Expectation factors over a finite product of mutually independent random variables

In the original finite-valued case, let $J\subseteq I$ and let $B_i\subseteq \mathbb{R}$ be Borel for $i\in J$. Write $A_i$ for the finite attained-value set of $X_i$. The event ${\bigcap }_{i\in J}\{X_i\in B_i\}$ is the disjoint union of the joint atoms indexed by ${\prod }_{i\in J}(A_i\cap B_i)$. By [L3] and [L4], its probability is ${\sum }_{(x_i)\in {\prod }_{i\in J}(A_i\cap B_i)}{\prod }_{i\in J}\mathbb{P}(X_i=x_i)={\prod }_{i\in J}{\sum }_{x_i\in A_i\cap B_i}\mathbb{P}(X_i=x_i)={\prod }_{i\in J}\mathbb{P}(X_i\in B_i)\mathrm{.}$ Factoring finite sums proves the middle equality, including empty factors. Thus the family is independent in the sense of [L1]. Each variable is bounded because its attained-value set is finite, hence integrable on a probability space.

Suppose $I\mathrm{\ne }\mathrm{\varnothing }$ and every $X_i$ is integrable. Enumerate the finite set $I$ by $0,\dots ,n-1$ and apply [L2] with $g_i(x)=x$, which is Borel. It gives product integrability and the asserted real factorization. Selecting one enumeration of one finite set uses no axiom of choice. Step 1.1 supplies the required independence and integrability in the original finite-valued case.

Suppose $I\mathrm{\ne }\mathrm{\varnothing }$ and every $X_i$ is nonnegative. Apply [L2] to the same enumeration with $g_i(x)=\max (x,0)$, a continuous nonnegative Borel function. Since $g_i(X_i)=X_i$, [L2] gives the asserted extended factorization with precisely its zero-factor convention.

For $I=\mathrm{\varnothing }$, the random product is the constant $1$, whose expectation is $1$ on a probability space, and the product of expectations is the empty product $1$. Steps 2.1 and 3.1 and this identity prove all stated cases.