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Let $A_1, A_2,\cdots$ be Lebesgue-measurable subsets of $[0, 1]$ of measure $1/2$, and let $A$ denote the set of points $x$ such that $x$ belongs to infinitely many of the sets $A_n$. How we can show that the Lebesgue measure of $A$ is at least $1/2$.

Segni
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  • Think of $\cap_k \cup_{n\geq k} A_k$, this should have points that belong to infinitely many of the sets $A_n$. Now use the monotone convergence theorem to say that to find the measure of that set, we can take the measure inside of the limits... – Juanito May 18 '17 at 21:03
  • If that hint is not enough, I am happy to elaborate more. Let me know. – Juanito May 18 '17 at 21:07
  • You can also forget about MCT if you have not done that. Just think of $B_k=\cup _{n\geq k} A_k$. Argue why all $B_k$s have measure of atleast .5, note that the $B_k$s are all nested. Now conclude about the measure of their intersection. – Juanito May 18 '17 at 21:12
  • I need more elaborate,thank u. – Segni May 18 '17 at 21:17
  • Do you know about MCT/ Fatou's Lemma? If not I can prove it without it.. – Juanito May 18 '17 at 21:21

2 Answers2

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Take $B=\cap _k \cup_{n\geq k} A_k $

Any x is present in an infinite number of $A_n$s, if and only if they are present in any $B_k=\cup_{n\geq k} A_k $ for all $n$. (You should be able to show both directions, or, see lim sup and lim inf of sequence of sets..)

Therefore, $\mu(B)=\mu(\cap_k B_k)=\mu (\limsup A_n)\geq \limsup \mu ( A_n)\geq .5$

The second last step following from Fatou's Lemma using characteristic functions of the sets $A_n$ involved. It can also be shown using the continuity of measures. The last step follows from the fact that all $A_n$s have measure .5.

Juanito
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First, $\bigcap_k\bigcup_{n>k} A_n$ is exactly the set of points which appear in infinitely many $A_k$s.

$\Rightarrow$: If $\exists i: x\in A_i$, then $\forall k < i:x\in \bigcup_{n>k}A_n$. If there are infinitely many such $i$, a choice would be available for any $k$. So $x$ is in every such union, hence in the infinite intersection.

$\Leftarrow$: If $x$ is in the infinite intersection, it is in every union $\bigcup_{n>k} A_n, \forall k$. So $\forall k:\exists i > k:x\in A_i$. The set of such $i$ has to be infinite. Otherwise it would have an upperbound $M$, and for $k > M$, no such $i$ exists that $x\in A_i$, since $M$ is the largest one possible.

Note that all $\bigcup_{n>k} A_n$ are measurable, and every measure is at least $1/2$, and that they form a non-increasing sequence of real numbers. Then we can use the infinite intersection property:

$$\mu\left(\bigcap_{k}^\infty \left(\bigcup_{n>k} A_n\right) \right) = \lim_{k\to\infty} \mu\left(\bigcup_{n>k} A_n\right)$$

which directly results from the $\sigma$-additivity. Note that the right limit exists, since they form a non-increasing, bounded sequence.

Colliot
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