Convergence in $L_p$ does not imply "a.s." convergence

Asked Sep 25 '15 at 18:36

Active Sep 25 '15 at 18:36

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I'm having a hard time understanding some examples about this that I found on stack exchange. So I'd like to ask for the simpler possible example where a sequence of random variables $\{T_n\}$ satisfies: $$ \lim_{n\to\infty}\mathbb{E}\left(||T_n-T||^d\right)=0 $$ but not $$ \mathbb{P}\left(\lim_{n\to\infty}||T_n-T||=0\right)=1 $$

Can someone give me a simple example where the above holds?

Thanks for helping!

asked Sep 25 '15 at 18:36

Guilherme Salomé

1,069

My solution here gives an example: http://math.stackexchange.com/questions/1378565/if-independent-r-v-converge-in-probability-to-a-constant-do-they-converge-almo/1378606#1378606 – Sep 25 '15 at 18:47
Can you link to the other examples, so that people don't repeat them? – Nate Eldredge Sep 25 '15 at 18:48
@NateEldredge specifically this one: http://math.stackexchange.com/questions/138043/does-convergence-in-lp-implies-convergence-almost-everywhere – Guilherme Salomé Sep 25 '15 at 19:00
I also only have proved Borel-Cantelli first lemma – Guilherme Salomé Sep 25 '15 at 19:01

Convergence in $L_p$ does not imply "a.s." convergence

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