Slight adjustments.

src/fa/lp/ui.tex

@@ -144,7 +144,7 @@
 \label{corollary:dct-filter}
     Let $(X, \cm, \mu)$ be a measure space, $p \in [1, \infty)$, $E$ be a normed vector space over $K \in \RC$, $\fF \subset 2^{L^p(X; E)}$ be a filter, and $g, h \in L^p(X; \real)$ such that:
     \begin{enumerate}[label=(\alph*)]
-        \item $\fF \to g$ pointwise and locally in measure.
+        \item $\fF \to g$ locally in measure.
         \item There exists $F \in \fF$ such that $|f| \le h$ for all $f \in F$. 
     \end{enumerate}
     
@@ -156,4 +156,3 @@
 \begin{proof}
     By \autoref{theorem:vitali-convergence}. 
 \end{proof}
-

src/measure/measurable-maps/in-measure.tex

@@ -18,6 +18,7 @@
     and $f_\alpha \to f$ \textbf{locally in measure} if $f_\alpha \to f$ in measure on every set of finite measure. 
 \end{definition}
 
+
 \begin{definition}[Cauchy in Measure]
 \label{definition:cauchy-in-measure}
     Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a metric space, and $\fF$ be a filter of $(\cm, \cb_Y)$-measurable functions, then $\fF$ is \textbf{Cauchy in measure} if for every $\eps, \delta > 0$, there exists $A \in \fF$ such that $\mu(\bracs{d(f, g) > \delta}) < \eps$ for all $f, g \in A$. 
@@ -25,6 +26,35 @@
     Alternatively, if $\net{f}$ is a net of $(\cm, \cb_Y)$-measurable functions, then $\net{f}$ is \textbf{Cauchy in measure} if for every $\eps, \delta > 0$, there exists $\alpha_0 \in A$ such that for each $\alpha, \beta \in A$ with $\alpha, \beta \ge \alpha_0$, $\mu(\bracs{d(f_\alpha, f_\beta) > \delta}) < \eps$. 
 \end{definition}
 
+\begin{proposition}
+\label{proposition:convergence-in-measure}
+    Let $(X, \cm, \mu)$ be a semifinite measure space, $(Y, d)$ be a metric space, and $\fF$ be a filter of $(\cm, \cb_Y)$-measurable functions, then $\fF$ is Cauchy in measure if and only if:
+    \begin{enumerate}
+        \item[(L)] $\fF$ is locally Cauchy in measure. 
+        \item[(T)] For each $\eps, \delta > 0$, there exists $F \in \fF$ and $A \in \cm$ with $\mu(A) < \infty$ such that
+        \[
+        \sup_{f, g \in F}\mu(A^c \cap \bracs{d(f, g) > \delta}) < \eps
+        \]
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (L) + (T) $\Rightarrow$ (In Measure): Let $\eps, \delta > 0$. By (T) then there exists $F_1 \in \fF$ and $A \in \cm$ with $\mu(A) < \infty$ such that 
+    \[
+    \sup_{f, g \in F_1}\mu(A^c \cap \bracs{d(f, g) > \delta}) < \eps
+    \]
+
+    By (L), there exists $F_2 \in \fF$ with $F_2 \subset F_1$ such that
+    \[
+    \sup_{f, g \in F_2}\mu(A \cap \bracs{d(f, g) > \delta}) < \eps
+    \]
+
+    Therefore
+    \[
+    \sup_{f, g \in F_2}\mu\bracs{d(f, g) > \delta} < 2\eps
+    \]
+\end{proof}
+
+
 \begin{lemma}
 \label{lemma:ae-in-measure}
     Let $(X, \cm, \mu)$ be a finite measure space, $(Y, d)$ be a metric space, and $\seq{f_n}$ and $f$ be Borel measurable functions from $X$ to $Y$ such that $f_n \to f$ almost everywhere, then $f_n \to f$ in measure.

