Prove or disprove the existence of a measurable set 'equally' distributed in [0,1]. Is there a measurable set E in $[0,1]$ such that for every open interval $I$ in $[0,1]$, we have $m(E\cap I)=m(I)/2$ where $m$ denotes the Lebesgue measure?
Intuitively I think such a set exists and is 'equally' distributed in $[0,1]$, but I don't know how to describe more precisely this set. 
 A: A standard approach to this problem is to apply the Lebesgue differentiation theorem. In particular, consider the function $$f(x)=\int_0^x \chi_E\,dx$$
where $\chi_E$ is the indicator function of $E$. One may check that for $b>a$ we have $$f(b)-f(a)=m(E\cap (a,b)).$$
If $E$ satisfies your property, we would have that dividing both sides by $b-a$ gives:
$$\frac{f(b)-f(a)}{b-a}=\frac{1}2.$$
However, if we, say, fix $b$ and send $a$ to $b$ as a limit, we get
$$f'(b)=\frac{1}2$$ 
which contradicts the Lebesgue differentiation theorem, since $f'$ must equal $\chi_E$ almost everywhere, but $\chi_E$ takes on the value $\frac{1}2$ nowhere.
A: Suppose there is such an $E.$ Clearly $m(E) > 0.$ Let $\epsilon>0.$ Then there are pairwise disjoint open intervals $I_1,I_2, \dots$ such that $E \subset \cup_n I_n$ and $\sum_n m(I_n) < m(E) +\epsilon.$ Thus
$$m(E) = m(E\cap (\cup_n I_n)) = \sum_n m(E\cap I_n) = \sum_n \frac{m(I_n)}{2} < \frac{m(E)+\epsilon}{2}.$$
This implies $m(E) <\epsilon.$ Since $\epsilon$ is arbitrarily small, $m(E) = 0,$ contradiction.
