For arbitrary $\{X_n\}$ if $\sum_{n=1}^{\infty}{ \mathbb{E}(|{X_n}|)} <\infty$ then $\sum_{n=1}^{\infty}{ {X_n}} $ converges absolutely $a.e$ For arbitrary  $\{X_n\}$ if
$$\sum_{n=1}^{\infty}{ \mathbb{E}(|{X_n}|)} <\infty$$
then $\sum_{n=1}^{\infty}{ {X_n}} $ converges absolutely $a.e$
MY SOLUTION: For Three series of Kolmogorov,
Series 1 We must verify that $\sum_{n\ge 1} \mathbb{E}(|X_n| > 1) < +\infty$. Observe
\begin{eqnarray}
\sum_{n\ge 1} \mathbb{P}(|X_n| > 1) &\le& \sum_{n\ge 1} \mathbb{E} (|X_n|)  < \infty.\\
\end{eqnarray}
thus series 1 ok.
Series 2 We need to show $\sum_{n\ge 1} \mathbb{E}(|X_n| 1_{|X_n|\le 1})$ converges. by the hypothesis we have.
$$
\ \ \ \ \ \ \sum_{n\ge 1} \mathbb{E}(|X_n| 1_{|X_n|\le 1}) \le  \sum_{n\le 1}^{}{ \mathbb{E}(|{X_n}|)} <\infty.
$$
thus series 2 ok.
Series 3 We must show $\sum_{n\ge 1} \mathrm{Var}(|X_n| 1_{|X_n|\le 1}) < \infty$. Given $(*)$ above, this is easy:
\begin{eqnarray}
(*)\,\,\,\,\,\,\,\,\,\,\sum_{n\ge 1} \mathrm{Var}(|X_n| 1_{|X_n|\le 1}) &\le& \sum_{n\ge 1} \mathbb{E}(|X_n|^2 1_{|X_n|\le 1}) < \infty.
\end{eqnarray}
$(*)\,\,$ why this$\sum_{n\ge 1} \mathbb{E}(|X_n|^2 1_{|X_n|\le 1})$ series converges?
I do not know, thanks for your help.

It's much simpler than that.
Just note by the Monotone Convergence Theorem that $$\Bbb E\bigg[ \sum_n |X_n|\bigg] = \sum_n \Bbb E[|X_n|]$$
and one side is finite iff the other is. Hence it follows that $\Bbb E \big[ \sum_n |X_n| \big]<\infty$.
But if $Y$ is any nonnegative random variable such that $\Bbb E[Y] < \infty$ then it must follow that $Y< \infty$ almost surely. Applying this to $Y=\sum_n |X_n|$, we see that $\sum_n |X_n|<\infty$ almost surely, i.e. the series for $|X_n|$ converges absolutely a.s.
