Prove or disprove the existence of a measurable set 'equally' distributed in [0,1].

Question

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.

Please learn to typeset the math correctly; it's not that hard. — zhw., Jun 05 '16 at 15:46
This question has been asked and answered here many times before. Let me see if I can find one of the previous questions. — Nate Eldredge, Jun 05 '16 at 20:32
Possible duplicate of Is there a measurable set $A$ such that $m(A \cap B) = \frac12 m(B)$ for every open set $B$? — Nate Eldredge, Jun 05 '16 at 20:36
See also: http://math.stackexchange.com/questions/1647113/existence-of-a-subset-s-subset-mathbb-r-s-t-forall-ab-s-cap-a-b-has/1647126#1647126 http://math.stackexchange.com/questions/596676/lebesgue-density-theorem-in-the-line/629721#comment1257338_596676 http://math.stackexchange.com/questions/664704/let-a-subset-mathbbr-a-measurable-and-bounded-set-show-that-exists-for-each?rq=1 — Nate Eldredge, Jun 05 '16 at 20:38

score 11 · Accepted Answer · answered Jun 05 '16 at 15:45

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.

score 3 · Answer 2 · answered Jun 05 '16 at 18:42

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.

Prove or disprove the existence of a measurable set 'equally' distributed in [0,1].

2 Answers2