Constructing an Equidistributed set

Question

I want to construct a subset $A \subset [0,1]$ of measure $0<\epsilon <1$, such that for any interval $J \subset [0,1]$ we have $m(A \cap J)=m(A)m(J)$. If anyone has any hints it would be great.

Answer 1

We can also get this without the LDT: Suppose there exists such a set $A.$ Because $m(A) < 1,$ there exists $U$ open in $[0,1]$ such that $A\subset U$ and $m(U) < 1.$ Now $U$ is the pairwise dijoint union of intervals $I_n, n=1,2,\dots $ So we have $A =A\cap U = \cup_n (A\cap I_n).$ Because the sets $A\cap I_n$ are pairwise disjoint,

$$m(A) = \sum_n m(A\cap I_n) = \sum_n m(A)m(I_n) = m(A)[\sum_n m(I_n)] = m(A)m(U).$$

Since $m(A)>0,$ the above implies $1 = m(U),$ contradiction.

Answer 2

Of course $A=[0,1]$ and $A=\emptyset$ are trivial examples, or they were examples until you added the condition $0<m(A)<1$. With that condition added there are no examples. This follows for example from the Lebesgue Differentiation Theorem applied to $\chi_A$: If $A$ is equidistributed in the sense you specify then LDT shows that $\chi_A=m(A)$ almost everywhere, hence $m(A)=0$ or $m(A)=1$.