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\begin{itemize}
  \item $f \ge 0$, $\mu^*(f) = \sup \{\mu(g),
    \text{$g$ \gls{simple}}, g \le f\}$. For $f$
    \gls{simple}, $\ge 0$, is $\mu^*(f) = \mu(f)$?
    $\mu^*(f) \ge \mu(f)$ is easy. For any $g$
    simple, $g \le f$, $\mu(g) \le \mu(f)$, taking
    supremum of LHS, we get $\mu^*(f) \le \mu(f)$.
  \item $f$ measurable, $f \ge 0$, $f$ bounded,
    then $2^{-n} \left\lfloor 2^n f \right\rfloor$
    is \gls{simple}. For unbounded $f$, we
    truncate to $f_n = 2^{-n} \left\lfloor 2^n f
    \right\rfloor \wedge n$ (which is
    \gls{simple} and has $f_n \uparrow f$).
\end{itemize}
A general question we begun to explore last time:
when can we say that
\[ \lim \int f_n \dd \mu = \int \lim f_n \dd \mu ?\]

\begin{example*}
  $f_n = \indicator{(n, n + 1)}$, $f_n \ge 0$,
  $f_n \to 0$ as $n \to \infty$. But $\lim_{n \to
  \infty} \lambda(f_n) = 1 > \lambda(0) = 0$.
\end{example*}

\begin{flashcard}[fatous-lemma]
\begin{lemma*}[Fatou's Lemma]
  \refstepcounter{customlemma}
  \label{fatous_lemma}
  Let $f_n : (E, \mathcal{E}, \mu) \to \RR$ be
  measurable, non-negative. Then
  \[ \mu(\liminf f_n) \cloze{\le} \liminf \mu(f_n) \]
\end{lemma*}
\fcscrap{
Recall that
\[ \boxed{\liminf x_n = \sup_m \inf_{k \ge m} x_k} \]
}

\begin{proof}
  \cloze{
  For $k \ge n$, $\inf_{m \ge n} f_m \le f_k$. So
  \[ \mu(\inf_{m \ge n} f_m) \le \mu(f_k) \quad
  \forall k \ge n \]
  i.e.
  \[ \label{lec11_l46_eq} \mu(\inf_{m \ge n} f_m) \le \inf_{k \ge n}
  \mu(f_k) \le \liminf \mu(f_k) \tag{$*$} \]
  Let $g_n = \inf_{m \ge n} f_m$. Then $g_n \ge
  0$, and $g_n \uparrow \sup_n g_n = \sup_n
  \inf_{m \ge n} f_m = \liminf f_n$. So by
  \nameref{mconv_thm}, $\mu(g_n) \uparrow
  \mu(\liminf f_n)$. Taking limit in
  \eqref{lec11_l46_eq}, we get
  \[ \mu(\liminf f_n) \le \liminf \mu(f_n)
  \qedhere \]
  }
\end{proof}
\end{flashcard}

\begin{flashcard}[dominated-conv-thm]
\begin{theorem*}[Dominated Convergence Theorem]
  \label{dconv_thm}
  Let \cloze{$(f_n) : (E, \mathcal{E}, \mu) \to \RR$ be
  measurable. Suppose $|f_n| \le g$ for all $n$,
  for some integrable function $g$ (i.e. $\mu(g) <
  \infty$). Also suppose $f_n \to f$ as $n \to
  \infty$ on $E$.} Then $f$, $f_n$ are integrable
  and
  \[ \mu(f_n) \to \mu(f) \]
\end{theorem*}

\begin{proof}
  \cloze{
  $f$ is measurable since it is a limit of
  measurable functions. Also, taking limits of
  $|f_n| \le g$, we get $|f| \le g$. So, $|f_n|
  \le g \implies \mu(|f_n|) \le \mu(g) < \infty$,
  $\mu(|f|) \le \mu(g) < \infty$. So $f_n, f$ are
  integrable.

  We have $0 \le g \pm g_n \to g \pm f$, so since
  $g \pm f_n$ are non-negative, by
  \nameref{fatous_lemma}, we have
  \[ \mu(g \pm f) \le \liminf \mu(g \pm f_n) \]
  i.e.
  \[ \mu(g) + \mu(f) \le \liminf (\mu(g) +
  \mu(f_n)) \le \mu(g) + \liminf \mu(f_n) \]
  \[ \mu(g) - \mu(f) \le \liminf (\mu(g) -
  \mu(f_n)) \le \mu(g) - \limsup \mu(f_n) \]
  As $\mu(g) < \infty$, we get
  \[ \mu(f) \le \liminf \mu(f_n) \le \limsup
  \mu(f_n) \le \mu(f) \]
  i.e. $\mu(f_n) \to \mu(f)$.
  }
\end{proof}
\end{flashcard}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item The theorem is still true if we change
      all the conditions to hold \gls{al_ev}
      (instead of everywhere).
    \item In fact, $\mu(|f_n - f|) \to 0$ (recall
      that $\mu(|g|) \ge |\mu(g)|$, which in this
      case shows $\mu(|f_n - f|) \ge |\mu(f_n) -
      \mu(f)|$).

