Let $(X,d)$ be a locally compact metric space and let $\mathscr{B}$ be the Borel sets. Let $\mu_n,\mu:\mathscr{B}\to [0,\infty]$ be regular measures over $(X,\mathscr{B})$. We say $\mu_n$ converges vaguely to $\mu$ (or $\mu_n \stackrel{\textrm{v}}{\to}\mu$) if $\int_Xud\mu_n\to \int_Xud\mu$ for all compactly supported continuous $u$ (i.e. $u \in C_c(X))$.
As part of a proof of a portmanteau theorem, I read that
$$\mu_n(B)\to\mu(B),\,\forall B \in \mathscr{B}\textrm{ relatively compact s.t. }\mu(\partial B)=0\implies\mu_n \stackrel{\textrm{v}}{\to}\mu$$
In the proof of this passage, it is first shown that $\{t>0:\mu(\partial\{u\geq t\})>0\}$ has Lebesgue measure zero. Then, a DCT is applied as following
$$\begin{aligned}\lim_{n \to \infty}\int_{(0,\infty)}\mu_n\{u\geq t\}dt&=\lim_{n \to \infty}\int_{(0,\|u\|_\infty)}\mu_n\{u\geq t\}dt\color{red}{\stackrel{(?)}{=}}\\
&=\int_{(0,\|u\|_\infty)}\lim_{n \to \infty}\mu_n\{u\geq t\}dt=\\
&=\int_{(0,\|u\|_\infty)}\mu\{u\geq t\}dt\end{aligned}$$
While the rest seems to me clear, I do not understand the $\color{red}{(?)}$ passage and how DCT has been applied. For example, if $M:=\sup_n\mu_n(X)<\infty$ we would have
$$\mu_n\{u\geq t\}\mathbf{1}_{(0,\|u\|_\infty)}(t)\leq M\mathbf{1}_{(0,\|u\|_\infty)}(t)\in L^1(dt)$$
and we could use DCT and take the limit over $\{t>0:\mu(\partial\{u\geq t\})=0\}$. However, I don't see how DCT is used seemingly without assumptions on $\mu_n$ with the exception of regularity. Is $(X,d)$ being locally compact involved here? Help appreciated.
Best Answer
Regular measures are finite on compact sets. Note that $\{x: u(x) >t\}$ is a subset of the support $K$ of $u$. We can choose an open set $U$ such that $K \subseteq U \subseteq \overline U$ and $\overline U$ is compact. Further we can choose $U$ to be a $\mu-$ continuity set. It now follows that $\mu_n (\{x: u(x)>t\})\leq \mu_n (K) \leq \mu_n (\overline U) \to \mu (\overline U)<\infty$. Hence, DCT can be applied.