Radon-Nikodým (write the density as a limit) 

Let $\mu$ be a probability measure and $\nu$ a $\sigma$-finite measure on $(\mathbb{R},\mathcal{B})$ with $\nu\ll\mu$. Show that it is $\mu$-a.s.
    $$
\lim_{h\to 0}\frac{\nu [x-h,x+h)}{\mu [x-h,x+h)}=\frac{d\nu}{d\mu}(x).
$$


I do not have many ideas...
Set $f:=\frac{d\nu}{d\mu}$. Then $\nu [x-h,x+h)=\int_{[x-h,x+h)}f\, d\mu$ and
$$
\lim_{h\to 0}\frac{\nu [x-h,x+h)}{\mu [x-h,x+h)}=\lim_{h\to 0}\frac{\int_{[x-h,x+h)}f\, d\mu}{\mu [x-h,x+h)}.
$$
But that does not really help.
Can you help me, please?
Miro

This is theorem 5.4 page 115 from Theory of the integral, S. Saks. It is proved via Vitali's covering theorem.
