Added regularity properties for Radon measures.

src/measure/measure/radon.tex

@@ -44,3 +44,94 @@
     \mu(K) = \inf_{U \in \cn^o(K)}\mu(U) \ge \inf_{U \in \cn^o(K)}\int f_U d\mu \ge \mu(K)
     \]
 \end{proof}
+
+\begin{proposition}[{{\cite[Proposition 7.5]{Folland}}}]
+\label{proposition:radon-regular-sigma-finite}
+    Let $X$ be a LCH space and $\mu: \cb_X \to [0, \infty]$ be a Radon measure, then
+    \begin{enumerate}
+        \item $\mu$ is inner regular on all its $\sigma$-finite sets.
+        \item If $X$ is $\sigma$-compact or $\mu$ is $\sigma$-finite, then $\mu$ is regular.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Let $E \in \cb_X$ with $\mu(E) < \infty$ and $\eps > 0$, then there exists
+    \begin{itemize}
+    \item $U \in \cn^o(E)$ with $\mu(U \setminus E) < \eps/2$.
+    \item $K \subset U$ compact with $\mu(K) > \mu(U) - \eps/2$.
+    \item $V \in \cn^o(U \setminus E)$ with $\mu(V) < \eps/2$.
+    \end{itemize}
+    In which case, $K \setminus V \subset E$ is compact with
+    \[
+    \mu(K \setminus V) \ge \mu(K) - \mu(V) \ge \mu(U) - \eps/2 - \eps/2 \ge \mu(E) - \eps
+    \]
+    so $\mu$ is inner regular on $E$.
+
+    Now suppose that $E$ is $\sigma$-finite, then there exists $\seq{E_n} \subset \cb_X$ with $\mu(E_n) < \infty$ for each $n \in \natp$ and $E = \bigsqcup_{n \in \natp}E_n = E$. Let $N \in \natp$ and $\seqf{K_n} \subset 2^X$ compact with $K_n \subset E_n$ for each $n \in \natp$, then $\bigcup_{n = 1}^N K_n \subset E$ is compact. Hence
+    \[
+    \sup_{\substack{K \subset E \\ \text{compact}}}\mu(K) \ge \sum_{n = 1}^N\sup_{\substack{K_n \subset E_n \\ \text{compact}}}\mu(K_n) = \sum_{n = 1}^N \mu(E_n)
+    \]
+    As the above holds for all $N \in \natp$,
+    \[
+    \sup_{\substack{K \subset E \\ \text{compact}}}\mu(K) \ge \sum_{n \in \natp}\mu(E_n) = \mu(E)
+    \]
+\end{proof}
+
+\begin{proposition}[{{\cite[Proposition 7.7]{Folland}}}]
+\label{proposition:radon-measurable-description}
+    Let $X$ be a LCH space, $\mu: \cb_X \to [0, \infty]$ be a $\sigma$-finite Radon measure, and $E \in \cb_X$, then
+    \begin{enumerate}
+        \item For every $\eps > 0$, there exists $U \in \cn^o(E)$ and $F \subset E$ closed such that $\mu(U \setminus F) < \eps$.
+        \item There exists a $F_\sigma$ set $A$ and a $G_\delta$ set $B$ such that $A \subset E \subset B$ and $\mu(B \setminus A) = 0$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Let $\seq{E_n} \subset \cb_X$ such that $\mu(E_n) < \infty$. By outer regularity, there exists $\seq{U_n}$ open such that $U_n \in \cn^o(E_n)$ and $\mu(U_n) < \mu(E_n) + \eps/2^n$ for all $n \in \natp$. In which case,
+    \[
+    \mu\paren{\bigcup_{n \in \natp}U_n \setminus E} \le \sum_{n \in \natp}\mu(U_n \setminus E_n) < \eps
+    \]
+    so $U = \bigcup_{n \in \natp}U_n$ is the desired open set.
+
+    Applying the above result to $E^c$, there exists $V \in \cn^o(E^c)$ such that $\mu(V \setminus E^c) < \eps$. Let $F = V^c$, then $F \subset E$ is closed and
+    \[
+    \mu(E \setminus F) = \mu(E \cap F^c) = \mu(E \cap V) = \mu(V \setminus E^c) < \eps
+    \]
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:finite-compact-regular}
+    Let $X$ be a LCH space and $\mu: \cb_X \to [0, \infty]$ be a Borel measure such that:
+    \begin{enumerate}
+        \item[(a)] For any $U \subset X$ open, $U$ is $\sigma$-compact.
+        \item[(b)] For any $K \subset X$ compact, $\mu(K) < \infty$.
+    \end{enumerate}
+    then $\mu$ is a regular measure on $X$.
+\end{proposition}
+\begin{proof}
+    By assumption (b), $f \mapsto \int f d\mu$ is a positive linear functional on $C_c(X; \real)$. By the Riesz Representation Theorem (\ref{theorem:riesz-radon}), there exists a Radon measure $\nu: \cb_X \to [0, \infty]$ such that for any $f \in C_c(X; \real)$, $\int f d\mu = \int f d\mu$.
+
+    Let $U \subset X$ be open, then by \ref{proposition:radon-measure-cc},
+    \[
+    \nu(U) = \sup_{\substack{f \in C_c(X; [0, 1]) \\ f \le \one_U}} \int f d\nu = \sup_{\substack{f \in C_c(X; [0, 1]) \\ f \le \one_U}} \int f d\mu \le \mu(U)
+    \]
+    By assumption (a), there exists $\seq{K_n} \subset 2^X$ compact such that $K_n \upto U$. By Urysohn's lemma (\ref{lemma:lch-urysohn}), there exists $\seq{f_n} \subset C_c(X; [0, 1])$ such that $\one_{K_n} \le f_n \le \one_U$ for all $n \in \natp$, and $f_n \upto f$ pointwise. Using the Monotone Convergence Theorem (\ref{theorem:mct}),
+    \[
+    \mu(U) = \limv{n} \int f_n d\mu \le \sup_{\substack{f \in C_c(X; [0, 1]) \\ f \le \one_U}}\int f d\mu = \nu(U)
+    \]
+    Therefore $\mu(U) = \nu(U)$.
+
+    Now let $E \in \cb_X$ be arbitrary and $\eps > 0$. By \ref{proposition:radon-measurable-description}, there exists $U \in \cn^o(E)$ and $V \subset E$ closed such that $\nu(U \setminus E), \nu(E \setminus V) < \eps$. In which case, since $U \setminus V$ is open,
+    \[
+    \mu(U \setminus E) \le \mu(U \setminus V) = \nu(U \setminus V) < \eps
+    \]
+    so $\mu(U) = \mu(U \setminus E) + \mu(E) \le \mu(E) + \eps$. Therefore
+    \[
+    \mu(E) = \inf_{U \in \cn^o(E)}\mu(U) = \inf_{U \in \cn^o(E)}\nu(U) = \nu(E)
+    \]
+    and $\mu = \nu$.
+
+    Since $X$ is $\sigma$-compact, $\mu$ is $\sigma$-finite, so $\mu$ is regular by \ref{proposition:radon-regular-sigma-finite}.
+    \[
+    \color{blue}{\int}\llap{\color{green}{\int}}
+    \]
+\end{proof}