14

Dense subsets of $[0,1]$ I know have Lebesgue measure $0$ or $1$, but, is there any dense, uniform subset of $[0,1]$ with meausre $1/2$?

What I mean with uniform: a subset $A$ of $[0,1]$ is uniform if $m(A\cap[a,b])=(b-a)m(A)$ for $0\le a\le b\le 1$. $m$ is Lebesgue's measure. The point is excluding examples like $\big([0,0.5]\cap\Bbb Q\big) \cup \big([0.5,1]\cap \Bbb I\big)$.

ajotatxe
  • 65,084

2 Answers2

18

Another twist of the answer is: uniform subsets $A$ are almost nothing or almost everything. $m(A\cap(a,b))=(b-a)\,m(A)=m(A)\,m((a,b))$ implies $m(A\cap B)=m(A)\,m(B)$ for every measurable set $B$, due the the same $\sigma$-additivity and regularity of the measure. Now with $B=A$, we obtain $m(A)=m(A)^2$, meaning $m(A)=0$ or $m(A)=1$.

  • Simply amazing. Amazingly simple. – ajotatxe Jun 02 '17 at 21:56
  • 1
    When you set $B = A$, you use the fact that $A$ is measurable, i.e. you show that if $A$ is measurable and "uniform", then it must have measure zero or 1. But you can construct a non-Lebesgue set $E$ that satisfies $m(E \cap B) = (1/2)m(B)$ for all Borel sets $B$. Roughly speaking this is explained here: https://terrytao.wordpress.com/2008/10/14/non-measurable-sets-via-non-standard-analysis/#more-717 – Thompson Jun 03 '17 at 00:30
  • Yes, but ajotatxe's definition of uniformity assumed measurability by writing $m(A)$ instead of $1/2$. –  Jun 03 '17 at 05:18
8

There certainly is no measurable example: if $E$ is measurable and of positive measure, and $0<\epsilon<1$, then there is some open interval $I$ such that $m(E\cap I)\ge\epsilon m(I)$ (this is an application of regularity, see this question). Now take $\epsilon>{1\over 2}$.

Noah Schweber
  • 245,398