I'm reading a proof of below theorem from [this paper](https://arxiv.org/pdf/2205.13207.pdf).



> 
> **Theorem A.3.** Let $\Omega$ be a locally compact normal Hausdorff space. Let $\left\{\mu\_n\right\} \cup\{\mu\} \subset \mathcal{M}(\Omega)$ and assume that $\underset{n \rightarrow \infty}{\operatorname{v-lim}} \mu\_n=\mu$. Then for any open set $\Theta \subset \Omega$,
> $$
> |\mu|(\Theta) \leq \liminf \_{n \rightarrow \infty}\left|\mu\_n\right|(\Theta) .
> $$
> In particular, $\|\mu\| \leq \liminf \_{n \rightarrow \infty}\left\|\mu\_n\right\|$.
> 
> 
> 


**Proof.** Let $\Theta \subset \Omega$ be open and $\varepsilon>0$. Since $\mu$ is inner regular and $\Omega$ is normal and locally compact, as a consequence of Urysohn's lemma [1, Lemma 2.46], there exists $f \in C\_c(\Omega)$ such that $|f| \leq 1, \operatorname{supp}(f) \subset \Theta$ and
$$
\int f \mathrm{~d} \color{red}{\mu} \geq \color{red}{|\mu|}(\Theta)-\varepsilon.
$$
Then by vague convergence of $\left\{\mu\_n\right\}$,
$$
|\mu|(\Theta)-\varepsilon \leq \int f \mathrm{~d} \mu=\lim \_{n \rightarrow \infty} \int f \mathrm{~d} \mu\_n \leq \liminf \_{n \rightarrow \infty} \int|f| \mathrm{d}\left|\mu\_n\right| \leq \liminf \_{n \rightarrow \infty}\left|\mu\_n\right|(\Theta)
$$
Now the result follows by letting $\varepsilon \downarrow 0$.




---


**My understading:** Below are the authors' related definitions. A finite signed measure $\mu$ is Radon if its variation $|\mu|$ is inner regular. In above proof, $\mu$ is Radon and thus $|\mu|$ is inner regular. So there is $f \in C\_c(\Omega)$ such that $f \le 1\_\Theta$ and
$$
\int f \mathrm{~d} \color{red}{|\mu|} \geq \color{red}{|\mu|}(\Theta)-\varepsilon.
$$


So I'm confused due to the appearance of $\color{red}{\mu}$ instead of $\color{red}{|\mu|}$ in the integral. Could you elaborate on my confusion?




---


**Authors' definitions:** Let


* $\Omega$ be a metric space and $\mathscr{B}(\Omega)$ its Borel $\sigma$-algebra.
* $C\_b(\Omega)$ the subspace of all real-valued bounded continuous functions on $\Omega$.
* $C\_0(\Omega)$ the subspace of all $f \in C\_b(\Omega)$ such that for any $\varepsilon>0$, there exists a compact set $K\_{\varepsilon}$ with $|f|<\varepsilon$ on $K\_{\varepsilon}^c$, and
* $C\_c(\Omega)$ the subspace of all $f \in C\_b(\Omega)$ such that $f$ has compact support.


For a signed measure $\mu$ on $(\Omega, \mathscr{B}(\Omega)$ ), we denote by $|\mu|$ its associated variation measure. A finite signed measure $\mu$ on $(\Omega, \mathscr{B}(\Omega))$ is called a finite signed Radon measure if **$|\mu|$ is inner regular**, i.e., for each $A \in \mathscr{B}(\Omega)$,
$$
|\mu|(A)=\sup \{|\mu|(K): K \in \mathscr{B}(\Omega), K \text { compact, } K \subset A\} .
$$
We denote the set of all finite signed Radon measures on $(\Omega, \mathscr{B}(\Omega))$ by $\mathcal{M}(\Omega)$ and the subset of all finite positive Radon measures by $\mathcal{M}^{+}(\Omega)$. We say that a sequence $\left\{\mu\_n\right\} \subset \mathcal{M}(\Omega)$ converges to $\mu \in \mathcal{M}(\Omega)$


* (a) weakly if $\int\_\Omega f \mathrm d \mu\_n \to \int\_\Omega f \mathrm d \mu$ for all $f \in C\_b(\Omega)$ and we write $\underset{n \rightarrow \infty}{\operatorname{w-lim}} \, \mu\_n= \mu$;
* (b) vaguely if $\int\_\Omega f \mathrm d \mu\_n \to \int\_\Omega f \mathrm d \mu$ for all $f \in C\_c(\Omega)$, and we write $\underset{n \rightarrow \infty}{\operatorname{v-lim}} \, \mu\_n=\mu$.



