Added the Vitali-Hahn-Saks theorem.

src/measure/measure/fn.tex

@@ -0,0 +1,16 @@
+\section{The Fréchet-Nikodym Metric}
+\label{section:frechet-nikodym}
+
+\begin{definition}[Fréchet-Nikodym Metric]
+\label{definition:frechet-nikodym}
+    Let $(X, \cm, \mu)$ be a finite measure space, then the \textbf{Fréchet-Nikodym metric} with respect to $\mu$ is the mapping
+    \[
+    d: \cm \times \cm \to [0, \infty) \quad (A, B) \mapsto \mu(E \Delta B)
+    \]
+
+    For each $A, B \in \cm$, $A$ and $B$ are \textbf{essentially equal} if $\mu(E \Delta B) = 0$. The set $\cm$ modulo essential equality is a complete metric space. 
+\end{definition}
+\begin{proof}
+    By the completeness of $L^1(X, \cm, \mu; \real)$. 
+\end{proof}
+

src/measure/measure/index.tex

@@ -10,3 +10,4 @@
 \input{./lebesgue-stieltjes.tex}
 \input{./product.tex}
 \input{./kolmogorov.tex}
+\input{./fn.tex}

src/measure/vector/ac.tex

@@ -35,3 +35,52 @@
     so (1) does not hold. 
 \end{proof}
 
+\begin{definition}[Uniformly Absolutely Continuous]
+\label{definition:u-ac}
+    Let $(X, \cm)$ be a measurable space, $\mu$ be a positive measure on $(X, \cm)$, $E$ be a normed space over $K \in \RC$, and $\mathcal{U} \subset M(X, \cm; E)$ be a family of $E$-valued vector measures on $(X, \cm)$, then $\mathcal{U}$ is \textbf{uniformly absolutely continuous} with respect to $\mu$ if for every $\eps > 0$, there exists $\delta > 0$ such that for all $A \in \cm$ with $\mu(A) < \delta$, $\norm{\nu(A)}_E < \eps$ for all $\nu \in \mathcal{U}$. 
+\end{definition}
+
+\begin{theorem}[Vitali-Hahn-Saks]
+\label{theorem:vitali-hahn-saks}
+    Let $(X, \cm, \mu)$ be a measurable space, $\mu$ be a positive measure on $(X, \cm)$, $E$ be a normed vector space over $K \in \RC$, and $\seq{\nu_n} \subset M(X, \cm; E)$ be $E$-valued vector measures on $(X, \cm)$ such that: 
+    \begin{enumerate}[label=(\alph*)]
+        \item For each $n \in \natp$, $\nu$ is absolutely continuous with respect to $\mu$. 
+        \item For each $A \in M(X, \cm; E)$, $\nu(A) = \limv{n}\nu_n(A)$ exists. 
+    \end{enumerate}
+
+    then
+    \begin{enumerate}
+        \item $\seq{\nu_n}$ is uniformly absolutely continuous with respect to $\mu$. 
+        \item The mapping $\nu$ is a $E$-valued vector measure absolutely continuous with respect to $\mu$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem II.2.1]{Yosida}}}. ]
+    Let $\cm_0$ be the euivalence classes of essentially equal sets in $\cm$, equipped with the \hyperref[Fréchet-Nikodym metric]{definition:frechet-nikodym} with respect to $\mu$. For each $n \in \natp$, $\nu_n \ll \mu$ by assumption (a), and $\nu_n \in UC(\cm_0; E)$. 
+
+    (1): Let $\eps > 0$. For each $N \in \natp$, let
+    \[
+    A_N = \bracs{A \in \cm_0 \bigg | \sup_{n \ge N}\norm{\nu_n(A) - \nu_N(A)} \le \eps}
+    \]
+
+    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 $\mu(A \Delta B) \le \delta$, $\norm{\nu_m(B) - \nu_n(B)}_E \le \eps$. 
+
+    Let $B \in \cm$ with $\mu(B) = \mu(\emptyset \Delta B) \le \delta$ and write $B = (A \cup B) \setminus (A \setminus B)$, then $\mu((A \cup B) \Delta B) \le \delta$ and $\mu((A \setminus B) \Delta B) \le \delta$. Thus
+    \begin{align*}
+        \norm{\nu_m(B) - \nu_n(B)}_E &= \norm{(\nu_m - \nu_n)(A \cup B) - (\nu_m - \nu_n)(A \setminus B)}_E \\
+        &\le 2\eps
+    \end{align*}
+
+    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 $N \in \natp$, 
+    \[
+        \norm{\nu(A) - \sum_{n = 1}^N \nu(A_n)}_E = 
+        \nu\paren{A \setminus \bigsqcup_{n = 1}^N A_n} =
+        \limv{k}\nu_k\paren{A \setminus \bigsqcup_{n = 1}^N A_n}
+    \]
+    
+    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}
+
+