Consider independent random variables $X_n\sim\operatorname{\mathsf{Ber}}\left(\frac1n\right)$. The problem is whether $$X_n\xrightarrow{\mathrm{a.s.}}0$$ is true or not. I tried to use the definition $$\Pr\left(\lim_{n\to\infty}X_n=0\right)=1$$ But I do not understand what the limit of those mean. I read that it can be rewritten like this $$\Pr\left(\omega:\lim_{n\to\infty}X_n(\omega)=0\right)=1$$ and it does not help because I am not aware of what $\Omega$ is. So, what does the limit above mean and how do we find the $\Omega$ like this for given $X_n$? For context, I have seen this question, but I do not know Borel-Cantelli or Lebesgue theory.

My friend and I were discussing and he suggested the following. Let $\Omega$ be the set of all sequences $(x_1,x_2,\dots)$ with each $x_i$ chosen uniformly from $[0,1]$ (that is the probability assignment). Then, define $X_n$ to be the indicator of $x_i\leq\frac1n$. Then, for the sequences of the form $x_k=1/k^{\alpha}$ for some $\alpha>1$, we have that $X_n$ does not converge to $0$. However, we do not know if this set has nonzero probability, which would imply that the almost sure convergence is not true.

  • 3
    $\begingroup$ Try using the second Borel-Cantelli lemma, noting that $\sum_{n=1}^{\infty}1/n = \infty$ along the way... $\endgroup$
    – jbowman
    Nov 17, 2023 at 18:09
  • $\begingroup$ @jbowman: I don't understand how to use it here. What are the $E_n$s? Is there a proof or intuition directly based on the definition of a.s. convergence? $\endgroup$ Nov 18, 2023 at 14:12
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    $\begingroup$ $E_n$ is the $n^{th}$ event whose probability you are looking at, e.g., $X_n = 1$. 2BC tells us that $X_n = 1$ will occur infinitely often. See if you can relate that to the definition of a.s. convergence. $\endgroup$
    – jbowman
    Nov 18, 2023 at 15:30


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