MAJOR RETRACTION IN UNIFORMITY DEFINING PROPOSITION.

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

@@ -3,8 +3,92 @@
 
 \begin{definition}[In Measure]
 \label{definition:in-measure}
-    Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a metric space, then the uniform
+    Let $(X, \cm, \mu)$ be a measure space and $(Y, d)$ be a metric space. For each $\eps, \delta > 0$, let
+    \[
+    U(\delta, \eps) = \bracs{(f, g) \in \mathscr{M}(X; Y)| \mu\bracs{d(f, g) > \delta} < \eps}
+    \]
+
+    then
+    \[
+    \fB = \bracs{U(\delta, \eps)|\eps, \delta > 0}
+    \]
+
+    forms a fundamental system of entourages for a uniformity. The uniformity induced by $\fB$ is the \textbf{uniform structure of convergence in measure} on $\mathscr{M}(X; Y)$.
 \end{definition}
+\begin{proof}
+    It is sufficient to check the conditions of \autoref{proposition:fundamental-entourage-criterion}:
+    \begin{enumerate}
+        \item[(FB1)] For each $\eps, \eps', \delta, \delta' > 0$,  
+        \[
+        U(\delta \wedge \delta', \eps \wedge \eps') \subset U(\delta, \eps) \cap U(\delta', \eps')
+        \]
+        \item[(UB3)] For each $\eps, \delta > 0$ and $f, g, h \in \mathscr{M}(X; Y)$, 
+        \[
+        \bracs{d(f, h) > \delta} \subset \bracs{d(f, g) > \delta} \cup \bracs{d(g, h) > \delta}
+        \]
+
+        so $U(\delta/2, \eps/2) \circ U(\delta/2, \eps/2) \subset U(\delta, \eps)$. 
+    \end{enumerate}
+\end{proof}
+
+\begin{definition}[Ky Fan Metric]
+\label{definition:ky-fan}
+    Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a metric space, and
+    \[
+    \alpha: \mathscr{M}(X; Y)^2 \to [0, \infty) \quad (f, g) \mapsto \inf\bracs{\eps > 0| \mu\bracs{d(f, g) > \eps} \le \eps}
+    \]
+
+    then:
+    \begin{enumerate}
+        \item $\alpha$ is a metric on $\mathscr{M}(X; Y)$ modulo almost everywhere equality. 
+        \item $\alpha$ induces the uniform structure of convergence in measure on $\mathscr{M}(X; Y)$. 
+    \end{enumerate}
+
+    The mapping $\alpha$ is the \textbf{Ky Fan metric} on $\mathscr{M}(X; Y)$. 
+\end{definition}
+\begin{proof}
+    (1): Let $f, g, h \in \mathscr{M}(X; Y)$, then
+    \begin{enumerate}
+        \item[(M)] If $\alpha(f, g) = 0$, then by \hyperref[continuity from above]{proposition:measure-properties}, 
+        \[
+        \mu\bracs{d(f, g) > 0} = \limv{n}\mu\bracs{d(f, g) > 1/n} \le \limv{n}\frac{1}{n} = 0
+        \]
+
+        so $f = g$ almost everywhere. 
+        \item[(PM3)] For each $\eps > 0$, 
+        \[
+        \bracs{d(f, h) > \eps} \subset \bracs{d(f, g) > \eps/2} \cup \bracs{d(g, h) > \eps/2}
+        \]
+
+        so $\alpha(f, h) \le \alpha(f, g) + \alpha(g, h)$. 
+    \end{enumerate}
+
+    so $\alpha$ is a metric on $\mathscr{M}(X; Y)$, modulo almost everywhere equality. 
+
+    (2): Let $f, g \in \mathscr{M}(X; Y)$. For any $\eps, \delta > 0$, if $\alpha(f, g) < \eps \wedge \delta$, then there exists $r \in (0, \eps \wedge \delta]$ such that $\mu\bracs{d(f, g) > r} \le r$. Thus
+    \[
+    \bracs{(f, g) \in \mathscr{M}(X; Y)|\alpha(f, g) < \eps \wedge \delta} \subset 
+    \bracs{(f, g) \in \mathscr{M}(X; Y)|\mu\bracs{d(f, g) >\delta} < \eps}
+    \]
+
+    On the other hand, if $\mu\bracs{d(f, g) > \eps} \le \eps$, then $d(f, g) \le \eps$. Therefore $\alpha$ induces the uniform structure of convergence in measure. 
+\end{proof}
+
+\begin{definition}[Locally In Measure]
+\label{definition:locally-in-measure}
+    Let $(X, \cm, \mu)$ be a measure space and $(Y, d)$ be a metric space. For each $\eps, \delta > 0$ and $A \in \cm$ with $\mu(A) < \infty$, let
+    \[
+    U(A, \delta, \eps) = \bracs{(f, g) \in \mathscr{M}(X; Y)| \mu(A \cap \bracs{d(f, g) > \delta}) < \eps}
+    \]
+
+    then
+    \[
+    \fB = \bracs{U(A, \delta, \eps)|\eps, \delta > 0, A \in \cm, \mu(A) < \infty}
+    \]
+
+    forms a fundamental system of entourages for a uniformity. The uniformity induced by $\fB$ is the \textbf{uniform structure of local convergence in measure} on $\mathscr{M}(X; Y)$.
+\end{definition}
+