In Section 8.1 Derivatives of Measures in Rudin's Real and Complex Analysis:

Suppose $\mu$ is a complex Borel measure on $\mathbb{R}^n$ and $m$ is the Lebesgue measure on $\mathbb{R}^n$, $\Omega$ is a substantial family, $x \in \mathbb{R}^n$, and $A$ is a complex number. If to each $\epsilon > 0$ there corresponds a $ \delta> 0$ such that $|\frac{\mu(E)}{m{E}} - A| < \epsilon$ for every $E \in \Omega$ with $x \in E$ and $diam(E) < \delta$, then we say $\mu$ is differentiable at $x$, and write $(D\mu)(x)=A$

I was wondering

  1. if and when $D \mu(x)$ and the Radon-Nikodym derivative of the continuous part of $\mu$ wrt the Lebesgue measure $m$ at $x$ are the same?
  2. why to distinguish the two above concepts?
  3. can $D \mu$ be generalized from $(\mathbb{R}^n, \mathcal{B(\mathbb{R}^n)}, m)$ to more general measure space $(\Omega, \mathcal{F}, \nu)$?

Thanks and regards!

  • 3
    $\begingroup$ Doesn't Rudin addresses 1. in his book? I thought so. 2. The Radon-Nikodym derivative makes sense on a general measure space (assume $m$ to be $\sigma$-finite), while the derivative requires some metric structure, which leads me to 3: Yes this is possible, but you need a metric structure on the measure space and some condition that the metric and the measure $m$ interact nicely, for example a doubling condition, i.e., if you double the radius of the ball the measure only grows in some controlled way. Moreover, you need some niceness condition on the interaction $\mu$ and $m$ as well (I guess). $\endgroup$ – t.b. May 12 '11 at 12:18
  • 1
    $\begingroup$ Thanks! As to 1, I just found Rudin mentioned it in the corollary of Theorem 8.6. My bad. $\endgroup$ – Tim May 12 '11 at 12:40

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