TITLE: Probability measures equipped with weak topology, is $P\mapsto P(A)$ measurable?
QUESTION [7 upvotes]: Let $X$ be a compact metric space and $\mathcal{M}(X)$ be the set of probability measures on $X$ equipped with the topology of weak convergence. So $\mathcal{M}(X)$ itself can be viewed as a compact metrizable space. Let $A\subseteq X$ be any measurable set in $X$, is the map
$$\mathcal{M}(X)\rightarrow [0,1]~,~~P\mapsto P(A)$$
measurable?

REPLY [3 votes]: The map $P\mapsto \int f \,dP$ from $(M(X),{\cal B}(M(X)))$ to $(\mathbb{R},{\cal B}(\mathbb{R}))$ is measurable for all bounded, Borel $f$ on $X$. In particular, for $f={\bf 1}_A,$ the map $P\mapsto P(A)$ is measurable.
Proof: If $f$ is a continuous function, then  $P\mapsto \int f \,dP$ is continuous by 
the definition of weak convergence of measures, and hence is also measurable. 
Now apply the functional Monotone Class Theorem. We use Theorem 2 at the link, with 
$${\cal H}=\left\{f\in {\cal B}_b(X): P\mapsto \int f\,dP\mbox{ is measurable}\right\},$$
and conclude that ${\cal H}={\cal B}_b(X).$