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

Since $\nu$ is a $\sigma$- finite measure, you can find a sequence of disjoint sets $E_n$ such that $\Bbb{R} = \cup_n E_n$ and $\nu(E_n) < \infty.$ Define now $$\nu_n(C)= \nu(C \cap E_n).$$ Note that $\nu(C) = \sum_n \nu_n(C)$
Let $f_n = \frac{d \nu_n}{d\mu} $ and consider the $\sigma$ algebra $\mathcal{B}_k$ generated by the dyadic decomposition of order k $$\mathcal{B}_k = \sigma \bigg(\bigg(\frac{i-1}{2^k},\frac{i}{2^k}\bigg]; i \in \mathbb{Z}\bigg)$$ 
 Note that this $\sigma$- algebra contains atoms (indivisible sets) of the form $\bigg(\frac{i-1}{2^k},\frac{i}{2^k}\bigg]$. Define now the conditional expectation of $f_n$ with respect to $\mathcal{B}_k$
$$f_n^k : = \Bbb{E} \big[f_n \big\vert \mathcal{B}_k\big] $$ Note that $$\nu_n \bigg(\bigg(\frac{i-1}{2^k},\frac{i}{2^k}\bigg]\bigg) = \int_{\bigg(\frac{i-1}{2^k},\frac{i}{2^k}\bigg]}  \, d\nu_n =\int_{\bigg(\frac{i-1}{2^k},\frac{i}{2^k}\bigg]}f_n  \, d\mu_n =\int_{\bigg(\frac{i-1}{2^k},\frac{i}{2^k}\bigg]}f_n^k  \, d\mu_n  = f^k_n(i 2^{-k})\int_{\bigg(\frac{i-1}{2^k},\frac{i}{2^k}\bigg]}  \, d\mu_n  = f^k_n(i 2^{-k})\mu_n\bigg(\bigg(\frac{i-1}{2^k},\frac{i}{2^k}\bigg]\bigg) $$
Therefore 
$$\frac{\nu_n \bigg(\bigg(\frac{i-1}{2^k},\frac{i}{2^k}\bigg]\bigg)}{\mu_n\bigg(\bigg(\frac{i-1}{2^k},\frac{i}{2^k}\bigg]\bigg)} = f_n^k(i 2^{-k}) $$
Since $f_n^k$ is $\mathcal{B}_k$ measurable and $f(x) = f_n^k(i 2^{-k}) $ for every $x \in \bigg(\frac{i-1}{2^k},\frac{i}{2^k}\bigg]$.
Now note that $(f^k_n,\mathcal{B}_k)_k$ is a martingale.
Indeed we only need to check that
$$  \Bbb{E} \big[f^{k+1}_n \big\vert \mathcal{B}_k\big]  = f^k_n $$
consider the set $\big(\frac{i-1}{2^k},\frac{i}{2^k}\big]$ and evaluate
$$\Bbb{E}\bigg[{1_{\big(\frac{i-1}{2^k},\frac{i}{2^k}\big]}\Bbb{E} \big[f^{k+1}_n \big\vert \mathcal{B}_k\big]}\bigg] =
 \Bbb{E}\bigg[{1_{\big(\frac{2i-2}{2^{k+1}},\frac{2i-1}{2^{k+1}}\big]}\Bbb{E} \big[f^{k+1}_n \big\vert \mathcal{B}_k\big]}\bigg]  + \Bbb{E}\bigg[{1_{\big(\frac{2i-1}{2^{k+1}},\frac{2i}{2^{k+1}}\big]}\Bbb{E} \big[f^{k+1}_n \big\vert \mathcal{B}_k\big]}\bigg] = \nu_n \bigg(\big(\frac{2i-2}{2^{k+1}},\frac{2i-1}{2^{k+1}}\big]\bigg) + \nu_n \bigg(\big(\frac{2i-1}{2^{k+1}},\frac{2i}{2^{k+1}}\big]\bigg)  = \nu_n \bigg(\big(\frac{i-1}{2^{k}},\frac{i}{2^{k}}\big]\bigg) = \Bbb{E}\bigg[{1_{\big(\frac{i-1}{2^k},\frac{i}{2^k}\big]}f^{k}_n\bigg] } $$
Now you only need to apply the martingale convergence theorem and note that $\bigvee_k \mathcal{B}_k = \sigma(\cup_k \mathcal{B}_k) = \mathcal{B}$ to conclude that almost surely $f_n^k \to f_n $ since there are only countable null sets where the convergence fails, you can consider them together and get
$f^k(x):=\sum_n f_n^k(x)  \to \sum_n f_n(x) = f(x) = \frac{d \nu}{d \mu}$
finaly note that for every $x \in \mathbb{R}$ that has not the form $i 2^{-k}$  you can find $h_k, \tilde{h}_k \to 0$ $ \frac{\nu_n \bigg(\bigg(\frac{i-1}{2^k},\frac{i}{2^k}\bigg]\bigg)}{\mu_n\bigg(\bigg(\frac{i-1}{2^k},\frac{i}{2^k}\bigg]\bigg)} =\frac{\nu_n \bigg(\big(x - h_k,x + \tilde{h}_k\big]\bigg)}{\mu_n\bigg(\big(x - h_k,x + \tilde{h}_k\big]\bigg)}= f^k(x) \to f(x)$
I hope this is helpful
remark: you can adapt this argument to the general case with $h \to 0$ and every $x \in \mathbb{R}$
