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](https://mathoverflow.net/a/433987/477203) 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](https://mathoverflow.net/questions/434000/complex-borel-measures-relation-between-the-total-variation-norm-of-a-measure-a), we have
$$
|\mu|(\Theta) \le |\mu^1|(\Theta) + |\mu^2|(\Theta).
$$


As such,
$$
|\mu|(\Theta) \le \liminf \_{n \rightarrow \infty}\left|\mu^1\_n\right|(\Theta) + \liminf \_{n \rightarrow \infty}\left|\mu^2\_n\right|(\Theta) \le \liminf \_{n \rightarrow \infty} \big (\left|\mu^1\_n\right|(\Theta) + \left|\mu^2\_n\right|(\Theta) \big ).
$$



Complex Borel measures: does $\mu_n \to \mu$ weakly imply $|\mu|(\Theta) \le \liminf_n |\mu_n|(\Theta)$ for every open subset $\Theta$?
$\newcommand{\Om}{\Omega}\newcommand{\Th}{\Theta}\newcommand{\B}{\mathscr B}\newcommand{\M}{\mathcal M}\newcommand\ep\varepsilon\newcommand{\de}{\delta}\newcommand{\R}{\mathbb R}$By the [polar decomposition of complex measures](https://en.wikipedia.org/wiki/Complex_measure#Variation_of_a_complex_measure_and_polar_decomposition), 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](https://en.wikipedia.org/wiki/Urysohn%27s_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.


