Complex Borel measures: does $\mu_n \to \mu$ weakly imply $|\mu|(\Theta) \le \liminf_n |\mu_n|(\Theta)$ for every open subset $\Theta$? Let

*

*$\Omega$ be a metric space,

*$C_b(\Omega)$ the space of all real-valued bounded continuous functions on $\Omega$, and

*$\mathcal{M}(\Omega)$ the space of all finite signed Borel measures on $\Omega$.

For $\mu \in \mathcal{M}(\Omega)$, we denote by $|\mu|$ its associated variation measure. We say that a sequence $\left\{\mu_n\right\} \subset \mathcal{M}(\Omega)$ converges to $\mu \in \mathcal{M}(\Omega)$ 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 $\underset{n \rightarrow \infty}{\operatorname{w-lim}} \, \mu_n= \mu$;
It is proved in this answer that

Theorem Let $\mu_n,\mu\in \mathcal{M}(\Omega)$ such that that $\underset{n \rightarrow \infty}{\operatorname{w-lim}} \, \mu_n=\mu$. Then for any open subset $\Theta$ of $\Omega$,
$$
|\mu|(\Theta) \leq \liminf _{n \rightarrow \infty}\left|\mu_n\right|(\Theta) .
$$

The essential part of the proof is that given $\varepsilon>0$ there is $f \in C_b(\Omega)$ such that
$$
|f| \le 1_\Theta
\quad \text{and} \quad
\int_\Omega f \mathrm d\mu\ge|\mu|(\Theta)-\varepsilon.
$$
Then by weak convergence of $(\mu_n)$ we have
$$
|\mu|(\Theta)-\varepsilon \leq \int f \mathrm{~d} \mu=\lim _{n \rightarrow \infty} \int f \mathrm{~d} \mu_n \leq \liminf _{n \rightarrow \infty} \int|f| \mathrm{d}\left|\mu_n\right| \leq \liminf _{n \rightarrow \infty}\left|\mu_n\right|(\Theta).
$$
The result then follows by taking the limit $\varepsilon \to 0^+$.

My understanding: To have above inequalities, we use the fact that $\mu,\mu_n$ are real-valued.
Question: Can the above theorem be extended to the setting of complex Borel measures?

