# Explain why $P(\liminf X_n \leq x) = 1=P(\limsup X_n \geq y)$ implies $P(\liminf X_n \leq x \text{ and } \limsup X_n \geq y) = 1$

I have a question about the following post

$P[X_n\:\mathrm{converges}] = 0$ for iid non-constant RVs

As far as I can understand, we have by Borell-Cantelli Lemma that $$P(X_n \leq x \text{ infinitely often }) = 1= P(X_n \geq y \text{ infinitely often })$$ from which it is clear that $$P(\liminf X_n \leq x) = 1 = P(\limsup X_n \geq y)$$

But, then how does one get

$$P(\liminf X_n \leq x \text{ and } \limsup X_n \geq y) = 1\ ?$$

• That is the one of the most vague titles possible. – Akiva Weinberger Oct 2 '15 at 1:00
• @AkivaWeinbergercolumbus I will try to come up with a better title – user74261 Oct 2 '15 at 1:03
• @pm021 To-do list: 1. Switch to nonempty titles. 2. When asking about details of a proof on a given page, mention the new question on the old page, if only by correction. – Did Oct 2 '15 at 6:35

If $P(A)=P(B)=1$, then $P(A\cap B)=1$. To check this notice that $P((A\cap B)^c)=P(A^c\cup B^c)\le P(A^c)+P(B^c)$.