I can't solve the following exercise, even if I find it quite intuitive.
Let $\nu, \mu$ be Radon measures on a metric space $(X,d)$. Suppose that: 1) $w\in L^1(X,\mu), w\geq 0$ $\mu$ a.e.; 2) $w$ is continuous in $x_0$ and $\mu(B(x_0,r))>0$ for any $r>0$; 3) $\nu(A)=\int_Aw d\mu$ for any $A\in \mathcal M$.
Show that there exists $$\lim_{r\to 0} {\nu(B(x_0,r))\over \mu(B(x_0,r))}=w(x_0).$$
We know that $$\lim_{r\to 0} {\nu(B(x_0,r))\over \mu(B(x_0,r))}=w(x_0) \iff \lim_{r\to 0} {\nu(B(x_0,r))\over \mu(B(x_0,r))}-w(x_0)=0$$ and $$\lim_{r\to 0} {\nu(B(x_0,r))\over \mu(B(x_0,r))}-w(x_0)=0 \iff \lim_{j\to \infty} {\nu(B(x_0,r_j))\over \mu(B(x_0,r_j))}-w(x_0)=0 ,$$ for any sequence $r_j\rightarrow 0$.
So I've thought to prove the latter one. Let $a_j={\nu(B(x_0,r_j))\over \mu(B(x_0,r_j))}-w(x_0)$ for any $j$, so I have to prove that (*) $\forall \epsilon>0 \exists N\in \Bbb N: j>j_N \implies |a_j|<\epsilon.$
Thus, let $\epsilon>0$ be given. Since $w$ is continuous in $x_0$ we have that $\exists N\in \Bbb N: j>j_N \implies |w(x_j)-w(x_0)|<\epsilon$ for any sequence $x_j\rightarrow x_0$.
And now? I don't know if this is the right approach to solve this exercise and, also if it were, I've lost myself in all this limits!
Can someone help me?
HINT:
$$ \frac{\int w(x) d\mu}{\mu(B(x_{0},r))} - w(x_{0}) = \frac{\int w(x) d\mu}{\mu(B(x_{0},r))} - \frac{\int w(x_{0}) d\mu}{\mu(B(x_{0},r))} = \frac{\int (w(x)- w(x_{0})) d\mu}{\mu(B(x_{0},r))}$$