Suppose $\{X_{n}\}$ is a sequence of non-negative but not necessarily independent random variables, $S_n$ is the sum of the first $n$ random variables, and $S_{n}$ converges to $S$ in probability. I would like to show in fact that $S_{n}$ converges to $S$ almost surely. I tried mimicking the argument in how to show convergence in probability imply convergence a.s. in this case? but I believe since we don't have independence here this won't work. I tried using the fact that $X_{n}$ is Cauchy in probability to try and show that set where $S_{n}$ goes to infinity has probability 0 but couldn't figure out the details. Appreciate if someone could help out.


Independence is not necessary. Since $X_n \geq 0$ it follows that $(S_n)$ is increasing. So $T=\lim S_n $ exists (but may be $\infty$). But $S_n \to S$ in probability and this implies there is a subsequence $(S_{n_k})$ which converges a.s to $S$. [Theorem 4.2.3 of K L Chung's "A Course in Probability Theory']. Now $(S_{n_k})$ converges to $T$ as well as $S$ so $S=T$ a.s.. Hence $S_n \to T=S$ with probability $1$.