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\begin{theorem*}[DCT for \gls{al_surely} convergence]
  $(\Omega, \mathcal{F}, \PP)$ is a probability
  space and $(X_n)$, $X$ are random variables.
  Suppose $X_n \to X$ $\PP$ \gls{al_surely} and
  $|X_n| \le Y$ $\forall n$ for some integrable
  random variable $Y$ (i.e. $\EE Y < \infty$),
  then
  \[ \EE|X_n - X| \to 0 \]
  as $n \to \infty$.
\end{theorem*}

\begin{theorem*}[DCT for in $\PP$ convergence]
  $(\Omega, \mathcal{F}, \PP)$ is a probability
  space and $(X_n)$, $X$ are random variables.
  SUppose $X_n \to X$ in $\PP$ probability, and
  $|X_n| \le Y$ $\forall n$ for some integrable
  random variable $Y$. Then
  \[ \EE|X_n - X| \to 0 \]
  as $n \to \infty$.
\end{theorem*}

\begin{proof}
  Suppose $\EE |X_n - X| \not\to 0$. Then there
  exists a subsequence $(n_k)$ such that
  $\EE|X_{n_k} - X| > \eps$ for all $k$, for some
  $\eps > 0$. Now, $X_n \convP X$ implies $X_{n_k}
  \convP X$. Hence $\exists (n_{k_l})$ such that
  $X_{n_{k_l}} \convas X$ and $|X_{n_{k_l}}| \le
  Y$. But then by \nameref{dconv_thm},
  $\EE|X_{n_{k_l}}| \to 0$, but that contradicts
  the constructed property of $(n_k)$.
\end{proof}

\begin{theorem*}[BCT for in $\PP$ convergence]
  $X_n \convP X$ and $|X_n| \le M$ for some
  constant $M > 0$, $\forall n \ge 0$. Then
  $\EE|X_n - X| \to 0$.
\end{theorem*}

\begin{theorem*}[Differentiation under the integral
  sign]
  Let $U \subseteq \RR$ be open and $f : U \times
  E \to \RR$ such that
  \begin{enumerate}[(i)]
    \item $x \mapsto f(t, x)$ is integrable for
      all $t \in U$.
    \item $t \mapsto f(t, x)$ is differentiable
      $\forall x \in E$.
    \item $\exists g : E \to \RR$ integrable such
      that $\forall x \in E$, $\forall t \in U$,
      \[ \left| \pfrac{f}{t} (t, x) \right| \le
      g(x) .\]
  \end{enumerate}
  THen $x \mapsto \pfrac{f}{t}(t, x)$ is
  integrable $\forall t$ and $F : U \to \RR$
  defined by
  \[ F(t) = \int_E f(t, x) \mu(\dd x) \]
  is differentiable and
  \[ \dfrac{}{t} F(t) = \int \pfrac{f}{t} (t, x)
  \mu(\dd x) .\]
\end{theorem*}

\begin{proof}
  For $h_n \to 0$, set
  \[ g_n(x) = \frac{f(t + h_n, x) - f(t, x)}{h_n}
  - \pfrac{f}{t} \]
  For any fixed $t$, $g_n(x) \to 0$ for all $x \in
  E$ (by (ii)), and
  \[ |g_n(x)| =  \left| \ub{\pfrac{f}{t} (\tilde{t},
  x)}_{\text{MVT}} - \pfrac{f}{t}(t, x) \right|
  \stackrel{\text{(iii)}}{\le}\le 2g(x) \]
  and $2g$ is integrable. Hence, $\mu(g_n) \to 0$
  by \nameref{dconv_thm}, i.e.
  \[ \int \frac{f(t + h_n) - f(t, x)}{h_n} \mu
  (\dd x) - \int \pfrac{f}{t} (t, x) \mu(\dd x)
  \to 0 \]
  i.e.
  \[ \ub{\lim_{n \to \infty} \frac{F(t + h_n) -
  F(t)}{h_n}}_{= F'(t)} = \int \pfrac{f}{t}(t, x) \mu(\dd x)
  . \qedhere \]
\end{proof}

\subsubsection*{Integrals and image measures}

Let $f : (E, \mathcal{E}, \mu) \to (G,
\mathcal{G})$ be measurable with the image
measure $\nu = \mu \circ f^{-1}$ on $(G,
\mathcal{G})$. If $g : (G, \mathcal{G}) \to \RR$
measurable, $\ge 0$, then
\[ \nu(g) = \int_G g(x) \dd \nu(x) = \int_G g \dd
\mu \circ f^{-1} \stackrel{?}{=} \int_E g(f(x))
\dd \mu(x) \]
(the $?$ equality is an exercise on \es{2}). Then
\[ \boxed{\mu \circ f^{-1}(g) = \mu(g \circ f)} \]
In part, for $X : (\Omega, \mathcal{F}, \PP) \to
\RR$ measurable, $X \ge 0$, we have
\[ \EE(g(X)) = \int_\Omega g(X(\omega)) \dd
\PP(\omega) = \int g(x) \dd \mu_X(x) \]
where $\mu_X = \PP \circ X^{-1}$ is the \emph{law} of
$X$.

