Added existence of the Haar measure.

src/measure/lcg/haar.tex

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+\section{Haar Measures}
+\label{section:haar}
+
+\begin{definition}[Haar Measure]
+\label{definition:haar-measure}
+    Let $G$ be a locally compact group and $\mu: \cb_G \to [0, \infty]$ be a non-zero Radon measure, then $\mu$ is a \textbf{left Haar measure} if
+    \begin{enumerate}
+        \item[(LH)] For each $g \in G$ and $A \in \cb_G$, $\mu(gA) = \mu(A)$. 
+    \end{enumerate}
+
+    and a \textbf{right Haar measure} if
+    \begin{enumerate}
+        \item[(RH)] For each $g \in G$ and $A \in \cb_G$, $\mu(Ag) = \mu(A)$. 
+    \end{enumerate}
+\end{definition}
+
+\begin{lemma}
+\label{lemma:lc-sigma-compact}
+    Let $G$ be a locally compact group, then there exists an open and closed subgroup $H$ that is $\sigma$-compact. 
+\end{lemma}
+\begin{proof}
+    Let $K \in \cn_G(1)$ and $K^{(1)} = K$. For each $n \in \natp$, let $K^{(n+1)} = KK^{(n)}$, then $K^{(n+1)}$ is compact by \autoref{proposition:compact-extensions} with $K^{(n+1)} \in \cn_G(K^{(n)})$. Let $H = \bigcup_{n \in \natp}K^{(n)}$, then $H$ is a subgroup of $G$, which is open by \autoref{lemma:openneighbourhood}. Since $H$ admits an exhaustion by compact sets, it is $\sigma$-compact. 
+
+    Finally, since
+    \[
+    G \setminus H = G \setminus \bigcup_{n \in \natp} K^{(n)} = \bigcap_{n \in \natp}G \setminus K^{(n)}
+    \]
+
+    and $K^{(n)}$ is closed for each $n \in \natp$ by \autoref{proposition:compact-closed}, $G \setminus H$ is closed, and hence $H$ is open. 
+\end{proof}
+
+
+\begin{definition}[Covering Ratio]
+\label{definition:lcg-covering-ratio}
+    Let $G$ be a locally compact group and $f, g \in C_c^+(G)$, then
+    \[
+    (f: g) = \inf\bracs{\sum_{j = 1}^n c_j \bigg | \seqf{c_j} \subset [0, \infty), \seqf{x_j} \subset G, f \le \sum_{j = 1}^n c_j L_{x_j}g}
+    \]
+
+    is the \textbf{covering ratio} of $f$ by $g$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:covering-ratio-gymnastics}
+    Let $G$ be a locally compact group and $f, h, g \in C_c^+(G)$, then:
+    \begin{enumerate}
+        \item If $g \ne 0$, then $(f: g) < \infty$. 
+        \item $(f, h: g) \le (h: g) + (h: g)$. 
+        \item For each $\lambda \ge 0$, $(\lambda f: g) = \lambda(f: g)$. 
+        \item If $f \le h$, then $(f: g) \le (h: g)$. 
+        \item $(f: g) \le (f: h)(h: g)$. 
+        \item $(f: g) \ge \norm{f}_u/\norm{g}_h$. 
+        \item For each $x \in G$, $(L_xf: g) = (f: g)$. 
+    \end{enumerate}
+\end{proposition}
+% Proof omitted due to obviousness. 
+
+\begin{lemma}
+\label{lemma:haar-approx}
+    Let $G$ be a locally compact group, $f, f'\in C_c^+(G)$, and $\eps > 0$, then there exists $V \in \cn_G(1)$ such that for any $g \in C_c^+(V)$ with $g \ne 0$, 
+    \[
+    (f: g) + (f': g) \le (f + f': g) + \eps
+    \]
+\end{lemma}
+\begin{proof}[Proof, {{\cite[Lemma 2.18]{FollandHarmonic}}}. ]
+    By \hyperref[Urysohn's Lemma]{lemma:lch-urysohn}, there exists $\eta \in C_c^+(G; [0, 1])$ such that $\eta|_{\supp{f} \cup \supp{f'}} = 1$. 
