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 $\mu\_n^B\implies\mu^B$? 



Weak convergence of conditional probabilities
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](https://en.wikipedia.org/wiki/Convergence_of_measures), we have $\mu\_n^B\Longrightarrow\mu^B$. 


