The derivative of a measure

Question

Let $\mu$, $\nu$ be two Radon Measure on $\mathbb{R}^n$. How can I prove that $D_{\mu}{\nu}=\lim_{r \to 0} \frac{\nu(B(x,r)}{\mu(B(x,r))}$ is in $L^1_{loc}(\mathbb{R}^n,\mu)$?

Answer 1

Can you prove that if $r \le D_\mu \nu(x)$ for all $x$ in some Borel set $B$, then $r \mu(B) \le \nu(B)$?

Once you have that fact, on any compact set $K$ and for any $t > 1$ you have $$\int_K D_\mu \nu \, d\mu = \int_{K \cap \{D_\mu \nu > 0\}} D_\mu \nu \, d\mu = \sum_{j=-\infty}^\infty \int_{K \cap \{t^j < D_\mu \nu \le t^{j+1}\}} D_\mu \nu \, d\mu$$

Of course $$\int_{K \cap \{t^j < D_\mu \nu \le t^{j+1}\}} D_\mu \nu \, d\mu \le t^{j+1} \mu(K \cap \{t^j < D_\mu \nu \le t^{j+1}\})$$ and according to the first observation you have $$t^{j+1} \mu(K \cap \{t^j < D_\mu \nu \le t^{j+1}\}) \le t \nu(K \cap \{t^j < D_\mu \nu \le t^{j+1}\}).$$

Thus $$\int_K D_\mu \nu \, d\mu \le \sum_{j=-\infty}^\infty t \nu(K \cap \{t^j < D_\mu \nu \le t^{j+1}\}) \le t \nu(K) < \infty.$$