src/measure/radon/radon.tex

@@ -51,7 +51,7 @@
 
 \end{proof}
 
-\begin{proposition}[{{\cite[Proposition 7.5]{Folland}}}]
+\begin{proposition}
 \label{proposition:radon-regular-sigma-finite}
     Let $X$ be a LCH space and $\mu: \cb_X \to [0, \infty]$ be a Radon measure, then
     \begin{enumerate}
@@ -59,7 +59,7 @@
         \item If $X$ is $\sigma$-compact or $\mu$ is $\sigma$-finite, then $\mu$ is regular.
     \end{enumerate}
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[Proposition 7.5]{Folland}}}. ]
     (1): Let $E \in \cb_X$ with $\mu(E) < \infty$ and $\eps > 0$, then there exists
     \begin{itemize}
     \item $U \in \cn^o(E)$ with $\mu(U \setminus E) < \eps/2$.
@@ -85,7 +85,7 @@
 
 \end{proof}
 
-\begin{proposition}[{{\cite[Proposition 7.7]{Folland}}}]
+\begin{proposition}
 \label{proposition:radon-measurable-description}
     Let $X$ be a LCH space, $\mu: \cb_X \to [0, \infty]$ be a $\sigma$-finite Radon measure, and $E \in \cb_X$, then
     \begin{enumerate}
@@ -93,7 +93,7 @@
         \item There exists a $F_\sigma$ set $A$ and a $G_\delta$ set $B$ such that $A \subset E \subset B$ and $\mu(B \setminus A) = 0$.
     \end{enumerate}
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[Proposition 7.7]{Folland}}}. ]
     (1): Let $\seq{E_n} \subset \cb_X$ such that $\mu(E_n) < \infty$. By outer regularity, there exists $\seq{U_n}$ open such that $U_n \in \cn^o(E_n)$ and $\mu(U_n) < \mu(E_n) + \eps/2^n$ for all $n \in \natp$. In which case,
     \[
     \mu\paren{\bigcup_{n \in \natp}U_n \setminus E} \le \sum_{n \in \natp}\mu(U_n \setminus E_n) < \eps
@@ -194,7 +194,7 @@
         \item If $E = \complex$ and $f$ is bounded, then $\phi$ can be taken such that $\norm{\phi}_u \le \norm{f}_u$.
     \end{enumerate}
 \end{theorem}
-\begin{proof}[Proof {{\cite[Theorem 7.10]{Folland}}}. ]
+\begin{proof}[Proof, {{\cite[Theorem 7.10]{Folland}}}. ]
     First assume that $f$ is bounded.
 
     (1, bounded): If $f$ is bounded, then $f \in L^1(X; E)$. By \autoref{proposition:radon-cc-dense}, there exists $\seq{\phi_n} \subset C_c(X)$ such that $\phi_n \to f$ in $L^1(\mu)$. Since $\phi_n \to f$ in $L^1(\mu)$, $\phi_n \to f$ in measure by \autoref{proposition:lp-in-measure}. By taking a subsequence using \ref{proposition:cauchy-in-measure-limit}, assume without loss of generality that $\phi_n \to f$ almost everywhere. 
@@ -226,14 +226,14 @@
     (2, unbounded): By (1) applied to $g$, there exists $\phi \in C_c(X; \complex)$ such that $\mu\bracs{\phi \ne g} < \eps/2$, and $\mu(\bracs{\phi \ne f}) < \eps$.
 \end{proof}
 
-\begin{proposition}[Monotone Convergence Theorem for Lower Semicontinuous Functions]
+\begin{proposition}[Monotone Convergence Theorem (LSC)]
 \label{proposition:mct-radon}
     Let $X$ be a LCH space, $\net{f}$ and $f: X \to [0, \infty]$ be non-negative lower semicontinuous functions such that $f_\alpha \upto f$, then for any Radon measure $\mu$ on $X$, 
     \[
     \int f d\mu = \sup_{\alpha \in A}\int f_\alpha d\mu
     \]
 \end{proposition}
-\begin{proof}[Proof {{\cite[Proposition 7.12]{Folland}}}. ]
+\begin{proof}[Proof, {{\cite[Proposition 7.12]{Folland}}}. ]
     Assume without loss of generality that $\int f d\mu < \infty$. By \autoref{proposition:semicontinuous-properties}, $f$ is Borel measurable, so $f \ge f_\alpha$ for all $\alpha \in A$ implies that
     \[
     \int f d\mu \ge \sup_{\alpha \in A}\int f_\alpha d\mu