This is a question on a study guide to prepare for a final. I wanted to make sure that my solution is correct.
Let $\{X_n\}$ be a sequence of random variables defined on the space $(\Omega, S,P)$ where $f_n\sim exp(\lambda_n)$. Show if $\sum_{n=1}^{\infty}\frac{1}{\lambda_n}$ converges then $\sum_{n=1}^{\infty}X_n $ converges a.e.
Attempt:
Will attempt to show that $S_n=\sum_{i=1}^{n}X_i$ is cauchy a.e. Consider for a given $\epsilon > 0$, $P(\cap_{n=1}^{\infty}\cup_{k\ge n}|S_k-S_n|> \epsilon)=\lim_{n\to\infty}P(\cup_{k\ge n}|S_k-S_n|> \epsilon)$ by continuity from below. Then by countable subadditivity:
$\lim_{n\to\infty}P(\cup_{k\ge n}|S_k-S_n|> \epsilon)\le\lim_{n\to\infty}\sum_{k\ge n}P(|S_k-S_n|>\epsilon)\le\lim_{n\to\infty} \sum_{k\ge n}\frac{E[S_k-S_n]}{\epsilon}$ by markov. Since $\sum_{n=1}^{\infty}\frac{1}{\lambda_n}=\sum_{n=1}^{\infty}E[X_n]$ then $lim_{n\to\infty} \sum_{k\ge n}\frac{E[S_k-S_n]}{\epsilon}=0$ and $S_n$ is cauchy a.e and as such $\sum_{n=1}^{\infty}X_n$ converges a.e. Any help would be appreciated.
Your proof is right. Note also that $X_i$ take only no-negative values, so for possibly non-proper r.v. $Y=\sum_{n=1}^\infty X_n$ $$ 0<\mathbb E[Y]=\sum_{n=1}^\infty \mathbb E[X_n]<\infty \Rightarrow 0<Y<\infty \text{ a.s.} $$ It means that sum converges a.s.