Update: Below is my failed attempt. It would be great if it can be fixed into a valid proof. I could not prove that
$$
\liminf _{n \rightarrow \infty} \big (\left|\mu^1_n\right|(\Theta) + \left|\mu^2_n\right|(\Theta)  \big ) \le \liminf_n \left|\mu_n\right|(\Theta) .
$$
My attempt: Let $\mu_n, \mu$ are complex Borel measures on $\Omega$ such that $\underset{n \rightarrow \infty}{\operatorname{w-lim}} \, \mu_n= \mu$. Assume $\mu_n, \mu$ are decomposed into $\mu_n =\mu_n^1 + i \mu_n^2$ and  $\mu =\mu^1 + i \mu^2$ where $i$ is the imaginary unit and $\mu_n^1, \mu_n^2, \mu^1, \mu^2$ are finite signed Borel measures. We have
$$
\begin{align*}
\underset{n \rightarrow \infty}{\operatorname{w-lim}} \, \mu_n = \mu &\iff \int_\Omega f \mathrm d \mu_n^1 + i \int_\Omega f \mathrm d \mu_n^2 \to \int_\Omega f \mathrm d \mu^1 + i \int_\Omega f \mathrm d \mu^2 \quad \forall f \in C_b(\Omega) \\
&\iff \int_\Omega f \mathrm d \mu_n^1 \to \int_\Omega f \mathrm d \mu^1 \quad \text{and} \quad \int_\Omega f \mathrm d \mu_n^2 \to \int_\Omega f \mathrm d \mu^2 \quad \forall f \in C_b(\Omega) \\
&\iff \underset{n \rightarrow \infty}{\operatorname{w-lim}} \, \mu^1_n = \mu^1 \quad \text{and} \quad \underset{n \rightarrow \infty}{\operatorname{w-lim}} \, \mu^2_n = \mu^2 \\
&\implies |\mu^1|(\Theta) \leq \liminf _{n \rightarrow \infty}\left|\mu^1_n\right|(\Theta) \quad \text{and} \quad |\mu^2|(\Theta) \leq \liminf _{n \rightarrow \infty}\left|\mu^2_n\right|(\Theta) 
\end{align*}
$$
for all open subsets $\Theta$ of $\Omega$. From this question, we have
$$
|\mu|(\Theta) \le |\mu^1|(\Theta) + |\mu^2|(\Theta).
$$
As such,
$$
|\mu|(\Theta) \le \liminf _{n \rightarrow \infty}\left|\mu^1_n\right|(\Theta)  + \liminf _{n \rightarrow \infty}\left|\mu^2_n\right|(\Theta) \le \liminf _{n \rightarrow \infty} \big (\left|\mu^1_n\right|(\Theta) + \left|\mu^2_n\right|(\Theta)  \big ).
$$
 A: $\newcommand{\Om}{\Omega}\newcommand{\Th}{\Theta}\newcommand{\B}{\mathscr B}\newcommand{\M}{\mathcal M}\newcommand\ep\varepsilon\newcommand{\de}{\delta}\newcommand{\R}{\mathbb R}$By the polar decomposition of complex measures, there is a Borel function $g\colon\Om\to[0,2\pi)$ such that
\begin{equation*}
    |\mu|(\Th)=\int_\Om 1_\Th e^{ig}\,d\mu. \tag{1}\label{1}
\end{equation*}
Take any real $\ep>0$. Next, take any natural
\begin{equation*}
    m>\frac{2\pi |\mu|(\Th)}{\ep/2} \tag{2}\label{2}
\end{equation*}
and any
\begin{equation*}
    \de\in\Big(0,\frac\ep{2(2m+1)}\Big). \tag{3}\label{3}
\end{equation*}
For $j\in[m]:=\{1,\dots,m\}$, let
\begin{equation*}
    I_j:=[\tfrac{2\pi(j-1)}m,\tfrac{2\pi j}m),\ A_j:=\Th\cap g^{-1}(I_j), \tag{4}\label{4}
\end{equation*}
so that the $A_j$'s are Borel sets forming a partition of $\Th$.
Since $\Om$ is a metric space and $|\mu|$ is a Borel measure, $|\mu|$ is regular. So, for each $j\in[m]$ there exist a closed set $F_j$ and an open set $G_j$ such that
\begin{equation*}
    F_j\subseteq A_j\subseteq G_j\text{ and }|\mu|(G_j\setminus F_j)<\de, \tag{5}\label{5}
\end{equation*}
so that the $F_j$'s are (pairwise) disjoint and for
\begin{equation*}
    F:=\bigcup_{j\in[m]}F_j\text{ and }G:=\Th\setminus F \tag{6}\label{6}
\end{equation*}
we have
\begin{equation*}
    |\mu|(G)=\sum_{j\in[m]}|\mu|(A_j\setminus F_j)<m\de. \tag{8}\label{8}
\end{equation*}
All metric spaces are normal. So, by Urysohn's lemma, for each $j\in[m]$ there exists a continuous function $h_j\colon\Om\to\R$ such that
\begin{equation*}
    h_j=1\text{ on }F_j,\ h_j=0\text{ on }G_j^c:=\Om\setminus G_j,\ 0\le h_j\le1. \tag{9}\label{9}
\end{equation*}
Let
\begin{equation*}
    h:=\sum_{j\in[m]} \frac{2\pi j}m\,h_j. \tag{10}\label{10}
\end{equation*}
Then, by \eqref{6}, \eqref{10}, \eqref{9}, and \eqref{4}, on $F$ we have $0\le h-g\le\frac{2\pi}m$, and hence
\begin{equation*}
    |e^{ih}-e^{ig}|\le\frac{2\pi}m\quad \text{on}\ F. \tag{11}\label{11}
\end{equation*}
Again by the regularity of $|\mu|$ and Urysohn's lemma, there exist a closed set $F_0$ and a continuous function $h_0\colon\Om\to\R$ such that
\begin{equation*}
    F_0\subseteq\Th,\ |\mu|(\Th\setminus F_0)<\de, \tag{12}\label{12}
\end{equation*}
\begin{equation*}
    h_0=1\text{ on }F_0,\ h_0=0\text{ on }\Th^c,\ 0\le h_0\le1. \tag{13}\label{13}
\end{equation*}
So, by \eqref{1}, \eqref{13}, \eqref{6}, \eqref{11}, \eqref{12}, \eqref{8}, \eqref{2}, \eqref{3},
\begin{equation*}
\begin{aligned}
    &\Big||\mu|(\Th)-\int_\Om h_0 e^{ih}\,d\mu\Big| \\ 
    =&\Big|\int_\Th e^{ig}\,d\mu-\int_\Th h_0 e^{ih}\,d\mu\Big| \\ 
    \le&\int_\Th |1-h_0|\,d|\mu|+\int_\Th|e^{ig}-e^{ih}|\,d|\mu| \\ 
    =&\int_\Th |1-h_0|\,d|\mu|+\int_F|e^{ig}-e^{ih}|\,d|\mu| +\int_G|e^{ig}-e^{ih}|\,d|\mu| \\ 
    \le&|\mu|(\Th-F_0)+ \frac{2\pi}m\,|\mu|(F)+2|\mu|(G) \\ 
    \le&\de+ \frac{2\pi}m\,|\mu|(\Th)+2m\de<\ep. 
\end{aligned}
\end{equation*}
So,
\begin{equation*}
\begin{aligned}
    |\mu|(\Th)&=\Re|\mu|(\Th) \\ 
    &\le\ep+\Re\int_\Om h_0 e^{ih}\,d\mu \\ 
    &=\ep+\lim_n\Re\int_\Om h_0 e^{ih}\,d\mu_n \\ 
    &=\ep+\liminf_n\Re\int_\Om h_0 e^{ih}\,d\mu_n \\   
    &=\ep+\liminf_n\Re\int_\Th h_0 e^{ih}\,d\mu_n \\   
    &\le\ep+\liminf_n|\mu_n|(\Th).  
\end{aligned}
\end{equation*}
Letting $\ep\downarrow0$, we conclude that
\begin{equation*}
    |\mu|(\Th)\le\liminf_n|\mu_n|(\Th),
\end{equation*}
as desired.
