Kolmogorov's Three-Series Theorem

$\quad$ Proof. Necessity. The sequence $\{ S_n , n \geq 1 \}$, converges a.s., if and only if it is fundamental a.s.. The sequence $\{ S_n , n \geq 1 \}$, is fundamental a.s. if and only if$$\begin{equation}\mathrm{P}\left\{\sup _{k \geq 1}\left|S_{n+k}-S_n\right| \geq \varepsilon\right\} \rightarrow 0, \quad n \rightarrow \infty .\end{equation}$$From $\;$ Kolmogorov's inequality, we get$$\begin{align*}\mathrm{P}\left\{\sup _{k \geq 1}\left|S_{n+k}-S_n\right| \geq \varepsilon\right\} & =\lim _{N \rightarrow \infty} \mathrm{P}\left\{\max _{1 \leq k \leq N}\left|S_{n+k}-S_n\right| \geq \varepsilon\right\} \\& \leq \lim _{N \rightarrow \infty} \frac{\sum_{k=n}^{n+N} \mathrm{E} X_k^2}{\varepsilon^2}=\frac{\sum_{k=n}^{\infty} \mathrm{E} X_k^2}{\varepsilon^2} .\end{align*}$$Therefore (2) is satisfied if $\sum_{k=1}^{\infty} E X_k^2<\infty$, and consequently $S_n$ is Cauchy sequence with a.s. and hence $\lim_{n \rightarrow \infty} S_n(\omega)$ exists a.s.

Sufficiency. Now, let $\sum_k X_k$ converge. Then, by (2), for sufficiently large $n$,$$\begin{equation}\mathrm{P}\left\{\sup _{k \geq 1}\left|S_{n+k}-S_n\right| \geq \varepsilon\right\}<\frac{1}{2} .\end{equation}$$By the second part of $\;$ Kolmogorov's inequality,$$\begin{equation*}\mathbf{P}\left\{\sup _{k \geq 1}\left|S_{n+k}-S_n\right| \geq \varepsilon\right\} \geq 1-\frac{(c+\varepsilon)^2}{\sum_{k=n}^{\infty} \mathrm{E} X_k^2} .\end{equation*}$$Therefore if we suppose that $\sum_{k=1}^{\infty} E X_k^2=\infty$, we obtain$$\begin{equation*}\mathrm{P}\left\{\sup _{k \geq 1}\left|S_{n+k}-S_n\right| \geq \varepsilon\right\}=1,\end{equation*}$$which contradicts (3).This completes the proof of the theorem. $\Box$

$\quad$ Proof. To prove it, let $\mu_n = \mathrm{E} X_n^c$. Convergence of $\sum \mathrm{V} X_n^c$ and Theorem of Kolmogorov and Khinchin imply that $\sum_{n=1}^{\infty} \left( X_n^c - \mu_n \right)$ converges a.s. Convergence of $\sum \mu_n$ now gives that $\sum X_n^c$ converges a.s.But if $\sum \mathrm{P}\left(\left|X_n\right| \geq c\right)<\infty$, then by the Borel-Cantelli lemma there is number $M$ such that $\forall m > M, \; X_m \leq c$ . Therefore $X_n=X_n^c$ for all $n$ starting from $M$. Therefore $\sum X_n$ also converges a.s..

If $\sum X_n$ converges a.s. then $X_n \rightarrow 0$ a.s., and therefore, for every $c>0$, at most a finite number of the events $\left\{\left|X_n\right| \geq c\right\}$ can occur a.s.. Therefore $\sum I\left(\left|X_n\right| \geq c\right)<\infty$ a.s., and, by the second part of the Borel-Cantelli lemma, $\sum \mathrm{P}\left(\left|X_n\right|>c\right)<\infty$. Moreover, the convergence of $\sum X_n$ implies the convergence of $\sum X_n^c$. To prove of convergence of the both series $\sum \mathrm{E} X_n^c$ and $\sum \mathrm{V} X_n^c$ we will use symmetrization method. In addition to the sequence$X_1^c, X_2^c, \ldots$, we consider a different sequence, $\hat{X_1^c}, \hat{X_2^c}, \ldots$ of independent random variable such $X_n^c$ has the same distribution as $\hat{X_n^c}, \: n\geq n$.
Then if $\sum_nX_n^c$ converges a.s., the series $\sum_n\hat{X_1^c}$ also converges, and hence so does $\sum_n \left(X_n^c-\hat{X_n^c}\right)$. But $\mathrm{E}\left(X_n-\hat{X_n^c}\right)=0$ and $\mathrm{P} \left( \left| X_n^c-\hat{X_n^c} \right| \leq 2 c\right)=1$. Therefore $\sum_n\mathrm{V}\left(X_n^c-\hat{X_n^c}\right)<\infty$ by Theorem of Kolmogorov and Khinchin. In addition,$$\begin{equation}\sum_n\mathrm{V} X_n=\frac{1}{2} \sum_n\mathrm{V}\left(X_n^c-\hat{X_n^c}\right)<\infty .\end{equation}$$Consequently, by Theorem of Kolmogorov and Khinchin, $\sum\left(X_n^c-\mathrm{E} \hat{X_n^c}\right)$ converges witha.s. , and therefore $\sum_n\mathrm{E} X_n^c$ converges.
Thus if $\sum_nX_n^c$ converges a.s. (and $\mathrm{P}\left(\left|X_n^c\right| \leq c\right)=1, n \geq 1$ ) it follows that both $\sum_n\mathrm{E} X_n^c$ and $\sum_n\mathrm{V} X_n^c$ converge. $\Box$