Can one measurably partition the interval $[0,1]=A\cup B$ with mass equally distributed between $A$ and $B$? Can one find measureable sets $A$ and $B$ such that $A\cap B=\emptyset$, $A\cup B=[0,1]$, and for any interval $I=[x,y]$, with $0\leq x<y\leq1$, the following holds.
1.) $m(A\cap I)=\dfrac{y-x}{2}$.
2.) $m(B\cap I)=\dfrac{y-x}{2}$.
(Note: obviously either of the items imply the other.)
Edit: If the answer is yes, is the analogous statement true for any measure on $\mathbb{R}$, or further, on any measure space?
 A: If it were possible then the bounded measurable function $\chi_A - 1/2$ would have zero integral over any interval so it would have to be zero almost everywhere. No, it's not possible.
A: Look at this exercise, if a measurable proper subset $E \subset [0,1]$ satisfies that 
$m(E\cap I)\geq \alpha m(I)$ for some $0 \leq \alpha < 1$ and all intervals $I$ than $m(E)=0$. So in your case you have $\alpha =1/2$ . So your decomposition is not possible. 
Edit:
Prove of the exercise:
Suppose by contradiction that $1>m(E)>0$ for your case you want that $m(E^c)\neq 0$ then $m(E^c)>0$ since we don't want $m(E)=1$, now take $x$ a point of density of $E^c$ , this means
that given $\epsilon $ exists $I$ interval centred at $x$ shuch that $\frac{m(E^c\cap I)}{m(I)}> 1-\alpha$. On the other hand we by mensuarability of $E$ we have that
$m(E\cap I)+m(E^c\cap I)=m( I)$ and by hypotesis,$m(E\cap I)\geq \alpha m(I)$, than
$m(I)=m(E\cap I)+m(E^c\cap I)\geq m(E^c\cap I)+\alpha m(I)>(1-\alpha)m(I)+\alpha m(I)=m(I)$  a contradiction.
