From 1ec807a9ca14f430ef3d8327af4033e3da183f6a Mon Sep 17 00:00:00 2001
From: Bokuan Li <bokuan.li@mail.mcgill.ca>
Date: Sun, 14 Jun 2026 21:29:15 -0400
Subject: [PATCH] Removed pretentiousness.

---
 src/measure/vector/rn.tex | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/src/measure/vector/rn.tex b/src/measure/vector/rn.tex
index 6473e27..566b9a9 100644
--- a/src/measure/vector/rn.tex
+++ b/src/measure/vector/rn.tex
@@ -37,7 +37,7 @@
 
     then
     \begin{enumerate}
-        \item By definition, $\nu = \nu_a + \nu_s$. 
+        \item Since $X = A \sqcup B$, $\nu = \nu_a + \nu_s$. 
         \item Let $E \in \cm$ with $\mu(E) = 0$, then $(1 - g)|_E = 0$ $\lambda$-almost everywhere. Thus $E \subset B$ modulo a $\lambda$-null set, and
         \[
             \nu_a(E) = \nu(E \cap A) \le \lambda(A \cap B) = 0