac{1}{n}\}.$$