\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 $(f + h: g) \le (f: g) + (h: g)$. 
        \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 relatively compact 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}, $\prod_{f \in C_c^+(G)}\ol{\mathcal{I}(f)}$ is compact, and $\bigcap_{V \in \cn_G(1)}\ol{E_V} \ne \emptyset$. 

    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 $f, f' \in C_c^+(G)$, $I(f + f') \le I(f) + I(f')$. 
    \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, $I(f) + I(f') \le I(f + f') + 3\eps$. Since this holds for all $\eps > 0$, 
    \begin{enumerate}[start=4, label=(\roman*)]
        \item For each $f, f' \in C_c^+(G)$, $I(f + f') \ge I(f) + I(f')$. 
    \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 $C_c(G; \real)$ in $L^1(\mu; \real)$]{proposition:radon-cc-dense} and (LH), $\mu$ is a left Haar measure. 

    (2): Let $f, g \in C_c^+(G) \setminus \bracs{0}$ and $\eps > 0$, then by \autoref{proposition:lcg-cc-uc}, there exists a symmetric neighbourhood $V \in \cn_G(1)$ such that for any $x \in G$ and $y \in V$, 
    \[
    |f(xy) - f(yx)|, |g(xy) - g(yx)| < \eps
    \]

    By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $h \in C_c^+(V) \setminus \bracs{0}$ such that $h(x) = h(x^{-1})$ for all $x \in V$. Since $f$ and $h$ are both compactly supported and $\mu, \nu$ are locally finite, by \hyperref[Tonelli's Theorem]{theorem:fubini-tonelli},
    \begin{align*}
        \paren{\int f d\mu}\paren{\int h d\nu} &= \iint f(x)h(y) \mu(dx)\nu(dy) \\
        &= \iint f(yx)h(y) \nu(dx)\mu(dy) \\
    \end{align*}

    Similarly, by symmetry of $h$, 
    \begin{align*}
        \paren{\int h d\mu}\paren{\int f d\nu} &= \iint h(x)f(y) \mu(dx)\nu(dy) \\
        &= \iint h(y^{-1}x)f(y) \mu(dx)\nu(dy) \\
        &= \iint h(x^{-1}y)f(y) \nu(dy)\mu(dx) \\
        &= \iint h(y)f(xy) \nu(dy)\mu(dx) \\
        &= \iint h(y)f(xy) \mu(dx)\nu(dy)
    \end{align*}

    Thus there exists $C > 0$ such that
    \begin{align*}
    &\abs{\paren{\int f d\mu}\paren{\int h d\nu} - \paren{\int h d\mu}\paren{\int f d\nu}} \\ 
    &\le \abs{\iint h(y)[f(xy) - f(yx)]\mu(dx)\nu(dy)} \le C\eps
    \end{align*}

    and
    \[
    \abs{\paren{\int g d\mu}\paren{\int h d\nu} - \paren{\int h d\mu}\paren{\int g d\nu}}
    \le C\eps
    \]

    so there exists $C' > 0$ such that
    \[
    \abs{\frac{\int h d\nu}{\int h d\mu} - \frac{\int g d\nu}{\int g d\mu}}
    \le C'\eps
    \]

    and
    \[
    \abs{\frac{\int f d\nu}{\int f d\mu} + \frac{\int h d\nu}{\int h d\mu}}
    \le C'\eps
    \]

    Therefore
    \[
    \abs{\frac{\int f d\nu}{\int f d\mu} - \frac{\int g d\nu}{\int g d\mu}} \le 2C' \eps
    \]

    As the above holds for all $\eps > 0$, $\int f d\nu/\int f d\mu = \int g d\nu/\int g d\mu$. By uniqueness from the \hyperref[Riesz Representation Theorem]{section:riesz-radon}, there exists $\lambda > 0$ such that $\mu = \lambda \nu$. 
\end{proof}

\begin{proposition}
\label{proposition:haar-measure-charge}
    Let $G$ be a locally compact group and $\mu: \cb_G \to [0, \infty]$ be a left/right Haar measure, then:
    \begin{enumerate}
        \item For each $U \subset G$ open with $U \ne \emptyset$, $\mu(U) > 0$. 
        \item For each $f \in C_c^+(G) \setminus \bracs{0}$, $\int \phi d\mu > 0$. 
    \end{enumerate}
\end{proposition}
\begin{proof}[Proof, for left Haar measures. ]
    Since $\mu \ne 0$, there exists $g \in C_c^+(G) \setminus \bracs{0}$ such that $\int g d\mu > 0$. By compactness of $\supp{g}$, there exists $\seqf{x_j} \subset G$ such that:
    \[
    \supp{g} \subset \bigcup_{j = 1}^n x_j^{-1}U \quad g \le \sum_{j = 1}^n L_{x_j}f
    \]

    Thus $0 < \int g d\mu \le n\mu(U)$ and $0 < \int g d\mu \le n \int f d\mu$. 
\end{proof}

\begin{proposition}
\label{proposition:haar-translation}
    Let $G$ be a locally compact group, $\mu: \cb_G \to [0, \infty]$ be a left Haar measure, $p \in [1, \infty)$, and $E$ be a normed vector space over $K \in \RC$, then the mapping
    \[
    G \times L^p(\mu; E) \quad (x, f) \mapsto L_xf
    \]

    is jointly continuous. Similarly, if $\nu: \cb_G \to [0, \infty]$ is a right Haar measure, then
    \[
    G \times L^p(\mu; E) \quad (x, f) \mapsto R_xf
    \]

    is also jointly continuous.
\end{proposition}
\begin{proof}[Proof of the left case. ]
    Let $\eps > 0$, $x, y \in G$, and $f, g \in L^p(\mu; E)$, then
    \begin{align*}
        \norm{L_xf - L_yg}_{L^p(\mu; E)} &\le \norm{L_xf - L_yf}_{L^p(\mu; E)} + \norm{L_yf - L_y g}_{L^p(\mu; E)} \\
        &= \norm{L_xf - L_yf}_{L^p(\mu; E)} + \norm{f - g}_{L^p(\mu; E)}
    \end{align*}

    By \autoref{proposition:radon-cc-dense}, there exists $\phi \in C_c(G)$ such that $\norm{\phi - f}_{L^p(\mu; E)} < \eps$. In which case, 
    \begin{align*}
        \norm{L_xf - L_yf}_{L^p(\mu; E)} &\le \norm{L_xf - L_x \phi}_{L^p(\mu; E)} + \norm{L_x\phi - L_y\phi}_{L^p(\mu; E)} \\
        &+ \norm{L_yf - L_y \phi}_{L^p(\mu; E)} \\
        &= 2\norm{f - \phi}_{L^p(\mu; E)} + \norm{L_x\phi - L_y\phi}_{L^p(\mu; E)} \\
        &\le 2\eps + \normn{L_{x^{-1}y}\phi - \phi}_u\mu\bracs{\phi \ne 0}
    \end{align*}

    By \autoref{proposition:lcg-cc-uc}, there exists $V \in \cn_G(1)$ such that if $x^{-1}y \in V$, then $\norm{L_{x^{-1}y}\phi - \phi}_u < \eps/\mu\bracs{\phi \ne 0}$. Thus if $x^{-1}y \in V$, then
    \[
    \norm{L_xf - L_yf}_{L^p(\mu; E)} \le 3\eps + \norm{f - g}_{L^p(\mu; E)}
    \]
\end{proof}