Added missing justification.

src/measure/vector/ac.tex

@@ -77,11 +77,18 @@
         \norm{\nu(A) - \sum_{n = 1}^N \nu(A_n)}_E &= 
         \norm{\nu\paren{A \setminus \bigsqcup_{n = 1}^N A_n}}_E \\
         &=
-        \norm{\limv{k}\nu_k\paren{A \setminus \bigsqcup_{n = 1}^N A_n}}_E
+        \limv{k}\norm{\nu_k\paren{A \setminus \bigsqcup_{n = 1}^N A_n}}_E
     \end{align*}
     
+    Since $\seq{\nu_k}$ is uniformly absolutely continuous, 
+    \[
+    \lim_{N \to \infty}\sup_{k \in \natp}\norm{\nu_k\paren{A \setminus \bigsqcup_{n = 1}^N A_n}}_E = 0
+    \]
     
-    so $\nu$ is a vector measure. Absolute continuity is equivalent to uniform continuity as a mapping $\cm_n \to E$, and uniform absolute continuity is equivalent to uniform equicontinuity as mappings $\cm \to E$. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, $\nu \in UC(\cm_0; E)$, so $\nu \ll \mu$. 
+    so $\nu$ is a vector measure. 
+    
+    
+    Absolute continuity is equivalent to uniform continuity as a mapping $\cm_n \to E$, and uniform absolute continuity is equivalent to uniform equicontinuity as mappings $\cm \to E$. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, $\nu \in UC(\cm_0; E)$, so $\nu \ll \mu$. 
     
 \end{proof}