Show that if $\mathbb{P}(X_{k}\neq 0\ \text{i.o.})=0$, then $\sum_{k=1}^{n}X_{k}$ converges almost surely.

Question

By i.o., I mean the infinity often. As asked in the title, I am working on the statement below:

Let $X_{k}$ be mutually independent random variable and set $S_{n}:=\sum_{k=1}^{n}X_{k}$, show that if $\mathbb{P}(X_{k}\neq 0\ \text{i.o.})=0$, then $S_{n}$ converges a.s.

My idea is to use $\mathbb{P}(X_{k}\neq 0\ \text{i.o.})=0$ which may imply $$\mathbb{P}(|X_{k}|>\epsilon\ \text{i.o.})=0,\ \text{for all} \ \epsilon>0.$$

Therefore, $X_{k}\rightarrow 0$ a.s. But this will give us $n^{-1}\sum_{k=1}^{n}X_{k}\rightarrow 0\ \text{a.s}$

How could I prove $S_{n}$ converges a.s.?

Thank you!

Answer 1

Let $A:=\{X_k=0\text{ ev.}\}$. By the stated assumption, $\mathsf{P}(A)=1$. For each $\omega\in A$ there exists $N_{\omega}\in \mathbb{N}$ s.t. $X_k(\omega)=0$ for all $k\ge N_{\omega}$ which means that $\lim_{n\to\infty}S_n(\omega)=\sum_{k=1}^{N_{\omega}}X_k(\omega)$.