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I searched but couldn't find an answer similar to this topic, apologies if I missed it. Here goes;

Let $A_j \subset [0,1], j=1,2,...$ such that each $A_j$ has Lebesgue measure $\mu(A_j)\geq 1/2$. Show that there exists a measurable set $S \subset [0,1]$ such that $\mu(S) \geq 1/2$ and each element $x \in S$ is contained in infinitely many of the $A_j's$.

I have been able to accomplish this in certain cases but the general solution has thus far evaded me. Any help would be appreciated. Cheers!

brawbro
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    Looks like you want to define $S$ as the $\limsup$ of the $A_j$ sets, which is precisely the set of points that are in infinitely many of those sets. Can you finish the rest? See also here: http://math.stackexchange.com/questions/107931/lim-sup-and-lim-inf-of-sequence-of-sets – Michael Jul 28 '16 at 19:35
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    I believe so. Since $A_j \subset [0,1] $ and $\mu([0,1]) =1 < \infty$ it follows that $\mu(\limsup A_j) \geq \limsup \mu(A_j) \geq 1/2$ – brawbro Jul 28 '16 at 19:59
  • @donkey_pills To check. For that last part, you implicitly used Fatou's lemma (is that right)? – Clement C. Jul 28 '16 at 20:06
  • Nope, just a theorem in the textbook I am working through. (Yeh J. Theory of Measure and Integration thm 1.28) – brawbro Jul 28 '16 at 20:24

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As suggested, you may as well make $S$ as large as possible by defining it to be the set of all points which lie in infinitely many $A_n$. The question then is how to show $\mu(S)\ge1/2$. A hint for that:

The point $x$ lies in infinitely many $A_n$ if and only if for every $N$ there exists $n\ge N$ with $a\in A_n$. That says that $$S=\bigcap_{N=1}^\infty\bigcup_{n=N}^\infty A_n.$$

(Note that however you complete this had better use the fact that $[0,1]$ has finite measure; the result is clearly false with $\Bbb R$ in place of $[0,1]$.)

David C. Ullrich
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If $x\in S^c$ it is in finitely many of the $A_j$'s only, so there must be $n=n(x)$ for which $x\in A_j^c$ (complement) for every $j\geq n$. Let $E_n= \cap_{j\geq n} A_j^c$ be the intersection of such sets. We have $E_1 \subset E_2 \subset ...$ so $E_n$ is an increasing sequence of measurable subsets and $S^c=\cup_{n\geq 1} E_n$. We have $\mu(E_n)\leq \mu(A_n^c)=1/2$ and by Lebesgue $1-\mu(S)=\mu(S^c)= \lim_n \mu(E_n)\leq 1/2$ so $\mu(S)\geq 1/2$.

H. H. Rugh
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