Incremental update.

src/measure/measure/outer.tex

@@ -148,9 +148,9 @@
     \end{align*}
     so $\nu(E) = \mu(E)$ whenever $\mu(E) < \infty$.
 
-    Finally, if $\mu$ is $\sigma$-finite, then there exists $\seq{E_n} \subset \cm$ such that $E_n \upto X$ and $\mu(E_n) < \infty$ for all $n \in \natp$. For each $n \in \natp$, there exists $\bracsn{E_{n, k}}_{k = 1}^\infty \subset \alg$ such that $E_n \subset \bigcup_{k \in \natp}E_{n, k}$ and $\mu\paren{\bigcup_{k \in \natp}E_{n, k}} < \infty$. Let $F_n = \bigcup_{j = 1}^n \bigcup_{k \in \natp}E_{n, k}$, then $\seq{F_n} \subset \sigma(\alg) \subset \cm \cap \cn$ with $F_n \upto X$ and $\mu(F_n) < \infty$ for all $n \in \nat$. For any $E \in \cm \cap \cn$, by continuity from below (\ref{proposition:measure-properties}),
+    Finally, if $\mu$ is $\sigma$-finite, then there exists $\seq{E_n} \subset \cm$ such that $E_n \upto X$ and $\mu(E_n) < \infty$ for all $n \in \natp$. For each $n \in \natp$, there exists $\bracsn{E_{n, k}}_{k = 1}^\infty \subset \alg$ such that $E_n \subset \bigcup_{k \in \natp}E_{n, k}$ and $\mu\paren{\bigcup_{k \in \natp}E_{n, k}} < \infty$. Let $F_n = \bigcup_{j = 1}^n \bigcup_{k \in \natp}E_{j, k}$, then $\seq{F_n} \subset \sigma(\alg) \subset \cm \cap \cn$ with $F_n \upto X$ and $\mu(F_n) < \infty$ for all $n \in \natp$. For any $E \in \cm \cap \cn$, by continuity from below (\ref{proposition:measure-properties}),
     \[
     \nu(E) = \limv{n}\nu(E \cap F_n) = \limv{n}\mu(E \cap F_n) = \mu(E)
     \]
-    so $\nu_{\cm \cap \cn} = \mu_{\cm \cap \cn}$.
+    so $\nu|_{\cm \cap \cn} = \mu|_{\cm \cap \cn}$.
 \end{proof}

src/measure/measure/regular.tex

@@ -23,7 +23,7 @@
 \end{definition}
 
 \begin{theorem}[{{\cite[Theorem 7.8]{Folland}}}]
-\label{theorem:sigma-finite-regular-measure}
+\label{theorem:sigma-compact-regular-measure}
     Let $X$ be a topological space and $\mu: \cb_X \to [0, \infty]$ be a Borel measure. If:
     \begin{enumerate}
         \item[(a)] $X$ is a LCH space.