Bounded convergence theorem

inputs/lecture_19.tex

@@ -136,10 +136,10 @@ However, some subsets can be easily described, e.g.
     Let $Y = \bE[X | \cG]$ for some sub-$\sigma$-algebra $\cG$.
     Then, by \autoref{cjensen}, $|Y| \le  \bE[ |X| | \cG]$.
     Hence $\bE[|Y|] \le  \bE[|X|]$.
-    It follows that $\bP[|Y| > k] < \delta$
-    for $k$ suitably large,
-    since $\bE[|X|] \le \infty$.
-    Note that $\{Y > k\}  \in \cG$.
+    By Markov's inequality,
+    it follows that $\bP[|Y| > k] < \delta$
+    for $k > \frac{\bE[|X|]}{\delta}$.
+    Note that $\{|Y| > k\}  \in \cG$.
     We have
     \begin{IEEEeqnarray*}{rCl}
         \bE[|Y| \One_{\{|Y| > k\} }] &<& \epsilon
@@ -200,7 +200,7 @@ However, some subsets can be easily described, e.g.
     \end{IEEEeqnarray*}
     for all $\delta > 0$ and suitable $k$.
 
-    Hence $\bP[|X_n| \ge k] < \delta$ by Markov's inequality.
+    Hence $\bP[|X_n| > k] < \delta$ by Markov's inequality.
     Then by \autoref{lec19f4} part (a) it follows that
     \[
     \int_{|X_n| > k} |X_n| \dif \bP \le  \underbrace{\int |X - X_n| \dif \bP}_{< \epsilon} + \int_{|X_n| > k} |X| \dif \bP \le  2 \epsilon.
@@ -232,7 +232,7 @@ Let $(\Omega, \cF, \bP)$ as always and let $(\cF_n)_n$ always be a filtration.
 \begin{theorem}
     \label{ceismartingale}
     Let $X \in L^p$ for some $p \ge 1$
-    and $\bigcup_n \cF_n \to \cF$.
+    and $\bigcup_n \cF_n \to \cF \supseteq \sigma(X)$.
     Then $X_n \coloneqq  \bE[X | \cF_n]$ defines a martingale which converges
     to $X$ in $L^p$.
 \end{theorem}

inputs/lecture_20.tex

@@ -45,7 +45,8 @@
         \int | X - X'|^p \dif \bP &=& \int_{\{|X| > M\} } |X|^p \dif \bP \xrightarrow{M \to \infty} 0
     \end{IEEEeqnarray*}
     as $\bP$ is regular,
-    i.e.~$\forall \epsilon > 0 . ~\exists  k . ~\bP[|X|^p \in [-k,k] \ge  1-\epsilon$.
+    i.e.~$\forall \epsilon > 0 . ~\exists  k . ~
+      \bP[|X|^p \in [-k,k]] \ge  1-\epsilon$.
 
     Take some $\epsilon > 0$ and $M$ large enough such that
     \[
@@ -111,7 +112,7 @@ we need the following theorem, which we won't prove here:
     \int_A X \dif \bP
     &=& \lim_{k \to \infty} \int_A X_{n_k} \dif \bP\\
     &=& \lim_{k \to \infty} \bE[X_{n_k} \One_A]\\
-    &\overset{\text{for }n_k \ge m}{=}& \int_{k \to \infty} \bE[X_m \One_A].
+    &\overset{\text{for }n_k \ge m}{=}& \lim_{k \to \infty} \bE[X_m \One_A].
     \end{IEEEeqnarray*}
     Hence $X_n = \bE[X | \cF_m]$ by the uniqueness of conditional expectation
     and by \autoref{ceismartingale},

inputs/prerequisites.tex

@@ -215,6 +215,23 @@ from the lecture on stochastic.
     \end{subproof}
 \end{refproof}
 
+\begin{theorem}[Bounded convergence theorem]
+    \label{thm:boundedconvergence}
+    Suppose that $X_n \xrightarrow{\bP} X$ and there exists
+    some $K$ such that $|X_n| \le K$ for all $n$.
+    Then $X_n \xrightarrow{L^1} X$.
+\end{theorem}
+\begin{proof}
+    Note that $|X| \le K$ a.s.~since
+    \[\bP[|X| \ge  K + \epsilon] \le \bP[|X_n - X| > \epsilon]
+        \xrightarrow{n \to \infty} 0.\]
+    Hence
+    \begin{IEEEeqnarray*}{rCl}
+        \int |X_n - X| \dif \bP &\le& \int_{|X_n - X| \ge \epsilon} |X_n - X| \dif \bP + \epsilon\\
+                                &\le &2K\bP[|X_n - X| \ge \epsilon] +\epsilon.
+    \end{IEEEeqnarray*}
+\end{proof}
+
 
 \subsection{Some Facts from Measure Theory}
 \begin{fact}+[Finite measures are {\vocab[Measure]{regular}}, Exercise 3.1]