I learned Radon-Nikodym theorem in class and I know what exactly it is. But I am not sure about how to compute Radon-Nikodym derivative... Any reference does not explicitly say about how to compute Radon-Nikodym derivative..

Can anybody help me about how to compute it or provide some useful thm regarding it?

  • $\begingroup$ You mean, in general? $\endgroup$ – Did Nov 6 '12 at 14:33
  • $\begingroup$ @did I hope so.. But I also thanks for any comment or advice to compute it! $\endgroup$ – Detectives Nov 6 '12 at 14:39
  • $\begingroup$ I am unaware of any general method for computing the Radon–Nikodym derivative. In most cases where you can, it is because the construction of the measures involves practically invites it. $\endgroup$ – Harald Hanche-Olsen Nov 6 '12 at 14:41
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    $\begingroup$ That would be the trivial case. Perhaps $\mu$ and $\nu$ are both defined in terms of densities with respect to a third measure, possibly with some limit involved. In which case the computation tends to boil down to chasing densities through the construction, making sure that nothing escapes into a null set along the way. Sorry if this is vague, I can't think of a good example. $\endgroup$ – Harald Hanche-Olsen Nov 6 '12 at 15:08
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    $\begingroup$ As @did alluded to there is no general methodology. But that shouldn't be too surprising in the sense that by now, you have been exposed to a number of mathematical entities with the same property. (i.e. integrals, ODE's, PDE's) $\endgroup$ – John Nov 6 '12 at 15:46

If $d\mu = f \, dm$, where $m$ is the Lebesgue measure on $\mathbb{R}^n$, then there is a concrete way of realizing the differentiation of measures; in particular, for almost every $x \in \mathbb{R}^n$, $$ \lim_{r \rightarrow 0} \frac{\mu(B(r,x))}{m(B(r,x))} = f(x)$$

In principle, a similar result holds if $d\mu = f \, d\nu$, but the issue is that then we don't want to use the sets $B(r,x)$ because we don't know how those behave under the measure $\nu$; so ultimately you have to know a lot about the measures explicitly if you want to do any computation.

  • $\begingroup$ Excuse me, but have you forgotten the integral sign? Should it be $d\mu=\int f dm$? $\endgroup$ – Pedro Gomes Feb 19 '17 at 18:46
  • $\begingroup$ That would not be standard notation as far as I'm aware. $\endgroup$ – Christopher A. Wong Mar 1 '17 at 21:59

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