# Convergence in Probability - Definition in Billingsley

In defining convergence in probability on p. 70, Billingsley considers the complement of the event $\{\omega \in \Omega : \lim_{n \to \infty} X_n(\omega) = X(\omega)\}$ and writes this as $$\left\{ \lim_{n \to \infty} X_n = X\right\}^c = \bigcup_{\epsilon > 0} \left\{ |X_n - X| \geq \epsilon \,\text{ i.o.}\right\}.$$ This is fine, but he then states,

...the union can be restricted to rational (positive) $\epsilon$ because the set in the union increases as $\epsilon$ decreases.

I'm trying to work out why this is so. For any $\epsilon > 0$ let $A_\epsilon := \left\{ |X_n - X) \geq \epsilon \,\text{ i.o.}\right\}$. Then for any $\epsilon_1 < \epsilon_2$ we have $A_{\epsilon_1} \supset A_{\epsilon_2}$, and I suppose want to use this with the fact that $(0,\infty) \cap \mathbb{Q}$ is dense in $(0,\infty)$, but I'm having trouble seeing the way forward.

Updated attempt: Select any $\epsilon > 0$ and the corresponding $A_{\epsilon}$. Then there is an $n \in \mathbb{N}$ such that $\frac{1}{n} < \epsilon$. Thus $A_\epsilon \subset A_{1/n} \subset \cup_n A_{1/n}$. Now select any $n \in \mathbb{N}$ and the corresponding $A_{1/n}$. Then there is an $\epsilon > 0$ such that $\epsilon < \frac{1}{n}$. Thus $A_{1/n} \subset A_\epsilon \subset \cup_\epsilon A_\epsilon$.

• We don't need density, it would suffice to take $\epsilon_n = 1/n$, or any other sequence converging to $0$. Given an arbitrary $\epsilon > 0$, there is an $\epsilon_k < \epsilon$, so $A_{\epsilon} \subset A_{\epsilon_k} \subset \bigcup_n A_{\epsilon_n}$. – Daniel Fischer May 6 '16 at 20:10
• @DanielFischer I updated my attempt at showing this based on your comments. Would you mind checking if it seems correct? – bcf May 6 '16 at 21:28