I've just encountered the following exercise:
Let $(X,\mathcal{M})$ be a locally compact metric space. Let $\nu$, $\mu$ be two Radon measures on $X$. Suppose that:
(i) $w \in L^1(X,\mu)$, $w \ge 0 $ a.e on $X$.
(ii) $w$ is continuous at $x_0$.
(iii) $\nu(A)=\int_{A} w \, d \mu$ for every $A \in \mathcal{M}$, that is $w=\frac{d \nu}{d \mu}$.
Prove that:
$$ \lim_{r \rightarrow 0} \frac{\nu(B(x_0,r))}{\mu(B(x_0,r))}=w(x_0)$$.
The statement seems very familiar when $X=\mathbb{R}^n$ $\mu \equiv \mathcal{L}$ (Lebesgue measure) since it's the Lebesgue differentiation theorem (or even the FTC since $w$ is continuous, if $n=1$). But in this general case I cannot quite get to the point: I've tried a standard $\epsilon$-$\delta$ argument that seems not working due to a $\frac{1}{\mu(B(x_0,r))}$ I cannot control just with the continuity of $w$ at $x_0$. How to attack the problem in this general case?
My attempt:
Fix $\epsilon > 0$. By the continuity of $w$ at $x_0$ we can find a $\delta > 0$ such that $|w(x)-w(x_0)| \le \epsilon $ whenever $ x \in B(x_0, \delta)$. It follows:
$$ \left| \frac{\nu(B(x_0,r))}{\mu(B(x_0,r))}-w(x_0) \right| \le \frac{1}{\mu(B(x_0,r)} \int_{B(x_0,r)} |w(x)-w(x_0)| \, d \mu \le \frac{1}{\mu(B(x_0,r)} \int_{B(x_0,\delta)} |w(x)-w(x_0)| \, d \mu \le \epsilon \frac{\mu(B(x_0,\delta))}{\mu(B(x_0,r))} $$, provided $|r| < \delta$.