# Convergence a.e. of a sequence of random variables (an equivalence is to show)

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

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

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

So

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

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.

• 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 $\varepsilon>0$? – math12 Dec 27 '13 at 13:01
• Yes, proving "for all $\epsilon > 0$..." is the same as proving "for all $\epsilon_n$...". – Michael Dec 27 '13 at 13:10
• I cant imagine how the sequence $(\varepsilon_n)$ looks like. Are this positive real numbers from $>0$ to $\infty$? – math12 Dec 27 '13 at 13:12
• Take $\epsilon_n = \frac{1}{n}$. – Michael Dec 27 '13 at 13:12
• Isn't it continuity from below here? – math12 Dec 27 '13 at 13:44