3
$\begingroup$

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!

  • 0
    "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 $n$.2011-04-09
  • 0
    @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 $m$?2011-04-09
  • 0
    @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 :)2011-04-09
  • 0
    @Tim (for (2) and (3)): by geometry I meant metric, but you can also think of topology. (3) yes. Isn't it surprising?2011-04-09
  • 0
    @user8268: Regarding (3), are there some reference (book, link, ...)? Thanks!2011-04-09
  • 0
    @Tim(regarding (3)): Sure there are many, but I don't know:) Here is a hint for isomorphism of $I=(0,1)$ and $I^2$. Write any number $a\in I$ in binary, $a=0.a_1a_2a_3\dots$, and associate to it a point in $I^2$, $(0.a_1a_3a_5\dots,\,0.a_2a_4\dots)$. That's a measure-preserving map (but you need to fight a bit with numbers that have non-unique binary expansion to make it a bijection)2011-04-09

1 Answers 1

2

regarding question 1:

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.

  • 0
    Thanks! I was wondering what "the derivative of the measure" is referred to?2011-05-11
  • 0
    @Tim: the first definition, "differentiation of integrals"2011-05-12
  • 0
    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 $(D\mu)(x)=A$"?2011-05-12
  • 0
    I don't know what a substantial family is, but this looks very much like the definition you referred to in the question. (The difference is using arbitrary sets rather than balls. I guess that a "substantial family" is some restriction on the volume in relation to the diameter)?2011-05-12