TITLE: Are dense subsets almost nothing or almost everything?
QUESTION [14 upvotes]: 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 $\big([0,0.5]\cap\Bbb Q\big) \cup \big([0.5,1]\cap \Bbb I\big)$.

REPLY [18 votes]: 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 $m(A)=1$.