+
+    Let $\delta > 0$, and define
+    \[
+    H = f + f' + \delta \eta \quad h = \frac{f}{H} \quad h' = \frac{f'}{H}
+    \]
+    
+    By \autoref{proposition:lcg-cc-uc}, there exists $V \in \cn_G(1)$ such that for any $x, y \in G$ with $x^{-1}y \in V$, 
+    \[
+    |h(x) - h(y)|, |h'(x) - h'(y)| < \delta
+    \] 
+
+    Let $g \in C_c^+(V)$, $\seqf{c_j} \subset [0, \infty)$, and $\seqf{x_j} \subset G$ such that $H \le \sum_{j = 1}^n c_j L_{x_j}\phi$, then for each $x \in G$, 
+    \begin{align*}
+        f(x) &= H(x)h(x) \le \sum_{j = 1}^n c_j L_{x_j}g(x)h(x) = \sum_{j = 1}^n c_jg(x_j^{-1}x)h(x) \\
+        &\le \sum_{j = 1}^n c_j[h(x_j) + \delta] \cdot  L_{x_j}g(x)
+    \end{align*}
+
+    Likewise, 
+    \[
+    f'(x) \le \sum_{j = 1}^n c_j[h'(x_j) + \delta] \cdot L_{x_j}g(x)
+    \]
+
+    As $h + h' \le 1$, 
+    \[
+        (f: g) + (f': g) \le \sum_{j = 1}^n c_j[h(x_j) + h'(x_j) + 2\delta]
+    \]
+    
+    Since the above holds for all such $\seqf{c_j} \subset [0, \infty)$ and $\seqf{x_j} \subset G$, 
+    \begin{align*}
+    (f: g) + (f': g) &\le (1 + 2\delta)(H: g) \\
+    &\le (1 + 2\delta)[(f + f': g) + \delta(\eta: g)]
+    \end{align*}
+\end{proof}
+
+\begin{theorem}[Haar]
+\label{theorem:haar}
+    Let $G$ be a locally compact group, then:
+    \begin{enumerate}
+        \item There exists a left/right Haar measure on $G$. 
+        \item For any two left/right Haar measures $\mu$ and $\nu$ on $G$, there exists $\lambda > 0$ such that $\mu = \lambda \nu$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 2.10, 2.20]{FollandHarmonic}}}. ]
+    (1): Fix $h \in C_c^+(G)$ with $h \ne 0$. For each $g \in C_c^+(G)$ with $g \ne 0$, let
+    \[
+    I_g: C_c^+(G) \to [0, \infty) \quad f \mapsto \frac{(f: g)}{(h: g)}
+    \]
+
+    then by (5) of \autoref{proposition:covering-ratio-gymnastics}, for each $f \in C_c^+(G)$ with $f \ne 0$, 
+    \[
+    \frac{1}{(h: f)} = \frac{(f: g)}{(h: f)(f: g)} \le I_g(f) \le \frac{(f: h)(h: g)}{(h: g)} = (f: h)
+    \]
+
+    Thus $\mathcal{I}(f) = \bracs{I_g(f)|g \in C_c^+(G) \setminus \bracs{0}}$ is precompact for each $f \in C_c^+(G)$. 
+
+    For each $V \in \cn_G(1)$, let $E_V = \bracs{I_g|g \in C_c^+(V) \setminus \bracs{0}}$, then $\fF = \bracs{E_V|V \in \cn_G(1)}$ is a filter on the product space $\prod_{f \in C_c^+(G)}\ol{\mathcal{I}(f)}$. By \hyperref[Tychonoff's Theorem]{theorem:tychonoff}, there exists $\bigcap_{V \in \cn_G(1)}\ol{E_V} \ne \emptyset$. 
+
+    Let $I \in \bigcap_{V \in \cn_G(1)}\ol{E_V}$, then by continuity, 
+    \begin{enumerate}[label=(\roman*)]
+        \item For each $f \in C_c^+(G) \setminus \bracs{0}$, $I(f) \in [(h: f)^{-1}, (f: h)]$. 
+        \item For every $\lambda \ge 0$ and $f \in C_c^+(G)$, $I(\lambda f) = \lambda I(f)$.
+        \item For any $x \in G$, $I(L_xf) = I(f)$. 
+        \item For each $f, f' \in C_c^+(G)$, $I(f + g) \le I(f) + I(g)$. 
+    \end{enumerate}
+    
+    Let $f, f' \in C_c^+(G)$ and $\eps > 0$. By \autoref{lemma:haar-approx}, there exists $V \in \cn_G(1)$ such that for each $g \in E_V$, 
+    \begin{enumerate}[label=(\alph*)]
+        \item $|I_g(f) - I(f)|, |I_g(f') - I(f')| < \eps$. 
+        \item $I_g(f) + I_g(f') \le I_g(f + f') + \eps$. 
+    \end{enumerate}
+
+    In which case, $I(f) + I(f') \le I(f + f') + \eps$. Since this holds for all $\eps > 0$, 
+    \begin{enumerate}[start=4, label=(\roman*)]
+        \item For each $f, f' \in C_c^+(G)$, $I(f + g) \ge I(f) + I(g)$. 
+    \end{enumerate}
+
+    Using \autoref{lemma:positive-functional-extension}, $I$ extends to a positive linear functional on $C_c(G; \real)$, with $I(f) > 0$ for all $f \in C_c^+(G) \setminus \bracs{0}$, and
+    \begin{enumerate}
+        \item[(LH)] For each $f \in C_c(G)$ and $x \in G$, $I(L_xf) = I(f)$. 
+    \end{enumerate}
+    
+    By (i) and the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon}, there exists a unique non-zero Radon measure $\mu: \cb_G \to [0, \infty]$ such that for each $f \in C_c^+(G)$, $I(f) = \int_G f d\mu$. Finally, by \hyperref[density of $C_c(\mu; \real)$ in $L^1(\mu; \real)$]{proposition:radon-cc-dense} and (LH), $\mu$ is a left Haar measure. 
+\end{proof}
+
+
+