 Solution

If the variables are independent random variables, apply the defining product rule first to indicator functions, then to simple functions, and finally use bounded measurable approximation to obtain the displayed identity. Conversely, choosing $f_n=1_{A_n}$ giveswhich is independence.

Let $S_m={\sum }_{n\le m}{X}_n$. Independence and zero means imply orthogonality in $L^2$, soIf the variance series converges, $(S_m)$ is Cauchy in the complete space $L^2$ and hence converges there. Conversely, $L^2$ convergence makes $(S_m)$ Cauchy, so every tail of the nonnegative variance series tends to zero; therefore ${\sum }_n{\sigma }_n^2<\infty$.