      We prove this by noting $|f_n - f| \le |f_n| + |f| \le g
      + g = 2g$ and $2g$ is integrable, so
      applying \nameref{dconv_thm} we get that
      $\mu(|f_n - f|) \to 0$.
    \item If $(E, \mathcal{E}, \mu) = (\Omega,
      \mathcal{F}, \PP)$, $X_n \to X$, $\PP$
      \gls{al_surely}, and $|X_n| \le Y$ with $\EE
      Y < \infty$. Then $\EE X_n \to \EE X$ and
      $\EE|X_n - X| \to 0$. In particular, if
      $|X_n| \le M ~\forall n$, for some constant
      $M > 0$, $M \in \RR$, then $\EE|X_n - X| \to
      0$. (Bounded Convergence Theorem).
    \item $f_n$ measurable on $[0, 1]$ and $|f_n|
      \le 1$, and $f_n \to f$ pointwise, then
      $\int f_n \dd x \to \int f \dd x$ ($\int f_n
      \dd \lambda(x)$). So stronger than Riemann
      integral as it requires uniform convergence.
  \end{enumerate}
\end{remark*}

\subsubsection*{Comparisons with Riemann integral}

\begin{enumerate}[(a)]
  \item FTC:
    \begin{enumerate}[(1)]
      \item Let $f : [a, b] \to \RR$ be
        continuous and set $F(t) = \int_a^t f(x) \dd
        x$. Then $F$ is differentiable on $[a, b]$
        with $F' = f$.
      \item Let $F : [a, b] \to \RR$ be
        differentiable and $F'$ is continuous,
        then $\int_a^b F'(x) \dd x = F(b) - F(a)$.
    \end{enumerate}
    Proofs are the same as before (just use
    $\int_t^{t + h} \dd x = h$).
  \item[(a$'$)] If $f : [a, b] \to \RR$ is Lebesgue
    integrable and $F(t) = \int_a^t f(x) \dd x$.
    Then $\lim_{h \to 0} \frac{F(t + h) -
    F(t)}{h} = \lim_{h \to 0} \frac{\int_t^{t + h}
    f(x) \dd x}{h} = f(t)$ \gls{al_ev} (Lebesgue
    differentiation Theorem in Analysis of
    Functions).
  \item Substitution formula: Let $\varphi : [a, b] \to
    \RR$, $\varphi$ strictly increasing and continuously
    differentiable. Then for all $g$ Borel
    measurable, $g \ge 0$ on $[\varphi(q), \varphi(b)]$,
    \[ \label{lec11_l156_eq}
    \int_{\varphi(a)}^{\varphi(b)} g(y) \dd y =
    \int_a^b g(\varphi(x)) \varphi'(x) \dd x
    \tag{$*$} \]
    \begin{proof}
      Let $\mathcal{V}$ be the set of all
      measurable functions $g$ for which
      \eqref{lec11_l156_eq} this holds. Then by
      lineariy of integrals, $\mathcal{V}$ is a
      vector space.
      \begin{itemize}
        \item $\indicator{} \in \mathcal{V}$ by FTC(2).
          holds. Also, $\indicator{(c, d]} \in
          \mathcal{V}$ by FTC(2).
        \item If $f_n \in \mathcal{V}$, $f_n
          \uparrow f$, $f_n \ge 0$, then by
          \nameref{mconv_thm}, $f \in
          \mathcal{V}$.
      \end{itemize}
      Hence by \nameref{monotone_class_thm},
      \eqref{lec11_l156_eq} holds $\forall g \ge
      0$ measurable.
    \end{proof}
  \item A (bounded) Riemann integrable (RI)
    function $f : [a, b] \to \RR$ is Lebesgue
    integrable in the following sense. If $f$ is
    bounded on $[a, b]$, $f$ is RI if and only if
    $\mathcal{D} = \{x \in [a, b] : \text{$f$ is
    not continuous at $x$}\}$ has
    $\lambda(\mathcal{D}) = 0$ (Lebesgue 1904),
    i.e. $f$ is continuous \gls{al_ev}. Such an
    $f$ need not be Borel (but is Lebesgue
    measurable), and can be modified on a Lebesgue
    measure $0$ set to make it Borel, i.e.
    $\exists \tilde{f}$ Borel such that $\tilde{f}
    = f$ on $A$ and $\lambda(A^c) = 0$, and $\int
    \tilde{f} \dd x = \int f \dd x$ (where we use
    Lebesgue integral on the left, and Riemann
    integral on the right).
  \item $\indicator{\QQ}$ on $[0, 1]$.
    $\mathcal{D} = [0, 1]$, $\lambda(\mathcal{D})
    \neq 0$, so $\indicator{\QQ}$ is not Riemann
    integrable. But $\indicator{\QQ}$ is Lebesgue
    integrable and $\indicator{\QQ} = 0$ $\lambda$
    \gls{al_surely}, so $\lambda(\indicator{\QQ}) =
    0$.
\end{enumerate}