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.

  • $\begingroup$ 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


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 $m$, $$ P\left\{\sup_{n\ge m}X_n\le 1/2\right\}=\lim_{N\to\infty}\prod_{n=m}^{N}(1-n^{-1})=0. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.