Measure Theory – Locally Compact Separable Space and Weak Convergence

general-topologymeasure-theorymetric-spacesweak-convergence

Let

  • $X$ be a metric space,
  • $\mathcal C_b(X)$ the space of real-valued bounded continuous functions on $X$,
  • $\mathcal C_c(X)$ the space of real-valued continuous functions on $X$ with compact supports,
  • $\mathcal M (X)$ the space of all complex Borel measures on $X$,
  • $\mathcal M_{\mathbb R}(X)$ the space of all finite signed Borel measures on $X$, and
  • $\mathcal M_+ (X)$ the space of all finite non-negative Borel measures on $X$.

For $\nu \in \mathcal M(X)$, let $|\nu| \in \mathcal M_+ (X)$ be its variation and $[\nu] := |\nu| (X)$ its total variation norm. Then $(\mathcal M(X), [\cdot])$ is a real Banach space. For $\mu_n,\mu \in \mathcal M(X)$, we define weak convergence by
$$
\mu_n \rightharpoonup \mu \overset{\text{def}}{\iff} \int_X f \mathrm d \mu_n \to \int_X f \mathrm d \mu \quad \forall f \in \mathcal C_b(X),
$$
and weak$^*$ convergence by
$$
\mu_n \overset{*}{\rightharpoonup} \mu \overset{\text{def}}{\iff} \int_X f \mathrm d \mu_n \to \int_X f \mathrm d \mu \quad \forall f \in \mathcal C_c (X).
$$

It has been shown that

Theorem Let $\mu_n,\mu\in \mathcal{M}(X)$ such that $\mu_n \rightharpoonup \mu$. Then for any open subset $O$ of $X$,
$$
|\mu|(O) \leq \liminf _{n \rightarrow \infty}\left|\mu_n\right|(O) .
$$

I would like to apply the same technique to prove that

Theorem Let $X$ be locally compact separable and $\mu_n,\mu\in \mathcal{M}(X)$ such that $\mu_n \overset{*}{\rightharpoonup} \mu$. Then for any open subset $O$ of $X$,
$$
|\mu|(O) \leq \liminf _{n \rightarrow \infty}\left|\mu_n\right|(O) .
$$

My proof seems to be much simpler than the original one. It's likely that I made some subtle mistakes. Could you please have a check on my attempt?

Proof: We need the following lemmas, i.e.,

  • Lemma 1 Let $X$ be a locally compact separable metric space. Then $X$ is a Radon space.

  • Lemma 2 Let $X$ be a locally compact Hausdorff space. Let $K$ be a compact subset of $X$ and $U$ an open subset of $X$ such that $K \subset U$. Then there is an open subset $V$ of $X$ such that $K \subset V \subset \overline V \subset U$ and $\overline V$ is compact.

Let $\mathcal X$ be the Borel $\sigma$-algebra of $X$. By polar decomposition, there is a measurable $g:X \to [0, 2\pi)$ such that
$$
|\mu| (B) = \int_B e^{i g} \mathrm d \mu \quad \forall B \in \mathcal X.
$$

Fix $\varepsilon>0$. By Lemma 1, there is a compact subset $K$ of $O$ such that $|\mu| (O \setminus K) < \varepsilon$. By Lemma 2, there is an open subset $V$ of $X$ such that $\overline V$ is compact and $K \subset V \subset \overline V \subset O$. By Urysohn's lemma, there is a continuous $h:X \to [0, 1]$ such that $f(K)=1$ and $f(X \setminus V) = 0$. This implies the support of $f$ is contained in $\overline V$, so $f \in \mathcal C_c (X)$. Also, $|f| \le 1_{O}$. We have
$$
\begin{align}
|\mu| (O) – \operatorname{Re} \int_O fe^{i g} \mathrm d \mu &= \operatorname{Re} \int_O (1-f)e^{i g} \mathrm d \mu \\
&\le \left | \int_O (1-f)e^{i g} \mathrm d \mu \right | \\
&\le \int_O |(1-f)e^{i g}| \mathrm d |\mu| \\
&= \int_O |1-f| \mathrm d |\mu| \\
&\le |\mu| (O \setminus K) < \varepsilon.
\end{align}
$$

