Added Riesz.

src/measure/measure/index.tex

@@ -10,4 +10,5 @@
 \input{./src/measure/measure/outer.tex}
 \input{./src/measure/measure/lebesgue-stieltjes.tex}
 \input{./src/measure/measure/radon.tex}
+\input{./src/measure/measure/riesz.tex}
 \input{./src/measure/measure/kolmogorov.tex}

src/measure/measure/radon.tex

@@ -10,3 +10,37 @@
         \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 Urysohn's lemma (\ref{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 Urysohn's lemma (\ref{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}

src/measure/measure/regular.tex

@@ -13,7 +13,7 @@
 \label{definition:outer-regular}
     Let $X$ be a topological space, $\mu: \cb_X \to [0, \infty]$ be a Borel measure, and $E \in \cb_X$, then $\mu$ is \textbf{outer regular} on $E$ if
     \[
-    \mu(E) = \sup\bracs{\mu(U)| U \supset A, U \text{ open}}
+    \mu(E) = \sup\bracs{\mu(U)| U \in \cn^o(A)}
     \]
 \end{definition}
 

src/measure/measure/riesz.tex

@@ -1,2 +1,165 @@
-\section{Riesz Representation Theorem}
+\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 a 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 Urysohn's lemma (\ref{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, {{\cite[Theorem 7.2]{Folland}}}]
+\label{theorem:riesz-radon}
+    Let $(X, \topo)$ be a 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}
+    (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 \ref{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 Carathéodory's Extension Theorem (\ref{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 Urysohn's lemma (\ref{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 Urysohn's lemma (\ref{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 \ref{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}