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.



How fine is the Borel $\sigma$-algebra induced by the weak topology?
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](https://en.wikipedia.org/wiki/Analytic_set) subset of $\mathcal{P}(\mathcal{X})$. Analytic sets are [universally measurable](https://en.wikipedia.org/wiki/Universally_measurable_set), 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 $$P\circ t^{-1}(B)=P\circ t^{-1}\Big(t\big(t^{-1}(A)\big)\Big)=P\circ t^{-1}(A).$$


