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I want to prove the following: If $A_n,n\ge 1$ are Borel sets on Lebesgue space $([0,1],B(0,1),m)$, and there is $\epsilon >0$ s.t. $\forall n, m(A_n)\ge\epsilon$, then there is at least one point that belongs to infinitely many sets $A_n$.

I know that if there $\epsilon > 0$ such that $m(A_n)\ge\epsilon$ for all $n$, then

$$m(\cup_{n\in\mathbb{N}}A_n)=\sum_{n\in\mathbb{N}}m(A_n)\ge \sum\epsilon=\infty$$

Then by Borel-Cantelli lemma 2, $m(A_n,i.o.)=1$.

What does it mean by "at least one point that belongs to infinitely many sets $A_n$"? Is that $m(x\in A_n, i.o.)=1$?

Amit Kumar Gupta
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lightfish
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    What does $m(A_n, i.o.) = 1$ mean? That $m(A_n) = 1$ infinitely often? "At least one point belongs to infinitely many sets $A_n$" means there is some $x$ such that for infinitely many $n$, $x \in A_n$. – Amit Kumar Gupta Feb 10 '14 at 04:40
  • You wrote $m(\bigcup_{n\in\Bbb N} A_n) = \sum_{n\in\Bbb N} m(A_n)$, but that is not a correct statement, since we're not assuming that the $A_n$ are pairwise disjoint (indeed they can't be under these assumptions). – Greg Martin Feb 10 '14 at 05:10
  • @GregMartin, is it because $m$ is not measurable so we don't have countable additivity? – lightfish Feb 10 '14 at 05:12
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    "$m$ is not measurable" doesn't make sense - $m$ is the measure! The property called "countable additivity" holds only for disjoint sets - for the same reason that $m(A\cup B) = m(A) + m(B)$ is guaranteed only for disjoint sets (take $A=B$ for example). – Greg Martin Feb 10 '14 at 05:13

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Define $B_m = \bigcup_{n=m}^\infty A_n$ and $C = \bigcap_{m=1}^\infty B_m$. $C$ is sometimes called the lim sup of the sets $A_n$, and it equals exactly the set of points that are in infinitely many of the $A_n$. So you're trying to prove that $C$ is nonempty.

In fact you can prove that $m(C)\ge\epsilon$. The $B_m$ form a nested decreasing sequence of sets, each with measure at least $\epsilon$, and they all live in a bounded measure space. You should have a measure theory lemma that allows you to conclude the lower bound of $m(C)$ and thus finish the problem (once you've verfiied the other assertions I've made).

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Greg Martin
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  • Isn't $m(C)$ here in the way you defined it the same as $m(A_n, i.o.)=m(\limsup_n A_n)$ that came from the Borel-Cantelli lemma? i'm a bit confused. – lightfish Feb 10 '14 at 06:10
  • Quite possibly, yes. (I've never seen the notation $m(A_n,i.o.)$ before.) But you derived the inequality $m(A_n,i.o.)\ge\infty$, which is definitely false (the entire measure space only has measure $1$); the derivation in this answer is correct, I believe. – Greg Martin Feb 10 '14 at 08:34
  • I showed that $m(A_n, i.o.)=m(\limsup A_n)=1$, is that ok? thank you – lightfish Feb 10 '14 at 14:49
  • Or would $m(\limsup A_n)=1$ mean that there are infinitely many points in $A_n$, which is too strong? Do you use Fatou's lemma for the last part? – lightfish Feb 10 '14 at 15:09
  • $m(\limsup A_n)=1$ is also false in general. Take for example the special case where all the $A_n$ are identical. – Greg Martin Feb 11 '14 at 00:28