better proof of convergence in L^1 => convergence in measure

inputs/prerequisites.tex

@@ -46,6 +46,7 @@ from the lecture on stochastic.
 \end{definition}
 % TODO Connect to AnaIII
 
+\pagebreak
 \begin{theorem}+
     \label{thm:convergenceimplications}
     \vspace{10pt}
@@ -109,20 +110,14 @@ from the lecture on stochastic.
     $X_n \xrightarrow{L^1} X \implies X_n\xrightarrow{\bP} X$
     \end{claim}
     \begin{subproof}
-    Let $\bE[|X_n - X|] \to  0$.
-    Suppose there exists an $\epsilon > 0$ such that
-    $\limsup\limits_{n \to \infty} \bP[|X_n - X| > \epsilon] = c > 0$.
-    W.l.o.g.~$\lim_{n \to \infty} \bP[|X_n - X| > \epsilon] = c$,
-    otherwise choose an appropriate subsequence.
-    We have
+        Suppose $\bE[|X_n - X|] \to  0$.
+        Then for every $\epsilon > 0$
         \begin{IEEEeqnarray*}{rCl}
-        \bE[|X_n - X|] &=& \int_\Omega |X_n - X | \dif\bP\\
-                       &=& \int_{|X_n - X| > \epsilon} |X_n - X| \dif\bP
-                        + \underbrace{\int_{|X_n - X| \le \epsilon}|X_n-X|\dif\bP}_{\ge 0}\\
-                       &\ge& \epsilon \int_{|X_n -X | > \epsilon} \dif\bP\\
-                       &=& \epsilon \cdot c > 0 \lightning
+            \bP[|X_n - X| \ge \epsilon]
+              &\overset{\text{Markov}}{\ge}& \frac{\bE[|X_n - X|]}{\epsilon}
+              &\xrightarrow{n \to \infty} & 0,
         \end{IEEEeqnarray*}
-    \todo{Improve this with Markov}
+        hence $X_n \xrightarrow{\bP} X$.
     \end{subproof}
     \begin{claim} %+
         $X_n \xrightarrow{\bP} X \implies X_n \xrightarrow{\text{dist}} X$.
@@ -142,13 +137,15 @@ from the lecture on stochastic.
         we have
         $\bP[|X_n - X| > \delta] < \frac{\epsilon}{2}$.
         It is
-        \[|F_n(t) - F(t)|
-            = |\bP[X_n \le t] - F(t)|
-            \le \max(|\frac{\epsilon}{2} + \bP[X \le t + \delta] - F(t)|,
+        \begin{IEEEeqnarray*}{rCl}
+            |F_n(t) - F(t)|
+            &=& |\bP[X_n \le t] - F(t)|\\
+            &\le& \max(|\frac{\epsilon}{2} + \bP[X \le t + \delta] - F(t)|,
                 |\bP[X \le  t -\delta] - F(t)|)\\
-            \le \max(|\frac{\epsilon}{2} + F(t + \delta) - F(t)|, |F(t-\delta) -F(t)|
-            \le \epsilon,
-        \]
+            &\le& \max(|\frac{\epsilon}{2} + F(t + \delta) - F(t)|,
+               |F(t-\delta) -F(t)|)\\
+            &\le& \epsilon,
+        \end{IEEEeqnarray*}
         hence $F_n(t) \to F(t)$.
     \end{subproof}
     \begin{claim}