Fixed typos.

src/measure/vector/ac.tex

@@ -59,7 +59,7 @@
 
     (1): Let $\eps > 0$. For each $N \in \natp$, let
     \[
-    A_N = \bracs{A \in \cm_0 \bigg | \sup_{n \ge N}\norm{\nu_n(A) - \nu_N(A)} \le \eps}
+    A_N = \bracs{A \in \cm_0 \bigg | \sup_{n \ge N}\norm{\nu_n(A) - \nu_N(A)}_E \le \eps}
     \]
 
     then since $\seq{\nu_n} \subset UC(\cm_0; E)$, $A_N \subset \cm_0$ is closed. By assumption (b), $\cm_0 = \bigcup_{N \in \natp}A_N$. By the \hyperref[Baire Category Theorem]{theorem:baire}, there exists $N \in \natp$, $A \in \cm_0$, and $\delta > 0$ such that for every $m, n \ge N$ and $B \in \cm_0$ with $\mu(A \Delta B) \le \delta$, $\norm{\nu_m(B) - \nu_n(B)}_E \le \eps$. 
@@ -73,11 +73,13 @@
     and $\seq{\mu_n}$ is uniformly absolutely continuous with respect to $\mu$. 
 
     (2): Let $\seq{A_n} \subset \cm$ and $A \in \cm$ with $A = \bigsqcup_{n \in \natp}A_n$, then for each $N \in \natp$, 
-    \[
-        \norm{\nu(A) - \sum_{n = 1}^N \nu(A_n)}_E = 
-        \nu\paren{A \setminus \bigsqcup_{n = 1}^N A_n} =
-        \limv{k}\nu_k\paren{A \setminus \bigsqcup_{n = 1}^N A_n}
-    \]
+    \begin{align*}
+        \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
+    \end{align*}
+    
     
     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 $\nu \ll \mu$.