# Borel measure absolutely continuous wrt Lebesgue integral

Is there a non-zero Borel measure $\mu$ on $\mathbb R$, absolutely continuous with respect to Lebesgue measure, such that $\mu (U)$ is integer for every open set $U$? Counterexample or Proof?

I have no idea how to approach.

• Welcome to math SE! I edited your first post a little bit, but I leave it to you to formulate your request ("If yes, give an example. If no, give a proof.") to the community less mandatory, by clicking on 'edit' in the lower left. – Hanno Dec 7 '17 at 8:03

If $\mu$ is absolutely continuous with respect to Lebesgue measure then $\mu (E)$ approaches 0 as Lebesgue measure of E approaches 0. Being integer valued it follows that all sufficiently small intervals have measure 0 under $\mu$. Hence $\mu$ is the zero measure.
• Yes, cover $\mathbb R$ by a sequence of intervals of small length and apply countable additivity – Kavi Rama Murthy Dec 7 '17 at 9:48
• $[n\epsilon, (n+1)\epsilon)$ n varying over all integers, for example. – Kavi Rama Murthy Dec 11 '17 at 8:01