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.


2 Answers 2


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 $\infty$.

  • 1
    $\begingroup$ 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
    Commented Mar 10, 2015 at 23:39
  • 1
    $\begingroup$ 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
    Commented Mar 11, 2015 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
    Commented May 10, 2019 at 22:19
  • $\begingroup$ @Rdrr Then you should have no trouble demonstrating the $X_n$ are not independent. $\endgroup$
    – whuber
    Commented May 11, 2019 at 13:43

A little late to the party but here's my attempt. Stumbled upon this when doing a similar problem, but came up with a different solution.

Usually, questions of this sort will involve some kind of dependent random variable (since the converse of the Borel-Cantelli lemma fails when there's non-independence).

Consider the martingale $S_n = \sum_{i=0}^n X_i$, which starts at $S_0 = X_0 = 1$, with increments $P(X_k = \pm 1) = \frac{1}{2}$.

Define the stopping time $\tau = \inf \{n: S_n = 0\}$. Since we are dealing with a simple symmetric random walk, $\tau < \infty$ almost surely, and hence, $S_{n \wedge \tau} = 0$ almost surely as $n \rightarrow \infty$.

Before we provide the counter example to show almost sure convergence does not imply complete convergence, lets first calculate $\mathbb{P} (n < \tau)$. The strategy used here is similar to the gambler ruin's problem, namely, $$ 1 = \mathbb{E}[S_{n \wedge \tau}] \leq n (\mathbb{P} (n < \tau)) + 0(\mathbb{P} (n \geq \tau)), $$ where the inequality is from the observation that $S_n \leq n$ and therefore, $\mathbb{P} (n < \tau) \geq \frac{1}{n}$.

Now, we proceed to demonstrate by counterexample that $S_{n \wedge \tau} \stackrel{a.s.}{\rightarrow} 0 $ does not imply that $S_{n \wedge \tau}$ converges completely to 0. By the definition of complete convergence, $$ \begin{split} \sum_{n=1}^\infty \mathbb{P}(|S_{n \wedge \tau}-0|>\epsilon) &= \sum_{n=1}^\infty \mathbb{P}(|S_{n \wedge \tau}|>\epsilon) \\ &= \sum_{n=1}^\infty \mathbb{P}(S_{n \wedge \tau} \neq 0) \\ &= \sum_{n=1}^\infty \mathbb{P}(n < \tau) \\ &\geq \sum_{n=1}^\infty \frac{1}{n} \\ &= \infty, \end{split} $$ and therefore, a.s. convergence does not necessarily imply complete convergence.


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