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Let $A_1,A_2,A_3, A_4$ be measurable subsets of $[0,1]$, such that $\displaystyle\sum_{k=1}^{4}m(A_k)>3$. Prove that
$$ m\left(\bigcap_{k=1}^{4}A_k\right)>0. $$

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Do you have any thoughts? – Ross Millikan Jul 25 '11 at 17:32
Assuming that the measure of the intersection is zero, you get a contradiction by inclusion-exclusion. – Mark Jul 25 '11 at 18:03
@Mark: Care to elaborate? I don't see how you can do that without some annoyances. – Eric Naslund Jul 25 '11 at 19:05
up vote 7 down vote accepted

Let the superscript $c$ denote the complement in $[0,1]$. Recall that $m(A)=1-m(A^c)$. Then $$ m\left(\bigcap_{i=1}^4A_i\right)=1-m\left(\bigcup_{i=1}^4A_i^c\right)\geq 1-\sum_{i=1}^4 m(A_i^c)$$ $$=\sum_{i=1}^4 m(A_i)-3>0.$$

The last inequality follows from our starting asumption.

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@ Eric: I do not think that you can use the intersection of [0,1] and A is the same as A complement and the union of A_i complement is not contained in [0,1] so I do not think that you can use the excision property. Would you mine to explain more in detail? – Yeonjoo Yoo Jul 26 '11 at 3:03
@Yeonjoo, when he wrote $A_i^c$, he meant complement with respect to $[0,1]$. He is considering $[0,1]$ as the whole space throughout his argument. – user1736 Jul 26 '11 at 4:19

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