\section{The Riesz Representation Theorem}
\label{section:riesz-radon}

\begin{definition}[Positive Linear Functional on $C_c$]
\label{definition:positive-linear-functional-cc}
    Let $X$ be a topological space and $I \in \hom(C_c(X; \real); \real)$, then $\phi$ is \textbf{positive} if for any $f \in C_c(X; [0, \infty))$, $\dpb{f, I}{C_c(X; \real)} \ge 0$.
\end{definition}

\begin{proposition}
\label{proposition:positive-linear-functional-cc-property}
    Let $X$ be an LCH space and $I \in \hom(C_c(X; \real); \real)$ be a positive linear functional, then
    \begin{enumerate}
        \item For any $f, g \in C_c(X; \real)$ with $f \le g$, $\dpb{f, I}{C_c(X; \real)} \le \dpb{g, I}{C_c(X; \real)}$.
        \item For any $K \subset X$ compact, there exists $C_K \ge 0$ such that for all $f \in C_c(X; \real)$ with $\supp{f} \subset K$, $|{\dpb{f, I}{C_c(X; \real)}}| \le \norm{f}_u$.
    \end{enumerate}
\end{proposition}
\begin{proof}
    (1): $\dpb{g - f, I}{C_c(X; \real)} \ge 0$.

    (2): By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $g \in C_c(X; [0, 1])$ such that $g|_K = 1$. In which case,
    \[
    -\norm{f}_u\dpn{g, I}{C_c(X; \real)} \le \dpn{f, I}{C_c(X; \real)} \le \norm{f}_u\dpn{g, I}{C_c(X; \real)}
    \]

    so $C_K = \dpn{g, I}{C_c(X; \real)}$ is a desired constant.
\end{proof}


\begin{theorem}[Riesz Representation Theorem]
\label{theorem:riesz-radon}
    Let $(X, \topo)$ be an LCH space and $I \in \hom(C_c(X; \real); \real)$ be a positive linear functional, then there exists a Borel measure $\mu: \cb_X \to [0, \infty]$ such that:
    \begin{enumerate}
        \item For any $U \subset X$ open, $\mu(U) = \sup_{f \prec U}\dpb{f, I}{C_c(X; \real)}$.
        \item For any $K \subset X$ compact, $\mu(K) = \inf_{f \in C_c(X; [0, 1]), f \ge \one_K}\dpb{f, I}{C_c(X; \real)}$.
        \item For any $f \in C_c(X; \real)$, $\int f d\mu = \dpb{f, I}{C_c(X; \real)}$.
        \item $\mu$ is a Radon measure.
        \item[(U)] If $\nu: \cb_X \to [0, \infty]$ is a Borel measure that satisfies (3) and (4), then $\mu = \nu$.
    \end{enumerate}
\end{theorem}
\begin{proof}[Proof {{\cite[Theorem 7.2]{Folland}}}. ]
    (1): For any $U \in \topo$ and $f \in C_c(X; \real)$, denote $f \prec U$ if $f \in C_c(X; [0, 1])$ and $\supp{f} \subset U$. Let
    \[
    \mu_0: \topo \to [0, \infty] \quad U \mapsto \sup_{f \prec U}\dpb{f, I}{C_c(X; \real)}
    \]

    and
    \[
    \mu^*: 2^X \to [0, \infty] \quad E \mapsto \inf\bracsn{\mu_0(U)|U \in \cn^o(E)}
    \]

    then
    \begin{enumerate}
        \item[(OM1)] Since $\emptyset \in \topo$ and $\mu_0(\emptyset) = 0$, $\mu^*(\emptyset) = 0$.
        \item[(OM2)] Let $E, F \subset X$ with $E \subset F$, then $\cn^o(E) \subset \cn^o(F)$, so $\mu^*(E) \le \mu^*(F)$.
        \item[(OM3)] Let $\seq{E_n} \subset 2^X$, $E = \bigcup_{n \in \natp}E_n$, $U \in \cn^o(E)$, and $\seq{U_n} \subset \topo$ such that $U_n \in \cn^o(E_n)$ for each $n \in \natp$.

