Let
- $X$ be a metric space,
- $\mathcal M(X)$ the space of all finite signed Borel measures on $X$,
- $\mathcal M_+(X)$ the space of all finite nonnegative Borel measures on $X$, and
- $\mathcal C_b(X)$ be the space of real-valued bounded continuous functions on $X$.
Then $\mathcal C_b(X)$ is a real Banach space with supremum norm $|\cdot|_\infty$. For $\mu \in \mathcal M(X)$, let $(\mu^+, \mu^-)$ be its Jordan decomposition and $|\mu| := \mu^+ + \mu^-$ its variation. We endow $\mathcal M(X)$ with the total variation norm $[\cdot]$ where $[\mu] := |\mu|(X)$. Then $(\mathcal M(X), [\cdot])$ is a Banach space. For $\mu_n,\mu \in \mathcal M(X)$, we define the weak convergence $$ \mu_n \rightharpoonup \mu \overset{\text{def}}{\iff} \int_X f \mathrm d \mu_n \to \int_X f \mathrm d \mu \quad \forall f \in \mathcal C_b(X). $$
I have mimicked the proof in this answer for below theorem, i.e.,
Theorem: Let $\mu_n,\mu \in \mathcal M(X)$ such that $\mu_n \rightharpoonup \mu$ and $[\mu_n] \to [\mu]$, then $|\mu_n| \rightharpoonup |\mu|$.
Could you have a check on my attempt?
Proof: By Portmanteau theorem for $\mathcal M_+(X)$, it suffices to prove that $$ |\mu|(\Theta) \leq \liminf _{n \rightarrow \infty}\left|\mu_n\right|(\Theta) $$ for every open subset $\Theta$ of $X$. Fix $\varepsilon>0$ and $\delta := \varepsilon/4$.
Let $(D^+, D^-)$ be the Hahn decomposition $X$ w.r.t. $\mu$, i.e., $\mu$ is positive on $D^+$ and negative on $D^-$. Let $A^\pm:=\Theta\cap D^\pm$.
- Since $|\mu|$ is inner regular, there are closed subsets $C^\pm$ of $X$ such that $C^\pm \subset A^\pm$ and $|\mu|(A^\pm \setminus C^\pm) < \delta$.
- Since $X$ is normal, there are open subsets $O^\pm$ of $X$ such that $C^\pm \subset O^\pm$ and $O^+ \cap O^- = \emptyset$. WLOG, we assume $O^\pm \subset \Theta$.
- By Urysohn's lemma, there is $f^\pm:X \to [0, 1]$ continuous such that $f^\pm (C^\pm) =1$ and $f^\pm (X \setminus O^\pm) =0$.
Notice that $f^\pm$ is supported on $O^\pm$, so $f^+ f^- =0$. Let $f:=f^+-f^-$. Then $f$ is continuous and $|f| \le 1$. Hence $|f| \in \mathcal C_b(X)$. It follows from $O^\pm \subset \Theta$ that $|f| \le 1_\Theta$. Let's prove that $$ \int_X f \mathrm d \mu \ge |\mu|(\Theta) - \varepsilon. $$
We have $$ \begin{aligned} |\mu|(O^-\setminus C^-) &\le |\mu|((\Theta\setminus O^+)\setminus C^-) \\ &= |\mu|(\Theta)-|\mu|(O^+)-|\mu|(C^-) \\ &\le |\mu|(\Theta)-|\mu|(C^+)-|\mu|(C^-) \\ &<|\mu|(\Theta)-|\mu|(A^+)-|\mu|(A^-)+2\delta \\ &= 2\delta. \end{aligned} $$
Then $$ \begin{aligned} \int_X f\mathrm d\mu&=\int_{O^+} f^+\mathrm d\mu-\int_{O^-} f^-\mathrm d\mu \\ &\ge\int_{C^+} f^+\mathrm d\mu-\int_{C^-} f^-\mathrm d\mu -\int_{O^-\setminus C^-} f^-\mathrm d\mu \\ &\ge \mu(C^+)-\mu(C^-) -\int_{O^-\setminus C^-} f^-\mathrm d\mu \\ &\ge\mu(C^+)-\mu(C^-) -|\mu|(O^-\setminus C^-) \\ &>\mu(A^+)-\delta-\mu(A^-)-\delta -2\delta \\ &=|\mu|(\Theta)-4\delta=|\mu|(X)-\varepsilon. \end{aligned} $$
Finally, $$ \begin{align} \int_X f \mathrm d \mu = \lim_n \int_X f \mathrm d \mu_n \le \liminf_n \int_X |f| \mathrm d |\mu_n| \le \liminf_n \int_X 1_{\Theta} \mathrm d |\mu_n| \le \liminf_n |\mu_n|(\Theta). \end{align} $$
This completes the proof.
Update: In above proof, we basically prove the following lemma, i.e.,
Lemma 1: If $X$ is a metric space and $\mu_n,\mu \in \mathcal M(X)$ such that $\mu_n \rightharpoonup \mu$, then $$ |\mu|(\Theta) \leq \liminf _{n \rightarrow \infty}\left|\mu_n\right|(\Theta) $$ for every open subset $\Theta$ of $X$.
Let $\mathcal C_c(X)$ be the space of real-valued continuous functions on $X$ with compact supports. For $\mu_n,\mu \in \mathcal M(X)$, the weak$^*$ convergence is defined by $$ \mu_n \overset{*}{\rightharpoonup} \mu \overset{\text{def}}{\iff} \int_X f \mathrm d \mu_n \to \int_X f \mathrm d \mu \quad \forall f \in \mathcal C_c (X). $$
With stronger assumption on $X$, we can get stronger result
Lemma 2: If $X$ is a locally compact separable metric space and $\mu_n,\mu \in \mathcal M(X)$ such that $\mu_n \overset{*}{\rightharpoonup} \mu$, then $$ |\mu|(\Theta) \leq \liminf _{n \rightarrow \infty}\left|\mu_n\right|(\Theta) $$ for every open subset $\Theta$ of $X$.
For the proof of Lemma 2, we need the following results, i.e.,
Lemma 3 Let $X$ be a locally compact separable metric space. Then $X$ is a Radon space.
Lemma 4 Let $X$ be a locally compact Hausdorff space. Let $K$ be a compact subset of $X$ and $U$ an open subset of $X$ such that $K \subset U$. Then there is an open subset $V$ of $X$ such that $K \subset V \subset \overline V \subset U$ and $\overline V$ is compact.
Lemma 3 allows us to get compact subsets $K^\pm$ instead of just closed subsets $C^\pm$. Lemma 4 allows us to construct an $f \in \mathcal C_c(X)$ instead of just $f \in \mathcal C_b(X)$.
