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\begin{example*}
  If $X_n \convP X$, then $X_n \convdist X$. $X_n
  \convP X$ but $X_n \not\to X$ \gls{al_surely}
  ($(0, 1]$, $\mathcal{B}$, $\lambda$).
  \[ f_1 = \indicator{(0,\half]}, \quad f_2 =
  \indicator{(\half, 1]} \]
  \[ f_3 =
  \indicator{(0, \third]}, \quad f_4 =
  \indicator{(\third, \frac{2}{3}]}, \quad f_5 =
  \indicator{(\frac{2}{3}, 1]} \]
  \[ f_6 = \indicator{(0, \frac{1}{4}]}, \ldots \]
  Then $f_n \to 0$ in $\lambda$-measure, but $f_n
  \not\to 0$ $\lambda$ \gls{al_ev}. For any $x \in
  (0, 1]$, $(f_n(x))$ has an infinite sequence of
  $1$s, hence $f_n(x) \not\to 0$.
\end{example*}
\vspace{-1em}

Recall, for $f$ simple, i.e. $f = \sum_{k = 1}^m
a_k \indicator{A_k}$, $a_k \ge 0$, $A_k \in
\mathcal{E}$, then
\[ \mu(f) = \int f \dd\mu = \sum_{k = 1}^m a_k
\mu(A_k) \]
(recall the
\nameref{lec9_int_properties} given
\lastlecture).

\begin{flashcard}[measure-of-function-defn]
\begin{definition*}[Measure of function]
  \cloze{
  For $f : E \to \RR$ measurable, $f \ge 0$,
  define
  \[ \mu(f) = \sup \{\mu(g) : \text{$g$ simple}, g
  \le f\} \]
  }
\end{definition*}
\end{flashcard}

Clearly, if $0 \le f_1 \le f_2$, then $\mu(f_1)
\le \mu(f_2)$.

For general $f : E \to \RR$ measurable, $f = f^+ -
f^-$, where $f^+ = \max(f, 0)$, $f^- = \max(-f,
0)$ and $|f| = f^+ + f^-$.

\begin{flashcard}[integrabl-defn]
\begin{definition*}[Integrable function]
  \glsadjdefn{integ_fn}{integrable}{function}
  \cloze{
  We say $f$ is
  \emph{integrable} if $\mu(|f|) < \infty$ and
  then define
  \[ \mu(f) = \mu(f^+) - \mu(f^-) \]
  ($\mu(|f|) = \mu(f^+) + \mu(f^-)$, hence
  $|\mu(f)| \le \mu(|f|)$).
  }
\end{definition*}
\end{flashcard}
\vspace{-1em}

If one of $\mu(f^+)$ or $\mu(f^-)$ is $\infty$ and
the other is finite, we define $\mu(f)$ to be
$\infty$ or $-\infty$ respectively (though $f$ is
not \gls{integ_fn}).
\begin{itemize}
  \item $x_n \uparrow x$ to mean $x_n \le x_{n +
    1} ~\forall n$, $x_n \to x$
  \item $f_n \uparrow f$ to mean $f_n(x) \le f_{n
    + 1}(x) ~\forall x \in E$ and $f_n(x) \to
    f(x)$.
\end{itemize}

\begin{flashcard}[monotone-convergence-thm]
\begin{theorem*}[Monotone Convergence Theorem]
  \label{mconv_thm}
  \cloze{
  Let $(f_n)_n, f : (E, \mathcal{E}, \mu) \to \RR$
  measurable and \fcemph{non-negative}, and
  suppose $f_n \uparrow f$. Then $\mu(f_n)
  \uparrow \mu(f)$.
  }
\end{theorem*}
\end{flashcard}

\begin{proof}
  Recall $\mu(f) = \sup\{\mu(g) : g \le f,
  \text{$g$ simple}\}$. Let $M = \sup_n
  (\mu(f_n))$, then $\mu(f_n) \uparrow M$. Need to show
  that $M = \mu(f)$.

  Since $f_n \le f$, $\mu(f_n) \le \mu(f)$, so by
  taking $\sup$, $M \le \mu(f)$.

  Now need to show $M \ge \mu(f)$. So it is enough
  to show $M \ge \mu(g)$ for all $g$ \gls{simple}, $g
  \le f$. Let $g = \sum_{k = 1}^m a_k
  \indicator{A_k} \le f$. Assume without loss of
  generality, that $(A_k)$s are disjoint. Define
  the approximation $g_n$ as
  \[ g_n(x) = (2^{-n} \left\lfloor 2^n f_n(x)
  \right\rfloor) \ub{\wedge}_{\min} g(x) \]
  So $g_n$ is \gls{simple}, $g_n \le \ol{f_n} \le
  f_n \uparrow f$, so $g_n = \ol{f_n} \wedge g
  \uparrow f \wedge g = g$, i.e. $g_n \uparrow g$,
  $g_n$ \gls{simple}, $g_n \le f_n$.

