# Weak convergence of conditional probabilities

Suppose $$\mu_n\implies\mu$$, i.e. $$\mu_n$$ converges weakly to $$\mu$$ where $$\mu_n$$, $$\mu$$ are probability measures on some metric space $$(X,d)$$. Given a Borel set $$B$$, define $$\mu^B$$ to be the conditional probability given $$B$$, i.e. $$\mu^B(A)=\mu(A\cap B)/\mu(B)$$, and similarly for $$\mu_n^B$$.

Under what conditions (e.g. on $$X$$, $$\mu$$, and/or $$B$$) does it follow that $$\mu_n^B\implies\mu^B$$?

For $$\mu_n^B\Longrightarrow\mu^B$$, it is enough that $$\mu(\partial B)=0$$ (and $$\mu(B)>0$$), where $$\partial B$$ denotes the boundary of $$B$$.
Indeed, then for any Borel set $$A$$ such that $$\mu(\partial A)=0$$ we have $$\mu(\partial (A\cap B))=0$$, because $$\partial(A\cap B)\subseteq(\partial A)\cup(\partial B)$$. So, by the Portmanteau theorem, we have $$\mu_n^B\Longrightarrow\mu^B$$.
A sufficient condition is that $$B$$ is a $$\mu$$-continuity set, namely, $$\mu(\partial B)=0$$.
To see this, recall weak convergence can be characterized by $$\mu_n(E)\rightarrow \mu(E)$$ for any $$\mu$$-continuity Borel set $$E$$. Then to establish the desired weak convergence of conditional probability measures, it suffices to note that for any Borel $$A$$ with $$\mu(\partial A)=0$$, we have $$\mu(\partial(A\cap B))=0$$ as well since $$\partial(A\cap B)\subset \partial A \cup \partial B$$.