Show: For a sequence $(X_n)_{n\in\mathbb{N}}$ of random variables it is: $$ (X_n)_{n\in\mathbb{N}}\text{ converges a.e.}\Leftrightarrow\forall~\varepsilon>0:~\lim_{n\to\infty}\mathbb{P}\left(\bigcup_{j,k\geq n}\left\{\lvert X_j-X_k\rvert>\varepsilon\right\}\right)=0 $$

Hello, concerning "$\Rightarrow$" here is my previous idea:

It is for any $\varepsilon > 0$ $$ \bigcup_{j,k\geq n}\left\{\lvert X_j-X_k\rvert > \varepsilon\right\}\subset \left\{\sup_{j,k\geq n}\lvert X_j-X_k\rvert>\varepsilon\right\}, $$ so it follows $$ \mathbb{P}(\bigcup_{j,k\geq n}\left\{\lvert X_j-X_k\rvert > \varepsilon\right\})\leq\mathbb{P}\left(\left\{\sup_{j,k\geq n}\lvert X_j-X_k\rvert>\varepsilon\right\}\right). $$ Moreover, it is $$ \left\{\sup_{j,k\geq n}\lvert X_j-X_k\rvert>\varepsilon\right\}\supseteq \left\{\sup_{j,k\geq n+1}\lvert X_j-X_k\rvert>\varepsilon\right\},~~~~~\mathbb{P}\left(\left\{\sup_{j,k\geq n}\lvert X_j-X_k\rvert>\varepsilon\right\}\right)<\infty $$ so that I can apply the $\sigma$-continuity of $\mathbb{P}$, getting $$ \mathbb{P}\left(\left\{\sup_{j,k\geq n}\lvert X_j-X_k\rvert>\varepsilon\right\}\right)\to \mathbb{P}\left(\left\{\lim_{n\to\infty}\sup_{j,k\geq n}\lvert X_j-X_k\rvert>\varepsilon\right\}\right). $$ I am not totally sure, but I think that $$ \left\{\lim_{n\to\infty}\sup_{j,k\geq n}\lvert X_j-X_k\rvert>\varepsilon\right\}\subset\left\{\lvert X_j-X_k\rvert>\varepsilon\text{ for infinite many }(j,k)\right\}\\\subset\left\{\omega\in\Omega: (X_n)\mbox{ does not converge }\right\}~~~(*), $$ what means that $$ \mathbb{P}\left(\left\{\lim_{n\to\infty}\sup_{j,k\geq n}\lvert X_j-X_k\rvert>\varepsilon\right\}\right)\leq\mathbb{P}(\left\{\omega\in\Omega: (X_n)\mbox{ does not converge }\right\})=0. $$ Put all together I get $$ \mathbb{P}(\bigcup_{j,k\geq n}\left\{\lvert X_j-X_k\rvert > \varepsilon\right\})\to 0. $$

As I said, I am not totally clear about the first inclusion of $(*)$.

Is that inclusion right? Is that proof right?

What about "$\Leftarrow$"? Can you give me a hint, please?


Let the underlying probability space be $\Omega$ and $\{ \epsilon_n\}$ a strictly positive sequence that goes to $0$. Then

$$ \{ \omega \in \Omega: X_n \; \mbox{does not converge at}\; \omega\} $$

$$ = \bigcup_{\epsilon_n} \bigcap_{n = 1}^{\infty} \bigcup_{j,k \geq n} \{ \lvert X_j-X_k \rvert >\epsilon_n \}. $$


$$ \mathbb{P} \left( \{ \omega \in \Omega: X_n \; \mbox{does not converge at}\; \omega\} \right) = 0 $$

if and only if, for every $\epsilon_n$,

$$ \mathbb{P} \left( \bigcap_{n = 1}^{\infty} \bigcup_{j,k \geq n} \{ \lvert X_j-X_k \rvert >\epsilon_n \} \right) = 0. $$

By continuity from above,

$$ \mathbb{P} \left( \bigcap_{n = 1}^{\infty} \bigcup_{j,k \geq n} \{ \lvert X_j-X_k \rvert >\epsilon_n \} \right) = \lim_{n\to\infty}\mathbb{P}\left(\bigcup_{j,k\geq n}\left\{\lvert X_j-X_k\rvert>\epsilon_n\right\}\right), $$

which proves your claim.

  • $\begingroup$ Ok, you are working with the limsup of sets. The only thing that is not clear to me is that thing with the sequence $(\varepsilon_n)$... is that enough to handle any $\varepsilon>0$? $\endgroup$ – math12 Dec 27 '13 at 13:01
  • $\begingroup$ Yes, proving "for all $\epsilon > 0$..." is the same as proving "for all $\epsilon_n$...". $\endgroup$ – Michael Dec 27 '13 at 13:10
  • $\begingroup$ I cant imagine how the sequence $(\varepsilon_n)$ looks like. Are this positive real numbers from $>0$ to $\infty$? $\endgroup$ – math12 Dec 27 '13 at 13:12
  • 1
    $\begingroup$ Take $\epsilon_n = \frac{1}{n}$. $\endgroup$ – Michael Dec 27 '13 at 13:12
  • $\begingroup$ Isn't it continuity from below here? $\endgroup$ – math12 Dec 27 '13 at 13:44

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