Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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
$$ m\left(\bigcap_{k=1}^{4}A_k\right)>0. $$

share|cite|improve this question
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.

share|cite|improve this answer
@ 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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.