Exotic Borel Set Let $A\subseteq[0,1]$ be a Borel set such that $0<m(A\cap I)<m(I)$ for every interval $I\subseteq [0,1]$. What can we say about measure of such a set $A$?  Can it be any number between $0$ and $1$. 
It is obvious that measure of $A$ can not be $1$. (taking $I=[0,1]$ contradicts with property of set $A$ above, note that inequalities above are strict)
Thank in advance  for any help and comments
 A: First of all, by varying the construction of a fat Cantor set $C_1$ one can make $m(C_1)$ to be any real number between $0$ and $1$. Iteratively assigning a new fat Cantor set $C_i$ inside one of the largest remaining empty intervals, we obtain the desired set $A$. The only thing you have to be careful with is that you are really able to obtain any value for $m(A) = \sum_{i=1}^\infty m(C_i)$.
A: Suppose $0\lt\alpha\lt1$. Construct an $F_\sigma$-set $S$ such that $0\lt m(S\cap I)\lt m(I)$ for every finite interval $I\subset\mathbb R$, and $m(S)\lt\infty$. [This is Quite Easily Done; e.g., using "fat Cantor sets", simultaneously construct two disjoint $F_\sigma$-sets of finite measure, each of which has non-null intersection with every rational interval.] Let $c$ be a Lebesgue density point of $S$. The function $$f(x)=\frac{m(S\cap[c-x,c+x])}{2x}$$is continuous on $(0,\infty)$, with $f(0+)=1$ and $f(\infty)=0$, so there is an $x\in(0,\infty)$ such that $f(x)=\alpha$. Let $B=S\cap[c-x,c+x]$ and let $A=\dfrac{B-(c-x)}{2x}$.
