# Existence of a measurable set

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 $$x\in [0,1]$$?

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.

• If we consider $\chi_E$, is the antiderivative given by $x\cdot\chi_E$? – Habagat Maliksi Apr 26 at 19:16
• @HabagatMaliksi It's given by $\int_0^x \chi_E(t)dt$. – eyeballfrog Apr 26 at 19:18
• No big deal, but usually antiderivatives exist throughout an entire interval. – zhw. Apr 26 at 20:48
• @zhw. $f(x) = \int_0^x\chi_E(t)dt$ is defined on all of $x\in[0,1]$, though perhaps I was imprecise in defining that. – eyeballfrog Apr 26 at 20:54

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 $$\sum m(I_n) Therefore $$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.