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This is a question that my teacher is having us do for homework but I think there might be a typo in it. I was hoping if someone clear this up for me.

The sequence $\{X_n\}$ of random variables converges almost everywhere to $X$ if and only if for every $\epsilon > 0$ we have $ \lim_{m \rightarrow \infty} P(|X_n-X|\le \epsilon \; for \; all \; n \ge m) =1 $ or equivalently $ \lim_{m \rightarrow \infty} P(|X_n-X|> \epsilon \; for \; some \; n \ge m) =0 $.

I think that $ \lim_{m \rightarrow \infty} P(|X_n-X|\le \epsilon \; for \; all \; n \ge m) =1 $ is the same thing as saying $ \lim_{n \rightarrow \infty} P(|X_n-X|\le \epsilon ) =1 $ which is then the same as saying $ \lim_{n \rightarrow \infty} P(|X_n-X|> \epsilon ) =0 $ which means convergence in probability. And this is where I think there is a mistake because convergence in probability does not imply almost everywhere convergence.

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  • $\begingroup$ They are not the same as $$ P\left\{\bigcap_{n\ge m}\{|X_n-X|\le \epsilon\}\right\}=P\left\{\sup_{n\ge m}|X_n-X|\le \epsilon\right\}\to 1 $$ is stronger than requiring $P\left\{|X_n-X|\le \epsilon\right\}\to 1$. $\endgroup$
    – user140541
    Apr 19, 2016 at 2:55
  • $\begingroup$ @d.k.o. could you please expand on what you said, I don't understand. $\endgroup$
    – alpastor
    Apr 19, 2016 at 3:19
  • $\begingroup$ I said that your assertion in the third paragraph is incorrect... $\endgroup$
    – user140541
    Apr 19, 2016 at 3:43
  • $\begingroup$ @d.k.o. yeah I got that part, could you explain where the intersection and sup came in. That's what I don't understand $\endgroup$
    – alpastor
    Apr 19, 2016 at 15:01
  • $\begingroup$ proofwiki.org/wiki/… $\endgroup$
    – user140541
    Apr 19, 2016 at 15:36

1 Answer 1

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Here is a (standard) counterexample: consider a sequence $\{X_n\}$ of independent r.v.s. with $$ P\{X_n=1\}=n^{-1} \text{ and }P\{X_n=0\}=1-n^{-1}. $$ Then $X_n\xrightarrow{p} 0$ because $\forall \epsilon>0$, $$ P\{X_n\le\epsilon\}\ge 1-n^{-1}\to 1 \text{ as }n\to \infty. $$ However, $X_n\not\xrightarrow{a.s.} 0$ because for any $m$, $$ P\left\{\sup_{n\ge m}X_n\le 1/2\right\}=\lim_{N\to\infty}\prod_{n=m}^{N}(1-n^{-1})=0. $$

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