Can we show $\frac{\nu\left(B_\varepsilon(x)\right)}{\mu\left(B_\varepsilon(x)\right)}\xrightarrow{\varepsilon\to0+}\frac{{\rm d}\nu}{{\rm d}\mu}(x)$?

Question

Let $(E,\mathcal E,\mu)$ be a finite measure space and $\nu$ be a finite signed measure on $(E,\mathcal E)$ with $\nu\ll\mu$. By the Radon-Nikodým theorem, $$\nu=f\mu\tag1$$ for some $f\in L^1(\mu)$.

If $(E,d)$ is a metric space and $\mathcal E=\mathcal B(E)$, are we able to show that $$g_\varepsilon(x):=\left.\begin{cases}\displaystyle\frac{\nu\left(B_\varepsilon(x)\right)}{\mu\left(B_\varepsilon(x)\right)}&\text{, if }\mu\left(B_\varepsilon(x)\right)>0\\0&\text{, otherwise}\end{cases}\right\}\;\;\;\text{for }x\in E$$ is convergent in a suitable mode of convergence (e.g. in $\mu$-almost everywhere or in $L^1(\mu)$ as $\varepsilon\to0+$ and the limit $g$ is a version of $f$?

If someone isn't aware of a rigorous proof, but can give an intuitive explanation of the basic idea, I'd appreciate that as well.

Answer 1

Note that $\frac{\nu(B_{\epsilon}(x))}{\mu(B_{\epsilon}(x))}=\frac{1}{\mu(B_{\epsilon}(x))}\int_{B_{\epsilon}(x)} f(y) \, d\mu(y).$

By the Lebesgue differentiation theorem, we know that $\frac{1}{\mu(B_{\epsilon}(x))}\int_{B_{\epsilon}(x)} f(y) \, d\mu(y)\to f(x)$ for $\mu$ a.e. $x,$ as $\epsilon\to 0.$

This completes the proof.