Let $X$ be a compact metric space and $M(X)$ the set of all Borel probability measures on $X$. It is know that $M(X)$ is a convex compact metric space endowed with the weak-* topology i.e. $(\mu_n)_n \subseteq M(X)$ converges to $\mu \in M(X)$ iff for all continuous function $f \in C(X)$ $\int_X f d\mu_n \to \int_X f d\mu$. With this definition the linear functions $\mu \to \int_X f d\mu $ are continuous and hence $\mathcal{B}(M(X))-$measurable.

There is a nice characterization of the set $\mathcal{B}(M(X))$?

I think that $\mathcal{B}(M(X))$ is the smaller $\sigma-$algebra that makes the maps $\mu\to \mu(A)$ measurable for all Borel sets $A$ but I don't know how to prove this.

My first attempt is to take a close set $A \subseteq X$ and a sequence of continuous functions $f_n : X \to [0,1]$ such that $f_n \to 1_A$ point-wise (this can be made using the Urysohn lemma and is important that $A$ is close). Since $|f_n| \leq 1 \in L^1(\mu)$ for every $\mu \in M(X)$ then the dominated convergence theorem tell us that $$\mu \to \mu(A) = \int_X 1_A d\mu = \lim_{n\to \infty} \int_X f_n d\mu $$ An then the function $\mu\to \mu(A)$ is a point-wise limit of continuous functions and then is measurable. Then if $A\subseteq X$ is open $\mu \to \mu(A) = 1 -\mu(A^c)$ and then the function is measurable.

I try to use the regularity of every measure in $M(X)$ to prove that $\mu \to \mu(A)$ is measurable for every Borel set $A$. Why I try this? for every measure $\mu$ and a Borel set $A$ exists a sequence of open sets $\theta_n^\mu$ such that $\mu \to \mu(A)=\lim_{n\to \infty}\mu(\theta_n^\mu)$. The problem here is that the open sets depends on the measure and then the functions $\mu \to \mu(\theta^\mu_n) $ are not clearly measurable by the arguments that I give above.

Any help will be appreciated

Sorry for my english! :)

Let $X$ be a compact metric space and $M(X)$ the set of all Borel probability measures on $X$. It is know that $M(X)$ is a convex compact metric space endowed with the weak-* topology i.e. $(\mu_n)_n \subseteq M(X)$ converges to $\mu \in M(X)$ iff for all continuous function $f \in C(X)$ $\int_X f d\mu_n \to \int_X f d\mu$. With this definition the linear functions $\mu \to \int_X f d\mu $ are continuous and hence $\mathcal{B}(M(X))$-measurable.

There is a nice characterization of the set $\mathcal{B}(M(X))$?

I think that $\mathcal{B}(M(X))$ is the smaller $\sigma$-algebra that makes the maps $\mu\to \mu(A)$ measurable for all Borel sets $A$ but I don't know how to prove this.

My first attempt is to take a close set $A \subseteq X$ and a sequence of continuous functions $f_n : X \to [0,1]$ such that $f_n \to 1_A$ point-wise (this can be made using the Urysohn lemma and is important that $A$ is close). Since $|f_n| \leq 1 \in L^1(\mu)$ for every $\mu \in M(X)$ then the dominated convergence theorem tell us that $$\mu \to \mu(A) = \int_X 1_A d\mu = \lim_{n\to \infty} \int_X f_n d\mu $$ An then the function $\mu\to \mu(A)$ is a point-wise limit of continuous functions and then is measurable. Then if $A\subseteq X$ is open $\mu \to \mu(A) = 1 -\mu(A^c)$ and then the function is measurable.

I try to use the regularity of every measure in $M(X)$ to prove that $\mu \to \mu(A)$ is measurable for every Borel set $A$. Why I try this? for every measure $\mu$ and a Borel set $A$ exists a sequence of open sets $\theta_n^\mu$ such that $\mu \to \mu(A)=\lim_{n\to \infty}\mu(\theta_n^\mu)$. The problem here is that the open sets depends on the measure and then the functions $\mu \to \mu(\theta^\mu_n) $ are not clearly measurable by the arguments that I give above.

Any help will be appreciated.