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 $j$th 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

References:

[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.