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\begin{theorem*}
  If $f_n \stackrel{\mu}{\longrightarrow} 0$, then
  $\exists$ subsequence $(n_k)$ such that
  $f_{n_k} \to 0$ \glsref[al_ev]{$\mu$.a.e}.
\end{theorem*}

\begin{proof}
  Suppose $f_n \stackrel{\mu}{\longrightarrow} 0$.
  Choosing $\eps = \frac{1}{k}$, then $\mu\left(|f_n| >
  \frac{1}{l} \right) \to 0$ as $n \to \infty$. We can get
  $n_k$ such that $\mu\left(|f_{n_k}| > \frac{1}{k}\right) <
  \frac{1}{k^2}$. We can choose $n_{k + 1}$ in the
  same way (i.e. $\mu \left( |f_{n_{k + 1}}| >
  \frac{1}{k + 1} \right) < \frac{1}{(k + 1)^2}$,
  and such that $n_{k + 1} > n_k$).

  So we get a subsequence $(n_k)$ such that
  \[ \mu \left( |f_{n_k}| > \frac{1}{k} \right) <
  \frac{1}{k^2} \]
  Also, $\sum_j \frac{1}{k^2} < \infty$.

  So $\sum_k \mu \left( |f_{n_k}| > \frac{1}{k}
  \right) < \infty$. So by
  \nameref{borel_cantelli_1},
  \[ \mu \left( \ub{|f_{n_k}| > \frac{1}{k} \text{
  \gls{io}}}_{f_{n_k} \not\to 0} \right) = 0 \]
  \[ \implies \mu(f_{n_k} \not\to 0) = 0 \]
  i.e. $f_{n_k} \to 0$ \glsref[al_ev]{$\mu$.a.e}.
\end{proof}

\begin{remark*}
  Going to a subsequence is necessary,  i.e.
  convergence in $\mu$ measure $\not\implies$
  \glsref[al_ev]{$\mu$.a.e} convergence.

  For example, let $(A_n)_{n \in \NN}$ be
  \gls{indep_event} events such that $\PP(A_n) =
  \frac{1}{n}$. Let $X_n = \indicator{A_n}$. Then
  $X_n \stackrel{\PP}{\longrightarrow} 0$ (as
  $\PP(|X_n| > \eps) = \PP(A_n) = \frac{1}{n} \to
  0 ~\forall \eps > 0$). But $\sum_n \PP(A_n) =
  \infty$, i.e. $\sum_n \PP(|X_n| > \eps) =
  \infty$ and $\{|X_n| > \eps\}$ are
  \gls{indep_rv}. So by \nameref{borel_cantelli_2}
  $\implies$ $\PP(|X_n| > \eps \text{\gls{io}}) =
  1$, hence $X_n \not\to 1$ \gls{al_ev}.
\end{remark*}
\vspace{-1em}

\begin{flashcard}[conv-dist-defn]
\begin{definition*}[Convergence in distribution]
  \glsnoundefn{conv_dist}{converges in
  distribution}{N/A}
  \glsnoundefn{conv_dist}{converges in
  distribution}{converges in distribution}
  \cloze{
  For $X, (X_n)_n$ a sequence of random variables,
  we say $X_n \convdist X$ (\emph{$X_n$ converges to $X$
  in distribution}), if
  \[ F_{X_n}(t) \stackrel{n \to
  \infty}{\longrightarrow} F_X(t) ~\forall
  \text{$t$ such that $F_X(t)$ is continuous} \]
  (this definition does not require $(X_n)$ to be
  defined  on the same probability space).
  }
\end{definition*}
\end{flashcard}

\begin{remark*}
  If $X_n \stackrel{\PP}{\longrightarrow} X$, then
  $X_n \convdist X$ (proof is on \es{2}).
\end{remark*}

\begin{example*}
  $(X_n)_{n \in \NN}$ be IID $\Exp(1)$, i.e.
  $\PP(x_n > x) = e^{-x} ~\forall n \in \NN, x \ge
  0$. Find a deterministic function $g : \NN \to
  \RR$ such that \gls{al_surely}
  \[ \limsup \frac{X_n}{g(n)} = 1 \]
  For $\alpha > 1$, $\PP(X_n > \alpha \log n) =
  e^{-\alpha \log n} = n^{-\alpha}$. So $\sum_n
  \PP(X_n > \alpha \log n) < \infty$ if and only
  if $\alpha > 1$.

  For any $\eps > 0$, $\sum_n \PP(X_n > (1 + \eps)
  \log n) < \infty$, so by
  \nameref{borel_cantelli_1},
  \[ \PP \left( \frac{X_n}{\log n} > 1 + \eps
  \text{ \gls{io}} \right) = 0 \]
  Also, $\sum_n \PP(X_n > \log n) = \infty$, also
  $\{X_n > \log n\}$ are \gls{indep_event} events (as
  $(X_n)$ \gls{indep_rv}), so by
  \nameref{borel_cantelli_2},
  \[ \PP \left( \frac{X_n}{\log n} > 1 \text{
  \gls{io}} \right) = 1 \]
  So
  \[ \PP \left( \limsup \frac{X_n}{\log n} = 1
  \right) = 1 .\]
\end{example*}

\begin{flashcard}[tail-event-defn]
\begin{definition*}[Tail events]
  \glsnoundefn{tail_sig}{tail
  $\sigma$-algebra}{tail $\sigma$-algebras}
  \glssymboldefn{tail_sig}{tail
  $\sigma$-algebra}{$\tau$}
  \cloze{
  Let $(X_n : n \in \NN)$ be a sequence of random
  variables. Define
  \[ \tau_n = \sigma\{X_{n + 1}, X_{n + 2}, \ldots\} \]
  and define $\tau = \bigcap_{n \in \NN} \tau_n$.

