# Weak star convergence of Borel probability measures on a metric space

Let $$(X,\rho)$$ be a compact metric space and let $$P(X)$$ be the set of Borel probability measures on the Borel $$\sigma$$-algebra of $$X$$. Suppose $$\mu_n,\mu \in P(X)$$ for $$n \in \mathbb{N}$$ such that $$\mu_n \to \mu$$ in the weak star topology, i.e. $$\lim_{n\to\infty}\int_X f(x) \hspace{1mm} d\mu_n(x) = \int_X f(x) \hspace{1mm} d\mu(x)$$ for all continuous functions $$f \colon X \to \mathbb{C}$$. If $$E \subseteq X$$ is a Borel set satisfying $$\mu(E) = \mu(\textrm{Int}(E))$$, then using the regularity of Borel measures on metric spaces, given $$\epsilon > 0$$ we can produce compact sets $$K_1 \subseteq \textrm{Int}(E), K_2 \subseteq E$$ and an open set $$U \supseteq E$$ such that $$\mu(U) - \mu(E), \mu(E) - \mu(K_1),\mu(E) - \mu(K_2) < \epsilon.$$ By Urysohn's lemma there exist continuous functions $$f \colon X \to [0,1]$$ and $$g \colon X \to [0,1]$$ such that $$\textrm{supp}(f) \subseteq U, \textrm{supp}(g) \subseteq \textrm{Int}(E)$$ and $$f(x) = g(y) = 1$$ if $$x \in K_2, y \in K_1$$. We have $$\int_X g(x) \hspace{1mm} d\mu_n(x) \leq \mu_n(E) \leq \int_X f(x) \hspace{1mm} d\mu_n(x)$$ for all $$n \in \mathbb{N}$$ by the construction of $$f$$ and $$g$$. By the choice of the sets $$K_1,K_2,U$$ and the weak star convergence of $$\{\mu_n\}_{n=1}^{\infty}$$ it follows that $$\mu(E) - \epsilon \leq \int_X g(x) \hspace{1mm} d\mu(x) \leq \liminf_{n\to\infty}\mu_n(E) \leq \limsup_{n\to\infty}\mu_n(E) \leq \int_X f(x) \hspace{1mm} d\mu_n(x) \leq \mu(E) + \epsilon.$$ This shows that $$\lim_{n\to\infty}\mu_n(E)$$ exists and is equal to $$\mu(E)$$.

My question is what additional assumptions do we need to conclude that $$\lim_{n\to\infty}\mu_n(E) = \mu(E)$$ holds for all Borel subsets $$E \subseteq X$$.

• I'm confused by your question. You gave an assumption on $E$ that guarantees $\lim_n \mu_n(E) = \mu(E)$, but you end up asking what conditions on $(\mu_n)_n$ and $\mu$ we need to ensure that $\lim_n \mu(E_n) = \mu(E)$ for all $E$? I'm not exactly sure what kind of answer you are looking for. Also, I think you want $X$ to be compact, so that $\int_X gd\mu$ always makes sense. – mathworker21 Mar 21 at 5:31