        Let $f \prec U$, then by compactness of $\supp{f}$, there exists $N \in \natp$ such that $\supp{f} \subset \bigcup_{n = 1}^N U_n$. By \autoref{proposition:lch-partition-of-unity}, there exists a partition of unity $\seqf{\phi_j} \subset C_c(X; [0, 1])$ on $K$ subordinate to $\seqf[N]{U_n}$. In which case,
        \[
        \dpb{f, I}{C_c(X; \real)} = \sum_{n = 1}^N \dpb{\phi_n f, I}{C_c(X; \real)} \le \sum_{n = 1}^N \mu_0(U_n) \le \sum_{n \in \natp}\mu_0(U_n)
        \]

        Since this holds for all $f \prec U$,
        \[
        \mu^*(E) \le \mu_0(U) \le \sum_{n \in \natp}\mu_0(U_n)
        \]


        Let $\eps > 0$, then there exists $\seq{U_n} \subset \topo$ such that $U_n \supset E_n$ and $\mu_0(U_n) \le \mu^*(E_n) + \eps/2^n$ for each $n \in \natp$. In which case,
        \[
        \mu^*(E) \le \sum_{n \in \natp}\mu_0(U_n) \le \eps + \sum_{n \in \natp}\mu^*(E_n)
        \]

        As this holds for all $\eps > 0$, $\mu^*(E) \le \sum_{n \in \natp}\mu^*(E_n)$.
    \end{enumerate}

    Therefore $\mu^*: 2^E \to [0, \infty]$ is an outer measure.

    To see that all Borel sets are $\mu^*$-measurable, let $E \subset X$ and $U \in \topo$. First suppose that $E$ is open. Let $f \prec E \cap U$, then for any $g \prec E \setminus \supp{f}$, $\supp{f} \cap \supp{g} = \emptyset$ and $f + g \prec E$, so
    \[
    \dpb{f, I}{C_c(X; \real)} + \dpb{g, I}{C_c(X; \real)} \le \mu_0(E)
    \]

    Since this holds for all $g \prec E \setminus \supp{f}$,
    \[
    \dpb{f, I}{C_c(X; \real)} + \mu_0(E \setminus \supp{f}) \le \mu_0(E)
    \]

    As $E \setminus \supp{f} \supset E \setminus U$,
    \[
    \dpb{f, I}{C_c(X; \real)} + \mu_0(E \setminus U) \le \mu_0(E)
    \]

    Finally, the above holds for all $f \prec E \cap U$,
    \[
    \mu_0(E \cap U) + \mu_0(E \setminus U) \le \mu_0(E)
    \]

    Now suppose that $E$ is arbitrary. Let $V \in \cn^o(E)$, then
    \[
    \mu^*(E \cap U) + \mu^*(E \setminus U) \le \mu_0(V \cap U) + \mu_0(V \setminus U) \le \mu_0(V)
    \]

    As this holds for all such $V$,
    \[
    \mu^*(E) = \mu^*(E \cap U) + \mu^*(E \setminus U)
    \]


    By \hyperref[Carathéodory's Extension Theorem]{theorem:caratheodory-extension}, there exists a Borel measure $\mu: \cb_X \to [0, \infty]$ such that for all $U \in \topo$, $\mu(U) = \mu_0(U)$.

    (2): Let $K \subset X$ be compact and $U \in \cn^o(K)$. By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $f \prec U$ with $f \ge \one_K$. In which case,
    \[
    \inf_{\substack{f \in C_c(X; \real) \\ f \ge \one_K}}\dpb{f, I}{C(X; \real)} \le \inf_{U \in \cn^o(K)}\mu(U) = \mu(K)
    \]

    On the other hand, let $f \in C_c(X; [0, 1])$ with $f \ge \one_K$. For any $r \in (0, 1)$, let $g \prec \bracs{f > r}$, then $r^{-1}f \ge g$ and
    \[
    \dpb{g, I}{C_c(X; \real)} \le r^{-1}\dpb{f, I}{C_c(X; \real)}
    \]

