Narrow convergence and support of the limit measure Let $(X,d)$ be a Polish metric space and $\{\mu_n\}_{n\in\mathbb{N}}$ a sequence of probability measures such that $\mu_n\rightarrow\mu$ narrowly (i.e. $\int_Xf\,\mathrm{d}\mu_n\rightarrow\int_Xf\,\mathrm{d}\mu$ for any $f$ continuous and bounded). If there exists a compact set $K$ that contains the supports of the $\mu_n$'s then does it contain also the support of the narrow limit $\mu$?
 A: Since $(X,d)$ is a Polish space, any finite Borel measure on $X$ is tight (Ulam's theorem). Hence, for any $n\in\mathbb{N}$
$$
1=\mu_n\big(\text{supp}(\mu_n)\big)\leq\mu_n(K)$$
The Portmanteau theorem implies that
$$\mu(K)\geq \limsup_n\mu_n(K)=1$$
Therefore $\mu(K)=1$, and the from the definition of support, $\text{supp}(\mu)\subset K$.
A: Ok, I tried like that:
Assume that $K\supset\bigcup_n\mathrm{supp}\,\mu_n$ and let $x\in\mathrm{supp}\,\mu$. I want to prove that $K\supset\mathrm{supp}\,\mu$.
So let $x\in\mathrm{supp}\,\mu$. By definition of support of a measure there exists an open nhood $\mathcal{N}_x$ of $x$ such that $\mu(\mathcal{N}_x)>0$. Using the Portmanteau's theorem$$\liminf_{n\rightarrow\infty}\mu_n(\mathcal{N}_x)\geq\mu(\mathcal{N}_x)>0.$$So $\exists\bar{n}\in\mathbb{N}$ such that $\forall n\geq\bar{n}$ we have $\mu_n(\mathcal{N}_x)>0$. But now $x\in\mathrm{supp}\,\mu_n\subset K$ and we're done.
Remark: compactness was needed for a bigger theorem I needed to prove, and a "minimal" $K$ need to be closed. As counterexample, pick $\mu_n=\delta_{\frac{1}{n}}$ over the real line $\mathbb{R}$ with the Euclidian topology. We have that 1) $\mathrm{supp}\,\mu_n=\frac{1}{n}$ and 2) that $\delta_{x_n}\rightharpoonup\delta_{x_0}\Leftrightarrow x_n\rightarrow x_0$. In this case if we take $K=\bigcup_n\mathrm{supp}\,\mu_n=(0,1]$ we wouldn't have $\mathrm{supp}\,\mu\subset K$ being $\mathrm{supp}\,\mu=\{0\}$ so here a "minimal" $K$ should be closed.
