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Let $X$ be a separable metric space, $\mu_{n}$ a sequence of Borel probability measures and $\mathcal{C}$ be a family of sets that is closed under finite unions and interections, and that contains all the balls. If $\mu_{n}(A)$ converges for every $A\in\mathcal{C}$, does there exists a Borel measure $\mu_{\infty}$ such that $\mu_{\infty}(E)=\lim\mu_{n}(E)$ for every $E\in\mathcal{C}?$

From Theorem 4.3 in this paper we can get this result when $X$ is locally compact. Here Sion makes an outer measure and then shows open sets are measurable. His proof definitively uses local compactness.

Does anybody know if the result is true when $X$ is not necessarily locally compact?

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    $\begingroup$ The referenced Theorem 4.3 does not imply the condition in the first paragraph, at least not as stated. The theorem is in terms of weak$^*$ convergence, i.e. $\int f \ d\mu_n \to \int f\ d\mu_\infty$ for all $f \in C_c(X)$ (continuous with compact support). This does not imply that $\mu_n(E) \to \mu_\infty(E)$ for $E\not\subset K$ for any $K$ compact (hence Andreas' counterexample below). The statement is true if you assume that all elements in $\mathcal C$ are relatively compact. $\endgroup$
    – D. Kelleher
    Commented Apr 20, 2013 at 19:33
  • $\begingroup$ Thanks for your comments. Theorem 4.3 talks about two kinds of limits. The other limit is the outer measure constructed in section 3. Theorem 3.3 mentions that the outer measure is Radon (hence every borel set is measurable) if every open set is the countable union of compact sets. $\endgroup$
    – FelipeG
    Commented Apr 21, 2013 at 17:18

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Why isn't the following a (locally compact) counterexample? Let $X$ be the set of natural numbers, with the metric where the distance between every two distinct points is 1. So the topology is discrete, and the only balls are the singletons and the whole space. Let $\mathcal C$ consists of the finite sets and the whole space. Let $\mu_n$ be the probability measure concentrated at the point $n$. Then for each finite set $A$ we have $\lim_n\mu_n(A)=0$, while for the whole space $X$ we have $\lim_n\mu_n(X)=1$. These limits are not the values of any countably additive measure $\mu_\infty$.

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The answer is no, and, without more conditions, I think that the best one can hope for is that $\mu_\infty$ is a finitely additive measure.

The idea from the paper of Maurice Sion that was cited is that finite measures are dual to $C_0(X)$, continuous functions vanishing at infinity. This technique doesn't work if $X$ is not locally compact -- if $x_0\in X$ is such that there is no relatively compact neighborhood of $x_0$, then $f(x_0)=0$ for all $f \in C_0(X)$. So a pointmass at $x_0$ is equivalent to the zero measure as elements of the dual of $C_0$!

As a work around, one could consider $L^\infty(X)$, (in particular, one would need an appropriate Borel reference measure). The dual of $L^\infty$ can be classified as finitely-additive measures. The condition $\mu_n(A)\to\mu_\infty(A)$ for all $A$ in a basis of $X$ means that the induced functionals converge on simple functions, and the limit is bounded because the measures were taken to be probabilities. So the limit should be a finitely additive measure.

I'm being a little fast and loose, so there may be things I'm forgetting to assume (regularity of the measures maybe?).

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