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We define $\mathcal{M}_{F}(X)$ for $X$ metric space to be the set of all finite measures defined on the borel sigma algebra.

Problem.

Assume $(X,d)$ is a metric space and $\mu_n,\mu \in \mathcal{M}_{F}(X)$. Then $\mu_n\rightarrow_{w} \mu$ if and only if $\mu(C)\geq \lim_n \sup_n \mu_n(C)$ for any closed set $C$ and $\lim_n \inf \mu_n(X)\geq \mu(X)$.

Forward direction. Assume $\mu_n$ converges weakly to $\mu$. Hence for each $n$, Let $f_n$ be a measurable function such that $0\leq f_n(x)\leq 1$ for all $x$ and $f_n|_{C}=1$ and $\lim_{n\rightarrow \infty}f_n(x)=\chi_{C}(x)$ for all $x\in X$. Now the result follows from Fatou's lemma and the dominated convergence theorem. The reverse inclusion is also true.

Why does the backwards direction follow?

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Here is a sketch:

The condition implies that if the boundary of a Borel set $E$ has $\mu$-measure zero, then $\lim_n \mu_n(E)=\mu(E)$.

Let $f:X\to\mathbb{R}$ be a continuous function. The idea is now to suitably partition the range of $f$ into small intervals and check that the measures of the preimages converge. This, plus a bit of epsilontics, proves the result.

To get this right, we want these preimages to have boundaries of $\mu$-measure zero. The preimage of the interval $[a,b]$ under $f$ is a closed set whose boundary is a subset of $f^{-1}\big(\{a,b\}\big)$. Now, if $y\neq z$, then $f^{-1}(\{y\})$ and $f^{-1}(\{z\})$ are disjoint. A finite measure cannot put strictly positive mass on uncountable many disjoint sets, so if we avoid taking the endpoints of the intervals from the at most most countably many points $y$ such that $\mu(f^{-1}(\{y\}))>0$, everything works out well.

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