  Fix $\eps \in (0, 1)$ and consider the sets
  \[ A_k(n) = \{x \in A_k : g_n(x) \ge (1 - \eps)
  a_k\} \]
  Now, $g = a_k$ on the set $A_k$, and $g_n
  \uparrow g$, so $A_k(n) \uparrow A_k$, hence
  $\mu(A_k(n)) \uparrow \mu(A_k)$. Also,
  \[ \indicator{A_k(n)} (1 - \eps) a_k \le
  \indicator{A_k(n)} g_n \le \indicator{A_k} g_n \]
  So as $\mu(f)$ is increasing, we have,
  \begin{align*}
    \mu(\indicator{A_k(n)} (1 - \eps) a_k)
    &\le \mu(\indicator{A_k} g_n) \\
    \label{lec10_l123_eq}
    \implies (1 - \eps) a_k \mu(A_k(n))
    &\le
    \mu(\indicator{A_k} g_n) \tag{$*$}
  \end{align*}
  Finally, $g_n = \sum_{k = 1}^n \indicator{A_k}
  g_n$ ($g_n \le g$ and $g$ support on $\bigcup_{k
  = 1}^n A_k$ and $(A_k)$ are disjoint), so
  \begin{align*}
     \mu(g_n)
     &= \mu \left( \sum_{k = 1}^n \indicator{A_k}
     g_n \right) \\
     &= \sum_{k = 1}^n \mu(\indicator{A_k} g_n) \\
     &\ge \sum_{k = 1}^n (1 - \eps) a_k \mu(A_k(n))
     &&\text{by \eqref{lec10_l123_eq}} \\
     &\uparrow \sum_{k = 1}^n (1 - \eps) a_k
     \mu(A_k) \\
     &= (1 - \eps) \mu(g)
  \end{align*}
  Then
  \[ (1 - \eps) \mu(g) \le \lim_{n \to \infty}
  \mu(g_n) \ub{\le}_{g_n \le f_n} \lim_{n \to
  \infty} \mu(f_n) \le M \]
   i.e. $\mu(g) \le \frac{M}{1 - \eps}$ for all
   $\eps > 0$, hence $\mu(g) \le M$.
\end{proof}

\begin{theorem*}
  Let $(f, g) : (E, \mathcal{E}, \mu) \to \RR$ be
  measurable, non-negative. Then $\forall \alpha,
  \beta \ge 0$,
  \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} $\iff$ $\mu(f) = 0$.
  \end{enumerate}
\end{theorem*}

\begin{proof}
  \begin{enumerate}[(a)]
    \item Let $f_n = (2^{-n} \left\lfloor 2^n f
      \right\rfloor) \wedge n$, $g_n = (2^{-n}
      \left\lfloor 2^n g \right\rfloor) \wedge n$.
      Then, $f_n, g_n$ are \gls{simple} and $f_n
      \uparrow f$, $g_n \uparrow g$. Then $\alpha f_n
      + \beta g_n \uparrow \alpha f + \beta g$. So by
      \nameref{mconv_thm}, $\mu(f_n) \uparrow \mu(f)$,
      $\mu(g_n) \uparrow \mu(g)$, $\mu(\alpha f_n +
      \beta g_n) \uparrow \mu(\alpha f + \beta g)$.
    \item Obvious from the definition.
    \item If $f = 0$ \gls{al_ev}, then $0 \le f_n
      \le f$, so $f_n = 0$ \gls{al_ev}, but $f_n$
      \gls{simple} $\implies \mu(f_n) = 0$ and
      $\mu(f_n) \uparrow \mu(f)$ so $\mu(f) = 0$.
      \qedhere
  \end{enumerate}
\end{proof}

\begin{theorem*}
  Now, let $f, g : (E, \mathcal{E}, \mu) \to \RR$
  be \gls{integ_fn}. Then $\forall \alpha, \beta
  \in \RR$,
  \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}
\end{theorem*}

\begin{proof}
  Exercise. Set $f = f^+ - f^-$, $g = g^+ - g^-$,
  and use definition, $\mu(f) = \mu(f^+) -
  \mu(f^-)$. If $\mu(f^+) = \mu(f^-)$, then
  $\mu(f) = 0$ but $f$ need not be $0$
  \gls{al_ev}.
\end{proof}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item If $0 \le f_n \uparrow f$ \gls{al_ev},
      then $\mu(f_n) \uparrow \mu(f)$.
    \item \nameref{mconv_thm} $\implies \lim_n
      \int f_n \dd \mu = \int \lim_n f_n \dd \mu$
      for $0 \le f_n \uparrow \lim f_n = f$. If
      $g_n \ge 0$, then (writing $f_n = \sum_{k =
      1}^n g_k$, $f_n \uparrow f = \sum_{k =
      1}^\infty g_k$),
      \begin{align*}
        \implies \lim_n \int \sum_{k = 1}^n g_k
        \dd \mu
        &= \int \left( \sum_{k = 1}^\infty g_k
        \right) \dd \mu \\
        \implies \sum_{k = 1}^\infty \int g_k \dd
        \mu
        &= \int \sum_{k = 1}^\infty g_k \dd \mu
      \end{align*}
      i.e.
      \[ \boxed{\sum_{k = 1}^\infty \mu(g_k) = \mu
      \left( \sum_{k = 1}^\infty g_k \right)} \]
      This generalises the countable additivity of
      $\mu$ to integrals of non-negative
      functions. In part, if $g_k =
      \indicator{A_k}$ where $(A_k)$ disjoint,
      implies countable additivity of $\mu$.
  \end{enumerate}
\end{remark*}
