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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?

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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.

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