    As this holds for all $g \prec \bracs{f > r}$,
    \[
    \mu(K) \le \mu(\bracs{f > r}) \le r^{-1}\dpb{f, I}{C_c(X; \real)}
    \]

    Since the above holds for all $r \in (0, 1)$,
    \[
    \mu(K) \le \inf_{\substack{f \in C_c(X; \real) \\ f \ge \one_K}}\dpb{f, I}{C_c(X; \real)}
    \]


    (3): Let $f \in C_c(X; \real)$. Using linearity, assume without loss of generality that $f \in C_c(X; [0, 1])$. Let $N \in \natp$. For each $1 \le n \le N$, let
    \[
    f_n = \min\paren{\max\paren{f - \frac{n - 1}{N}, 0}, \frac{1}{N}}
    \]

    For any $x \in X$, there exists $1 \le n \le N$ such that $f(x) \in [(n-1)/N, n/N]$. In which case,
    \begin{itemize}
    \item For each $1 \le j < n$, $f_j(x) = 1/N$.
    \item $f_n(x) = f(x) - \frac{n - 1}{N}$.
    \item For each $n < j \le N$, $f_j(x) = 0$.
    \end{itemize}
    so $\sum_{j = 1}^Nf_j(x) = \frac{n-1}{N} + f(x) - \frac{n - 1}{N} = f(x)$.

    Let $K_0 = \supp{f}$, and for each $1 \le n \le N$, let $K_n = \bracs{f \ge n/N}$.
    \[
    \frac{1}{N}\mu\paren{K_n} \le \int f_n d\mu \le \frac{1}{N}\mu\paren{K_{n-1}}
    \]

    Since $\supp{f_n} \subset \bracs{f \ge (n-1)/N}$, for any $U \supset \bracs{f_n \ge (n - 1)/N}$, $f_n \prec U$, so
    \[
    \dpb{f_n, I}{C_c(X; \real)} \le \frac{1}{N}\mu\paren{K_{n-1}}
    \]

    On the other hand, $f_n \ge \frac{1}{N}\one_{K_n}$,
    \[
    \mu(K_n) \le \dpb{f_n, I}{C_c(X; \real)}
    \]

    so
    \[
    \frac{1}{N}\sum_{n = 1}^N \mu(K_n) \le \dpb{f, I}{C_c(X; \real)}, \int f d\mu \le \frac{1}{N}\sum_{n = 1}^N \mu(K_{n-1})
    \]


    Therefore
    \[
    \abs{\int f d\mu - \dpb{f, I}{C_c(X; \real)}} \le \frac{\mu(K_{0}) - \mu(K_N)}{N} \le \frac{\mu(\supp{f})}{N}
    \]

    As this holds for all $N \in \natp$, $\int f d\mu = \dpb{f, I}{C_c(X; \real)}$.

    (4):
    \begin{enumerate}
        \item[(R1)] For any $K \subset X$ compact, by \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $f \in C_c(X; [0, 1])$ with $f \ge \one_K$. In which case, $\mu(K) \le \int f d\mu = \dpb{f, I}{C_c(X; \real)}$.
        \item[(R2)] By definition of $\mu^*$, $\mu$ is outer regular.
        \item[(R3')] For any $U \in \topo$,
        \[
        \mu(U) = \sup_{f \prec U}\dpb{f, I}{C_c(X; \real)} \le \sup_{f \prec U}\mu(\supp{f}) \le \sup_{\substack{K \subset U \\ \text{compact}}}\mu(K) \le \mu(U)
        \]

    \end{enumerate}

    (U): By \autoref{proposition:radon-measure-cc}, $\nu$ also satisfies (2). Thus for any $E \in \cb_X$, by (R2) and (R3'),
    \[
    \nu(E) = \inf_{U \in \cn^o(E)}\sup_{\substack{K \subset U \\ \text{compact}}}\inf_{\substack{f \in C_c(X; [0, 1]) \\ f \ge \one_K}}\dpb{f, I}{C_c(X; \real)}
    \]

    so $\nu$ is uniquely determined by $I$.
\end{proof}
