I'm reading a proof of below theorem from this paper.

Theorem A.3. Let $\Omega$ be a locally compact normal Hausdorff space. Let $\left\{\mu_n\right\} \cup\{\mu\} \subset \mathcal{M}(\Omega)$ and assume that $\underset{n \rightarrow \infty}{\operatorname{v-lim}} \mu_n=\mu$. Then for any open set $\Theta \subset \Omega$, $$ |\mu|(\Theta) \leq \liminf _{n \rightarrow \infty}\left|\mu_n\right|(\Theta) . $$ In particular, $\|\mu\| \leq \liminf _{n \rightarrow \infty}\left\|\mu_n\right\|$.

Proof. Let $\Theta \subset \Omega$ be open and $\varepsilon>0$. Since $\mu$ is inner regular and $\Omega$ is normal and locally compact, as a consequence of Urysohn's lemma [1, Lemma 2.46], there exists $f \in C_c(\Omega)$ such that $|f| \leq 1, \operatorname{supp}(f) \subset \Theta$ and $$ \int f \mathrm{~d} \color{red}{\mu} \geq \color{red}{|\mu|}(\Theta)-\varepsilon. $$ Then by vague convergence of $\left\{\mu_n\right\}$, $$ |\mu|(\Theta)-\varepsilon \leq \int f \mathrm{~d} \mu=\lim _{n \rightarrow \infty} \int f \mathrm{~d} \mu_n \leq \liminf _{n \rightarrow \infty} \int|f| \mathrm{d}\left|\mu_n\right| \leq \liminf _{n \rightarrow \infty}\left|\mu_n\right|(\Theta) $$ Now the result follows by letting $\varepsilon \downarrow 0$.

My understading: Below are the authors' related definitions. A finite signed measure $\mu$ is Radon if its variation $|\mu|$ is inner regular. In above proof, $\mu$ is Radon and thus $|\mu|$ is inner regular. So there is $f \in C_c(\Omega)$ such that $f \le 1_\Theta$ and $$ \int f \mathrm{~d} \color{red}{|\mu|} \geq \color{red}{|\mu|}(\Theta)-\varepsilon. $$

So I'm confused due to the appearance of $\color{red}{\mu}$ instead of $\color{red}{|\mu|}$ in the integral. Could you elaborate on my confusion?

Authors' definitions: Let

  • $\Omega$ be a metric space and $\mathscr{B}(\Omega)$ its Borel $\sigma$-algebra.
  • $C_b(\Omega)$ the subspace of all real-valued bounded continuous functions on $\Omega$.
  • $C_0(\Omega)$ the subspace of all $f \in C_b(\Omega)$ such that for any $\varepsilon>0$, there exists a compact set $K_{\varepsilon}$ with $|f|<\varepsilon$ on $K_{\varepsilon}^c$, and
  • $C_c(\Omega)$ the subspace of all $f \in C_b(\Omega)$ such that $f$ has compact support.

For a signed measure $\mu$ on $(\Omega, \mathscr{B}(\Omega)$ ), we denote by $|\mu|$ its associated variation measure. A finite signed measure $\mu$ on $(\Omega, \mathscr{B}(\Omega))$ is called a finite signed Radon measure if $|\mu|$ is inner regular, i.e., for each $A \in \mathscr{B}(\Omega)$, $$ |\mu|(A)=\sup \{|\mu|(K): K \in \mathscr{B}(\Omega), K \text { compact, } K \subset A\} . $$ We denote the set of all finite signed Radon measures on $(\Omega, \mathscr{B}(\Omega))$ by $\mathcal{M}(\Omega)$ and the subset of all finite positive Radon measures by $\mathcal{M}^{+}(\Omega)$. We say that a sequence $\left\{\mu_n\right\} \subset \mathcal{M}(\Omega)$ converges to $\mu \in \mathcal{M}(\Omega)$

  • (a) 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 $\underset{n \rightarrow \infty}{\operatorname{w-lim}} \, \mu_n= \mu$;
  • (b) vaguely if $\int_\Omega f \mathrm d \mu_n \to \int_\Omega f \mathrm d \mu$ for all $f \in C_c(\Omega)$, and we write $\underset{n \rightarrow \infty}{\operatorname{v-lim}} \, \mu_n=\mu$.

2 Answers 2


$\newcommand{\Om}{\Omega}\newcommand{\Th}{\Theta}\newcommand{\B}{\mathscr B}\newcommand{\M}{\mathcal M}\newcommand\ep\varepsilon\newcommand{\de}{\delta}\newcommand{\R}{\mathbb R}$Take any $\mu\in\M(\Om)$, any open subset $\Th$ of $\Om$, and any real $\ep>0$. Let $\de:=\ep/4$.

