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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 $x\in [0,1]$?

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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.

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  • $\begingroup$ If we consider $\chi_E$, is the antiderivative given by $x\cdot\chi_E$? $\endgroup$ – Habagat Maliksi Apr 26 at 19:16
  • $\begingroup$ @HabagatMaliksi It's given by $\int_0^x \chi_E(t)dt$. $\endgroup$ – eyeballfrog Apr 26 at 19:18
  • $\begingroup$ No big deal, but usually antiderivatives exist throughout an entire interval. $\endgroup$ – zhw. Apr 26 at 20:48
  • $\begingroup$ @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. $\endgroup$ – eyeballfrog Apr 26 at 20:54
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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.

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