TITLE: Sets with equal positive measure in every interval
QUESTION [12 upvotes]: Hi,
I want to write a proof that relies on the fact that:
There are Borel Sets $A$ and $B$ contained in $\mathbb{R}$ such that
$A \cap B = \emptyset$ and $\lambda(A \cap (x,y)) = \lambda(B \cap (x,y)) > 0$.
Note that $x < y \in \mathbb{R}$ are arbitrary.
I'm fairly sure this is true, but am having trouble coming up with a construction of such sets and it's driving me up the wall.  Can anyone help?

REPLY [2 votes]: Let $\mu$ and $\nu$ be two measure on $(E,{\cal E})$. If :

$\mu(A)=\nu(A)$ for all $A$ in a subset $\pi$ of ${\cal E}$ such that $\pi$ is closed under finite intersection and the $\sigma-$algebra generated by $\pi$ is ${\cal E}$
there exists an increasing sequence of set $A_n$ in ${\cal E}$ such that $\mu(A_n)=\nu(A_n)<\infty$ and $\cup A_n = E$.

then $\mu=\nu$. (This is standard generating argument).
If you apply this result with $\mu=1_A \lambda$, $\nu=1_B \lambda$ and ${\cal B}$ the set of open interval, you get that if your condition on $\lambda(A\cap...)$ holds, then $A=B$ almost everywhere. With the other conditions you see that it is not possible.