I have a sequence of independent random variables $X_1, X_2,...$ such that $P(X_n = 1) = \frac{1}{n}$ and $P(X_n = 0) = 1 - \frac{1}{n}$. Using the second Borel-Cantelli Lemma, we have $\sum P(X_n \neq 0) = \sum P(X_n = 1) = \sum \frac{1}{n} = \infty$ implies that $X_n$ does not converge to 0 a.s.

But I feel like we can construct an example where almost convergence occurs. Let $\Omega = (0, 1)$, and let the probability measure be the Lebesgue measure restricted to $(0, 1)$. Finally, let $X_n = \mathbb{I}\{(0, \frac{1}{n})\}$. Then for any $\omega \in \Omega$, $\underset{n \to \infty}{\lim} X_n(\omega) = 0$, so $X_n$ converges to 0 a.s.

I don't understand why I'm getting contradicting conclusions. Any help would be much appreciated. Thank you!


The random variables $X_n := 1_{(0,1/n)}$ are not independent. This follows from the fact that

$$\lambda(X_n=1,X_m=1) = \lambda(X_m=1) = \frac{1}{m} \neq \frac{1}{n} \frac{1}{m} = \lambda(X_n=1) \lambda(X_m=1)$$

for any $m \leq n$. (Here $\lambda$ denotes the Lebesgue measure restricted to $(0,1)$.)

Consequence: The assumptions of the Borel-Cantelli Lema are not satisfied. Therefore, the almost surely convergence of the sequence is no contradiction to the Borel-Cantelli Lemma.

  • $\begingroup$ Thank you so much for the reply! I feel so stupid now. :P I'm sorry I couldn't up-vote because of my low reputation. $\endgroup$ – user207886 Jan 14 '15 at 18:16
  • $\begingroup$ @user207886 You are welcome. $\endgroup$ – saz Jan 14 '15 at 18:36

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