Why sets of nonzero measure contain almost all the point from Borel-cantelli Leems?

Question

The question referred from this discussion Borel-Cantelli Lemma "Corollary" in Royden and Fitzpatrick

I understand the proof of the Leems.But i still don't get "Then almost all x ∈ R belong to at most finitely many of the E_Ks"? i understood that if x belongs to infinitely E_k then the set contain all such x has measure 0 but how it implies finitely many x's belongs to that set? or infinitely many x ∈ R belong to at most finitely many of the Ek's? how i can get a contradiction?

THanks in advance for the help.

Answer 1

Let $E= \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty E_k$. Note that $x \in E$ if and only if $x$ belongs to infinitely many of the $E_k$.

The Borel-Cantelli lemma says that if $\sum m(E_k) < \infty$, then $m(E) = 0$. That is to say, almost all $x \in \mathbb{R}$ do not belong to $E$. So this means that almost all $x$ have the property that $x$ does not belong to infinitely many of the $E_k$, which is to say that each such $x$ belongs to only finitely many of the $E_k$.

Note that nothing in the statement is talking about "infinitely many $x$" or "finitely many $x$". You seem to have got that part mixed up.