\subsubsection*{Densities of measures}

For $f : (E, \mathcal{E}, \mu) \to \RR$
measurable, $\ge 0$, let
\[ \nu(A) = \mu(f \indicator{A}) ~\forall A \in
\mathcal{E} \qquad \left( = \int_A f \dd \mu
\right) \]
Then $\nu$ is a measure on $(E, \mathcal{E})$
(check).

For any $g$ measurable, $g \ge 0$ on $E$, $\nu(g)
= \mu(fg)$, i.e. $\int g \dd \nu = \int gf \dd
\mu$.

\begin{proof}
  Holds for indicators by definition, then holds
  for simple functions by additivity, and for
  non-negative measurable functions by MCT.
\end{proof}

We say that $\nu$ has the density $f$ with respect
to $\mu$. ($\mu(f \indicator{A}) = \mu(g
\indicator{A}) ~\forall A \in \mathcal{E}$
$\implies$ $f = g$ $\mu$ \gls{al_ev} (\es{2})).

IN part, for $\mu = \lambda$ (Lebesgue measure),
$\forall f$ Borel, if $\mu(f \indicator{A}) = 0$
for all $A$ in a \pisys\ generating $\mathcal{E}$,
then $f= 0$ \gls{al_ev}.

There exists a Borel measure $\nu$ on $\RR$ given
by $\nu(A) = \int_A f(x) \dd x$ and then $\forall
g$ Borel, $g \ge 0$, $\nu(g) = \int f(x) g(x) \dd
x$. We say that $\nu$ has density $f$. This $\nu$
is a probability measure on $(\RR, \mathcal{B})$
if and only if $\int f(x) \dd x = 1$. For $X :
(\Omega, \mathcal{F}, \PP) \to \RR$, if the law
$\mu_X$ (i.e. $\PP \circ X^{-1}$) has the density
$f_X$ (with respect to $\lambda$), we call $f_X$
the probability density of $X$. Then
\[ \PP(X \in A) = \PP \circ X^{-1}(A) = \mu_X(A) =
\int_A f_X(x) \dd x \qquad \forall A \in
\mathcal{B} \forall \text{$g$ Borel}, g \ge 0 \]
(taking $A = (-\infty, x]$, $F_X(x) =
\int_{-\infty}^\infty f_X(x) \dd x$. If $F_X' =
f_X$, then this holds).
\[ = \int g(x) \dd \mu_X(x) = \int g(x) f_X(x) \dd
x \]

\subsection{Product Measures}

\begin{flashcard}[product-measure-defn]
\begin{definition*}[Product measure]
  \glssymboldefn{prod_measure}{$\otimes$}{$\otimes$}
  \cloze{
  Let $(E_1, \mathcal{E}_1, \mu_1)$ and $(E_2,
  \mathcal{E}_2, \mu_2)$ be 2 \emph{finite}
  measure spaces. On the Cartesian product $E
  \defeq E_1 \times E_2$, we consider the set of
  `rectangles'
  \[ \mathcal{A} = \{A_1 \times A_2 : A_1 \in
  \mathcal{E}_1, A_2 \in \mathcal{E}_2\} \]
  Then $\mathcal{A}$ is a \pisys. Define the
  product \sigalg\ $\mathcal{E} = \mathcal{E}_1
  \otimes \mathcal{E}_2 \defeq
  \sigma(\mathcal{A})$.
  }
\end{definition*}
\end{flashcard}
\vspace{-1em}

One can show that if $E_i$ are topological spaces
with a countable basis, then
\[ \mathcal{B}(E_1 \times E_2) = \mathcal{B}(E_1)
\pmeas \mathcal{B}(E_2) \]
(where $E_1 \times E_2$ is the product topology on
$E_1 \times E_2$; see Dudley for a proof).

\textbf{Goal:} To construct a product measure on
$(E_1 \times E_2, \mathcal{E} = \mathcal{E}_1
\pmeas \mathcal{E}_2)$.