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Do empirical measures converge weakly to the measure almost surely? In particular suppose $\mu$ is a Borel probability measure on $\mathbb R^d$ and that $X_1,X_2,\dots$ are IID drawn from $\mu$. Let $\hat\mu_N=\frac{1}{N}\sum_{i=1}^N\delta_{X_i}$. The Strong Law of Large numbers Says that $\mathbb P(\hat\mu_N(A)\to\mu(A))=1$ for every $A$. Almost sure weak convergence, on the other hand, would be that $\mathbb P(\hat\mu_N(A)\to\mu(A)\text{ for every continuity set $A$})=1$. Is that statement true?

Note that this is very different from the stronger statement $\mathbb P(\sup_{\text{continuity set $A$}}\left|{\hat\mu_N(A)-\mu(A)}\right|\to0)=1$, which would require continuity sets to constitute a Glivenco-Cantelli class, something that only happens in $d=1$ as far as I understand.

Applying the Hewitt–Savage zero–one law (https://en.wikipedia.org/wiki/Hewitt%E2%80%93Savage_zero-one_law) it's clear that $\mathbb P(\hat\mu_N(A)\to\mu(A)\text{ for every continuity set $A$})$ is either $1$ or $0$ and nowhere in between. Can we argue it cannot be $0$?

If this is not true in general are there reasonable sufficient conditions on $\mu$ that would guarantee almost sure weak convergence of $\hat\mu_N$? My feeling is that something like compactness of support should do it.

Thanks.

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The result is true.

Theorem 11.4.1 in Real Analysis and Probability by R.M.Dudley explains why the empirical measures converge almost surely for a Borel probability measure $\mu$ on a separable metric space $(S,d)$. I can provide more detail if you can't find this reference.

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