\section{Radon Measures}
\label{section:radon-measure}

\begin{definition}[Radon Measure]
\label{definition:radon-measure}
    Let $X$ be a LCH space and $\mu: \cb_X \to [0, \infty]$ be a Borel measure, then $\mu$ is a \textbf{Radon measure} if:
    \begin{enumerate}
        \item[(R1)] For any $K \subset X$ compact, $\mu(K) < \infty$.
        \item[(R2)] $\mu$ is outer regular on all Borel sets.
        \item[(R3')] $\mu$ is inner regular on all open sets.
    \end{enumerate}
\end{definition}

\begin{proposition}
\label{proposition:radon-measure-cc}
    Let $X$ be a LCH space and $\mu: \cb_X \to [0, \infty]$ be a Radon measure, then
    \begin{enumerate}
        \item For any $U \subset X$ open,
        \[
        \mu(U) = \sup\bracs{\int f d\mu \bigg | f \in C_c(X; [0, 1]), \supp{f} \subset U}
        \]

        \item For any $K \subset X$ compact,
        \[
        \mu(K) = \inf\bracs{\int f d\mu \bigg | f \in C_c(X; [0, 1]), f \ge \one_K}
        \]

    \end{enumerate}
\end{proposition}
\begin{proof}
    (1): Let $U \subset X$ be open. By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, for any $K \subset U$ compact, there exists $f_K \in C_c(X; [0, 1])$ such that $f_K|_K = 1$ and $\supp{f_K} \subset U$. In which case,
    \[
    \mu(K) \le \int f_K d\mu \le \mu(U)
    \]

    By (R3'),
    \[
    \mu(U) = \sup_{K \subset U \text{ compact}}\mu(K) \le \sup_{K \subset U \text{ compact}}\int f_K d\mu \le \mu(U)
    \]


    (2): Let $K \subset X$ be compact and $U \in \cn^o(K)$. By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $f_U \in C_c(X; [0, 1])$ such that $f_U|_K = 1$ and $\supp{f_U} \subset U$. In which case,
    \[
    \mu(K) \le \int f_U d\mu \le \mu(U)
    \]

    By (R2),
    \[
    \mu(K) = \inf_{U \in \cn^o(K)}\mu(U) \ge \inf_{U \in \cn^o(K)}\int f_U d\mu \ge \mu(K)
    \]

\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 \hyperref[Riesz Representation Theorem]{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 \autoref{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 \hyperref[Urysohn's lemma]{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 \hyperref[Monotone Convergence Theorem]{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 \autoref{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 \autoref{proposition:radon-regular-sigma-finite}.
\end{proof}

\begin{proposition}
\label{proposition:radon-cc-dense}
    Let $X$ be a LCH space, $\mu: \cb_X \to [0, \infty]$ be a Radon measure, $E$ be a normed vector space, and $p \in [1, \infty)$, then $C_c(X; E)$ is dense in $L^p(X; E)$.
\end{proposition}
\begin{proof}
    By \autoref{proposition:lp-simple-dense}, $\Sigma(X, \cm; E) \cap L^p(X; E)$ is dense in $L^p(X; E)$. Using linearity, it is sufficient to approximate indicator functions of Borel sets with finite measure.

    Let $A \in \cb_X$ and $\eps > 0$. By \autoref{proposition:radon-regular-sigma-finite}, there exists $U \in \cn^o(A)$ and $K \subset A$ compact such that $\mu(U \setminus A), \mu(A \setminus K) < \eps/2$. By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $f \in C_c(X; [0, 1])$ such that $f|_K = 1$ and $\supp{f} \subset U$. In which case, for any $y \in E$,
    \[
    \norm{x \cdot \one_A - x \cdot f}_{L^p(X; E)} \le \norm{x}_E \mu(\bracs{f \ne \one_A})^{1/p} \le \norm{x}_E\mu(U \setminus K)^{1/p} < \eps^{1/p}\norm{x}_E
    \]
\end{proof}

