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 $E \subseteq X$.

  • $\begingroup$ 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. $\endgroup$ – mathworker21 Mar 21 at 5:31

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