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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.

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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 $(\Omega, \mathfrak{F}, \mu)$.

Figure

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 $\infty$.

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    $\begingroup$ Thanks a lot!. Two comments, is there a reason to define $$X_n(\omega) = 2 + (-1)^n \text{ when } \omega \le 1/n$$ instead of $$X_n(\omega) = 1 \text{ when } \omega \le 1/n$$? second, should it be $\sum\text{Pr}\left(X_n > \epsilon\right)$? $\endgroup$ – Manuel Mar 10 '15 at 23:39
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    $\begingroup$ 1. No good reason. While I was thinking this through, I used the $\pm 1$ term as a reminder that there might not be convergence at such points. 2. I fixed the $\lt$ typo, thanks. $\endgroup$ – whuber Mar 11 '15 at 3:10
  • $\begingroup$ Are the $X_n$ independent? They seem to be to me, which by the Second Borel Cantelli lemma would imply the convergence is not almost sure. $\endgroup$ – Rdrr May 10 '19 at 22:19
  • $\begingroup$ @Rdrr Then you should have no trouble demonstrating the $X_n$ are not independent. $\endgroup$ – whuber May 11 '19 at 13:43

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