# Differentiation theorem for Radon measures

I have trouble to understand a detail in the proof of the following Theorem:

Theorem: Let $\nu, \mu: \mathcal P(\mathbb R^n) \to [0, \infty]$ be outer Radon measures, such that $\nu \ll \mu$. Then we have $$\nu(A) = \int_A \mathrm D_\mu \nu \, \mathrm d \mu$$ for every $\mu$-measurable set $A \subset \mathbb R^n$.

Now in the proof of this theorem, the author considers the set $$Z := \left\{ x \in \mathbb R^n \; | \; \mathrm D_\mu \nu(x) = 0 \right\} \, .$$ Since for each $\alpha > 0$ the inclusion $$Z \subset \left\{ x \in \mathbb R^n \; | \; {\mathrm D^-}_{\mu} \nu(x) \leq \alpha \right\} \; ,$$ holds, we can conclude from a Lemma that $$\nu(Z) \leq \alpha \mu(Z) \; .$$ Now the author concludes, that $\nu(Z) = 0$. But this conclusion is only possible, if $\mu(Z) < \infty$, right? So why is $\mu(Z) < \infty$?

• By the way, it's the Theorem 2 on page 40 of "Evans / Gariepy - Measure Theory and Fine Properties of Functions". – aexl Jan 7 '15 at 3:16
• I had been studying this book, too, and I think most of the proofs are quite quick and need some refinements. If you are interested in how I tried to to this, let me know. But for this particular problem, see my answer below. – sranthrop Jan 7 '15 at 3:22

Since $\mu$ and $\nu$ are Radon measures, you can assume that $\mu$ and $\nu$ are finite measures. To be precise: Let \begin{align*} Z:=\{x\in A\ |\ \mathrm D_\mu\nu(x)=0\}. \end{align*} We would like to show that $\nu(Z)=0$. To do this consider \begin{align*} Z_n:=Z\cap\{x\in\mathbb R^n\ |\ ||x||\leq n\}. \end{align*} Then $Z_n$ is a subset of a compact set and hence has finite measure. Therefore, we can apply the Lemma and obtain $\nu(Z_n)\leq\alpha\mu(Z_n)$ for each $\alpha>0$ and $n\in\mathbb N$. Thus, $\nu(Z_n)=0$ for each $n\in\mathbb N$. Finally, \begin{align*} \nu(Z)=\nu\left(\bigcup_{n\geq 1}Z_n\right)\leq\sum_{n\geq 1}\nu(Z_n)=0 \end{align*} and $\nu(Z)=0$.