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\begin{proposition*}
  Let $(X_n)$ be independent with $\EE X_n = \mu$
  and $\EE X_n^4 \le M$ for all $n$. Then
  \[ \frac{S_n}{n} \to \mu \]
  \gls{al_surely} as $n \to \infty$ (where $S_n =
  X_1 + \cdots + X_n$).
\end{proposition*}

\begin{proof}
  Let $X_n' = X_n - \mu$. Then
  \[ \EE X_n'^4 \le 2^4(\EE X_n^4 + \mu^4) \le
  \ub{2^4(M + \mu^4)}_{=M'} \qquad \forall n .\]
  So we assume that $\mu = 0$ (without loss of
  generality). Then using independence, for
  distinct indices $i, j, k, l$,
  \[ 0 = \EE(X_i X_j^3) = \EE(X_i X_j X_k^2) =
  \EE(X_i X_j X_k X_l) \]
  Hence,
  \begin{align*}
    \EE(S_n^4)
    &= \EE(X_1 + \cdots + X_n)^4 \\
    &= \EE \left( \sum_{1 \le i \le n} X_i^4 + 6
    \sum_{1 \le i < j \le n} X_i^2 X_j^2 \right)
    \\
    &\le nM + 6\frac{n(n - 1)}{2} \\
    &\le nM + 3n(n - 1) \\
    &\le 3n^2 M
  \end{align*}
  But
  \[ \EE(X_i^2 X_j^2) \stackrel{\text{C-S}}{\le} \sqrt{\EE X_i^4 X_j^4}
  \le M ,\]
  i.e.
  \[ \EE \left( \left( \frac{S_n}{n} \right)^4
  \right) \le \frac{3M}{n^2} \]
  Now,
  \[ \EE \left( \sum_{n = 1}^\infty \left(
  \frac{S_n}{n} \right)^4 \right)
  \stackrel{\text{\nameref{mconv_thm}}}{=}
  \sum_{n = 1}^\infty \EE \left( \frac{S_n}{n}
  \right)^4 \le \sum_n \frac{3M}{n^2} < \infty .\]
  So,
  \[ \sum_{n = 1}^\infty \left( \frac{S_n}{n}
  \right)^4 < \infty \]
  \gls{al_surely}. Hence $\left( \frac{S_n}{n}
  \right)^4 \to 0$ \gls{al_surely}, i.e.
  $\frac{S_n}{n} \to 0$ \gls{al_surely}.
\end{proof}

\newpage
\section{Ergodic Theory}

\begin{flashcard}[measure-preserving-defn]
\begin{definition*}[Measure preserving map]
  \glsadjdefn{mp}{measure preserving}{map}
  \cloze{Let $(E, \mathcal{E}, \mu)$ be a $\sigma$-finite measure
  space. A measurable map $\theta : E \to E$ is
  called ($\mu$-)measure-preserving (m.p.) if
  \[ \mu \circ \theta^{-1} = \mu ,\]
  i.e. $\mu(\theta^{-1}(A)) = \mu(A)$ for all $A
  \in \mathcal{E}$.
  In this case, for all $f \in \Lp[1]$,
  \[ \int_E f \dd \mu = \int_E f \circ \theta \dd
  \mu = \int f \dd \mu \circ \theta^{-1} \]}
\end{definition*}
\end{flashcard}

\begin{flashcard}[theta-invariant]
\begin{definition*}[$\theta$-invariant]
  \glsadjdefn{tinv}{$\theta$-invariant}{set}
  \cloze{A set $A \in \mathcal{E}$ is
  \emph{$\theta$-invariant} if $\theta^{-1}(A) =
  A$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[theta-invariant-f-defn]
\begin{definition*}[$\theta$-invariant function]
  \glsadjdefn{tinv_fn}{$\theta$-invariant}{function}
  \cloze{A measurable function $f$ is
  \emph{$\theta$-invariant} if $f = f \circ
  \theta$.}
\end{definition*}
\end{flashcard}
\vspace{-1em}

The space of all \gls{tinv} sets
$\mathcal{E}_\theta$ is a \sigalg{} and $f$ is
\gls{tinv_fn} if and only if $f$ is
$\mathcal{E}_\theta$-measurable (exercise on \es{4}).

\begin{flashcard}[ergodic-map-defn]
\begin{definition*}[Ergodic map]
  \glsadjdefn{ergodic}{ergodic}{map}
  \cloze{The map $\theta$ is called \emph{ergodic} if
  $\mathcal{E}_\theta$ is $\mu$-trivial, i.e.
  $\forall A \in \mathcal{E}_\theta$, $\mu(A) = 0$
  or $\mu(A^c) = 0$. ``well-mixed''.}
\end{definition*}
\end{flashcard}
\vspace{-1em}

Boltzman (1880) Ergodic hypothesis for dynamical
systems
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/25f3edf71d4947c4.png}
\end{center}
``space filling''. For Markov chains, ergodicity
$\iff$ irreducibility.

