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Let $\{A_n\}$ be a collection of measurable sets in $\mathbb{R}$. If $A$ is the union of the collection and almost every $x\in A$ belongs to no more than $k$ of the $A_n$ then I need to show that $m(A)\geq\frac{\sum m(A_n)}{k}$.

My idea: For each such x, I would like to identify it with a sub-collection of $\{A_n\}$ in such a way that the sub-collections are disjoint. A starting point would be the collection of, say, j sets that $x$ belongs to: $\{A_{x_i}:\ x\in\bigcap_{i=1}^j A_{x_i},\ 1\leq j\leq k\}$. However, I can't see that such sub-collections must be disjoint and I am not sure where to go from here or if this will lead to a solution. Hints would be appreciated!

dsimo04
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1 Answers1

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Hint: Denote $\chi_B$ the characteristic function of a set $B$. Then $\sum_n\chi_{A_n}\leq k\cdot\chi_A$ a.e.

Luiz Cordeiro
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  • Thanks for your hint. I see that if $E={x\in A:\ x\text{belongs to at most k}\ A_n}$ then $x\in E$ implies your statement. Also, I see that $\sum_n \chi_{A_n}\leq k\cdot \chi_A$ gives a way of relating $A$, $A_n$ and $k$ with the desired inequality. But I don't see how to change the characteristic functions of the sets to their measures. Do I need to use that $\chi_B$ is measurable for any measurable set $B$? Thanks again. – dsimo04 Oct 28 '16 at 00:50
  • @dsimo04 You are right in your coment that the inequality is valid for $x\in E$ (I added "a.e." to my answer, so it is correct now). Now use the fact that $\int\chi_Bdm=m(B)$. – Luiz Cordeiro Oct 28 '16 at 01:15