# Vague convergence: confusion about the regularity of a signed Radon measure and that of its variation

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 $$\underset{n \rightarrow \infty}{\operatorname{v-lim}} \, \mu_n=\mu$$.


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{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} 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. $$\quad\Box$$

• 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? Nov 6, 2022 at 3:04
• @Analyst : The equality there holds because $f^\pm=1$ on $K^\pm$. So, of course, the inequality $\ge$ also holds. Nov 6, 2022 at 14:12
• 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... Nov 6, 2022 at 14:18
• @Analyst : I don't think your latter comment is wrong. But what is your question now, if any? Nov 6, 2022 at 15:30
• Could you please have a look at this related question? Thank you so much for your detailed answer here. 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 $$|f|\leq 1$$.

• Thank you so much for your dedicated support and detailed answers! Nov 7, 2022 at 5:35