2
$\begingroup$

Let $\mu$ be Lebesgue measure on $\mathbb{R}$ and $\nu$ Borel measure such that $\nu(A) > 0 \Rightarrow \mu(A) > 0$. Show that limit $$g(x) := \lim_{\varepsilon \to 0}\frac{\nu\left(B(x,\varepsilon)\right)}{\mu\left(B(x,\varepsilon)\right)} $$ exists almost everywhere and for every Borel set $A$ $$\nu(A)=\int_{A}g(x)dx.$$

I've find this question at problem set for preperation for calculus exam with an adnotation to not use Radon-Nikodym theorem in proof. I don't know how to start with this problem so any hint would be appreciated.

$\endgroup$
  • 1
    $\begingroup$ Is that integral meant to end in a $d\mu$? $\endgroup$ – Elchanan Solomon Jan 23 '14 at 3:05
  • $\begingroup$ is $\nu$ finite? $\endgroup$ – user28877 Jan 23 '14 at 5:46
  • $\begingroup$ @IsaacSolomon I think so, user710587 there is no such assumption. $\endgroup$ – Stephen Dedalus Jan 23 '14 at 10:40
  • $\begingroup$ @user710587 I have rethink your question. Condition $\mu(A)>0 \Rightarrow \nu(A)>0$ pretends to be absolute continuity of $\mu$ but to fullfill ac we need $\nu < \infty$, too. So let's suppose that $\nu$ is finite. $\endgroup$ – Stephen Dedalus Jan 23 '14 at 13:08

Your Answer

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

Browse other questions tagged or ask your own question.