  Then $\tau$ is a \sigalg\ called the \emph{tail
  \sigalg} (contains events that depend only on
  the ``limiting behaviour'' of the sequence).
  }
\end{definition*}
\end{flashcard}

\begin{flashcard}[kol-0-1-law]
\begin{theorem*}[Kolmogorov $0-1$ law]
  \label{kol_0_1}
  \cloze{
  Let $(X_n)_n$ be a sequence of \gls{indep_rv}
  random variables. Then for the \gls{tail_sig}
  $\tailsig$, if $A \in \tailsig$, then $\PP(A) =
  0$ or $\PP(A) = 1$. If $Y : (\Omega, \tailsig)
  \to \RR$ is measurable, then $Y$ is
  \gls{al_surely} convergent.
  }
\end{theorem*}

\begin{proof}
  Let $\mathcal{F}_n = \sigma(X_1, \ldots, X_n)$.
  Then $\mathcal{F}_n$ is generated by the
  \pisys\ of sets
  \[ A = \{X_1 \le x_1, X_2 \le x_2, \ldots, X_n
  \le x_n\} \qquad x_1, \ldots, x_n \in \RR .\]
  and $\tailsig_n = \sigma(X_{n + 1}, \ldots)$ is
  generated by the \pisys\ of events
  \[ B = \{X_{n + 1} \le x_{n + 1}, \ldots, X_{n +
  k} \le x_{n + k}\}, \qquad x_{n + 1}, \ldots,
  \in \RR, k \in \NN \]
  By by independence $\PP(A \cap B) = \PP(A)
  \PP(B)$ for all such $A$ and $B$. Hence, by an
  earlier theorem, $\mathcal{F}_n$ and
  $\tailsig_n$ are independent. But $\tailsig
  \subseteq \tailsig_n$, so $\mathcal{F}_n$ and
  $\tailsig$ are \gls{indep_rv} for all $n$.

  Now, consider $\bigcup_n \mathcal{F}_n$
  ($\mathcal{F}_1 \subseteq \mathcal{F}_2
  \subseteq \cdots$) is a \pisys\ that generates
  $\mathcal{F}_\infty \defeq \sigma(X_n, n \in
  \NN)$. But $\bigcup_n \mathcal{F}_n$ and
  $\tailsig$ are \gls{indep_sigma}, so by the
  theorem again, $\mathcal{F}_\infty$ and
  $\tailsig$ are \gls{indep_sigma}. But $\tailsig
  \subseteq \mathcal{F}_\infty$, so for any $A \in
  \tailsig$, $A \in \mathcal{F}_\infty$,
  \[ \PP(A) = \PP(\ub{A}_{\in \tailsig} \cap
  \ub{A}_{\in \mathcal{F}_\infty}) = \PP(A) \PP(A)
  = \PP(A)^2 ,\]
  i.e. $\PP(A) = 0$ or $\PP(A) = 1$.

  Finally, if $Y$ is $\tailsig$ measurably, for
  any $y \in \RR$, $\{Y \le y\} \in \tailsig$, so
  $\PP(Y \le y) = 0$ or $1$. Then $c = \inf\{y :
  \PP(Y \le y) = 1\}$, then $\PP(Y = c) = 1$.
  $X_i$ IID, $\EE X < \infty$, then
  \[ \limsup \frac{\sum_{i = 1}^n X_i}{n}, \qquad
  \liminf \frac{\sum_{i = 1}^n X_i}{n} \]
  are constants \gls{al_surely}.
\end{proof}
\end{flashcard}

\newpage
\section{Integration}

For $(E, \mathcal{E}, \mu)$ a measure space, $f :
T \to \RR$ measurable and $f \ge 0$, we shall
define the integral of $f$ and write it as
\[ \mu(f) = \int_E f \dd \mu = \int_E f(x) \dd
\mu(x) \]

When $(E, \mathcal{E}, \mu) = (\RR, \mathcal{B},
\lambda)$, we write it as $\int f(x) \dd x$.

For $(E, \mathcal{E}, \mu) = (\Omega, \mathcal{F},
\PP)$, and $X$ a random variable, we define its
\emph{Expectation}
\[ \EE(X) = \int_\Omega X \dd P = \int_\Omega =
X(\omega) \dd P(\omega) \]
\glsadjdefn{simple}{simple}{function}
To start, we say $f : E \to \RR$ is \emph{simple}
if $f = \sum_{k = 1}^m a_K \indicator{A_k}$, $0
\le a_k < \infty$, $A_k \in \mathcal{E} ~\forall
k, m  \in \NN$. Define for such simple $f$,
\[ \mu(f) = \sum_{k = 1}^m a_k \mu(A_k) \]
(where $0 \cdot \infty = 0$). This is well defined
(see \es{2}). Check for $f, g$ simple, $\alpha,
\beta \ge 0$,
\refsteplabel[properties]{lec9_int_properties}
\begin{enumerate}[(a)]
  \item $\mu(\alpha f + \beta g) = \alpha \mu(f) +
    \beta \mu(g)$
  \item $f \le g \implies \mu(f) \le \mu(g)$
  \item $f = 0$ \gls{al_ev} $\implies$ $\mu(f) =
    0$.
\end{enumerate}