Does there exist a measurable $A \subseteq [0,1]$ with $\lambda(A \cap [0,a]) = a/2$ $\forall a \in [0,1]$? does there exist a Borel set $A \in [0,1]$ such that, for any $a \in [0,1]$, the Lebesgue measure of the set $A \cap [0,a]$ is $a/2$?
Thanks

Let $f(a) = \int_0^a 1_A$, then the Lebesgue differentiation theorem gives
$f'(a) = 1_A(a)$ for ae. $a$. In particular, we can't have $f'(a) = {1 \over 2}$.
