# How fine is the Borel $\sigma$-algebra induced by the weak topology?

Let $$(\mathcal{X},\Sigma,P)$$ be a Polish probability measure space, and $$(\mathcal{X}^n,\Sigma^{\otimes n},P^n)$$ be the product of its $$n$$ copies. Let $$t: x^n \in \mathcal{X}^n \mapsto L_{x^n} \in \mathcal{P}(\mathcal{X})$$ be the empirical measure function, where $$L_{x^n}$$ is the empirical measure of the $$n$$-length sequence $$x^n$$, and $$\mathcal{P}(\mathcal{X})$$ is the set of probability measures on $$(\mathcal{X},\Sigma)$$. We consider two kinds of subsets of $$\mathcal{P}(\mathcal{X})$$:

1. A subset $$A$$ of $$\mathcal{P}(\mathcal{X})$$ such that $$t^{-1} (A)$$ is $$\Sigma^{\otimes n}$$-measurable.
2. A subset $$B$$ of $$\mathcal{P}(\mathcal{X})$$ such that $$B$$ is measurable with respect to the Borel $$\sigma$$-algebra induced by the weak topology.

Obviously, $$B$$ is a special case of $$A$$. My question is: For any $$A$$ above, can we find a sequence $$B_k$$ as the above such that $$P\circ t^{-1} (B_k)\to P\circ t^{-1} (A)$$ as $$k\to \infty$$? Are there references for this question? Thanks.

Yes, and you can even do the approximation by a single set. All the structure you need is that $$t$$ is a measurable function between Polish spaces.
The function $$t$$ is continuous and hence measurable, and the space $$\mathcal{P}(\mathcal{X})$$ is again Polish. Let $$A$$ satisfy the condition in 1. Then $$t^{-1}(A)$$ is a Borel set in $$\mathcal{X}^n$$ and, therefore, $$t\big(t^{-1}(A)\big)$$ an analytic subset of $$\mathcal{P}(\mathcal{X})$$. Analytic sets are universally measurable, so $$P\circ t^{-1}\Big(t\big(t^{-1}(A)\big)\Big)$$ is well defined. Since every measurable set in the completion is the union of a Borel set and a subset of a null set, there exists a Borel set $$B\subseteq t\big(t^{-1}(A)\big)$$ such that $$P\circ t^{-1}(B)=P\circ t^{-1}\Big(t\big(t^{-1}(A)\big)\Big)=P\circ t^{-1}(A).$$
• Thanks Michael! That is indeed what I want. In fact, my question was not clearly stated since it should additionally requires that $B_k \subseteq A$. (After seeing your answer I have realized this.) But you got my points even if the question is not clear.