0

I know by the Borel-Cantelli lemma that for all $\epsilon > 0, \ \sum_{n=1}^{\infty} P(|X_n|>\epsilon)<\infty \Rightarrow X_n \xrightarrow{a.e.} 0$. But, how can I prove that the if and only if condition holds, which means that also holds that if $X_n \xrightarrow{a.e.} 0 \Rightarrow$, for all $\epsilon > 0,\ \ \sum_{n=1}^{\infty} P(|X_n|>\epsilon)<\infty$. I know that by another proposition of this lemma holds that $X_n \xrightarrow{a.e.} 0 \Leftrightarrow$, $P(|X_n|>\epsilon)= 0$, but I need to add that the sum of this probabilities greater than $\epsilon$ is less than $\infty$?.

Thank you very much.

NHB
  • 1

1 Answers1

1

It's not true without further assumptions. If all you know is that $X_n \to 0$ almost surely, then $P(|X_n| > \epsilon)$ can go to $0$ arbitrarily slowly.

EDIT: For example, let $U$ have uniform distribution on $[0,1]$, and $X_n = 1$ if $U < t_n$, $0$ otherwise, where $t_n \to 0+$. Thus if $0 < \epsilon < 1$, $P(|X_n| > \epsilon) = t_n$. We do have $X_n \to 0$ almost surely, but $t_n$ can go to $0$ arbitrarily slowly, and in particular $\sum_{n=1}^\infty t_n$ need not converge.

EDIT: If the $X_n$ are independent, you can use the second Borel-Cantelli lemma.

Robert Israel
  • 448,999
  • could you be so kind as to explain yourself a little further, please? Thank you so much for your help. – NHB Nov 17 '14 at 19:52
  • Sorry, and if I assume that are independent, can I prove that if $X_n \xrightarrow{a.e.}0\ \Rightarrow \sum P(|X_n|>\epsilon)<\infty$??? thank you very much – NHB Nov 18 '14 at 17:35