Added the Vitali convergence theorem.

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

@@ -3,18 +3,15 @@
 
 \begin{definition}[Convergence in Measure]
 \label{definition:convergence-in-measure}
-    Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a metric space, and $\seq{f_n}$ and $f$ be Borel measurable functions from $X$ to $Y$, then $f_n \to f$ \textbf{in measure} if for every $\eps > 0$,
+    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, and $f$ be a $(\cm, \cb_Y)$-measurable function, then $\fF \to f$ \textbf{in measure} if for each $\eps > 0$, 
     \[
-    \lim_{n \to \infty}\mu(\bracs{d(f_n, f) > \eps}) = 0
+    \lim_{g, \fF}\mu(\bracs{d(f, g) > \eps}) = 0
     \]
 \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 $\seq{f_n}$ be Borel measurable functions from $X$ to $Y$, then $\seq{f_n}$ is \textbf{Cauchy in measure} if for every $\eps > 0$,
-    \[
-    \mu(\bracs{d(f_m, f_n) > \eps}) \to 0 \quad \text{as}\ m, n \to \infty
-    \]
+    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 $f, g \in A$. 
 \end{definition}
 
 \begin{lemma}

src/measure/vector/ac.tex

@@ -38,6 +38,8 @@
 \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 $\nu \in \mathcal{U}$. 
+
+    For any family $\cf$ of measurable functions, $\cf$ is \textbf{uniformly absolutely continuous} with respect to $\mu$ if the measures $\mathcal{U} = \bracs{fd\mu|f \in \cf}$ are uniformly absolutely continuous with respect to $\mu$. 
 \end{definition}
 
 \begin{theorem}[Vitali-Hahn-Saks]