Weak convergence of conditional probabilities

Question

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 $\mu_n^B\implies\mu^B$?

Answer 1

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 $\mu_n^B\Longrightarrow\mu^B$.

Answer 2

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 $\partial(A\cap B)\subset \partial A \cup \partial B$.