%! TEX root = PM.tex
% vim: tw=50
% 25/11/2023 10AM

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

\begin{flashcard}[maximal-ergodic-lemma]
\begin{lemma*}[Maximal Ergodic lemma]
  \label{max_erg_lemma}
  \cloze{For $f \in \Lp[1](\mu)$ and $S^* = \sup_{n \ge
  0} S_n(f)$, we must have
  \[ \int_{\{S^* > 0\}} f \dd \mu \ge 0 .\]
  }
\end{lemma*}
\end{flashcard}

\begin{proof}
  \[ \mu(|S_n|) \le \sum_{i = 0}^{n - 1} \mu(|f
  \circ \theta^i|) \stackrel{\text{$\theta$
  \gls{mp}}}{=} \sum_{i = 1}^{n - 1}
  \mu(|f|) = n |f| \]
  So $\mu \left( \left| \frac{S_n}{n} \right|
  \right) \le \mu(|f|)$. So
  \[ \mu(|\ol{f}|) = \mu \left(\liminf \left|
  \frac{S_n}{n} \right|\right)
  \stackrel{\text{\nameref{fatous_lemma}}}{\le}
  \liminf \mu \left( \left| \frac{S_n}{n} \right|
  \right) \le \mu(|f|) .\]
  Note that  $\frac{S_n \circ \theta}{n} =
  \frac{S_{n + 1} - f}{n + 1} \times \frac{n +
  1}{n}$, so $\limsup \frac{S_n}{n} \circ \theta =
  \limsup \frac{S_{n + 1}}{n + 1} = \limsup
  \frac{S_n}{n}$. Similarly, $\liminf
  \frac{S_n}{n} \circ \theta = \liminf
  \frac{S_n}{n}$. For $a < b$,
  \[ D = D(a, b) = \left\{ \liminf_n \left(
  \frac{S_n}{n} \right) < a < b < \limsup_n \left(
  \frac{S_n}{n} \right) \right\} \]
  is \gls{tinv}. Shall show $\mu(D) = 0$.
  \[ \Delta = \left\{ \liminf \left( \frac{S_n}{n}
  \right) < \limsup \left( \frac{S_n}{n} \right)
  \right\} = \bigcup_{\substack{a, b \in \QQ \\ a <
  b}} D(a, b) \]
  Hence if $\mu(D) = 0$ for all $a < b$, then
  \[ \mu \left( \bigcup_{\substack{a, b \in \QQ \\
  a < b}} D(a, b) \right) = 0 \implies \mu(\Delta)
  = 0 .\]
  Define
  \[ \ol{f} = \begin{cases}
    \liminf \left( \frac{S_n}{n} \right) = \limsup
    \left( \frac{S_n}{n} \right) & \text{on
    $\Delta^c$} \\
    0 & \text{on $\Delta$}
  \end{cases} \]
  Then $\frac{S_n}{n} \to \ol{f}$ $\mu$
  \gls{al_ev} and $\ol{f}$ is \gls{tinv_fn} (as
  $\liminf \frac{S_n}{n}$ is \gls{tinv_fn} and
  $\Delta$ is \gls{tinv}).

  Fix $a < b$. Note that
  $\theta : D \to D$ by invariance and $\theta$ is
  $\mu|_D$-\gls{mp}. Also, either $b > 0$ or $a <
  0$ (if $a < 0$, change $f$ to $-f$, $b$ to $-a$,
  $a$ to $-b$, then $b = -a > 0$). So assume $b >
  0$ without loss of generality. Shall apply
  \nameref{max_erg_lemma} on $D$ with $\mu|_D$.
  For any $B \subseteq D$ measurable and $\mu(B) <
  \infty$, let $g = f - b\indicator{B}$. Then $g
  \in \Lp[1]$, and on $D$,
  \[ S_n(g) = S_n(f) - bS_n(\indicator{B})
  \ge S_n(f) - nb > 0 \]
  for some $n$. Hence, $S^*(g) = \sup_{n \ge 0}
  S_n(g) > 0$ on $D$. Hence
  \[ \{S^*(g) . 0\} \cap D = D .\]
  Thus by \nameref{max_erg_lemma} on $D$,
  $\int_{\{S^*(g) > 0\} \cap D} g \dd \mu \ge 0$,
  i.e.
  \[ 0 \le \int_D (f - b \indicator{B}) \dd \mu =
  \int_D f \dd \mu - b \mu(B) \]
  hence $b\mu(B) \le \int_D f \dd \mu$. Since
  $\mu$ is $\sigma$-finite, there exists $(B_n)$
  measurable sets, $B_n \uparrow D$ and $\mu(B_n)
  < \infty$ for all $n$. Hence
  \[ b\mu(D) = b\mu(B_n) \le \int_D f \dd \mu <
  \infty \]
  (as $f \in \Lp[1]$). Hence $\mu(D) < \infty$. A
  similar argument applied to $(-f)$ and $(-a)$
  will give
  \[ (-a) \mu(D) \le \int_D(-f) \dd \mu \]
  (just take $D$ instead of $B$ now, $(-f) -
  (-a)\indicator{D}$). Hence
  \[ b\mu(D) \le \int_D f \dd \mu \le a \mu(D) \]
  But $a < b$. As $\mu(D) < \infty$, hence $\mu(D)
  = 0$.
\end{proof}

