1

Prove that there exists no Lebesgue-measurable subset $A$ of the real numbers such that for every interval $I$ the Lebesgue measure of $A\cap I$ is half the length of $I$.

My attempt:

assuming there exist such $A$ that works with some interval $I$, I wanted to show that there exists some interval in $I\setminus A$ but I couldn't show it.

Kareem Metwaly
  • 162

1 Answers1

0

Assume it exists and assume that has finite positive measure.

(If has zero measure then it is obvious that is not true)

Then $\forall \epsilon>0 $,exists a covering $\{I_n:n \in \Bbb{N}\}$ of $A$ such that $\sum_nm(I_n) \leq m(A)+\epsilon$

Thus $A=\bigcup_n(A \cap I_n)$ and $$m(A) \leq \sum_nm(A \cap I_n)=\sum_n\frac{m(I_n)}{2} \leq \frac{m(A)+\epsilon}{2}$$

Choose $\epsilon =\frac{m(A)}{4}$ and you have a contradiction.

if $A$ has infinite measure and we assume that has the property then set $A_m=A \cap[-m,m]$ will have the same property so we can restrict ourselves to every subinterval of $[-m,m]$ so by the previous case you can arrive to a contradiction again

Marios Gretsas
  • 21,906
  • The argument is more complicated when the measure of $A$ is infinity. – Kavi Rama Murthy Oct 30 '19 at 00:15
  • 1
    @KaboMurphy if $A$ has infinite measure and we assume that has the property then set $A_m=A \cap[-m,m]$ will have the same property so we can restrict ourselves to every subinterval of $[-m,m]$...i think – Marios Gretsas Oct 30 '19 at 00:23