\begin{theorem}[Lusin, {{\cite[Theorem 7.10]{Folland}}}]
\label{theorem:lusin}
    Let $X$ be a LCH space, $\mu$ be a Radon measure on $X$, $E$ be a normed vector space, and $f: X \to E$ be a measurable function with $\mu\bracs{f \ne 0} < \infty$, then for any $\eps > 0$,
    \begin{enumerate}
        \item There exists $A \subset \bracs{f \ne 0}$ such that $f|_A$ is continuous and $\mu(\bracs{f \ne 0} \setminus A) < \eps$
        \item If $E = \complex$, then there exists $\phi \in C_c(X; E)$ such that $\mu\bracs{f \ne \phi} < \eps$.
        \item If $E = \complex$ and $f$ is bounded, then $\phi$ can be taken such that $\norm{\phi}_u \le \norm{f}_u$.
    \end{enumerate}
\end{theorem}
\begin{proof}
    First assume that $f$ is bounded.

    (1, bounded): If $f$ is bounded, then $f \in L^1(X; E)$. By \autoref{proposition:radon-cc-dense}, there exists $\seq{\phi_n} \subset C_c(X)$ such that $\phi_n \to f$ in $L^1(\mu)$. Since $\phi_n \to f$ in $L^1(\mu)$, $\phi_n \to f$ in measure by \autoref{proposition:lp-in-measure}. By taking a subsequence using \ref{proposition:cauchy-in-measure-limit}, assume without loss of generality that $\phi_n \to f$ almost everywhere. 

    By \hyperref[Egoroff's Theorem]{theorem:egoroff}, there exists $A \subset \bracs{f \ne 0}$ such that $\phi_n \to f$ uniformly and $\mu(\bracs{f \ne 0} \setminus A) < \eps/3$. By \autoref{proposition:radon-regular-sigma-finite}, there exists $K \subset A$ compact such that $\mu(A \setminus K) < \eps/3$. 

    Let $\phi_0 = \lim_{n \to \infty}\phi_n$ on $K$, then $\phi_0 \in C(K; E)$ by \autoref{proposition:uniform-limit-continuous}, $\phi_0 = f$ almost everywhere on $K$, and $\mu(\bracs{f \ne 0} \setminus K) < \eps$.

    (2, bounded): By outer regularity, there exists $U \in \cn^o(\bracs{f \ne 0})$ such that $\mu(U \setminus \bracs{f \ne 0}) < \eps/3$. 

    By the \hyperref[Tietze Extension Theorem]{theorem:lch-tietze}, there exists $\phi \in C_c(X; \complex)$ such that $\phi|_K = \phi_0$ and $\supp{\phi} \subset U$. In which case, 
    \[
    \mu\bracs{\phi \ne f} \le \underbrace{\mu(\bracs{f \ne 0} \setminus K)}_{< 2\eps/3} + \underbrace{\mu(U \setminus \bracs{f \ne 0})}_{< \eps/3} < \eps
    \]

    (3): Let
    \[
    \psi: \complex \to \complex \quad z \mapsto \begin{cases}
        z &|z| \le \norm{f}_u \\
        z\norm{f}_u/|z| &|z| > \norm{f}_u
    \end{cases}
    \]
    then $\psi$ is continuous with $\bracs{\phi = f} = \bracs{\psi \circ \phi = f}$ and $\norm{\psi \circ \phi}_u \le \norm{f}_u$. 

    Now assume that $f$ is arbitrary.

    (1, unbounded): Since $\mu\bracs{f \ne 0} < \infty$, there exists $\alpha > 0$ such that $\mu\bracs{|f| > \alpha} < \eps/2$. Let $g \in L^\infty(X; E)$ such that $g = f$ on $\bracs{|f| \le \alpha}$. By (1) applied to $g$, there exists $A \subset \bracs{|f| \le \alpha}$ such that $f|_A$ is continuous and $\mu(\bracs{|f| \le \alpha} \setminus A) < \eps/2$. In which case, $\mu(\bracs{f \ne 0} \setminus A) < \eps$, as desired. 

