# Necessary and sufficient condition for convergence almost surely by Borel-Cantelli lemma

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.
• 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 $a_n\leq \frac{1}{n}$. – Keen-ameteur Aug 20 at 9:28
• $P(X_n>t)=\frac 1 n$ for any $t>0$ because $X_n \neq 0$ only when $\omega \in (0,\frac 1 n)$. – Kabo Murphy Aug 20 at 9:29
• But $a_n$ is not constant, aside from the fact that it is true only for $n$ large enough. – Keen-ameteur Aug 20 at 9:31
• Do you agree that $X_n >a_n$ implies $\omega \in (0,\frac 1 n)$ and the converse is also true as long as $n >a_n$? – Kabo Murphy Aug 20 at 9:33
• 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 $a_n=\frac{2}{n}$. – Keen-ameteur Aug 20 at 9:37