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](http://msp.org/pjm/1964/14-3/pjm-v14-n3-p23-p.pdf) 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?



Existence of limit measure
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 $\mu\_\infty$.


