The Borel-Cantelli Lemma in Royden and Fitzpatrick's "Real Analysis" seems to be a sort of "corollary" of the non-probabilistic ones I see online.
It says:
"Let $(E_k)_{k=1}^{\infty}$ be a countable collection of measurable sets for which $\sum_{k=1}^{\infty} m(E_k) < \infty$. Then almost all x $\epsilon \ \mathbb{R}$ belong to at most finitely many of the $E_k$'s"
I don't get three things:
- "Then almost all x $\in \ \mathbb{R}$ belong to at most finitely many of the $E_k$'s"
What does that mean? Since the Borel-Cantelli Lemma comes after the "almost everywhere" definition, I guess: $\exists \ Z \ \subset \mathbb{R}$ s.t. $m(Z) = 0$ and $\forall x \ \in \ \mathbb{R} \setminus Z$, x belongs to $(E_{j}, E_{j+1}, E_{l}, ..., E_{m})$.
- In the proof, one can see the $m(\limsup \ E_k) = 0$ part, stated as $m(\cap_{n=1}^{\infty} [\cup_{k=n}^{\infty} E_k]) = 0$. For some reason, $m(\cap_{n=1}^{\infty} [\cup_{k=n}^{\infty} E_k]) = 0$ implies that "Therefore almost all x $\in \ \mathbb{R}$ fail to belong to $\cap_{n=1}^{\infty} [\cup_{k=n}^{\infty} E_k]$ and therefore belong to at most finitely many $E_k$'s"
Okay, why? What is the contradiction if almost all $x \epsilon \ \mathbb{R}$ belong to all the $E_k$'s? Or a countably infinite subcollection?
- What is the significance of this? Why not just state the Borel-Cantelli Lemma as "Let $(E_k)_{k=1}^{\infty}$ be a countable collection of measurable sets for which $\sum_{k=1}^{\infty} m(E_k) < \infty$. $m(\cap_{n=1}^{\infty} [\cup_{k=n}^{\infty} E_k]) = 0$"?