Vague convergence: confusion about the regularity of a signed Radon measure and that of its variation
$\newcommand{\Om}{\Omega}\newcommand{\Th}{\Theta}\newcommand{\B}{\mathscr B}\newcommand{\M}{\mathcal M}\newcommand\ep\varepsilon\newcommand{\de}{\delta}\newcommand{\R}{\mathbb R}$Take any $\mu\in\M(\Om)$, any open subset $\Th$ of $\Om$, and any real $\ep>0$. Let $\de:=\ep/4$.


By the [Hahn decomposition theorem](https://en.wikipedia.org/wiki/Hahn_decomposition_theorem), there is a partition of $\Om$ into Borel sets $D^\pm$ such that $D^+$ is a positive set for $\mu$ and $D^-$ is a negative set for $\mu$.


Let
\begin{equation\*}
 A^\pm:=\Th\cap D^\pm. \tag{1}\label{1}
\end{equation\*}
Since $|\mu|$ is inner regular, there exist compact sets
\begin{equation\*}
 K^\pm\subseteq A^\pm\text{ such that }|\mu|(A^\pm\setminus K^\pm)<\de. \tag{2}\label{2}
\end{equation\*}
Since $\Om$ is normal, there exist open subsets $U^\pm$ of $\Th$ such that
\begin{equation\*}
 U^\pm\supseteq K^\pm\text{ and }U^+\cap U^-=\emptyset. \tag{3}\label{3}
\end{equation\*}
Since the sets $K^\pm$ are compact and $\Om$ is locally compact, without loss of generality the closures of the sets $U^\pm$ are compact.


By [Urysohn'slemma](https://en.wikipedia.org/wiki/Urysohn%27s_lemma), there exist continuous functions $f^\pm\colon\Om\to\R$ such that
\begin{equation\*}
 0\le f^\pm\le1,\quad f^\pm=1\text{ on }K^\pm,\quad f^\pm=0\text{ on }\Om\setminus U^\pm. \tag{4}\label{4}
\end{equation\*}
Let
\begin{equation\*}
 f:=f^+-f^-. 
\end{equation\*}
Then $f^+f^-=0$, whence $|f|\le1$. Also, $f=0$ on $\Om\setminus(U^+\cup U^-)$. So, recalling that the closures of the sets $U^\pm$ are compact, we see that $f\in C\_c(\Om)$. Also, since $U^\pm$ are subsets of $\Th$, we have $|f|\le1\_\Th$.


It remains to show that
\begin{equation\*}
 \int\_\Om f\,d\mu\ge|\mu|(\Th)-\ep. \tag{$\*$}\label{\*}
\end{equation\*}
To do this, note that, by \eqref{3}, \eqref{2}, and \eqref{1},  

\begin{equation}
\begin{aligned}
 |\mu|(U^-\setminus K^-)&\le|\mu|(\Th\setminus U^+\setminus K^-) \\ 
 &=|\mu|(\Th)-|\mu|(U^+)-|\mu|(K^-) \\ 
 &\le|\mu|(\Th)-|\mu|(K^+)-|\mu|(K^-) \\ 
 &<|\mu|(\Th)-|\mu|(A^+)-|\mu|(A^-)+2\de=2\de. 
\end{aligned}
\tag{5}\label{5}
\end{equation}
So, by \eqref{4}, \eqref{3}, \eqref{2}, \eqref{5}, and \eqref{1},
\begin{equation\*}
\begin{aligned}
 \int\_\Om f\,d\mu&=\int\_{U^+} f^+\,d\mu-\int\_{U^-} f^-\,d\mu \\ 
 &\ge\int\_{K^+} f^+\,d\mu-\int\_{K^-} f^-\,d\mu -\int\_{U^-\setminus K^-} f^-\,d\mu \\ 
 &=\mu(K^+)-\mu(K^-) -\int\_{U^-\setminus K^-} f^-\,d\mu \\ 
 &\ge\mu(K^+)-\mu(K^-) -|\mu|(U^-\setminus K^-) \\ 
 &>\mu(A^+)-\de-\mu(A^-)-\de -2\de \\ 
 &=|\mu|(\Th)-4\de=|\mu|(\Th)-\ep, 
\end{aligned}
\end{equation\*}
so that \eqref{\*} is proved. $\quad\Box$


