Suppose for $a<b$ we have $P(X_n < a$ i.o. and $X_n > b$ i.o.$) = 0$. Then $lim_{n \rightarrow \infty} X_n$ exists a.e. but may be infinite.

Here "i.o." means "infinitely often"; for any sequence of events $E_n$, $P(E_n$ i.o.$) = P(limsup_n E_n)$.

The problem comes with a hint to "consider all pairs of rational numbers (a, b) and take a union over them", which is a little ambiguous. I guessed that the antecedent of "them" is supposed to be the events $\{X_n < a$ i.o. and $X_n > b$ i.o.$\}$, and quickly found that $P(\cup_{i=1}^\infty \{X_n < a_i$ i.o. and $X_n > b_i$ i.o.$\}) \leq \sum_{i=1}^\infty P(X_n < a$ i.o. and $X_n > b$ i.o.$) = \sum_{i=1}^\infty 0 = 0$ for any countable indexing of pairs of rational numbers $(a_i, b_i)$ with $a_i < b_i$ at each index. I don't quite understand what this says about the limit of the $X_n$'s, however, since the set $\cup_{i=1}^\infty \{X_n < a_i$ i.o. and $X_n > b_i$ i.o.$\}$ is kind of strange, and I can't think of a way to show the existence of a limit other than constructing it or using Cauchy, the latter of which (as far as I know) only tends to work when the limit is finite.

If I've taken the wrong antecedent, and I'm actually meant to say something about the union of intervals $(a_i, b_i)$, then I really don't know how to relate that to a set of helpful probabilities at all, since the only one I have in hand is about what's outside of those intervals, not what's inside of them, and seemingly not in a way that gives rise to an obvious sort of complementation to go from one to the other, thanks to the infinitely often business.

  • $\begingroup$ Just note a small thing, don't forget to mention that $a, b$ are rationals otherwise the expression might not be measurable afterall~ $\endgroup$
    – Ran Wang
    Mar 2 '17 at 15:21

Hint: Let $(x_n)_{n \in \mathbb{N}}$ be a sequence of real numbers. Show that

$$\lim_{n \to \infty} x_n$$

does not exist if, and only if, there exist $a,b \in \mathbb{Q}$ such that for all $n \in \mathbb{N}$ there exist $n_1,n_2 \geq n$ such that

$$x_{n_1}<a \qquad \text{and} \qquad x_{n_2} > b.$$ To this end, recall that $\lim_{n \to \infty} x_n$ does not exist if (and only if) $$\liminf_{n \to \infty} x_n < \limsup_{n \to \infty} x_n.$$

  • $\begingroup$ Not sure if I understand what you are saying here. Let x_n=7 for all n. So lim x_n=7. But there exists an 'a' in Q such that a<x_n for infinitely many n. For example, a=2. So by your statement lim x_n does not exist. $\endgroup$
    – PeterR
    Oct 8 '14 at 19:22
  • $\begingroup$ @PeterR You are right, it didn't make sense; sorry. Should be fine now. $\endgroup$
    – saz
    Oct 8 '14 at 19:30
  • $\begingroup$ Should it be x_n_1<a? $\endgroup$
    – PeterR
    Oct 8 '14 at 19:36
  • $\begingroup$ @PeterR Argh... too many typos this evening :/. Thanks for your comments. $\endgroup$
    – saz
    Oct 8 '14 at 19:41
  • $\begingroup$ Now I'm happy! Thanks! $\endgroup$
    – PeterR
    Oct 8 '14 at 19:44

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