    (2, unbounded): By (1) applied to $g$, there exists $\phi \in C_c(X; \complex)$ such that $\mu\bracs{\phi \ne g} < \eps/2$, and $\mu(\bracs{\phi \ne f}) < \eps$.
\end{proof}

\begin{proposition}[Monotone Convergence Theorem for Lower Semicontinuous Functions, {{\cite[Proposition 7.12]{Folland}}}]
\label{proposition:mct-radon}
    Let $X$ be a LCH space, $\net{f}$ and $f: X \to [0, \infty]$ be non-negative lower semicontinuous functions such that $f_\alpha \upto f$, then for any Radon measure $\mu$ on $X$, 
    \[
    \int f d\mu = \sup_{\alpha \in A}\int f_\alpha d\mu
    \]
\end{proposition}
\begin{proof}
    Assume without loss of generality that $\int f d\mu < \infty$. By \autoref{proposition:semicontinuous-properties}, $f$ is Borel measurable, so $f \ge f_\alpha$ for all $\alpha \in A$ implies that
    \[
    \int f d\mu \ge \sup_{\alpha \in A}\int f_\alpha d\mu
    \]

    Let $\phi \in \Sigma^+(X, \cm)$ with $0 \le \phi < f$ and $\beta < \int \phi d\mu$. Let $\seqf{a_j} \subset (0, \infty)$ and $\seqf{E_j} \subset \cb_X$ such that $\phi = \sum_{j = 1}^n a_j \one_{E_j}$. Since $\int \phi d\mu < \infty$, for each $1 \le j \le n$, $\mu(E_j) < \infty$ by \hyperref[Markov's Inequality]{theorem:markov-inequality}. By \autoref{proposition:radon-regular-sigma-finite}, there exists compact sets $\seqf{K_j} \subset 2^X$ such that $K_j \subset E_j$ for each $1 \le j \le n$ and $\int \sum_{j = 1}^n a_j \one_{K_j}d\mu > \beta$. 

    Let $\psi = \sum_{j = 1}^n a_j \one_{K_j}$ and $K = \bigcup_{j = 1}^n K_j$, then $K$ is compact by \autoref{proposition:compact-extensions} and $-\psi$ is lower semicontinuous by \autoref{proposition:semicontinuous-properties}.
    
    For any $x_0 \in K$, since $f_\alpha \upto f$ and $\psi \le \phi < f$, there exists $\alpha(x_0) \in A$ such that $f_{\alpha(x_0)}(x_0) > \psi(x_0)$. By \autoref{proposition:semicontinuous-properties}, $f_{\alpha(x_0)} - \psi$ is lower semicontinuous, so
    \[
    \bracsn{\bracsn{f_{\alpha(x_0)} > \psi}|x_0 \in K}
    \]

    is an open cover of $K$. Let $\bracs{x_j}_1^N \subset K$ such that $K \subset \bigcup_{j = 1}^N \bracsn{f_{\alpha(x_j)} > \psi}$. Since $f_\alpha \upto f$, there exists $\alpha_0 \in A$ such that
    \[
    f_{\alpha_0} \ge \max_{1 \le j \le N}f_{\alpha(x_j)} \ge \psi
    \]
    
    so
    \[
    \int f_{\alpha_0} d\mu \ge \int \psi d\mu \ge \beta
    \]

    As such an $\alpha_0 \in A$ exists for all $\beta < \int \phi d\mu$, 
    \[
    \int \phi d\mu \le \sup_{\alpha \in A}\int f_\alpha d\mu
    \]

    Since the above holds for all $\phi \in \Sigma^+(X, \cm)$ with $0 \le \phi < f$, 
    \[
    \int f d\mu \le \sup_{\alpha \in A}\int f_\alpha d\mu
    \]

    by \autoref{lemma:lebesgue-non-negative-strict}.

\end{proof}



