Independent events and Kolmogorov Suppose we have a probability space $(\Omega, \mathfrak{F}, P)$, and independent events $(E_n)_n$. 

Consider $$M_n = \sum_{k=1}^n I_{E_k}$$ Is it correct to say that by the Kolmogorov $01$ law $M_n$ converges a.e or diverges a.e ?
Yes, it's a consequence of Kolmogorive 0-1 law. We can also derive it in the following way.

When $E(\sum_{k=1}^\infty I_{E_k}) &lt; +\infty$, $\sum_{k=1}^\infty I_{E_k}$ is finite almost surely, so $M_n$ converges almost surely.

When $E(\sum_{k=1}^\infty I_{E_k}) = +\infty$, we have

\begin{align}
E e^{-\sum_{k=1}^\infty I_{E_k}} &amp;= E\prod_{k=1}^\infty e^{-I_{E_k}}  \\
&amp;= \prod_{k=1}^\infty Ee^{-I_{E_k}}  \\
&amp;= \prod_{k=1}^\infty (1-P(E_k) + e^{-1}P(E_k))\\
&amp;= \prod_{k=1}^\infty(1 - (1-e^{-1})P(E_k))
\end{align}

Since $\sum_{k=1}^\infty P(E_k) = E(\sum_{k=1}^\infty I_{E_k}) = +\infty$, we have $\prod_{k=1}^\infty(1 - (1-e^{-1})P(E_k)) = 0$, (see this), i.e. $$E e^{-\sum_{k=1}^\infty I_{E_k}} = 0$$

So $e^{-\sum_{k=1}^\infty I_{E_k}} = 0$ almost surely, i.e. $M_n$ diverges almost surely.
A third proof: (First is by Kolmogorov 0-1 Law. Second is Petite Etincelle's answer)

When $E(\sum_{k=1}^\infty I_{E_k}) &lt; +\infty$, same as in Petite Etincelle's answer

When $E(\sum_{k=1}^\infty I_{E_k}) = +\infty$, we have for $A=1$,

$$\sum_{k=1}^\infty P(|I_{E_k}|\ge A) = +\infty$$

By Kolmogorov 3-Series Theorem, $\sum_{k=1}^\infty I_{E_k} = +\infty$