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.

  • $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ The conclusion follows from countable additivity? $\endgroup$ – Vergil Qu Dec 7 '17 at 8:19
  • $\begingroup$ Yes, cover $\mathbb R$ by a sequence of intervals of small length and apply countable additivity $\endgroup$ – Kavi Rama Murthy Dec 7 '17 at 9:48
  • $\begingroup$ Sry for my foolishness, but could you elaborate more on how to cover R b y a sequence of sufficiently small intervals? Thanks. $\endgroup$ – Vergil Qu Dec 7 '17 at 21:08
  • $\begingroup$ Could that be let An=(-nƐ,nƐ), Bn=An\An-1, and use {Bn} to cover R? $\endgroup$ – Vergil Qu Dec 7 '17 at 21:28
  • $\begingroup$ $[n\epsilon, (n+1)\epsilon)$ n varying over all integers, for example. $\endgroup$ – Kavi Rama Murthy Dec 11 '17 at 8:01

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