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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.

Biswajyoti Roy
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  • actually i think this is a sort of duplicate of my eponymous/namesake question because even though the answer says 'check this!', the answerer actually elaborates a little more in the comments? – BCLC Jan 06 '21 at 06:44

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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.

Nate Eldredge
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  • Thanks for the response Nate. I am still want to clarify why ∑m(Ek)<∞, then m(E)=0 implies almost all x∈R do not belong to E? If it because s set with positive measure should contain more element than set of measure 0?. Thanks again. I'm typing from mobile. Do please pardon any typing issue. Thanks – Biswajyoti Roy Jun 02 '20 at 19:27
  • That's just the definition of the phrase "almost all" - that it's true for every $x$ except those in some set of measure zero. The terminology is just meant to reinforce the informal idea that sets of measure zero are very small. – Nate Eldredge Jun 02 '20 at 19:30
  • actually i think this is a sort of duplicate of my eponymous/namesake question because even though the answer says 'check this!', the answerer actually elaborates a little more in the comments? – BCLC Jan 06 '21 at 06:45