From c3751d034f588d5e61db825e24999b6b0e9d82fd Mon Sep 17 00:00:00 2001
From: Bokuan Li <bokuan.li@mail.mcgill.ca>
Date: Tue, 23 Jun 2026 20:18:44 -0400
Subject: [PATCH] Added the modular function.

---
 src/measure/lcg/index.tex   |  1 +
 src/measure/lcg/lcg.tex     |  2 ++
 src/measure/lcg/modular.tex | 49 +++++++++++++++++++++++++++++++++++++
 3 files changed, 52 insertions(+)
 create mode 100644 src/measure/lcg/modular.tex

diff --git a/src/measure/lcg/index.tex b/src/measure/lcg/index.tex
index ceda358..32d87f2 100644
--- a/src/measure/lcg/index.tex
+++ b/src/measure/lcg/index.tex
@@ -3,3 +3,4 @@
 
 \input{./lcg.tex}
 \input{./haar.tex}
+\input{./modular.tex}
diff --git a/src/measure/lcg/lcg.tex b/src/measure/lcg/lcg.tex
index 963f07c..9f32a7c 100644
--- a/src/measure/lcg/lcg.tex
+++ b/src/measure/lcg/lcg.tex
@@ -13,3 +13,5 @@
 \begin{proof}
     By \autoref{lemma:lch-compact-neighbour}, there exists a compact neighbourhood $U \in \cn_G(\text{supp}(\phi))$. By \autoref{proposition:uniform-continuous-compact}, $\phi|_{U}$ and $\phi|_{\supp(\phi)^c}$ are both left and right uniformly continuous. Therefore $\phi$ is left and right uniformly continuous. 
 \end{proof}
+
+
diff --git a/src/measure/lcg/modular.tex b/src/measure/lcg/modular.tex
new file mode 100644
index 0000000..c215f11
--- /dev/null
+++ b/src/measure/lcg/modular.tex
@@ -0,0 +1,49 @@
+\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}
+
+