1 Product σ-Algebra
A product σ-algebra is the σ-algebra for the cartesian products of sets. We begin by giving the formal definition of a product σ-algebra.
Let \(\{ X_{\alpha} \}_{\alpha \in A}\) be an indexed collection of nonempty sets \(X=\prod_{\alpha \in A}^{}X_{\alpha}\) and \(\pi_{\alpha}:X \to X_{\alpha}\) the coordinate maps. If \(\mathcal{M}_{\alpha}\) is a σ-algebra on \(X_{\alpha}\) for each \(\alpha\), the product σ-algebra on \(X\) is the σ-algebra generated by \[ \{ \pi_{\alpha}^{-1}(E_{\alpha}):E_{\alpha}\in\mathcal{M}_{\alpha},~\alpha \in A \}. \] We denote this σ-algebra by \(\bigotimes_{\alpha \in A}\mathcal{M}_{\alpha}\). If \(A=\{ 1, \dots, n \}\) we also write \(\bigotimes_{1}^n\mathcal{M}_{j}\) or \(\mathcal{M}_{1}\otimes \cdots \otimes \mathcal{M}_{n}\).
The following is an alternative, and perhaps more intuitive, characterization of product σ-algebras in the case of countably many factors.
If \(A\) is countable, then \(\bigotimes_{\alpha \in A} \mathcal{M}_{\alpha}\) is the \(\sigma\)-algebra generated by \[ \left\{ \prod_{\alpha \in A}^{}E_{\alpha}:E_{\alpha}\in\mathcal{M}_{\alpha} \right\}. \]
PROOF: If \(E_{\alpha}\in\mathcal{M}_{\alpha}\), then \(\pi_{\alpha}^{-1}(E_{\alpha})=\prod_{\beta \in A}^{}E_{\beta}\) where \(E_{\beta}=X\) for \(\beta \neq \alpha\); on the other hand \(\prod_{\alpha \in A}^{}E_{\alpha}=\bigcap_{\alpha \in A}\pi_{\alpha}^{-1}(E_{\alpha})\). The result follows from properties of σ-algebra. \(\square\)
Suppose that \(\mathcal{M}_{\alpha}\) is generated by \(\mathcal{E}_{\alpha}\), \(\alpha \in A\). Then \(\bigotimes_{\alpha \in A}\mathcal{M}_{\alpha}\) is generated by \(\mathcal{F}_{1}=\{ \pi_{\alpha}^{-1}(E_{\alpha}):E_{\alpha}\in\mathcal{E}_{\alpha}, \alpha \in A \}\). If \(A\) is countable and \(X_{\alpha} \in \mathcal{E}_{\alpha}\) for all \(\alpha\), \(\bigotimes_{\alpha \in A}\mathcal{M}_{\alpha}\) is generated by \(\mathcal{F}_{2}=\left\{ \prod_{\alpha \in A}^{}E_{\alpha}:E_{\alpha}\in \mathcal{E}_{\alpha} \right\}.\)
PROOF: Obviously \(\mathcal{M}(\mathcal{F}_{1})\subset \bigotimes_{\alpha \in A}\mathcal{M}_{\alpha}\). On the other hand, for each \(\alpha\), the collection \(\{ E \subset X_{\alpha}:\pi_{\alpha}^{-1}(E)\in \mathcal{M}(\mathcal{F}_{1}) \}\) is easily seen to be a \(\sigma\)-algebra on \(X_{\alpha}\) that contains \(\mathcal{E}_{\alpha}\) and hence \(\mathcal{M}_{\alpha}\). In other words \(\pi_{\alpha}^{-1}(E)\in\mathcal{M}(\mathcal{F}_{1})\) for all \(E \in \mathcal{M}_{\alpha}\), \(\alpha \in A\), and hence \(\bigotimes_{\alpha \in A}\mathcal{M}_{\alpha}\subset \mathcal{M}(\mathcal{F}_{1})\). The second assertion follows from the first as in the proof above. \(\square\)
Let \(X_{1}, \dots, X_{n}\) be metric spaces and let \(X=\prod_{j=1}^{n}X_{j}\) equipped with the product metric. Then \(\bigotimes_{1}^n\mathcal{B}_{X_{j}}\subset \mathcal{B}_{X}\). If the \(X_{j}\)’s are separable, then \(\bigotimes_{1}^n \mathcal{B}_{X_{j}}=\mathcal{B}_{X}\).
PROOF:
\(\mathcal{B}_{\mathbb{R}^n}=\bigotimes_{1}^n \mathcal{B}_{\mathbb{R}}\).