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 \sigma-algebra on X_\alpha for each \alpha, the product \sigma-algebra on X is the \sigma-algebra generated by

\{\pi_\alpha^{-1}(E_\alpha): E_\alpha\in\mathcal M_\alpha, \alpha\in A\}.

Denote this \sigma-algebra by \bigotimes_{\alpha\in A}\mathcal M_\alpha. (If A=\{1, 2, ..., n\} we also write \bigotimes_1^n\mathcal M_j or \bigotimes_1\bigotimes...\bigotimes\mathcal M_n.)

Proposition 1. If A is countable then \bigotimes_{\alpha\in A}\mathcal M_\alpha is the \sigma-algebra generated by \{\prod_{\alpha\in A}E_\alpha: E_\alpha\in\mathcal M_\alpha\}.

Proof. If E_\alpha\in\mathcal M_\alpha, then \pi_\alpha^{-1}(E_\alpha)=\prod_{\beta\in A}E_\beta where E_\beta=X_\beta for \beta\not=\alpha; on the other hand, \prod_{\alpha\in A}E_\alpha=\cap_{\alpha\in A}\pi_\alpha^{-1}(E_\alpha). \Box

Proposition 2. Suppose that $latex\mathcal M_\alpha$ 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 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=\{\prod_{\alpha\in A}E_\alpha: E_\alpha\in\mathcal E_\alpha\}.

Proof. It’s easily seen that \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 E_\alpha and hence $latex\mathcal M_\alpha$. \Box

Proposition 3. Let X_1, ..., X_n be metric spaces and let X=\prod_1^nX_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 $latex\bigotimes_1^n\mathcal B_{X_j}=\mathcal B_X$.

Proof. By the Proposition 2, \bigotimes_1^n\mathcal B_{X_j} is generated by the sets \pi_j^{-1}(U_j), 1\le j\le n, where U_j is open in X_j. Since these sets are open in X, Lemma implies that \bigotimes_1^n\mathcal B_{X_j}\subset \mathcal B_X. Suppose now that C_j is a countable dense set in X_j, and let \mathcal E_j be the collection of balls in X_j with rational radius and center in C_j. Then every open set in X_j is a union of members of \mathcal E_j (a countable union, since E_j itself is countable). Moreover, the set of points in X whose jth coordinate is in C_j for all j is a countable dense subset of X, and the balls of radius r in X are merely products of balls of radius r in the X_j‘s. It follows that \mathcal B_{X_j} is generated by \mathcal E_j and \mathcal B_X is generated by \{\prod_1^n E_j: E_j\in\mathcal E_j\}. Therefore \mathcal B_X=\bigotimes_1^n\mathcal B_{X_j}. \Box

Corollary. \mathcal B_{\mathbb R^n}=\bigotimes_1^n\mathcal B_{\mathbb R}.

Definition. An elementary family is a collection \mathcal E of subsets of X such that:

1) \emptyset\in\mathcal E,

2) if E, F\in\mathcal E then E\cap F\in\mathcal E,

3) if E\in\mathcal E then E^c is a finite disjoint union of members of \mathcal E.

Proposition. If E is an elementary family, the collection \mathcal A of finite disjoint unions of members of \mathcal E is an algebra.

Proof. If A, B\in\mathcal E and B^c=\cup_1^J C_j (C_j\in\mathcal E, disjoint), then A\setminus B=\cup_1^J(A\cap C_j) and A\cup B=(A\setminus B)\cup B, where these unions are disjoint, so A\setminus B\in\mathcal A and A\cup B\in\mathcal A. It now follows by induction that if A_1, ..., A_n\in\mathcal E, then \cup_1^nA_j\in\mathcal A; By inductive hypothesis we may assume that A_1, A_2, ..., A_{n-1} are disjoint, and then \cup_1^n A_j=A_n\cup(\cup_1^{n-1}(A_j\setminus A_n)), which is a disjoint union. Suppose A_1, ..., A_n\in\mathcal E and A_m^c=\cup_{j=1}^{J_m}B_m^j with B_m^1, ..., B_m^{J_m} disjoint members of $latex\mathcal E$, then \mathcal A is closed under complements. \Box

“The power of imagination is incredible. Often we see athletes achieving unbelievable results and wonder how they did it. One of the tools they use is visualization or mental imagery… they made the choice to create their destinies and visualized their achievements before they ultimately succeeded.” ~ George Kohlrieser


[1] Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, 2ed, page 22-24.

[2] Purpose Fairy’s 21-Day Happiness Challenge, http://www.jrmstart.com/wordpress/wp-content/uploads/2014/10/Free+eBook+-+PurposeFairys+21-Day+Happiness+Challenge.pdf.