By the Hahn decomposition theorem, there is a partition of $\Om$ into Borel sets $D^\pm$ such that $D^+$ is a positive set for $\mu$ and $D^-$ is a negative set for $\mu$.

Let \begin{equation*} A^\pm:=\Th\cap D^\pm. \tag{1}\label{1} \end{equation*} Since $|\mu|$ is inner regular, there exist compact sets \begin{equation*} K^\pm\subseteq A^\pm\text{ such that }|\mu|(A^\pm\setminus K^\pm)<\de. \tag{2}\label{2} \end{equation*} Since $\Om$ is normal, there exist open subsets $U^\pm$ of $\Th$ such that \begin{equation*} U^\pm\supseteq K^\pm\text{ and }U^+\cap U^-=\emptyset. \tag{3}\label{3} \end{equation*} Since the sets $K^\pm$ are compact and $\Om$ is locally compact, without loss of generality the closures of the sets $U^\pm$ are compact.

By Urysohn'slemma, there exist continuous functions $f^\pm\colon\Om\to\R$ such that \begin{equation*} 0\le f^\pm\le1,\quad f^\pm=1\text{ on }K^\pm,\quad f^\pm=0\text{ on }\Om\setminus U^\pm. \tag{4}\label{4} \end{equation*} Let \begin{equation*} f:=f^+-f^-. \end{equation*} Then $f^+f^-=0$, whence $|f|\le1$. Also, $f=0$ on $\Om\setminus(U^+\cup U^-)$. So, recalling that the closures of the sets $U^\pm$ are compact, we see that $f\in C_c(\Om)$. Also, since $U^\pm$ are subsets of $\Th$, we have $|f|\le1_\Th$.

It remains to show that \begin{equation*} \int_\Om f\,d\mu\ge|\mu|(\Th)-\ep. \tag{$*$}\label{*} \end{equation*} To do this, note that, by \eqref{3}, \eqref{2}, and \eqref{1},
\begin{equation} \begin{aligned} |\mu|(U^-\setminus K^-)&\le|\mu|(\Th\setminus U^+\setminus K^-) \\ &=|\mu|(\Th)-|\mu|(U^+)-|\mu|(K^-) \\ &\le|\mu|(\Th)-|\mu|(K^+)-|\mu|(K^-) \\ &<|\mu|(\Th)-|\mu|(A^+)-|\mu|(A^-)+2\de=2\de. \end{aligned} \tag{5}\label{5} \end{equation} So, by \eqref{4}, \eqref{3}, \eqref{2}, \eqref{5}, and \eqref{1}, \begin{equation*} \begin{aligned} \int_\Om f\,d\mu&=\int_{U^+} f^+\,d\mu-\int_{U^-} f^-\,d\mu \\ &\ge\int_{K^+} f^+\,d\mu-\int_{K^-} f^-\,d\mu -\int_{U^-\setminus K^-} f^-\,d\mu \\ &=\mu(K^+)-\mu(K^-) -\int_{U^-\setminus K^-} f^-\,d\mu \\ &\ge\mu(K^+)-\mu(K^-) -|\mu|(U^-\setminus K^-) \\ &>\mu(A^+)-\de-\mu(A^-)-\de -2\de \\ &=|\mu|(\Th)-4\de=|\mu|(\Th)-\ep, \end{aligned} \end{equation*} so that \eqref{*} is proved. $\quad\Box$

  • $\begingroup$ May I ask if it should be "$\ge \mu(K^+)-\mu(K^-) -\int_{U^-\setminus K^-} f^-\,d\mu$" rather than "$=\mu(K^+)-\mu(K^-) -\int_{U^-\setminus K^-} f^-\,d\mu$" in the transformation at the end? $\endgroup$
    – Analyst
    Nov 6, 2022 at 3:04
  • $\begingroup$ @Analyst : The equality there holds because $f^\pm=1$ on $K^\pm$. So, of course, the inequality $\ge$ also holds. $\endgroup$ Nov 6, 2022 at 14:12
  • $\begingroup$ But we only know that $f^\pm (\Omega \setminus U^\pm) =0$ so it's possible that $f^\pm >0$ on $U^\pm \setminus K^\pm$. Maybe I'm wrong... $\endgroup$
    – Analyst
    Nov 6, 2022 at 14:18
  • 1
    $\begingroup$ @Analyst : I don't think your latter comment is wrong. But what is your question now, if any? $\endgroup$ Nov 6, 2022 at 15:30
  • $\begingroup$ Could you please have a look at this related question? Thank you so much for your detailed answer here. $\endgroup$
    – Analyst
    Nov 6, 2022 at 15:33

First we state here some general results about Radon measures following Bichteler, K. Integration: A Functional Approach, Birkhauser, 1991 and Bichteler, K. Integration theory: With Special Attention to Vector Measures, Lecture Notes in Mathematics, Springer 1973.

