Given a sequence of random variables $\{X_n \}_{n=1}^\infty$, the first Borel-Cantelli lemma tells us that if there is a positive sequence $\{ a_m \}_{m=1}^\infty$ for which:

$$ a_m \overset{m\rightarrow\infty}{\longrightarrow} 0 \quad \text{and} \quad \sum\limits_{n,m=1}^\infty \mathbb{P}\big( \vert X_n\vert> a_m \big) <\infty \tag{$\circledast$} $$ Then $X_n$ converges almost surely to $0$. My question is whether there is also a converse relation, i.e, $X_n\rightarrow0$ almost surely implies that there exists a positive sequence $\{a_m \}$ such that $\circledast$ holds?


If $a_n \to 0$ and $\sum_n P(|x_n| >a_n) <\infty$ then $X_n \to 0$ almost surely, by Borel Cantelli. Assuming that this is the sum you intended I will give a counterexample for the converse. Consider $(0,1)$ with Lebesgue measure and let $X_n(\omega)=n$ for $0 <\omega <\frac 1 n$, $X_n(\omega)=0$ for $\omega \geq \frac 1 n$. Then $X_n \to 0$ at every point but $P(X_n >a_n)=\frac 1 n$ for so $\sum_n P(X_n >a_n)=\infty$ no matter sequence $(a_n)$ you choose.

Of course this example also works for your double sum.

  • $\begingroup$ First, thank you for your answer, but I am not sure that $\mathbb{P}(X_n>a_n)=\frac{1}{n}$ in your example. For $a_n=\frac{1}{2n}$ this is zero constantly. It seems to me that $\mathbb{P}(X_n>a_n)=\frac{1}{n}-a_n$ when $a_n\leq \frac{1}{n}$. $\endgroup$ – Keen-ameteur Aug 20 '19 at 9:28
  • $\begingroup$ $P(X_n>t)=\frac 1 n$ for any $t>0$ because $X_n \neq 0$ only when $\omega \in (0,\frac 1 n)$. $\endgroup$ – Kavi Rama Murthy Aug 20 '19 at 9:29
  • $\begingroup$ But $a_n$ is not constant, aside from the fact that it is true only for $n$ large enough. $\endgroup$ – Keen-ameteur Aug 20 '19 at 9:31
  • $\begingroup$ Do you agree that $X_n >a_n$ implies $\omega \in (0,\frac 1 n)$ and the converse is also true as long as $n >a_n$? $\endgroup$ – Kavi Rama Murthy Aug 20 '19 at 9:33
  • $\begingroup$ I think that $X_n>a_n$ implies that $\omega\in \big( a_n, \frac{1}{n} \big)$. And I noticed that I miswrote the example, and meant to suggest $a_n=\frac{2}{n}$. $\endgroup$ – Keen-ameteur Aug 20 '19 at 9:37

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