s23-probability-theory/inputs/lecture_20.tex

\begin{refproof}{ceismartingale}
    By the tower property (\autoref{cetower})
    it is clear that $(\bE[X | \cF_n])_n$
    is a martingale.
    
    First step:
    Assume that $X$ is bounded.
    Then, by \autoref{cejensen}, $|X_n| \le  \bE[|X| | \cF_n]$,
    hence $\sup_{\substack{n \in \N \\ \omega \in  \Omega}} | X_n(\omega)| < \infty$.
    Thus $(X_n)_n$ is a martingale in $L^{\infty} \subseteq L^2$.
    By the convergence theorem for martingales in $L^2$ % TODO REF
    there exists a random variable $Y$,
    such that $X_n \xrightarrow{L^2} Y$.

    Fix $m \in \N$ and $A \in \cF_m$.
    Then
    \begin{IEEEeqnarray*}{rCl}
        \int_A Y \dif \bP
        &=& \lim_{n \to \infty} \int_A X_n \dif \bP\\
        &=& \lim_{n \to \infty} \bE[X_n \One_A]\\
        &=& \lim_{n \to \infty} \bE[\bE[X | \cF_n] \One_A]\\
        &\overset{A \in \cF_n}{=}& \lim_{\substack{n \to \infty\\n \ge m}} \bE[X \One_A]\\
    \end{IEEEeqnarray*}
    Hence $\int_A Y \dif \bP = \int_A X \dif \bP$ for all $m \in \N, A \in  \cF_m$.
    Since $\cF = \sigma\left( \bigcup \cF_n \right)$
    this holds for all $A \in  \cF$.
    Hence $X = Y$ a.s., so $X_n \xrightarrow{L^2} X$.
    Since $(X_n)_n$ is uniformly bounded, this also means
    $X_n \xrightarrow{L^p}  X$.


    Second step:
    Now let $X \in  L^p$ be general and define
    \[
    X'(\omega) \coloneqq  \begin{cases}
        X(\omega)& \text{ if }  |X(\omega)| \le  M,\\
        0&\text{ otherwise}
    \end{cases}
    \]
    for some $M  > 0$.
    Then $X' \in  L^\infty$ and
    \begin{IEEEeqnarray*}{rCl}
        \int | X - X'|^p \dif \bP &=& \int_{\{|X| > M\} } |X|^p \dif \bP \xrightarrow{M \to \infty} 0
    \end{IEEEeqnarray*}
    as $\bP$ is \vocab{regular}, \todo{Definition?}
    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
    % \[
    % \int |X - X'| \dif \bP < \epsilon.
    % \]

    % Let $(X_n')_n$ be the martingale given by $(\bE[X' | \cF_n])_n$.
    % Then $X_n' \xrightarrow{L^p} X'$ by the first step.

    % It is
    % \begin{IEEEeqnarray*}{rCl}
    %     \|X_n - X_n'\|_{L^p}^p  &=& \bE[\bE[X - X' | \cF_n]^{p}]\\
    %                             &\overset{\text{Jensen}}{\le}& \bE[\bE[(X- X')^p | \cF_n]\\
    %                             &=& \|X - X'\|_{L^p}^p\\
    %                             &<& \epsilon.
    % \end{IEEEeqnarray*}

    Hence
    \[
    \|X_n - X\|_{L^p} \le |X_n - X_n'|_{L^p} + |X_n' - X'|_{L^p} +  | X - X'|_{L^p} \le 3 \epsilon.
    \]
    Thus $X_n \xrightarrow{L^p} X$.
\end{refproof}

For the proof of \autoref{martingaleisce},
we need the following theorem, which we won't prove here:
\begin{theorem}[Banach Alaoglu]
    \label{banachalaoglu}
    Let $X$ be a  normed vector space and $X^\ast$ its
    continuous dual.
    Then the closed unit ball in $X^\ast$ is compact
    w.r.t.~the ${\text{weak}}^\ast$ topology.
\end{theorem}
\begin{fact}
    We have $L^p \cong (L^q)^\ast$ for $\frac{1}{p} + \frac{1}{q} = 1$
    via
    \begin{IEEEeqnarray*}{rCl}
        L^p &\longrightarrow & (L^q)^\ast \\
        f &\longmapsto & (g \mapsto \int g f \dif d\bP)
    \end{IEEEeqnarray*}
    
    We also have $(L^1)^\ast \cong L^\infty$,
    however $ (L^\infty)^\ast \not\cong L^1$.
\end{fact}

\begin{refproof}{martingaleisce}
    Since $(X_n)_n$ is bounded in $L^p$, by \autoref{banachalaoglu},
    there exists $X \in L^p$ and a subsequence
    $(X_{n_k})_k$ such that for all $Y \in L^q$ ($\frac{1}{p} + \frac{1}{q} = 1$ )
    \[
    \int X_{n_k} Y \dif \bP \to  \int XY \dif \bP
    \] 
    (Note that this argument does not work for $p = 1$,
    because $(L^\infty)^\ast \not\cong  L^1$).

    Let $A \in  \cF_m$ for some fixed $m$ and write
    $Y = \One_A$.
    Then
    \begin{IEEEeqnarray*}{rCl}
    \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].
    \end{IEEEeqnarray*}
    Hence $X_n = \bE[X | \cF_m]$ by the uniqueness of conditional expectation
    and by \autoref{ceismartingale},
    we get the convergence.
    
\end{refproof}