From 3d1e095e8217a5a13ae25a81310e5b3206351649 Mon Sep 17 00:00:00 2001
From: Bokuan Li <bokuan.li@mail.mcgill.ca>
Date: Sun, 28 Jun 2026 20:10:25 -0400
Subject: [PATCH] Fixed typos in the gluing lemma.

---
 src/measure/measurable-maps/approx.tex | 3 ++-
 src/measure/measure/localisable.tex    | 8 ++++----
 2 files changed, 6 insertions(+), 5 deletions(-)

diff --git a/src/measure/measurable-maps/approx.tex b/src/measure/measurable-maps/approx.tex
index 0e6ff97..440a320 100644
--- a/src/measure/measurable-maps/approx.tex
+++ b/src/measure/measurable-maps/approx.tex
@@ -38,6 +38,7 @@
     \begin{enumerate}
         \item $\seq{I_n}$ is an $\mathcal{A}$-admissible \hyperref[approximation of the identity]{definition:approximation-id-measure}. 
         \item For each $N \in \natp$, $I_N$ is Borel measurable with $I_N(X) \subset \bracsn{x_n|1 \le n \le N}$. 
+        \item For each $n \in \natp$ and $x \in X$, $d(x, I_{n+1}(x)) \le d(x, I_n(x))$. 
     \end{enumerate}
 \end{lemma}
 \begin{proof}
@@ -83,7 +84,7 @@
 
     Finally, for each $x \in X$ and $\eps > 0$, since $x \in \ol{\mathcal{A}(x)^o}$, there exists $N_0 \in \natp$ such that $x_{N_0} \in \mathcal{A}(x)$ and $d(x, x_{N_0}) < \eps$. In which case, for any $N \ge N_0$, $N_0 \in C_N(x)$ and $d(x, I_N(x)) \le d(x, x_{N_0}) < \eps$. Thus $I_N(x) \to x$ as $N \to \infty$, and $\seq{I_N}$ satisfies (AI1). 
     
-    Therefore $\seq{I_N}$ is an approximation of the identity satisfying (1) and (2). 
+    Therefore $\seq{I_N}$ is an approximation of the identity satisfying (1)-(3). 
 \end{proof}
 
 
diff --git a/src/measure/measure/localisable.tex b/src/measure/measure/localisable.tex
index 326f428..8bc6958 100644
--- a/src/measure/measure/localisable.tex
+++ b/src/measure/measure/localisable.tex
@@ -121,7 +121,7 @@
     (1): Let $A, B \in \cf$, then
     \begin{align*}
         \mu(E_A \setminus F_B^c) &= \mu(E_A \cap F_B) = \mu(E_A \cap F_B \cap A \cap B) \\
-        &\le \mu(E_A \cap F_A) + \mu((F_A \cap B) \Delta F_B \cap A \cap B) = 0
+        &\le \mu(E_A \cap F_A) + \mu((F_A \cap B) \Delta (F_B \cap A)) = 0
     \end{align*}
 
     so $F_B^c$ is an essential upper bound of $\bracs{E_A}_{A \in \cf}$. Since $E$ is an essential supremum of $\bracs{E_A}_{A \in \cf}$, $\mu(E \setminus F_B^c) = \mu(E \cap F_B) = 0$. 
@@ -182,15 +182,15 @@
         \item For each $A \in \cf$, $f_n|_A = f_{A, n}$ almost everywhere. 
     \end{enumerate}
     
-    Since $Y$ is Polish, \autoref{proposition:metric-measurable-limit} implies that $\bracsn{\limv{n}f_n \text{ exists}} \in \cm$. For each $A \in \cf$, 
+    Since $Y$ is Polish, \autoref{proposition:metric-measurable-limit} implies that $\bracsn{\limv{n}f_n \text{ exists}} \in \cm$. For each $A \in \cf$, since $\limv{n}f_{A, n}$ and $\limv{n}f_{n}|_A$ exist and are equal almost everywhere on $A$, 
     \[
-    \mu\paren{\bracs{\limv{n}f_n \text{ does not exist}} \cap A} = \mu\bracs{\limv{n}f_{A, n} \ne f_A} = 0
+    \mu\paren{\bracs{\limv{n}f_n \text{ does not exist}} \cap A} \le \mu\bracs{\limv{n}f_{A, n} \text{ does not exist}} = 0
     \]
 
     As $\mu$ is semifinite, $\mu\bracsn{\limv{n}f_n \text{ does not exist}} = 0$, so there exists $f \in \mathcal{L}^0(X; Y)$ such that $f = \limv{n}f_n$ almost everywhere. In which case, 
     \begin{enumerate}
         \item $f \in \mathcal{L}^0(X; Y)$. 
-        \item For each $A \in \cf$, $f|_A = \limv{n}f_n|A = \limv{n}f_{A, n} = f_A$ almost everywhere. 
+        \item For each $A \in \cf$, $f|_A = \limv{n}f_n|_A = \limv{n}f_{A, n} = f_A$ almost everywhere. 
         \item[(U)] For all $A \in \cf$, $f|_A = g|_A$ almost everywhere. Since $\mu$ is semifinite, $f = g$ almost everywhere. 
     \end{enumerate}
 \end{proof}