# Is there a measurable set $A$ such that $m(A \cap B) = \frac12 m(B)$ for every open set $B$?

Is there a measurable set $A$ such that $m(A \cap B)= \frac12 m(B)$ for every open set $B$?

Edit: (t.b.) See also A Lebesgue measure question for further answers.

-

Alternatively, approximate $A\cap[0,1]$ with a finite union of intervals.
On second thought, those hints are overly complicated. You can use the definition of Lebesgue measure to find an open set $B$ containing $A\cap[0,1]$ with measure close to that of $A\cap[0,1]$.
For Lebesgue density, think about what it says for the intersection of $A$ with small intervals. For the second hint, for every set of finite measure $E$ (here $E=A\cap[0,1]$) and every $\varepsilon>0$ there is a finite union of intervals $U$ such that the measure of the symmetric difference of $E$ and $U$ is less than $\varepsilon$. – Jonas Meyer Feb 3 '11 at 2:28