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Suppose that $(S,\Sigma)$ is a measurable space with $S$ Polish and $\Sigma$ its Borel sigma algebra. Let $\mathcal{C}$ be the collection of discrete probability measures on $S$ having countably infinite many distinct point masses. For any $P \in \mathcal{C}$ there exists a set $F_P \in \Sigma^{\infty}$ of full $P^{\infty}$ measure with the property that for any $(s_1,s_2,...) \in F_P$ we have that $s_i$ is a point mass of $P$ for all $i$, and that the empirical probability measures for $(s_1,s_2,...)$ converge setwise (sometimes called strongly) to $P$. Let

$$G = \bigcup_{P \in \mathcal{C}} F_P $$

My question is this: is $G \in \Sigma^{\infty}$?

Additionally, is $G \in \Sigma^{\infty}$ if we change $\mathcal{C}$ to be the collection of all discrete probability measures on $S$?

Many thanks for any help on this.

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  • $\begingroup$ Can you explain what you mean by setwise convergence of probability measures? $\endgroup$
    – Anthony Quas
    Commented Jun 22, 2017 at 21:24
  • $\begingroup$ @Anthony Quas: $P_n \rightarrow P$ setwise means that $P_n(A) \rightarrow P(A)$ for all $A \in \Sigma$. $\endgroup$
    – shanex
    Commented Jun 22, 2017 at 21:29
  • $\begingroup$ Can you explain what you mean by all discrete probability measures on $\mathcal S$ (how does this differ from the discrete probability measures on $\mathcal S$ having countably many distinct point masses?) $\endgroup$
    – Anthony Quas
    Commented Jun 22, 2017 at 22:29
  • $\begingroup$ @Anthony Quas: by all I mean those discrete $P$ having finite or countably many distinct point masses. $\endgroup$
    – shanex
    Commented Jun 22, 2017 at 22:32
  • $\begingroup$ OK. So when you say "countable", you mean countably infinite (as finite sets are generally also defined to be countable). My answer is fine in either case. $\endgroup$
    – Anthony Quas
    Commented Jun 22, 2017 at 22:41

1 Answer 1

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Define $\bar f_n(x)=\limsup_{N\to\infty} \frac 1N\#\{j\le N\colon x_j=x_n\}$, $\underline f_n(x)=\liminf_{N\to\infty} \frac 1N\#\{j\le N\colon x_j=x_n\}$, $g_n(x)=\prod_{j<n}\mathbf 1_{\{x_j\ne x_n\}}$.

Then define $F_n(x)=g_n(x)\cdot \mathbf 1_{\bar f_n(x)=\underline f_n(x)}\cdot \bar f_n(x)$. This function is the limiting density of terms that are equal to $x_n$ provided that (1) $x_n$ is the first time this point has appeared; (2) the limiting density of $x_n$'s exists.

Then $$ G=\left\{x\colon \sum_{n=1}^\infty F_n(x)=1\right\}\cap \bigcap_n\{x\colon \bar f_n(x)>0\}. $$

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  • $\begingroup$ Thanks very much. But wouldn't $\{x:\Sigma_{i=1}^{\infty} F_n(x) = 1\}$ contain tuples $x$ having entries that could repeat only finitely often? No tuple in any $F_P$ has this property. $\endgroup$
    – shanex
    Commented Jun 23, 2017 at 22:34
  • $\begingroup$ OK. I hadn't seen you wanted each $x_i$ to have positive mass. But then just intersect with $\bigcap_n\{x\colon \bar f_n(x)>0\}$. I will edit my answer to reflect this. $\endgroup$
    – Anthony Quas
    Commented Jun 23, 2017 at 23:36

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