If $S$ is compact then the separating classes are same as convergence determining classes Suppose $(S,\rho)$ is a metric space with Borel $\sigma-$algebra $\mathcal S$. 
A class of subsets $\mathcal A\subset\mathcal S$ is called a separating class if whenever two Borel probability measures $P$ and $Q$ satisfy $P(A)=Q(A)$ for all $A\in\mathcal A$ then $P=Q$ on $\mathcal S$.
A class of subsets $\mathcal A\subset\mathcal S$ is called a convergence determining class if whenever, for any sequence $P_n,P$ of Borel probability measures, $P_n(A)\to P(A)$ for all $P$-continuity sets $A\in \mathcal A$ implies $P_n\implies P$ i.e. $P_n(A)\to P(A)$ for all $P-$continuity sets $A\in\mathcal S$.
In general, a separating class need not be a convergence determining class although the converse is always true. However, I want to show that if $S$ is compact, then every separating class is also a convergence determining class.
Any help will be appreciated.

In the case of a compact metric space the space of all Borel probability measures is relatively compact (by Prohorov's Theorem). Let $A$ be a $P-$ continuity set in $\mathcal S$. It is enough to show that every subsequence of $\{P_n(A)\}$ has a further subsequence converging to $P(A)$. But every subsequence of $\{P_n\}$ has a further subsequence converging weakly to some probability measure $Q$. Hence $Q(A)=P(A)$ for every $P$ continuity set $A$. This finishes the proof. 
