TITLE: Cauchy convergence in probability implies the existence of a (finite a.e.) limit $X$
QUESTION [11 upvotes]: Cauchy convergence of a sequence $X_n$ of random variables in probability implies the existence of an $X$ (finite a.e.), such that $X_n$ converges to $X$ in probability.

The problem's hint suggests constructing a subsequence $n_k$ so that $\sum_{k=1}^\infty P\left(|X_{n_{k-1}} - X_{n_k}| > 1/2^k\right) < \infty$, and this I have accomplished, showing that in fact I have a subsequence with $\sum_{k=1}^\infty P\left(|X_{n_{k-1}} - X_{n_k}| > 1/2^k\right) \leq \sum_{k=1}^\infty 1/2^k = 1 < \infty$. But now that I have this subsequence, I'm not clear on how it implies that a limit random variable $X$ exists. I feel like I must be missing something obvious here, but I just can't put it together.

REPLY [6 votes]: By Borel-Cantelli Lemma $$ P\left(\cap_{i=1}^\infty \cup_{k=i}^\infty\left\{\mid X_{n_{k-1}} - X_{n_k}\mid>2^{-k}\right\}\right) =0, $$ so for all $\omega$, except for those belonging to an event of probability $0$, the sequence $X_{n_k}(\omega)$ is a Cauchy sequence of real numbers, which in turn must converge to a finite limit, that can be denoted $X(\omega)$. So $X_{n_k}$ converges almost surely to $X$.