Removed pretentiousness.

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