\section{The Modular Function}
\label{section:modular-function}

\begin{definition}[Modular Function]
\label{definition:modular-function}
    Let $G$ be a locally compact group and $\mu$ be a left Haar measure on $G$, then
    \begin{enumerate}
        \item For any $f, g \in C_c^+(G; \real) \setminus \bracs{0}$, $A, B \in \cb_G$ with $\mu(A), \mu(B) > 0$, and $y \in G$, 
        \[
        \Delta_G(y) = \frac{\int R_{y^{-1}} f d\mu}{\int f d\mu} = \frac{\int R_{y^{-1}} g d\mu}{\int g d\mu} = \frac{\mu(Ay)}{\mu(A)} = \frac{\mu(By)}{\mu(B)} > 0
        \]
        \item For each $y \in G$ and $A \in \cb_G$, denote $\mu_y(A) = \mu(Ay)$, then $\mu_y(dx) = \Delta_G(y)\mu(dx)$. 
        \item For any choice of $f \in C_c^+(G; \real) \setminus \bracs{0}$, the mapping $\Delta_G: G \to (0, \infty)$ defined by $y \mapsto \int R_{y^{-1}}f d\mu$ is a continuous group homomorphism. 
        \item For each $A \in \cb_G$, let $\nu(A) = \mu(A^{-1})$, then $\nu(dx) = \Delta(x^{-1})\mu(dx)$. 
    \end{enumerate}
    
    The homomorphism $\Delta_G: G \to (0, \infty)$ is the \textbf{modular function} of $G$. 
\end{definition}
\begin{proof}[Proof, {{\cite[Proposition 2.24, Proposition 2.31]{FollandHarmonic}}}. ]
    (1), (2): For each $y \in G$, $\mu_y$ is also a left Haar measure. By \hyperef[Haar's Theorem]{theorem:haar}, there exists $\lambda > 0$ such that $\mu_y = \lambda \mu$. In which case, 
    \[
        \Delta_G(y^{-1}) = \frac{\int R_y f d\mu}{\int f d\mu} = \frac{\int R_y g d\mu}{\int g d\mu} = \frac{\mu(Ay)}{\mu(A)} = \frac{\mu(By)}{\mu(B)}
    \]

    (3): By \autoref{proposition:haar-translation}, the mapping $y \mapsto \int R_y f d\mu$ is continuous. For any $x, y \in G$ and $A \in \cb_G$ with $\mu(A) > 0$, 
    \[
    \Delta_G(xy)\mu(A) = \mu(Axy) = \Delta_G(y)\mu(Ax) = \Delta_G(y)\Delta_G(x)\mu(A)
    \]

    (4): Let $f \in C_c^+(G)$ and $y \in G$, then
    \begin{align*}
        \int R_{y}f(x) \Delta_G(x^{-1})\mu(dx) &= \Delta_G(y)\int f(xy)\Delta_G[(xy)^{-1}]\mu(dx) \\
        &= \int f(x)\Delta(x^{-1})\mu(dx)
    \end{align*}

    so $\Delta_G(x^{-1})\mu(dx)$ is a right Haar measure. By Haar's Theorem, there exists $\lambda > 0$ such that $\nu(dx) = \lambda \Delta_G(x^{-1})\mu(dx)$. 

    If $\lambda \ne 1$, then there exists $U \in \cn_G(1)$ symmetric and compact such that $|\Delta_G(x^{-1}) - 1| \le 2^{-1}|\lambda - 1|$ on $U$. In which case, by symmetry, $\mu(U) = \nu(U)$, and
    \begin{align*}
        |\lambda - 1|\mu(U) &= |\lambda \nu(U) - \mu(U)| = \abs{\int_U \Delta_G(x^{-1})-1 \mu(dx)} \\
        &\le \frac{1}{2}|\lambda - 1|\mu(U)
    \end{align*}

    which contradicts the fact that $\lambda \ne 1$. Therefore $\lambda = 1$, and $\nu(dx) = \Delta_G(x^{-1})\mu(dx)$. 
\end{proof}

\begin{definition}[Unimodular]
\label{definition:lcg-unimodular}
    Let $G$ be a locally compact group, then $G$ is \textbf{unimodular} if $\Delta_G = 1$.
\end{definition}