\textbf{Fact:} If $f : E \to \RR$ is \gls{tinv_fn}, $\theta$
Is \gls{ergodic} if and only if $f = c$, a
constant, \gls{al_surely} (exercise on \es{4}).
($\mu(f^{-1}(-\infty, x)) = 0$ or $\mu(f^{-1}[x,
\infty)) = 0$).

\begin{example*}
  On $((0, 1], \mathcal{B}, \lambda_{(0, 1]})$ the
  maps
  \begin{enumerate}[(1)]
    \item $\theta_a(x) = x + a \pmod{1}$ (rotation
      of a circle).
    \item $\theta(x) = 2x \pmod{1}$
  \end{enumerate}
  are \gls{mp} and \gls{ergodic} unless $a \in
  \QQ$ (see \es{4}).
\end{example*}

\begin{theorem*}[Birkhoff's ergodic theorem (1931)]
  SUppose $(E, \mathcal{E}, \mu)$ is
  $\sigma$-finite and $f \in \Lp[1](E,
  \mathcal{E}, \mu)$ and $\theta : E \to E$
  \gls{mp}. Define $S_0 = 0$ and
  \[ S_n = S_n(f) = f
  + f \circ \theta + f \circ \theta^2 + \cdots + f
  \circ \theta^{n - 1} .\]
  Then there exists a \gls{tinv_fn} function
  $\ol{f}$ with $\mu(|\ol{f}|) \le \mu(|f|)$ such
  that $\frac{S_n(f)}{n} \to \ol{f}$ \gls{al_ev}
  as $n \to \infty$.
\end{theorem*}

\begin{remark*}
  If $\theta$ is \gls{ergodic}, $\ol{f} = c$
  \gls{al_ev}.
\end{remark*}

\begin{lemma*}[Maximal ergodic theorem]
  For $f \in \Lp[1](\mu)$, set $S^* = S^*(f) =
  \sup_{n \ge 0} S_n(f)$. Then
  \[ \int_{\{S^* > 0\}} f \dd \mu \ge 0 .\]
\end{lemma*}

\begin{proof}
  Set $S_n^* = \max_{0 \le m \le n} S_m$. Then for
  $m = 1, 2, \ldots, n + 1$,
  \[ S_m = f + S_{m - 1} \circ \theta \le f +
  S_n^* \circ \theta \]
  (as for $m = 1, \ldots, n + 1$, $S_{m - 1} \le
  S_n^*$, hence $S_{m - 1} \circ \theta \le S_n^*
  \circ \theta$). On $A_n \defeq \{S_n^* > 0\}$,
  we have
  \[ S_n^* = \max_{1 \le m \le n} S_m \le \max_{1
  \le m \le n + 1} S_m \le f + S_n^* \circ
  \theta \]
  So integrating,
  \[ \int_{A_n} S_n^* \dd \mu \le \int_{A_n} f \dd
  \mu + \int_{A_n} S_n^* \circ \theta \dd \mu
  \tag{1} \label{lec22_l172_eq} \]
  On $A_n^c$, we have $S_n^* = 0 \le S_n^* \circ
  \theta$ (as $S_n^* \ge 0$ since $S_0 = 0$).
  Thus,
  \[ \int_{A_n^c} S_n^* \dd \mu \le \int_{A_n^c}
  S_n^* \circ \theta \dd \mu \tag{2}
  \label{lec22_l177_eq} \]
  Then \eqref{lec22_l172_eq} +
  \eqref{lec22_l177_eq} gives
  \[ \int_E S_n^* \dd \mu \le \int_{A_n} f \dd \mu
  + \int_E S_n^* \circ \theta \dd \mu\]
  i.e.
  \[ \int S_n^* \dd \mu \le \int_{A_n} f \dd \mu
  + \int S_n^* \dd \mu \]
  Hence, (as $S_n^* \in \Lp[1]$), $\int_{A_n} f
  \dd \mu \ge 0$ ($*$) for all $n$.
  \[  A_n = \{S_n^* > 0\} = \{\max_{0 \le m \le n}
  S_m > 0\} = \bigcup_{m = 0}^n \{S_m > 0\}
  \uparrow \bigcup_{m = 0}^\infty \{S_m > 0\} =
  \{\ub{\sup S_m}_{=S^*} > 0\} .\]
  Hence $f \indicator{A_n} \to f\indicator{(S^* >
  0)}$. Hence (as $|f \indicator{A_n}| \le |f|$
  and $f \in \Lp[1]$) by \nameref{dct} (and using
  ($*$)),
  \[ 0 \le \int_{A_n}f \dd \mu \to \int_{(S^* > 0)} f \dd \mu \]
  Hence $\int f \dd \mu \ge 0$.
\end{proof}

\begin{remark*}
  Let $\mu$ be a finite measure. Then for $f \in
  \Lp[1]$ and any $\alpha > 0$, define $\ol{S_k} =
  \frac{S_k(f)}{k}$ and $\ol{S}^* = \sup_{k \ge 0}
  \ol{S_k}$, then
  \[ \ub{\mu(\ol{S}^* > \alpha)}
  _{\mu(\ub{\sup_{k \ge 0} \ol{S_k}}_{\ge S_1 = f} > \alpha)}
  \le \frac{1}{\alpha}
  \int_{\{\ol{S}^* > \alpha\}} f \dd \mu \le
  \frac{1}{\alpha} \int |f| \dd \mu \]
  \begin{proof}
    Exercise.
  \end{proof}
  $\mu(f > \alpha) \le \frac{\int |f|
  \dd \mu}{\alpha}$ is Markov.
\end{remark*}

\begin{remark*}
  For $\mu$ a probability maesure and $f \in
  \Lp[1](\mu)$, show that $\left\{
  \frac{S_n(f)}{n} : n \in \NN \right\}$ is
  \gls{ui} (exercise). Hence $\frac{S_n(f)}{n}
  \stackrel{\Lp[1]}{\longrightarrow} \ol{f}$. If
  $\theta$ \gls{ergodic}, $\ol{f} = \int f \dd
  \mu$ \gls{al_surely}.
\end{remark*}
