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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?

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  • $\begingroup$ 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). $\endgroup$ – Did Dec 17 '18 at 10:41
  • $\begingroup$ @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? $\endgroup$ – Ovi Dec 17 '18 at 10:52
  • $\begingroup$ 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. $\endgroup$ – Did Dec 17 '18 at 11:44
  • $\begingroup$ @Did Thank you, I completely got it now! $\endgroup$ – Ovi Dec 17 '18 at 11:54
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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.

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  • $\begingroup$ Thanks for the comment! Here is the link: nptel.ac.in/courses/108106083/lecture28_Convergence.pdf). Page 3, Example. I also think that, clearly, we can only talk about convergence in probability. $\endgroup$ – Ovi Dec 17 '18 at 10:39
  • $\begingroup$ See comment on main. $\endgroup$ – Did Dec 17 '18 at 10:41

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