26.2 Product $\sigma$-Algebras
Definition 26.2.1 (Product $\sigma$-Algebra).label 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 26.2.2.label 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.
- (P1)
For any $j \in I$, $\emptyset = \pi_{j}^{-1}(\emptyset) \in \ce$.
- (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
\[\bigcap_{j \in J}\pi_{j}^{-1}(E_{j}) \cap \bigcap_{j \in J'}\pi_{j}^{-1}(F_{j}) = \bigcap_{j \in J}\pi_{j}^{-1}(E_{j} \cap F_{j}) \in \ce\] - (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
\[\braks{\bigcap_{j \in J}\pi_j^{-1}(E_j)}^{c} = \bigcup_{j \in J}\pi_{j}^{-1}(E_{j}^{c}) = \bigsqcup_{\emptyset \subsetneq K \subset J}\underbrace{\paren{\bigcap_{k \in K}\pi_j^{-1}(E_k^c)} \cap \paren{\bigcap_{j \in J \setminus K}\pi_j^{-1}(E_k)}}_{\in \ce}\]
$\square$
Proposition 26.2.3.label Let $\seq{X_j}$ be topological spaces, $X = \prod_{n \in \natp}X_{n}$, then:
- (1)
$\bigotimes_{n \in \natp}\cb_{X_n}\subset \cb_{X}$.
- (2)
If $\seq{X_n}$ are separable and metrisable, then $\bigotimes_{n \in \natp}\cb_{X_n}= \cb_{X}$.
Proof, [Proposition 1.5, Fol99] and [Proposition 8.1.7, Coh13]. (2): For each $J \subset \natp$ finite, $x \in X$, and $\eps > 0$, define
By Proposition 5.5.7, $X$ is also separable. Let $\seq{x_n}\subset X$ be a countable dense subset, and
For each $x \in X$ and $U \in \cn_{X}(x)$, there exists $J \subset X$ finite and $k \in \natp$ such that $x \in B_{J}(x, 1/k) \subset U$. By density of $\seq{x_n}$, there exists $n \in \natp$ such that $x_{n} \in B_{J}(x, 1/(2k))$. In which case, $x \in B_{J}(x_{n}, 1/(2k)) \subset B_{J}(x, 1/k) \subset U$. Thus $\fB$ is a countable base for the product topology, consisting of cylinder sets in $\bigotimes_{n \in \natp}\cb_{X_n}$.
As such, every open set in $X$ is a countable union of elements of $\fB$. Therefore $\bigotimes_{n \in \natp}\cb_{X_n}\supset \cb_{X}$.$\square$
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