# Almost sure convergence and Borell - Cantelli Lemma 2

Suppose we have the following random variable:

$$X_n = n$$ with probability $$\frac{1}{n}$$ and $$0$$ with probability $$1-\frac{1}{n}$$. We we can define this variable on the probability space $$([0,1], \mathcal{B}[0,1], \lambda)$$ where $$\lambda$$ is the Lebesgue measure. Also, we have independence for all $$n\geq1$$.

It is said that this $$X_n$$ converges to $$0$$ almost surely (several sources). However, once we check this by the Borel - Cantelli Lemma 2, we get that $$\sum_{n=1}^\infty P(X_n=n)=\infty$$. Given that the events are independent, we know that $$X_n=n$$ infinitely often.

What could be the reason I am receiving this contradiction?

• The source is correct, and never asserts that $(X_n)$ is independent (a hypothesis which you seem to have invented by yourself and which does not hold in the setting of the notes you are referring to). – Did Dec 17 '18 at 10:41
• @Did I used the independence, because they do claim it in the same note later. It may have been a typo in the note. However, could you elaborate why the independence does not hold in this setting? – Ovi Dec 17 '18 at 10:52
• Because they explicitely build $(X_n)$ as $X_n=\mathbf 1_{(0,1/n)}$ on $[0,1]$ endowed with its Borel sigma-field and the Lebesgue measure, and these are not independent. "they do claim it in the same note later" Sure they do, but for completely different sequences $(X_n)$. No typo here. – Did Dec 17 '18 at 11:44
• @Did Thank you, I completely got it now! – Ovi Dec 17 '18 at 11:54

This is an answer to the question as posted here. The random variables are not defined explicitly in this question. It is just mentioned that they are defined on $$(0,1)$$ with Lebesgue measure. Explicit definition was given in the comments after I posted this answer.
I don't know what those sources are but we can only conclude that $$X_n \to 0$$ in probability. It need not converge almost surely.