0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.