Fixed a handful of typos.

src/fa/interpolation/complex.tex

@@ -40,12 +40,12 @@
         \item The mapping $C_\theta$ defined by $(E_0, E_1) \mapsto [E_0, E_1]_\theta$ is an interpolation functor of exact exponent $\theta$. 
     \end{enumerate}
 
-    and functor $C_\theta$ is the \textbf{method of complex interpolation}. 
+    and the functor $C_\theta$ is the \textbf{method of complex interpolation}. 
 \end{definition}
 \begin{proof}[Proof, {{\cite[Theorem 4.1.2]{BerghInterpolation}}}. ]
     (1): Let $\seq{x_n} \subset [E_0, E_1]_\theta$ with $\sum_{n \in \natp}\norm{x_n}_{[E_0, E_1]_\theta} < \infty$, then there exists $\seq{f_n} \subset \cf(E_0, E_1)$ such that for each $n \in \natp$, $f_n(\theta) = x_n$ and $\norm{f_n}_{\cf(E_0, E_1)} \le 2\norm{x_n}_{[E_0, E_1]_\theta}$. Since $\cf(E_0, E_1)$ is complete, there exists $f \in \cf(E_0, E_1)$ such that $f = \sum_{n = 1}^\infty f_n$. Let $x = f(\theta)$, then since $\sum_{n = 1}^N f_n \to f$ in $\cf(E_0, E_1)$ as $N \to \infty$, $\sum_{n = N}^\infty x_n \to x$ in $[E_0, E_1]_\theta$ as $N \to \infty$. Therefore $[E_0, E_1]_\theta$ is a Banach space by \autoref{lemma:banach-criterion}. 
 
-    For any $x \in E_0 \cap E_1$ and $\delta > 0$, let $f_\delta(z) = x_0 e^{(z - \theta)^2}$, then $f_\delta \in \cf(E_0, E_1)$ with $\norm{f_\delta}_{\cf(E_0, E_1)} \le e^\delta\norm{x}_{E_0 \cap E_1}$. Thus $x \in [E_0, E_1]_\theta$ with
+    For any $x \in E_0 \cap E_1$ and $\delta > 0$, let $f_\delta(z) = x_0 e^{\delta(z - \theta)^2}$, then $f_\delta \in \cf(E_0, E_1)$ with $\norm{f_\delta}_{\cf(E_0, E_1)} \le e^\delta\norm{x}_{E_0 \cap E_1}$. Thus $x \in [E_0, E_1]_\theta$ with
     \[
     \norm{x}_{[E_0, E_1]_\theta} \le \norm{f}_{\cf(E_0, E_1)} \le e^\delta\norm{x}_{E_0 \cap E_1}
     \]

src/measure/vector/ac.tex

@@ -88,7 +88,7 @@
     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 $\nu \ll \mu$. 
+    Absolute continuity is equivalent to uniform continuity as a mapping $\cm_0 \to E$, and uniform absolute continuity is equivalent to uniform equicontinuity as mappings $\cm_0 \to E$. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, $\nu \in UC(\cm_0; E)$, so $\nu \ll \mu$. 
     
 \end{proof}