# Cauchy convergence in probability implies the existence of a (finite a.e.) limit $X$

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.

You have $$\sum_{k=1}^\infty P(|X_{n_{k-1}} - X_{n_k}| > 1/2^k) < \infty$$, we are going to show almost surely $$\{X_{n_k}\}$$ is a Cauchy sequence in $$\mathbb{R}$$ (or $$\mathbb{C}$$ and any other complete metric space).

It's easy to see $$E\sum_{k=1}^\infty 1_{\{|X_{n_{k-1}} - X_{n_k}| > 1/2^k\}} = \sum_{k=1}^\infty P(|X_{n_{k-1}} - X_{n_k}| > 1/2^k) < \infty$$

which implies $$\sum_{k=1}^\infty 1_{\{|X_{n_{k-1}} - X_{n_k}| > 1/2^k\}}$$ is finite almost surely. This is to say that almost surely $$\exists N(\omega)$$ such that for all $$k > N(\omega)$$, we have $$|X_{n_{k-1}} - X_{n_k}| \leq \dfrac{1}{2^k}$$

For any $$\epsilon >0$$, take $$K$$ such that $$\dfrac{1}{2^K} < \epsilon$$ and $$M = \max\{K, N(\omega)\}$$. Then for any $$l>k >M$$, we have

$$|X_{n_l} - X_{n_k}| \leq \sum_{i=k+1}^{l}|X_{n_i} - X_{n_{i-1}}| \leq \sum_{i=k+1}^l \dfrac{1}{2^i} \leq \sum_{i=k+1}^{\infty} \dfrac{1}{2^i} = \frac{1}{2^k} < \frac{1}{2^K} <\epsilon.$$

So we see $$\{X_{n_k}\}$$ are almost surely a Cauchy sequence, define $$X$$ as its almost surely limit, of course $$X$$ is almost surely finite.

Then use the fact \begin{align} P(|X_n - X| > \epsilon) &\epsilon\right) \\ &< P\left(|X_n - X_{n_k}| > \frac{\epsilon}{2}) + P(|X_{n_k} - X|>\dfrac{\epsilon}{2}\right) \end{align}

and $$X_{n}$$'s Cauchy convergence in probability and $$X_{n_k} \to X$$ almost surely to conclude.

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 $X$.