Let $\Omega$ be a Polish space and $\mathcal{B}(\Omega)$ be its Borel $\sigma$-algebra. Let $\{\mu\_n\}$ be a sequence of probability measures on $\mathcal{B}(\Omega)$ such that $\mu\_n$ weak-converges to $\mu$. Then in general, $\mu\_n(E) \nrightarrow \mu(E)$ for $E \in \mathcal{B}(\Omega)$.


But if for some $F \in \mathcal{B}(\Omega)$ we have $\partial F = F$ and $\mu\_n(F) = 0$ for all $n$, can we say that $\mu\_n(F) \rightarrow \mu(F)$? If not, are there any conditions on $\Omega$ that would make this true?



Weak-convergence of probability measures implies the convergence of the measure of a continuity set
A trivial counterexample is $\Omega=\mathbb R$, $\mu\_n=\delta\_{1/n}, \mu=\delta\_0$, and $F=\{0\}$. A standard result is $\mu\_n(F)\to \mu(F)$ if $\mu(\partial F)=0$. This looks somehow opposite to what you want. 


