Does there exist a Lebesgue measurable subset $A$ of $R$ such that for every $a<b$ we have $m(A\cap(a,b))=(b-a)/2$?

Question

Does there exist a Lebesgue measurable subset $A$ of $R$ such that for every $a<b$ we have $m(A\cap(a,b))=(b-a)/2$?

I searched before posting and found a similar question here, but it isn't exactly the same as my question. I looked over the given answer there but I couldn't figure out if that answer was relevant to my question and how to adapt the answer to my problem if it is. I'm unsure at this point how to actually go about showing if such an $A$ exists or not.

Answer 1

The answer is NO, and the proof might uses Lebesgue differentiation theorem applying to the characteristic function of $A$.

Answer 2

Suppose there were such a set $A.$ Let $B=A\cap (0,1).$ Then $m(B)=1/2,$ and for every $(a,b)\subset (0,1),$ $m(B\cap(a,b))=(b-a)/2.$ Now there exist open intervals $I_n\subset (0,1)$ such that $B\subset \cup I_n$ and $\sum m(I_n)<3/4.$ So we have

$$B= B\cap (\cup I_n) = \cup (B\cap I_n),$$ which implies

$$m(B) \le \sum m(B\cap I_n) = \sum m( I_n)/2 = (1/2)\sum m( I_n) <3/8,$$

contradiction.

Answer 3

The following steps should give you an elemental solution of the problem:

1. We can rewrite the equality as $m(A\cap (a,b))=\frac{m(a,b)}{2}$.
2. Since every open set can be written as the union of countable many disjoint intervals we have (using the properties of the mesures)

$$m(A\cap G)=m\left(A\cap \bigcup_n (a_n,b_n)\right)=\sum_n m\left(A\cap (a_n,b_n)\right)=\sum_n \frac{m(a_n,b_n)}{2}=\frac{m(G)}{2}$$

1. Any mesurable set is the union of a null set and a $G_\delta$ (intersection of open sets containing the measurable set) then, since $A$ is measurable $A=N\cup G_\delta$ with $N$ null. So, if $A$ has finite measure (otherwise intersect it with a bounded open set, then it will preserve the desired property)

$$m(A)=m(N\cup G_\delta)=m(G_\delta)=m\left(\bigcap_n (G_n \cap A)\right)=\lim_{n\to \infty} m(G_n\cap A)=\lim_{n\to \infty} \frac{m(G_n)}{2} =\frac{m(A)}{2}$$.

This clearly gives us the contradiction.