Proof of Expectation of a Product of Independent Random Variables

Step 1 (indicators). For Borel sets $B_1,B_2$, the events $A=\{X\in B_1\}$ and $B=\{Y\in B_2\}$ are independent, so

the expectations of indicators being their probabilities directly from Simple Function and Its Integral.

Step 2 (nonnegative case). Suppose first $X\ge 0$ and $Y\ge 0$. Let $s_m$ be the dyadic approximating functions of Step 0(b) of the proof of Linearity and Monotonicity of the Lebesgue Integral, viewed as functions of the value: $s_m(X)=\varphi_m\circ X$ where $\varphi_m(t)=\min\{m,2^{-m}\lfloor 2^{m}t\rfloor\}$, so that $s_m(X)=\sum_i c_i\mathbf{1}_{\{X\in B_i\}}$ for finitely many values $c_i\ge 0$ and pairwise disjoint Borel sets $B_i$ (preimages under $\varphi_m$ of points, which are Borel: each is a finite union of intervals). Similarly $t_m(Y)=\sum_j d_j\mathbf{1}_{\{Y\in C_j\}}$. Expanding the product and using Step 1 with linearity (claim 1 of Linearity and Monotonicity of the Lebesgue Integral),

The products $s_m(X)t_m(Y)$ increase pointwise to $XY$, so by Monotone Convergence Theorem on the left and on each factor of the right (suprema of products of nondecreasing nonnegative real sequences multiply),

Step 3 (general case). Write $X=X^{+}-X^{-}$, $Y=Y^{+}-Y^{-}$ as in Integrable Function and the Lebesgue Integral. Each pair $(X^{\pm},Y^{\pm})$ satisfies the independence hypothesis of Step 2: $\{X^{+}\in B\}=\{X\in(\cdot)\}$ for a Borel set obtained from $B$ (the preimage of $B$ under $t\mapsto\max\{t,0\}$, a continuous, hence Borel measurable, map by the generator criterion of Measurable Function and Real-Valued Measurable Function), and similarly for the other parts; so the defining product formula for events transfers. By Step 2, $\mathbb{E}[X^{\pm}Y^{\pm}]=\mathbb{E}[X^{\pm}]\mathbb{E}[Y^{\pm}]$, all finite; in particular $\mathbb{E}[|XY|]=\mathbb{E}[|X|]\,\mathbb{E}[|Y|]<\infty$, so $XY$ is integrable. Expanding $XY=(X^{+}-X^{-})(Y^{+}-Y^{-})$ bilinearly and applying linearity (claim 2 of Linearity and Monotonicity of the Lebesgue Integral) to the four integrable products,

Step 4 (consequences). Let $X_1,\dots,X_r$ be independent with finite second moments, $\mu_i=\mathbb{E}[X_i]$. For $i\ne j$, the pair $(X_i,X_j)$ is independent (a subfamily), and centering preserves the independence hypothesis as in Step 3 (the map $t\mapsto t-\mu_i$ is continuous, hence Borel); products of square-integrable variables are integrable since $|uv|\le\tfrac12(u^2+v^2)$, so by Steps 1–3 and linearity,

Finally, expanding the square of the centered sum and using linearity (claim 2 of Linearity and Monotonicity of the Lebesgue Integral),

using $\mathbb{E}[\sum_i X_i]=\sum_i\mu_i$ and the variance definition. $\blacksquare$