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Iosif Pinelis
Iosif Pinelis
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$\newcommand{\Om}{\Omega}\newcommand{\Th}{\Theta}\newcommand{\B}{\mathscr B}\newcommand{\M}{\mathcal M}\newcommand\ep\varepsilon\newcommand{\de}{\delta}\newcommand{\R}{\mathbb R}$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}\quad F \tag{11}\label{11} \end{equation*}\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.

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

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

little mistake in an index. A function $h_0$ is defined but a property for this function was listed as $h_j$ ...
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edit approved Nov 8, 2022 at 1:59
Oliver Díaz
edit approved Nov 8, 2022 at 1:59
Oliver Díaz
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$\newcommand{\Om}{\Omega}\newcommand{\Th}{\Theta}\newcommand{\B}{\mathscr B}\newcommand{\M}{\mathcal M}\newcommand\ep\varepsilon\newcommand{\de}{\delta}\newcommand{\R}{\mathbb R}$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. \tag{11}\label{11} \end{equation*}\begin{equation*} |e^{ih}-e^{ig}|\le\frac{2\pi}m\quad \text{on}\quad 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_j=0\text{ on }\Th^c,\ 0\le h_0\le1. \tag{13}\label{13} \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.

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

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

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

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

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

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Iosif Pinelis
Iosif Pinelis
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Iosif Pinelis
Iosif Pinelis
  • 127.7k
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  • 107
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