We say $X_1, X_2, \ldots$ converge completely to $X$ if for every $\epsilon>0$ $\sum_{n=1}^\infty \text{P}\left(|X_n-X|>\epsilon\right) <\infty$.
With Borel Cantelli's lemma is straight forward to prove that complete convergence implies almost sure convergence.
I am looking for an example were almost sure convergence cannot be proven with Borel Cantelli. This is, a sequence of random variables that converges almost surely but not completely.
Best Answer
Let $\Omega=(0,1)$ with the Borel sigma-algebra $\mathfrak{F}$ and uniform measure $\mu$. Define
$$X_n(\omega) = 2 + (-1)^n \text{ when } \omega \le 1/n$$
and $X_n(\omega)=0$ otherwise. The $X_n$ are obviously measurable on the probability space $(\Omega, \mathfrak{F}, \mu)$.
For any $\omega\in\Omega$ and all $N \gt 1/\omega$ it is the case that $X_n(\omega)=0$. Thus, by definition, the sequence $(X_n)$ converges to $0$ (not just almost surely!).
However, whenever $0 \lt \epsilon \lt 1$, $\Pr(X_n\gt \epsilon) = \Pr(X_n \ne 0) = 1/n$, whence
$$\sum_{n=1}^\infty \Pr(X_n \gt \epsilon) = \sum_{n=1}^\infty \frac{1}{n},$$
which diverges to $\infty$.