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\begin{lemma*}[Lemma 1]
  Let $f : E \to \RR$ be $\mathcal{E}$-measurable.
  Then $\forall x_1 \in E_1$, the function
  \[ \label{lec13_l8_eq} x_2 \mapsto f(x_1, x_2) : E_2 \to \RR
  \tag{$*$} \]
  is $\mathcal{E}_2$-measurable.
\end{lemma*}

\begin{proof}
  $f = \indicator{A}$ where $A = A_1 \times A_2$.
  \[ \indicator{A} (x_1, x_2) = \indicator{A_1
  \times A_2} (x_1, x_2) = \indicator{A_1}(x_1)
  \cdot \indicator{A_2}(x_2) = \begin{cases}
    \indicator{A_2}(x_2) & \text{if $x_1 \in A$}
    \\
    0 & \text{otherwise}
  \end{cases} \]
  Let $\mathcal{V}$ be the set of all functions
  for which \eqref{lec13_l8_eq} holds. If $f_n \in
  \mathcal{V}$ for all $n$, $f_n \uparrow f$, then
  $f \in \mathcal{V}$. By
  \nameref{monotone_class_thm}, $\mathcal{V}$
  contains all bounded measurable functions $f$.
  $f_n = (-n) \vee f \wedge n$, $f_n \in
  \mathcal{V}$ and $f_n \to f$. So $f \in
  \mathcal{V}$.
\end{proof}

\begin{lemma*}[Lemma 2]
  Let $f : (E, \mathcal{E}) \to \RR$ be measurable
  and
  \begin{enumerate}[(i)]
    \item $f$ is bounded or
    \item $f$ is non-negative.
  \end{enumerate}
  Then
  \[ f_1(x_1) = \int_{E_2} f(x_1, x_2) \mu_2( \dd
  x_2), \quad x_1 \in E_1 \]
  is $\mathcal{E}_1$-measurable and (i) bounded or
  (ii) non-negative, taking values in $[0,
  \infty]$.
\end{lemma*}

\begin{remark*}
  A function $f$ taking values in $[0, \infty]$ is
  measurable means $f^{-1}(\infty) \in
  \mathcal{E}$, $f^{-1}(A) \in \mathcal{E}$ for
  all $A \in \mathcal{B}$.
\end{remark*}

\begin{proof}
  $f = \indicator{A_1 \times A_2}$ then $f_1(x_1)
  = \int \indicator{A_1 \times A_2} (x_1, x_2)
  \mu_2(\dd x_2) = \indicator{A_1}(x_1) \times
  \mu_2(A_2)$ is $\mathcal{E}_1$-measurable. $f_n
  \uparrow f$. Use
  \nameref{mconv_thm} and limit of
  measurable functions if measurable. Conclude
  using the \nameref{monotone_class_thm}.
\end{proof}

\begin{theorem*}
  \glssymboldefn{pmu}{$\otimes$}{$\otimes$}
  There exists a unique measure $\mu \defeq \mu_1
  \otimes \mu_2$ on $\mathcal{E}$ such that
  \[ \mu(A_1 \times A_2) = \mu_1(A_1) \mu_2(A_2)
  \qquad \forall A_1 \in \mathcal{E}_1, A_2 \in
  \mathcal{E}_2 \]
\end{theorem*}

\begin{proof}
  Uniqueness obvious as $\mathcal{A}$ is a
  \pisys\ generating $\mathcal{E}$ and $\mu$ is a
  finite measure. For existence, define the
  iterated integral
  \[ \mu(A) = \int_{E_1} \left( \int_{E_2}
  \indicator{A} (x_1, x_2) \mu_2(\dd x_2) \right)
  \mu_1(\dd x_1) \qquad A \in \mathcal{E} \]
  This definition makes sense by the previous two
  lemmas. Clearly, $\mu(\emptyset) = 0$, and
  $\mu(A_1 \times A_2) = \mu_1(A_1) \mu_2(A_2)$.
  $\mu$ is countably additive: if $(A_i)$
  disjoint, $A = \bigcup{i = 1}^\infty A_i$, then
  $\indicator{A} = \sum_{i = 1}^\infty
  \indicator{A_i}$, so apply \nameref{mconv_thm}
  twice.
\end{proof}

\begin{remark*}
  Note that
  \[ \mu(A) = \int_{E_2} \left( \int_{E_1}
  \indicator{A} (x_1, x_2) \mu_1(\dd x_1) \right)
  \mu_2(\dd x_2) \]
  by \nameref{dynkins_lemma}.
\end{remark*}

