diff --git a/src/measure/measure/localisable.tex b/src/measure/measure/localisable.tex
index 5712aaf..7bfb207 100644
--- a/src/measure/measure/localisable.tex
+++ b/src/measure/measure/localisable.tex
@@ -153,11 +153,13 @@
         \item For each $y \in Y$, $f^{-1}(y) = P(y)$, so $f \in \mathcal{L}^0(X; Y)$ is measurable. 
         \item Let $A \in \cf$, then for each $y \in Y$, $\mu(f_A^{-1}(y) \setminus P(y)) = 0$. On the other hand, 
         \begin{align*}
-         (P(y) \cap A) \setminus f_A^{-1}(y) &= P(y) \cap \bigcup_{z \in Y \setminus \bracs{y}}f_A^{-1}(z) \\
-         &\subset \braks{P(y) \cap \bigcup_{z \in Y \setminus \bracs{y}}P(z)} \cup \bigcup_{z \in Y \setminus \bracs{y}}f_A^{-1}(z) \setminus P(z)
+            \mu\braks{(P(y) \cap A) \setminus f_A^{-1}(y)} &\le \sum_{z \in Y \setminus \bracs{y}}\mu(P(y) \cap f_A^{-1}(z) ) \\
+            &\le \sum_{z \in Y \setminus \bracs{y}}\mu(P(y) \cap P(z)) \\
+            &+ \sum_{z \in Y \setminus \bracs{y}}\mu(f_A^{-1}(z) \setminus P(z)) \\
+            &= 0
         \end{align*}
-
-        is a null set. Therefore $f|_A = f_A$ almost everywhere. 
+        
+        so $f|_A = f_A$ almost everywhere on $A$. 
         \item[(U)] For all $A \in \cf$, $f|_A = g|_A$ almost everywhere. Since $\mu$ is semifinite, $f = g$ almost everywhere. 
     \end{enumerate}
     
@@ -182,12 +184,14 @@
         \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 $f_n|_A = f_{A, n}$ almost everywhere, $\limv{n}f_{A, n}$ and $\limv{n}f_{n}|_A$ exist and are equal almost everywhere on $A$. Thus
-    \[
-    \mu\paren{\bracs{\limv{n}f_n \text{ does not exist}} \cap A} \le \mu\bracs{\limv{n}f_{A, n} \text{ does not exist}} + 0 = 0
-    \]
+    Since $Y$ is Polish, \autoref{proposition:metric-measurable-limit} implies that $\bracsn{\limv{n}f_n \text{ exists}} \in \cm$. For each $A \in \cf$, $f_{A, n} \to f_n$ pointwise by (i). For each$n \in \natp$, $f_{A, n} = f_n|_A$ almost everywhere on $A$ by (2). Thus
+    \begin{align*}
+        \bracs{\limv{n}f_n \text{ exists}} \cap A &\subset \bigcap_{n \in \natp}\bracs{f_{A, n} = f_n|_A} \\ 
+        \mu\paren{\bracs{\limv{n}f_n \text{ exists}} \cap A} &= \mu\paren{\bigcap_{n \in \natp}\bracs{f_{A, n} = f_n|_{A}}} = \mu(A) \\
+        \mu\paren{\bracs{\limv{n}f_n \text{ does not exist}} \cap A} &= 0
+    \end{align*}
 
-    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, 
+    As $\mu$ is semifinite, $\mu\bracsn{\limv{n}f_n \text{ does not exist}} = 0$. By (1) and \autoref{proposition:metric-measurable-limit}, 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.