Let $\mathcal{K}$ denote the collection of compact subsets in $\Omega$ and $\mathcal{G}$ the collection of all open subsets in $\Omega$.

The Markov-Riesz representation theorem states that if $\Omega$ is a l.c. H. space and $I$ is a nonnegative linear functional on $\mathcal{C}_{00}(\Omega)$ then there is a order continuous and non-negative measure $\mu$ on a $\sigma$-algebra $\mathscr{M}(\Omega)$ containing the Borel sets such that

  • A. For any compact set $K\subset \Omega$ $$\mu(K)=\inf\{I(\phi): K\prec \phi\}<\infty$$

  • B. For any open set $G\subset \Omega$, $$\mu(G)=\sup\{I(\phi):0\leq \phi\prec G\}=\sup\{\mu(K): K \in\mathcal{K},\,K\subset G\}$$

  • C. For any $A\subset \Omega$, $$\mu^\bullet(A)=\inf\{\mu(G): G\in\mathcal{G},\, A\subset G\}$$

  • D. For any $A$ with $\mu(A)<\infty$, $$\mu(A)=\sup\{\mu(K): K\in\mathcal{K},\, K\subset A\}$$

Conditions B, D are called inner regularity; condition $C$ is called outer regularity.

The notation $K\prec f\prec G$, where $K\in\mathcal{K}$, $f\in\mathcal{C}_{00}(X)$, $G\in\mathcal{G}$ means that $\mathbb{1}_K\leq f\leq \mathbb{1}_G$, $\operatorname{supp}(f)\in\mathcal{K}$ and $\operatorname{supp}\subset G$.

A real (or complex) Radon measure $m$ is a linear functional on $\mathcal{C}_{00}(\Omega)$ that satisfies the property:
Property R: For any sequence $(\phi_n:n\in\mathbb{N})\subset\mathcal{C}_{00}(\Omega)$ whose supports are contained in a common compact set and which converges uniformly to some $\phi\in\mathcal{C}_{00}(\Omega)$, $\lim_n m(\phi_n)=m(\phi)$.

It is well known that

Theorem: A real linear functional on $\mathcal{C}_{00}(\Omega)$ satisfies property R iff and only if and $m$ has finite variation $|m|$, that is, for any $\psi\in\mathcal{C}^+_{00}(\Omega)$, \begin{align}|m|(\psi):=\sup\{m(\phi): \phi\in\mathcal{C}_{00}(\Omega), |\phi|\leq\psi\}<\infty\tag{1}\label{one}\end{align} The functional $|m|$ is linear and positive homogeneous on $\mathcal{C}_{00}^+(\Omega)$ and thus can be extended uniquely as a nonnegative linear functional on $\mathcal{C}_{00}(\Omega)$.

It follows from the Markov-rise representation theorem that if $m$ satisfies property R, then $|m|$ is a Radon measure in the sense that appears in many standard textbooks.

For any pair of measures $m$ and $n$ of finite variation, define $m\wedge n=\frac{m+n-|m-n|}{2}$ and $m\vee n=\frac{m+n+|m-n|}{2}$. $m\wedge n$ is the largest of all measures $\rho$ such that $\rho\leq m$ and $\rho\leq n$; similarly, $m\vee n$ is the smallest of all measures $\eta$ such that $m\leq \eta$ and $n\leq \eta$. Two measures $m$ and $n$ are orthogonal (mutually singular) iff $m\wedge n=0$.

For any real measure $m$ if finite variation, the measures $m_+=\frac{|m|+m}{2}$ and $m_-=\frac{|m|-m}{2}$ are orthogonal. Then the Hanh-Jordan decomposition theorem states that

Theorem: There is a measurable set $P$ such that \begin{align} m_+(\cdot)=m(\cdot\cap P)=|m|(\cdot \cap P),\qquad m_-(\cdot)=-m(\cdot\setminus P)=|m|(\cdot\setminus P)\tag{2}\label{two} \end{align}

For the rest of this posting, $\mathcal{M}(\Omega)$ denotes the space of (real) Radon measures $m$ that have finite total variation: $|m|(\Omega)<\infty$.

