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.



The Borel $\sigma$-algebra of the set of probability measures
Consider the set $\mathcal A$ of measurable subsets $A \subset X$, such that $\mu \mapsto \mu[A]$ is measurable. Obviously, $\mathcal A$ contains all open sets.


It's also easy to see that $\mathcal A$ contains an algebra of sets - say, sets $A \subset X$, such that their indicator is a pointwise limit of a sequence of continuous functions. Indeed, if $f\_n \to \mathsf{1} [A]$ and $0 \le f\_n \le 1$ (which can always be assumed, since we can replace $f\_n$ by $\max(\min(f\_n, 0), 1)$) then the function $\mu \mapsto \mu[A]$ is the pointwise limit of $\mu \mapsto \intop f\_n d\mu$.


A similar argument to the above shows that $\mathcal{A}$ is closed under [sequential limits of sets](http://en.wikipedia.org/wiki/Set-theoretic_limit). And anything that contains an algebra and is closed under sequential limits contains a $\sigma$-algebra (I would call that a version of the [monotone class theorem](http://en.wikipedia.org/wiki/Monotone_class_theorem)). Therefore, $\mathcal A \supset \mathcal{B} (X)$.


