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Consider the sequence $\{X_n\}$ given in Davide Giraudo's answer to this question.

As is explained in the answer, the lack of a.s. convergence comes from the Borel-Cantelli lemma.

Another way of proving it is by using the equivalence $$X_n\xrightarrow[n\rightarrow\infty]{a.s.} X\iff\forall\epsilon>0 \lim\limits_{n\rightarrow\infty}\Bbb P(\bigcup\limits_{m\ge n}\{|X_n-X|>\epsilon\})=0$$

and showing that for $0<\epsilon<1$ we have

$\Bbb P(\bigcap\limits_{m\ge n}\{|X_n-0|\le\epsilon\})=\prod\limits_{m\ge n}\Bbb P(X_m=0)=\prod\limits_{m\ge n}(1-{1\over m})=\lim\limits_{N\rightarrow\infty}\prod\limits_{m=n}^N {m-1\over N}=\lim\limits_{n\rightarrow\infty}{n-1\over N}=0$

1. Why can we say/conclude that by taking the complementary $$\lim\limits_{n\rightarrow\infty}\Bbb P(\bigcup\limits_{m\ge n}\{|X_m-0|>\epsilon\})=1$$

2. I understand both proofs (with the exception of 1.) but I don't see what $\Omega$ is and what kind positive measure subset $A\subset\Omega$ would satisfy $\Bbb P(\omega\in A : X_n(\omega)\nrightarrow X(\omega)\}\ $, where $X(\omega)=0\ \forall\omega\in\Omega$ is the limit variable.

John Cataldo
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  • If $\mathbb P(A)=0$ then $\mathbb P(A^c)=1$...

  • We don't care about what $\Omega $ is (the only thing important is that such a probability space exist)

    • – user657324 Apr 16 '19 at 16:51
  • @user657324 1) yes but why does it work for limits? 2) Yes but what if you had to describe it? Your comment is of no help at all – John Cataldo Apr 16 '19 at 17:48
  • I don't get your point ! $\mathbb P(\bigcup_{m\geq n}{|X_m|>\varepsilon })=1-\mathbb P(\bigcap_{m\geq n}{|X_m|\leq \varepsilon })=1-0=1.$ 2) You usually cannot describe it, especially when you work with stochastic processes. (sorry if it doesn't help, but it's hard to be more precise).
    • – user657324 Apr 16 '19 at 18:56