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Consider random variables $X_1, X_2, \dots$ and $X$ on $(\Omega, \mathcal F, \mathbb P)$. We say that $X_n$ converges to $X$ almost surely if $$\mathbb P\left(\lim_{n \to \infty} X_n =X\right)=1.$$ It is claimed that an equivalent condition is: for any $\epsilon >0$, $$\lim_{n \to\infty} \mathbb P \left( |X_m -X| < \epsilon, \mbox{for all } m \geq n \right)=1. (*)$$ I have never seen this condition before, is it generally true and widely used, please? Any reference to this formula? What does this formula mean exactly, please? In particular, the part "for all $m \geq n$"? We already have $\lim_{n\to\infty}$, does this mean $m$ goes to $\infty$ faster than $n$, please? Is there another way to write this which is understandable? The proof of this equivalence is as follows, which I also have questions about. It says that $$ \left\{ \omega: \lim_{n \to\infty} X_n(\omega) = X(\omega) \right\} = \cap_{\epsilon >0} \cup_{n=1}^\infty \left\{ \omega: |X_m(\omega)-X(\omega)| < \epsilon, \mbox{for all } m \geq n \right\}. (**) $$ The $\cap_{\epsilon>0}$ part does not seem right since it produces an UN-countable intersection of measurable set, which might not be measurable any more. I suppose it should be something like $\cap_{\epsilon \in \{1/2, 1/3, \dots\}}$. Right?

From $(**)$ it claims that we have $$ \left\{ \omega: \lim_{n \to\infty} X_n(\omega) = X(\omega) \right\} = \lim_{\epsilon \to 0} \lim_{n \to\infty} \left\{ \omega: |X_m(\omega)-X(\omega)| < \epsilon, \mbox{for all } m \geq n \right\}. (***) $$ I am not sure whether we can rewrite set operation in $(**)$ as limit operation in $(***).$ Is it correct?

Then by continuity of probability measure, $(***)$ implies that $$ \mathbb P(X_n \to X) =\lim_{\epsilon \to 0} \lim_{n \to\infty} \mathbb P\left\{ |X_m-X| < \epsilon, \mbox{for all } m \geq n \right\}, $$ which yields one direction of the equivalence. Likewide, $(**)$ implies for any $\epsilon >0$, $$ \mathbb P(X_n \to X) \leq \lim_{n \to\infty} \mathbb P\left\{ |X_m-X| < \epsilon, \mbox{for all } m \geq n \right\}, $$ which yields the other direction.

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  • $\begingroup$ You should spell out the quantifiers to try to make sense of the definition. As for the $\cap_{\epsilon > 0}$, it turns out to be exactly the same to intersect over $n \in \mathbb{N}$ and replace $\epsilon$ with $1/n$. $\endgroup$ – Ian Jul 14 '15 at 1:06
  • $\begingroup$ $(*)$ means that for any $\epsilon>0$, $P\{\sup_{m\ge n}|X_m-X|\ge \epsilon\}\rightarrow 0$ as $n\rightarrow \infty$ which also implies that $X\xrightarrow{a.s.}X$ if $\sum P\{|X_n-X|\ge \epsilon\}<\infty$ for every $\epsilon>0$. $\endgroup$ – d.k.o. Jul 14 '15 at 1:12
  • $\begingroup$ @Ian Why is $\cap_\epsilon$ the same as $1/n$, please? $\endgroup$ – LaTeXFan Jul 14 '15 at 1:42
  • $\begingroup$ @20824 It isn't always; the reason it is here is because the sets being intersected are the preimage of $(0,\epsilon)$. $\endgroup$ – Ian Jul 14 '15 at 1:45
  • $\begingroup$ @Ian I do not get it. So what? I guess it is dumb to ask. $\endgroup$ – LaTeXFan Jul 14 '15 at 1:57

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