The proof of Kolmogorovs two series theorem. I have a question about the proof of Kolmogorov's two-series theorem from Wikipedia, forund here:
https://en.wikipedia.org/wiki/Kolmogorov%27s_two-series_theorem
Here is the theorem:

Is this proof correct? What I do not understand is do we not need to take care of the cases of $\infty-\infty$, and $\infty-(-\infty)$ and $-\infty-(-\infty)$? Or are these cases taken care of in the proof?
 A: You  don't really need to worry about the infinite cases. Note that $\liminf_NS_N\le\limsup_NS_N$ almost surely, with equality if and only if the limit exists. If $\liminf_NS_N=\infty$ then equality is trivial, so assume $\liminf S_N<\infty$. Similarly if $\limsup_NS_N=-\infty$. The other cases are allowable sums: $\infty-(-\infty)=\infty$ and $-\infty-\infty=-\infty$. So there is no issue with looking instead at $\limsup_NS_N-\liminf_NS_N$ instead.
Having said that, this is a weird proof. (Wikipedia seems to make a habit of this for probability theory - the proof for Kolmogorov's inequality uses martingales, despite the fact that martingale convergence is less elementary and usually encountered later in a first probability theory course.) I believe a better approach is this:
We will show that $\{S_N\}$ is almost surely Cauchy. By Kolmogorov's inequality, we have
$$\mathbb P\left(\sup_{m<n\le k}|S_n-S_m|>\varepsilon\right)\le\varepsilon^{-2}\mathbb E|S_k-S_m|^2=\varepsilon^{-2}\sum_{n=m+1}^k\sigma_n^2\le\varepsilon^{-2}\sum_{n=m+1}^\infty\sigma_n^2$$
and so
$$L_{m,\varepsilon}:=\mathbb P\left(\sup_{n>m}|S_n-S_m|>\varepsilon\right)\le\varepsilon^{-2}\sum_{n=m+1}^\infty\sigma_n^2\to0$$
as $m\to\infty$. If you can recognize that this means $(S_n)$ is almost surely Cauchy, then you are done. If not, notice that $\sup_{m,n>N}|S_n-S_m|\le2\sup_{n>N}|S_n-S_N|$ and so
$$\mathbb P\left(\bigcap_N\bigcup_{m,n>N}\{|S_n-S_m|>\varepsilon\}\right)=\lim_{N\to\infty}\mathbb P\left(\sup_{m,n>N}|S_n-S_m|>\varepsilon\right)\le\limsup_NL_{N,\varepsilon/2}=0.$$
Now take a union over all rational $\varepsilon>0$ to deduce $\mathbb P(\{S_N\}\text{ is not Cauchy})=0$.