\begin{theorem*}[Fubini-Tonelli]
  \label{fubini_tonelli}
  Consider $(E, \mathcal{E}, \mu) = (E_1 \times
  E_2, \mathcal{E} \pmeas \mathcal{E}_2, \mu_1
  \pmu \mu_2)$, $\mu_i(E_i) < \infty$.
  \begin{enumerate}[(1)]
    \item Let $f : E \to \RR$ be measurable, $f
      \ge 0$. Then
      \[ \mu(f) \stackrel{(\dag)}{=} \int_{E_1}
      \left( \int_{E_2} f(x_1, x_2) \mu_2(\dd x_2)
      \right)\mu_1 (\dd x_1)
      \stackrel{(*)}{=} \int_{E_2} \left(
      \int_{E_1} f(x_1, x_2) \mu_1(\dd x_1)
      \right) \mu_2(\dd x_2) \]
    \item Let $f : E \to \RR$ be $\mu$-integrable
      (i.e. $\int |f| \dd \mu < \infty$). If we
      set
      \[ A_1 = \{x_1 \in E_1 : \int_{E_2} |f(x_1,
      x_2)| \dd \mu_2(x_2) < \infty\} \]
      and define $f_1 : E_1 \to \RR$ by $f_1(x_1)
      = \int_{E_2} f(x_1, x_2) \dd \mu_2(x_2)$ for
      all $x_1 \in A_1$ and $0$ otherwise. Then
      $\mu_1(A_1^c) = 0$ and $f_1$ is
      $\mu_1$-integrable and $\mu_1(f_1) =
      \mu(f)$.
  \end{enumerate}
\end{theorem*}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(1)]
    \item The identities ($\dag$) and ($*$) hold
      for $f = \indicator{A}$ for $A \in
      \mathcal{E}$, by definition of product
      measure $\mu$. Hence they extend to simple
      functions by linearity and for general
      functions $f \ge 0$ by \nameref{mconv_thm}
      and approximation by simple functions $f_n =
      2^{-n} \left\lfloor 2^n f \right\rfloor
      \wedge n$.
    \item Define $h : E_1 \to [0, \infty]$ as
      $h(x_1) = \int_{E_2} |f(x_1, x_2)| \mu_2(\dd
      x_2)$. By the Lemma 2, $h$ is
      measurable (as $|f| \ge 0$), is
      non-negative, so $A_1 \in \mathcal{E}_1$ (as
      $h^{-1}(\{\infty\}) \in \mathcal{E}_1$,
      $\{\infty\} = A_1^c$). So by (1),
      \[ \int_{E_1} \left( \int_{E_2} |f(x_1,
      x_2)| \mu_2(\dd x_2) \right)\mu_1(\dd x_1) =
      \mu(|f|) < \infty .\]
      Hence, $\mu_1(A_1^c) = 0$ (hence $f_1$
      integrable) (as $\mu(h) \ge \mu(h
      \indicator{A_1^c}) = \infty$ if $\mu(A_1^c)
      > \infty$). Setting
      \[ f_1^+(x_1) = \int_{E_2} f^+(x_1, x_2)
      \mu_2(\dd x_2), \qquad f_1^-(x) = \int_{E_2}
      f^-(x_1, x_2) \mu_2(\dd x_2) ,\]
      we see
      \[  f_1 = (f_1^+ - f_1^-) \indicator{A_1} =
      f_1^+ \indicator{A_1} - f_1^- \indicator{A_1} \]
      Also, $\mu_1(f_1^+) \stackrel{(1)}{=}
      \mu(f^+) < \infty$ and $\mu_1(f_1^-)
      \stackrel{(1)}{=} \mu(f^-) < \infty$, so,
      \[ \mu(f) = \mu(f^+) - \mu(f^-) =
      \mu_1(f_1^+) - \mu_1(f_1^-) = \mu_1(f_1) \qedhere \]
  \end{enumerate}
\end{proof}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item The proof of (2) is symmetric in
      $\mu_1$, $f_1$, so $\mu_1(f) = \mu(f) =
      \mu_2(f_2)$. So we can interchange the order
      of integrals whenever $f \ge 0$ or $f$
      integrable.
    \item The theorems extend to $\sigma$-finite
      measures $\mu$.
    \item Associativity is easy to check, i.e.
      $(\mathcal{E}_1 \pmeas \mathcal{E}_2) \pmeas
      \mathcal{E}_3 = \mathcal{E}_1 \pmeas
      (\mathcal{E}_2 \pmeas \mathcal{E}_3)$ and
      $\mu_1 \pmu (\mu_2 \pmu \mu_3) = (\mu_1 \pmu
      \mu_2) \pmu \mu_3$. So we can define the
      $n$-fold products $\bigotimes_{i = 1}^n
      \mu_i$ on $(E_1 \times \cdots \times E_n,
      \mathcal{E} \pmeas \cdots \pmeas
      \mathcal{E}_n)$ and $n$-fold integrals. In
      particular, when $E_i = \RR$ and $\mu_i =
      \lambda$, we get $\bigotimes_{i = 1}^n
      \lambda_i$ on $(\RR^n, \mathcal{B}(\RR^n))$.
  \end{enumerate}
\end{remark*}
