Borel-Cantelli Lemma "Corollary" in Royden and Fitzpatrick 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 $\sum_{k=1}^{\infty} m(E_k) < \infty$. $m(\cap_{n=1}^{\infty} [\cup_{k=n}^{\infty} E_k]) = 0$"?

 A: *

*I think you're getting a little mixed up (just think about your formulation, intuitively it seems false), but I agree that the conclusion is rather vaguely stated. Let me cite E. Stein's book Real Analysis, in which he poses the Borel-Cantelli lemma as Exercise 16, Chapter 1:

Suppose $\{E_k\}$ is a countable family of measurable subsets of $\mathbb{R}^n$ and that $$\sum_{k=1}^\infty m(E_k)<\infty.$$ Let $$E = \{x\in\mathbb{R}^n: x\in E_k~\text{for infinitely many}~k\} = \limsup_{k\to\infty} E_k.$$
  Then $E$ is measurable, and $m(E)=0$.


*You can (check this!) write $E = \cap_{n=1}^\infty \cup_{k=n}^\infty E_k$, so if this set is measure zero then almost every $x$ fails to belong to $E$. But if $x$ fails to belong to $E$, then it means precisely that $x$ is contained in at most finitely many $E_k$. This is the desired conclusion, so we're done!

*We could state it like that, but no one would know what it means. For its intuitive interpretation, the probabilistic version is very helpful. If you read the $E_k$-s as events and $m(E_k)$ as the probability of the event $E_k$ occurring, then $E$ is the event that infinitely many of the events $E_k$ occur simultaneously. The Borel-Cantelli lemma says that under a suitable decay condition on the probabilities of $E_k$ (namely convergence in the infinite series), the probability of the event $E$ is zero. Believable enough, I think. For its significance, it is an example of a zero-one law: the probability of infinitely many events occurring together is zero. Zero-one laws are of significant interest to probabilists; the page at http://en.wikipedia.org/wiki/Borel-Cantelli_lemma will mention other examples, some related.
