15.2 Product $\sigma$-Algebras
Definition 15.2.1 (Product $\sigma$-Algebra). Let $\bracs{(X_i, \cm_i)}_{i \in I}$ be measurable spaces, then the product $\sigma$-algebra $\bigotimes_{i \in I}\cm_{i}$ on $\prod_{i \in I}X_{i}$ is the $\sigma$-algebra generated by $\seqi{\pi_i}$.
Lemma 15.2.2. Let $\bracs{(X_i, \cm_i)}_{i \in I}$, let
then $\ce$ is an elementary family, and $\sigma(\ce) = \bigotimes_{i \in I}\cm_{i}$.
Proof.
For any $j \in I$, $\emptyset = \pi_{j}^{-1}(\emptyset) \in \ce$.
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
\[\bigcap_{j \in J}\pi_{j}^{-1}(E_{j}) \cap \bigcap_{j \in J'}\pi_{j}^{-1}(F_{j}) = \bigcap_{j \in J}\pi_{j}^{-1}(E_{j} \cap F_{j}) \in \ce\]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
\[\braks{\bigcap_{j \in J}\pi_j^{-1}(E_j)}^{c} = \bigcup_{j \in J}\pi_{j}^{-1}(E_{j}^{c}) = \bigsqcup_{\emptyset \subsetneq K \subset J}\underbrace{\paren{\bigcap_{k \in K}\pi_j^{-1}(E_k^c)} \cap \paren{\bigcap_{j \in J \setminus K}\pi_j^{-1}(E_k)}}_{\in \ce}\]
$\square$
Proposition 15.2.3. Let $\seqf{X_j}$ be topological spaces, $X = \prod_{j = 1}^{n} X_{j}$, then:
$\bigotimes_{j = 1}^{n} \cb_{X_j}\subset \cb_{X}$.
If $\seq{X_j}$ are separable, then $\bigotimes_{j = 1}^{n} \cb_{X_j}= \cb_{X}$
Proof. (2): By Proposition 4.5.7, $X$ 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
The open set $U$ is in $\bigotimes_{j = 1}^{n} \cb_{X_j}$.$\square$