Understanding probabilistic Borel-Cantelli Lemma "Corollary" in Royden and Fitzpatrick

Question

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

2. 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?

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

Answer 1

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.

Answer 2

What is the contradiction if almost all xϵ R belong to all the $E_k$'s?

This contradicts the hypothesis of the convergence of the series formed by the measures of each $E_k$.

My reasoning is as follows:

Let's assume that almost all $x \in\mathbb{R}$ belong to all $E_k$'s. This means that there exists a set $E_0\subseteq\mathbb{R}$ such that $m(E_0)=0$ and if $x\in\mathbb{R}-E_0$ then $x\in \cap_{k=1}^{\infty}E_k$. We then have that: $$\mathbb{R}-E_0 \subseteq\cap_{k=1}^{\infty}E_k\subseteq\cup_{k=1}^{\infty}E_k,$$ and by the countable monotonicity of the measure we have: $$m(\mathbb{R}-E_0)\leq m(\cup_{k=1}^{\infty}E_k)<\infty.$$ This contradicts the hypothesis.