The Borel $\sigma$-algebra of the set of probability measures

Question

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.

Answer 1

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. 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). Therefore, $\mathcal A \supset \mathcal{B} (X)$.

Answer 2

One direction is covered by Alexander Shamov's answer or my comment above. Here is the other direction.

Let $\mathcal{F}$ be the smallest $\sigma$-algebra such that, for every Borel $A \subset X$, $\mu \mapsto \mu(A)$ is $\mathcal{F}$-measurable. As mentioned, we have shown that $\mathcal{F} \subset \mathcal{B}(M(X))$.

Note further that for any bounded Borel $f : X \to \mathbb{R}$, the map $\mu \mapsto \int f\,d\mu$ is $\mathcal{F}$-measurable. (This is clear when $f$ is simple. Otherwise, approximate $f$ by a bounded sequence $f_n$ of simple functions and note that by dominated convergence, $\int f\,d\mu = \lim_{n \to \infty} \int f_n\,d\mu$, so that $\mu \mapsto \int f\,d\mu$ is a pointwise limit of $\mathcal{F}$-measurable functions.)

In particular, this is true when $f$ is continuous. So for any continuous $f$ and $a < b \in \mathbb{R}$, we have $$U_{f,a,b} := \left\{\mu : a < \int f\,d\mu < b\right\} \in \mathcal{F}.$$ But the collection of all $U_{f,a,b}$ is a sub-basis for the the weak-* topology. In particular, every weak-* open set in $M(X)$ is a union of finite intersections of such sets. Since $M(X)$ is second countable, we may say every open set is a countable union of finite intersections of $U_{f,a,b}$ sets, and hence every open set is in $\mathcal{F}$. So $\mathcal{B}(M(X)) \subset \mathcal{F}$ and your characterization is proved.