From caa710b2d257a3e83214555fda1bf0bf5f097a9d Mon Sep 17 00:00:00 2001
From: Bokuan Li <bokuan.li@mail.mcgill.ca>
Date: Wed, 7 Oct 2026 15:20:00 -0400
Subject: [PATCH] Thank you henry.

---
 src/measure/measurable-maps/product.tex | 25 +++++++++++++++----------
 1 file changed, 15 insertions(+), 10 deletions(-)

diff --git a/src/measure/measurable-maps/product.tex b/src/measure/measurable-maps/product.tex
index 0aa2e87..868afbc 100644
--- a/src/measure/measurable-maps/product.tex
+++ b/src/measure/measurable-maps/product.tex
@@ -37,21 +37,26 @@
 \end{enumerate}
 \end{proof}
 
-\begin{proposition}[{{\cite[Proposition 1.5]{Folland}}}]
+\begin{proposition}
 \label{proposition:product-sigma-algebra-metric}
-    Let $\seqf{X_j}$ be topological spaces, $X = \prod_{j = 1}^n X_j$, then:
+    Let $\seq{X_j}$ be topological spaces, $X = \prod_{n \in \natp} X_n$, then:
     \begin{enumerate}
-        \item $\bigotimes_{j = 1}^n \cb_{X_j} \subset \cb_{X}$.
-        \item If $\seqf{X_j}$ are separable, then $\bigotimes_{j = 1}^n \cb_{X_j} = \cb_{X}$
+        \item $\bigotimes_{n \in \natp} \cb_{X_n} \subset \cb_{X}$.
+        \item If $\seq{X_n}$ are separable and metrisable, then $\bigotimes_{n \in \natp} \cb_{X_n} = \cb_{X}$. 
     \end{enumerate}
 \end{proposition}
-\begin{proof}
-    (2): By \autoref{proposition:separable-product}, $X$ is also separable. Let $U \subset X$, then there exists $\seq{x_j} \subset X$ such that $\overline{\bracs{x_j|j \in \natp}} \supset U$. For each $j \in \nat$, let $\bracs{U_{j, k}}$ such that $U_{j, k} \in \cn_{X_k}(\pi_k(x_j))$ for each $1 \le k \le n$ and $x_j \in \bigcap_{k = 1}^n \pi_k^{-1}(U_{j, k}) \subset U$.
-
-    Since $\bigcap_{k = 1}^n \pi_k^{-1}(U_{j, k}) \in \bigotimes_{j = 1}^n \cb_{X_j}$ and
+\begin{proof}[Proof, {{\cite[Proposition 1.5]{Folland}}} and {{\cite[Proposition 8.1.7]{CohnMeasure}}}. ]
+    (2): For each $J \subset \natp$ finite, $x \in X$, and $\eps > 0$, define
     \[
-    U = \bigcup_{j \in \natp}\bigcap_{k = 1}^n \pi_k^{-1}(U_{j, k})
+    B_J(x, \eps) = \bigcap_{j \in J}\pi_j^{-1}(B_{X_j}(x_j, \eps))
     \]
 
-    The open set $U$ is in $\bigotimes_{j = 1}^n \cb_{X_j}$.
+    By \autoref{proposition:separable-product}, $X$ is also separable. Let $\seq{x_n} \subset X$ be a countable dense subset, and
+    \[
+    \fB = \bracsn{B_J(x_n, 1/k)|n, k \in \natp, J \subset \natp \text{ finite}}
+    \]
+
+    For each $x \in X$ and $U \in \cn_X(x)$, there exists $J \subset X$ finite and $k \in \natp$ such that $x \in B_J(x, 1/k) \subset U$. By density of $\seq{x_n}$, there exists $n \in \natp$ such that $x_n \in B_J(x, 1/(2k))$. In which case, $x \in B_J(x_n, 1/(2k)) \subset B_J(x, 1/k) \subset U$. Thus $\fB$ is a countable base for the product topology, consisting of cylinder sets in $\bigotimes_{n \in \natp} \cb_{X_n}$. 
+
+    As such, every open set in $X$ is a countable union of elements of $\fB$. Therefore $\bigotimes_{n \in \natp} \cb_{X_n} \supset \cb_{X}$.
 \end{proof}