Existence of a measurable set

Question

I don't know where to start so a hint would be greatly appreciated.

Use the fundamental theorem of calculus for absolutely continuous functions to answer:

Is there a measurable set $E\subset [0,1]$ such that $m(E\cap [0,x])= x/3$ for almost every $x\in [0,1]$?

Answer 1

Just for fun, here's an elementary proof. Suppose there is such an $E.$ Let $(a,b)\subset [0,1].$ Then

$$\tag 1 m(\chi_E\cap (a,b)) = m(\chi_E\cap (0,b)) - m(\chi_E\cap (0,a)) = b/3-a/3 = (b-a)/3.$$

Let $\epsilon>0.$ Then there are disjoint open intervals $I_n$ such that $E\subset \cup I_n$ and $\sum m(I_n)<m(E)+\epsilon.$ Therefore $m(E)= m(E\cap(\cup I_n)) = \sum m(E\cap I_n) = \sum m(I_n)/3 = (1/3)\sum m(I_n) <(1/3)(m(E)+\epsilon).$

This implies $(2/3)m(E) < \epsilon/3,$ or $m(E)<\epsilon/2.$ Since $\epsilon$ is arbitrary, $m(E)=0,$ contradiction.

Answer 2

Hint: If $E$ is measurable, its characteristic function $\chi_E$ is also measurable, and the Lebesgue integrable. Consider the antiderivative of this function, and what happens when you differentiate it.