# Almost everywhere convergence of random variables

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.

• 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 $P\left\{|X_n-X|\le \epsilon\right\}\to 1$.
– user140541
Apr 19, 2016 at 2:55
• @d.k.o. could you please expand on what you said, I don't understand. Apr 19, 2016 at 3:19
• I said that your assertion in the third paragraph is incorrect...
– user140541
Apr 19, 2016 at 3:43
• @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 Apr 19, 2016 at 15:01
• proofwiki.org/wiki/…
– user140541
Apr 19, 2016 at 15:36

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