So
$$
\begin{align}
|\mu| (O) – \varepsilon &< \operatorname{Re} \int_O fe^{i g} \mathrm d \mu \\
&\overset{(1)}{=} \operatorname{Re} \lim_n \int_O fe^{i g} \mathrm d \mu_n \\
&= \lim_n \operatorname{Re} \int_O fe^{i g} \mathrm d \mu_n \\
&\le \liminf_n \left| \int_O fe^{i g} \mathrm d \mu_n \right| \\
&\le \liminf_n \int_O |fe^{i g}| \mathrm d |\mu_n | \\
&= \liminf_n \int_O |f| \mathrm d |\mu_n | \\
&\le \liminf_n \int_O \mathrm d |\mu_n | \\
&= \liminf_n |\mu_n| (O).
\end{align}
$$

Notice that $(1)$ is true even if $fe^{i g}$ is complex-valued. The claim then follows by taking the limit $\varepsilon \to 0^+$.

Best Answer

Here is a short proof of the main Theorem in the OP. Let $G$ be an open set in $X$. Since $|\mu|$ is a Radon measure, for $\varepsilon>0$ there is $f\in\mathcal{C}_{00}(X)$ with $0\leq f\prec G$ such that $$|\mu|(G)-\tfrac{\varepsilon}{2}<|\mu|(f)=\int f\,d|\mu|$$ Since $|\mu|(f)=\sup\{|\mu(h)|:h\in\mathcal{C}_{00}(X;\mathbf{C}), |h|\leq f\}$ (see Proposition below), there is $g\in\mathbf{C}_{00}(X)$ such that $|g|\leq f$ and $$|\mu|(f)-\tfrac{\varepsilon}{2}<|\mu(g)|$$ Hence \begin{align} |\mu|(G)-\varepsilon&<|\mu(g)|=\lim_n|\mu_n(g)|\\ &\leq\liminf_n|\mu_n|(|g|)\\ &\leq \liminf_n|\mu_n|(G) \end{align} since $|g|\leq f\leq\mathbb{1}_G$.

This is to show that

Proposition: For any $f\in\mathcal{C}^+_{00}(X)$ $$|\mu|(f)=\sup\{|\mu(h)|:h\in\mathcal{C}_{00}(X;\mathbf{C}), |h|\leq f\}$$

By the Radon-Nykodym theorem (all measures here are finite) there is $\phi\in L_1(|\mu|;\mathbb{C})$ with $|\phi|=1$ such that $d\mu=\phi\cdot\,d|\mu|$. Notice that $|\mu(f\overline{\phi})|=\big||\mu|(f)\big|=|\mu|(f)$ and $|f\overline{\phi}|\leq|f|$. The only problem is that there is not reason for $\phi$ to be continuous. Let $\varepsilon>0$. Since $K_f=\operatorname{supp}(f)$ is compact, there is a relatively compact open set $U$ such that $K_f\subset U\subset \overline{U}$ with $|\mu|(U\setminus K_f)<\varepsilon$. Appealing to Lusin's theorem, there is $g\in\mathcal{C}_{00}(X;\mathbb{C})$ with $\operatorname{supp}(g)\subset U$ and such that $$|\mu|(\{f\overline{\phi}\neq g\})<\varepsilon$$ and $\|g\|_u\leq \|f\|_u$. Suppose $g=g_r+ ig_i$, $g_r,g_i\in\mathcal{C}_{00}(X;\mathbb{R})$, and set $$h=\max(-f,\min(g_r,f))+ i\max(-f,\min(g_i,f))$$ Then, $h\in\mathcal{C}_{00}(X;\mathbb{C})$, $|h|\leq f$ and $|\mu(h)-|\mu|(f)|=|\mu(h-f\overline{\phi})|\leq 2\|f\|_u\varepsilon$

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