2
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ @IosifPinelis I have tried but could not prove that $$ \liminf _{n \rightarrow \infty} \big (\left|\mu^1_n\right|(\Theta) + \left|\mu^2_n\right|(\Theta) \big ) \le \liminf_n \left|\mu_n\right|(\Theta) . $$ Please see my update. $\endgroup$
    – Analyst
    Commented Nov 6, 2022 at 21:01
  • $\begingroup$ Write $|\mu(U)|=e^{i\theta}\mu(U)=\nu (U)$, where $\nu$ is the real measure $Re\, ( e^{i\theta}\mu)$. Then take $|f| \leq \chi_U$ real such that $$|\nu(U)| \leq \epsilon \int fd\nu=\epsilon +Re \int f e^{i\theta} d\mu.$$ Then it should go on as in your post. $\endgroup$ Commented Nov 7, 2022 at 15:04
  • $\begingroup$ @GiorgioMetafune From this Wikipedia page, the polar form is $\mathrm d \mu=e^{i \theta} \mathrm d|\mu|$, whereas yours is $\mathrm d |\mu| = e^{i \theta} \mathrm d\mu$. Could you elaborate on this difference? $\endgroup$
    – Analyst
    Commented Nov 7, 2022 at 18:45
  • 1
    $\begingroup$ @Analyst: Notice that if $\mu(dx)=e^{i\theta(x)}\,|\mu|(dx)$, then since $|e^{i\theta(x)}|=1$, $|\mu|(dx)=e^{-i \theta}\mu(dx)$ $\endgroup$ Commented Nov 7, 2022 at 19:22

1 Answer 1

4
$\begingroup$

$\newcommand{\Om}{\Omega}\newcommand{\Th}{\Theta}\newcommand{\B}{\mathscr B}\newcommand{\M}{\mathcal M}\newcommand\ep\varepsilon\newcommand{\de}{\delta}\newcommand{\R}{\mathbb R}$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.

$\endgroup$
2
  • $\begingroup$ Ah! I see know, I forgot that you've taken care of the continuity of the character $e^{ig}$ by using a continuous approximation $e^{ih}$. Sorry! Everything is clear now...I'll remove my previous comments shortly. $\endgroup$ Commented Nov 9, 2022 at 22:27
  • $\begingroup$ Thank you again for your very detailed answer! $\endgroup$
    – Analyst
    Commented Nov 10, 2022 at 15:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.