We now prove that under the setting of the OP, for any open set $G$ in $\Omega$ and $\varepsilon>0$ there is $f\in\mathcal{C}_{00}(\Omega)$ with $|f|\prec G$ and such that $$\int f\,d\mu>|\mu|(G)-\varepsilon$$

  • A short proof can be obtained from the Markov-Riesz representation theorem and the definition of variation \eqref{one}: $|\mu|$ is a a Radon measure and so, for any $\varepsilon>0$ there is $g\in\mathcal{C}_{00}(\Omega)$ with $0\leq g\prec G$ such that $|\mu|(\Omega)-\varepsilon/2<|\mu|(g)=\int g\,d|\mu|$. By \eqref{one}, there is $f\in\mathcal{C}_{00}(\Omega)$ such that $|f|\leq g$ and $|\mu|(g)-\varepsilon/2<\mu(f)$. Putting things toguether $$|\mu|(\Omega)-\varepsilon<\mu(f).$$

  • An alternative approach is based on the Hahn-Jordan decomposition theorem: The liner operators $\mu_+=\frac{|\mu|+\mu}{2}$ and $\mu_-=\frac{|\mu|-\mu}{2}$ are positive linear functionals on $\mathbb{C}_{00}$ and so by the Markov-Riesz representation theorem they extend to measure satisfying the conditions in the bullets above. Since $\mu_+$ and $\mu_-$ are orthogonal, there are $\mathcal{M}(X)$ measurable sets $P$ and $N=X\setminus P$ such that $\mu_+(N)=0=\mu_-(P)$. Given $\varepsilon>0$ there are compact sets $K_p\subset P\cap G$ and $K_n\subset N\cap G$ such that \begin{align} \mu_+(G_p)=\mu_+(P\cap G)<\mu_+(K_p) + \varepsilon/4,&\qquad \mu_-(G)=\mu_-(N\cap G)<\mu_-(K_n)+\varepsilon/4 \end{align} Further, since $K_p$ and $K_n$ are disjoint compact sets, and $\mu_\pm$ are Radon, there are disjoint open sets $G_p$, and $G_n$ such that $K_p\subset G_p\subset G$ and $K_n\subset G_n\subset G$. Uryshohn's lemma yields function $f_p,f_n\in\mathcal{C}_{00}(\Omega)$ such that $$K_p\prec f_p\prec G_p,\qquad K_n\prec f_n\prec G_n$$ Then \begin{align} \int(f_p-f_n)\,d\mu&=\int f_p\,d\mu_+ -\int f_p\,d\mu_- -\int f_n\,d\mu_+ +\int f_n\,d\mu_-\\ &\geq \mu_+(K_p) + \mu_-(K_n)-\mu_-(G_p)-\mu_+(G_n)\\ &\geq \mu_+(G)-\varepsilon/4+ \mu_-(G)-\varepsilon/4-\mu_-(G_p)-\mu_+(G_n)\\ &=|\mu|(G)- \varepsilon/2-\mu_-(G_p)-\mu_+(G_n) \end{align} Since $K_p\subset P$ \begin{align} \mu_-(G_p)&= \mu_-(G_p\setminus K_p)\leq \mu_-((G\setminus G_n)\setminus K_p )=\mu_-(G)-(\mu_-(G_n)+\mu_-(K_p))\\ &= \mu_-(G)-\mu_-(G_n)\leq \mu_-(G) -\mu_-(K_n)<\varepsilon/2 \end{align} Similarly, since $K_n\subset N=\Omega\setminus P$ \begin{align} \mu_+(G_n)&= \mu_+(G_n\setminus K_n)\leq \mu_+((G\setminus G_p)\setminus K_n )=\mu_+(G)-(\mu_+(G_p)+\mu_+(K_n))\\ &= \mu_+(G)-\mu_+(G_p)\leq \mu_+(G) -\mu_+(K_p)<\varepsilon/2 \end{align} Putting things together we obtain that $$\int (f_p-f_n)\,d\mu\geq |\mu|(G)-\varepsilon$$ Since $G_p\cap G_n=\emptyset$, $f:f_p-f_n$ satisfies $f\in\mathcal{C}_{00}(\Omega)$, and $|f|\leq 1$.

  • $\begingroup$ Thank you so much for your dedicated support and detailed answers! $\endgroup$
    – Analyst
    Nov 7, 2022 at 5:35

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