TITLE: Kolmogorov's maximal inequality and convergence of random series.
QUESTION [6 upvotes]: Let $(X_n)_{n\ge 1}$ be a sequence of mutually independent random variables, on the same probability space, with expectation 0 and finite variance. Let $S_n = \sum_{l=1}^n X_l$. Prove that for any $\lambda > 0$,
$$\lambda^2 \displaystyle P \left(\max_{1\le k\le n} \big| S_k \big| \ge \lambda \right) \le Var(S_n)$$
Prove that if $\sum_l Var(X_l) < \infty$, then $(S_n)_{n\ge 1}$ converges almost surely. 
gf

REPLY [10 votes]: There's a more elementary proof which avoids the use of martingales$^1$. Observe that the event $\{\max_{1\le k\le n}|S_k|\ge\lambda\}$ is the disjoint union of the events $A_k=\{|S_k|\ge\lambda,|S_j|<\lambda\text{ for }j<k\}$, $k=1,\ldots,n$. Moreover, we have
\begin{align*}
\mathbb P(A_k) &\le\lambda^{-2}\mathbb E[S_k^2\mathbf1_{A_k}]\\
&\le\lambda^{-2}\mathbb E[\big(S_k^2+(S_n-S_k)^2\big)\mathbf1_{A_k}]\\
&=\lambda^{-2}\mathbb E[\big(S_k^2+2S_k(S_n-S_k)+(S_n-S_k)^2\big)\mathbf1_{A_k}],
\end{align*}
where the last line follows since $S_k\mathbf1_{A_k}$ and $S_n-S_k$ are independent and $\mathbb E[S_n-S_k]=0$. Thus we have
$$\mathbb P(A_k)\le\lambda^{-2}\mathbb E[\big(S_k+(S_n-S_k)\big)^2\mathbf1_{A_k}]=\lambda^{-2}\mathbb E[S_n^2\mathbf1_{A_k}]$$
and so summing over $k$ we find
$$\mathbb P\left(\max_{1\le k\le n}|S_k|\ge\lambda\right)\le\lambda^{-2}\mathbb E[S_n^2\mathbf1_{\max_{1\le k\le n}|S_k|\ge\lambda}]\le\lambda^{-2}\mathbb  E[S_n^2].$$

$1$. I have always found the use of martingales here unsatisfactory. Kolmogorov's inequality is often used to prove the Strong Law of Large Numbers which I imagine most would encounter before martingales in a first course in rigorous probability theory. Moreover, one of the existing answers here lifts directly from Wikipedia - aside from the blatant plagiarism, that proof has some issues. (E.g. it assumes both $\mathbb E[S_i]=0$ and $S_i\ge0$, it does not justify why $Z_n$ is a martingale, etc.)