# Are two interpretations of “differentiation of measures” related?

As Wikipedia mentioned, there are two interpretation of "differentiation of measures":

I was wondering

1. if they are related to each other, or unrelated concepts?
2. if there are other concepts that can also be viewed as differentiation of measures?

Thanks and regards!

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"differentiation of integrals" works in $\mathbb{R}^n$ or in similar spaces; it relies on the geometry of $\mathbb{R}^n$ (which is much more than the $\sigma$-algebra of measurable sets and the measure). Radon-Nikodym theorem is for any ($\sigma$-finite) measures, and uses no additional structure on the space. So the two concepts can only be compared on $\mathbb{R}^n$. (one clarifying thing - as measure spaces, all $\mathbb{R}^n$ (for all $n$'s, with Lebesgue measure) are isomorphic. Geometrically they are quite different - differentiation using balls depends strongly on $n$. – user8268 Apr 9 '11 at 13:00
@user8268: Thanks! (1) So are they unrelated? (2) Is the concept "gemoetry" generally equivalent to"metric"? (3) "as measure spaces, all $\mathbb{R}^n$ (for all n's, with Lebesgue measure) are isomorphic" do you mean that there always exists a bijective measure-preserving measurable mapping between $\mathbb{R}^n$ and $\mathbb{R}^m$ for any $n$ and $m$? – Tim Apr 9 '11 at 13:11
@Tim: I would rather say that the differentiation on $\mathbb{R}^n$ (which does give you the Radon-Nikodym derivative) is a connection between Radon-Nikodym theorem and the geometry of $\mathbb{R}^n$. That connection is (IIRC) quite difficult to prove. I believe that somebody more competent will enlighten you here :) – user8268 Apr 9 '11 at 13:12
@Tim (for (2) and (3)): by geometry I meant metric, but you can also think of topology. (3) yes. Isn't it surprising? – user8268 Apr 9 '11 at 13:15
@user8268: Regarding (3), are there some reference (book, link, ...)? Thanks! – Tim Apr 9 '11 at 13:15

If $\mu$ is a measure on $\mathbb{R}^n$, with a Radon-Nikodym derivative (of the continuous part), in respect to the Lesbeuge measure h ($\int_E\ h\ dm = \mu (E)$), then at every Lebesgue point of h, the derivative of the measure, is equal to h (practically straight from the definition). Since for every integrable function almost every point is a Lebesgue point, and the Radon-Nikodym derivative is defined up to a set of measure 0, then it is indeed the case that both "derivatives" come out the same.
Thanks! Is "the derivative of the measure" same as the one defined 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 $(D\mu)(x)=A$"? – Tim May 12 '11 at 11:42