\begin{flashcard}[von-neumann-erg-thm]
\begin{theorem*}[von Neumann's $\Lp$ Ergodic
  Theorem]
  \cloze{Let $\mu(E) < \infty$ and $1 \le p < \infty$.
  Then for $f \in \Lp(\mu)$,
  \[ \frac{S_n(f)}{n} \to \ol{f} \]
  in $\Lp$ as $n \to \infty$.}
\end{theorem*}

\begin{proof}
  \cloze{
  For any $f \in \Lp(\mu)$, $\normp\|f \circ
  \theta^n\|_p = \normp\|f\|_p$ as $\theta$ is
  $\mu$ \gls{mp}. So by Minkowski inequality,
  \[ \left\| \frac{S_n(f)}{n} \right\|_p \le
  \frac{1}{n} \sum_{i = 0}^{n - 1} \ub{
  \normp\|f \circ
  \theta^1\|_p}_{=\normp\|f\|_p} = \normp\|f\|_p \]
  Now, let $\eps > 0$ be given, then choose $K <
  \infty$ such that $\|f - f_K\|_p <
  \frac{\eps}{3}$ where $f_K = (-K) \vee f \wedge
  (K)$.
  \[ \left( \normp\|f - f_K\|_p^p =
  \int |f - f_K|^p \dd \mu \le \int |f|^p
  \indicator{|f| > K} \dd \mu
  \stackrel{\text{\nameref{dct}}}{\longrightarrow} 0 \right) \]
  But $f_K$ is bounded and $\mu$ a finite measure,
  so $f_K \in \Lp[1](\mu)$, hence by
  \nameref{birkhoff}, there exists $\ol{f_K} \in
  \Lp[1]$ such that $\frac{S_n(f_K)}{n} \to
  \ol{f_K}$ \gls{al_ev}. Again, $f \in \Lp(\mu)
  \subset \Lp[1](\mu)$ (as $\mu$ is a finite
  measure), so by Birkhoff, there exists $\ol{f}
  \in \Lp[1]$ such that $\frac{S_n(f)}{n} \to
  \ol{f}$ \gls{al_ev} as $n \to \infty$. Then
  \begin{align*}
    \normp\|\ol{f} - \ol{f_K}\|_p^p
    &= \int |\ol{f} - \ol{f_K}|^p \dd \mu \\
    &= \int \liminf_n \left| \frac{S_n(f)}{n} -
    \frac{S_n(f_K)}{n} \right|^p \dd \mu \\
    &= \int \liminf_n \left| \frac{S_n(f -
    f_K)}{n} \right|^p \dd \mu \\
    &\le \liminf \int \left| \frac{S_n(f -
    f_K)}{n} \right|^p \dd \mu \qquad
    \text{\nameref{fatous_lemma}} \\
    &= \liminf_n \left\| \frac{S_n(f - f_K)}{n} \right\|_p^p
  \end{align*}
  Then
  \begin{align*}
    \left\| \frac{S_n}{n} - \ol{f} \right\|_p
    &\stackrel{\text{\nameref{fatous_lemma}}}{\le}
    \left\| \frac{S_n(f)}{n} -
    \frac{S_n(f_K)}{n} \right\|_p + \|\ol{f} -
    \ol{f_K}\|_p + \left\| \frac{S_n(f_K)}{n} -
    \ol{f_K} \right\|_p \\
    &= 
    \ub{\left\| \frac{S_n(f - f_K)}{n}
    \right\|_p}_{= \normp\|f - f_K\|_p \le \eps/3}
    + \ub{\|\ol{f} -
    \ol{f_K}\|_p}_{\le \eps/3} + \ub{\left\| \frac{S_n(f_K)}{n} -
    \ol{f_K} \right\|_p}_{\to 0} \\
  \end{align*}
  so $\frac{S_n}{n} \to \ol{f}$ in $\Lp$ as $n \to
  \infty$ as desired.}
\end{proof}
\end{flashcard}
