# Measure limit, proof without Radon-Nikodym theorem

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.

• Is that integral meant to end in a $d\mu$? – Elchanan Solomon Jan 23 '14 at 3:05
• is $\nu$ finite? – user28877 Jan 23 '14 at 5:46
• @IsaacSolomon I think so, user710587 there is no such assumption. – Stephen Dedalus Jan 23 '14 at 10:40
• @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. – Stephen Dedalus Jan 23 '14 at 13:08