# The borel $\sigma-$algebra of the set of probability measures

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! :)

• To show that $\mu \mapsto \mu(A)$ is measurable for every Borel $A$, try a $\pi$-$\lambda$ or monotone class argument. Let $\mathcal{L}$ be the set of all $A \subset X$ such that $\mu \mapsto \mu(A)$ is measurable, and let $\mathcal{P}$ be the closed subsets of $X$. Then it is easy to check that $\mathcal{P}$ is closed under finite intersections, $\mathcal{L}$ is a $\lambda$-system, and $\mathcal{P} \subset \mathcal{L}$. The conclusion is that $\mathcal{L} \supset \sigma(\mathcal{P})$, i.e. the Borel sets. – Nate Eldredge May 22 '14 at 0:49
• You can find a proof in Stochastic Optimal Control: The Discrete-Time Case by Bertsekas and Shreve, where this is Proposition 7.25. – Michael Greinecker May 25 '14 at 11:08

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

• Perhaps it should be made clear that the algebra you mention needs to contain all the open sets. Certainly the indicator of an open set (or at least an open ball) can be written as a pointwise limit of continuous functions, but we are taking advantage of the metric here. – Nate Eldredge Apr 19 '16 at 4:12

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.