I'm reading a proof of below theorem from this paper.

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 $\underset{n \rightarrow \infty}{\operatorname{v-lim}} \, \mu_n=\mu$.