# 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:

1. "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 $(E_{j}, E_{j+1}, E_{l}, ..., E_{m})$.

1. 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?

1. 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 $\sum_{k=1}^{\infty} m(E_k) < \infty$. $m(\cap_{n=1}^{\infty} [\cup_{k=n}^{\infty} E_k]) = 0$"?

1. 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 $m(E)=0$.

2. 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!

3. 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.

• "But if x fails to belong to E, then it means precisely that x is contained in at most finitely many $E_k$" Why?
– BCLC
Apr 13, 2014 at 12:43
• Also I seem to have completely forgotten that math is extremely theoretical and abstract and often has no significance unless applied like in the probabilistic interpretation. XD
– BCLC
Apr 13, 2014 at 12:46
• Well, just think about what $E^C$ must be: if $E = \{x\in\mathbb{R}^n: \exists~\text{infinitely many}~E_k~\text{such that}~x\in E_k\}$, then just negate the appropriate quantifiers and you get exactly what I claimed. Apr 13, 2014 at 23:43
• You could also write $E$ as $\{x : f(x) < \infty \}$ where $f(x) = \sum_{n=1}^\infty \chi_{E_n}(x)$.
– user42761
May 27, 2014 at 13:02