\section{Product $\sigma$-Algebras}
\label{section:product-sigma-algebra}

\begin{definition}[Product $\sigma$-Algebra]
\label{definition:product-sigma-algebra}
    Let $\bracs{(X_i, \cm_i)}_{i \in I}$ be measurable spaces, then the \textbf{product} $\sigma$-algebra $\bigotimes_{i \in I}\cm_i$ on $\prod_{i \in I}X_i$ is the $\sigma$-algebra generated by $\seqi{\pi_i}$.
\end{definition}

\begin{lemma}
\label{lemma:rectangles}
    Let $\bracs{(X_i, \cm_i)}_{i \in I}$, let
    \[
    \ce = \bracs{\bigcap_{j \in J}\pi_j^{-1}(E_j) \bigg | J \subset I \text{ finite}, E_j \in \cm_j}
    \]

    then $\ce$ is an elementary family, and $\sigma(\ce) = \bigotimes_{i \in I}\cm_i$.
\end{lemma}
\begin{proof}
\begin{enumerate}
    \item[(P1)] For any $j \in I$, $\emptyset = \pi_j^{-1}(\emptyset) \in \ce$.
    \item[(P2)] Let
    \[
    \bigcap_{j \in J}\pi_j^{-1}(E_j), \bigcap_{j \in J'}\pi_j^{-1}(F_j) \in \ce
    \]

    Assume without loss of generality that $J = J'$, then
    \[
    \bigcap_{j \in J}\pi_j^{-1}(E_j) \cap \bigcap_{j \in J'}\pi_j^{-1}(F_j) = \bigcap_{j \in J}\pi_j^{-1}(E_j \cap F_j) \in \ce
    \]

    \item[(E)] For any $j \in I$, $\prod_{i \in I}X_i = \pi_j^{-1}(X_j) \in \ce$. Thus it is sufficient to show that $\ce$ is closed under complements. Let $\bigcap_{j \in J}\pi_j^{-1}(E_j) \in \ce$, then
    \[
    \braks{\bigcap_{j \in J}\pi_j^{-1}(E_j)}^c = \bigcup_{j \in J}\pi_j^{-1}(E_j^c) = \bigsqcup_{\emptyset \subsetneq K \subset J}
    \underbrace{\paren{\bigcap_{k \in K}\pi_j^{-1}(E_k^c)} \cap \paren{\bigcap_{j \in J \setminus K}\pi_j^{-1}(E_k)}}_{\in \ce}
    \]

\end{enumerate}
\end{proof}

\begin{proposition}[{{\cite[Proposition 1.5]{Folland}}}]
\label{proposition:product-sigma-algebra-metric}
    Let $\seqf{X_j}$ be topological spaces, $X = \prod_{j = 1}^n X_j$, then:
    \begin{enumerate}
        \item $\bigotimes_{j = 1}^n \cb_{X_j} \subset \cb_{X}$.
        \item If $\seq{X_j}$ are separable, then $\bigotimes_{j = 1}^n \cb_{X_j} = \cb_{X}$
    \end{enumerate}
\end{proposition}
\begin{proof}
    (2): By \autoref{proposition:separable-product}, $X$ is also separable. Let $U \subset X$, then there exists $\seq{x_j} \subset X$ such that $\overline{\bracs{x_j|j \in \natp}} \supset U$. For each $j \in \nat$, let $\bracs{U_{j, k}}$ such that $U_{j, k} \in \cn_{X_k}(\pi_k(x_j))$ for each $1 \le k \le n$ and $x_j \in \bigcap_{k = 1}^n \pi_k^{-1}(U_{j, k}) \subset U$.

    Since $\bigcap_{k = 1}^n \pi_k^{-1}(U_{j, k}) \in \bigotimes_{j = 1}^n \cb_{X_j}$ and
    \[
    U = \bigcup_{j \in \natp}\bigcap_{k = 1}^n \pi_k^{-1}(U_{j, k})
    \]

    The open set $U$ is in $\bigotimes_{j = 1}^n \cb_{X_j}$.
\end{proof}