= 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 $f_n=1_{A_n}$ gives
$$
\mathbb P(X_1\in A_1,\ldots,X_N\in A_N)=\prod_n\mathbb P(X_n\in A_n),
$$
which is independence.
Let $S_m=\sum_{n\leq m}X_n$. Independence and zero means imply orthogonality in $L^2$, so
$$
\|S_m-S_l\|_2^2=\sum_{n=l+1}^mσ_n^2.
$$
If 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 $\sum_nσ_n^2<∞$.
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