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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?

BCLC
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Detectives
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  • You mean, in general? – Did Nov 06 '12 at 14:33
  • @did I hope so.. But I also thanks for any comment or advice to compute it! – Detectives Nov 06 '12 at 14:39
  • 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. – Harald Hanche-Olsen Nov 06 '12 at 14:41
  • @HaraldHanche-Olsen you mean it's not easy to find Radon-Nikodym derivative in general. If computation is possible, it should be the way that by comparing both sides of $ v(E)= \int_E f d \mu $? – Detectives Nov 06 '12 at 14:56
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    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. – Harald Hanche-Olsen Nov 06 '12 at 15:08
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    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) – John Nov 06 '12 at 15:46
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    There is no constructive version of the Radon-Nikodym theorem known. A book that discusses cases in which one can compute the derivatives in detail is "Conditional Measures and Applications" by M.M. Rao, especially the second edition. But it is a very advanced book.For now, I would simply accept that this is a very non-constructive result. – Michael Greinecker Nov 07 '12 at 10:49

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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.

Christopher A. Wong
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  • Excuse me, but have you forgotten the integral sign? Should it be $d\mu=\int f dm$? – Pedro Gomes Feb 19 '17 at 18:46
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    That would not be standard notation as far as I'm aware. – Christopher A. Wong Mar 01 '17 at 21:59
  • Would this construction also work with $m$ taken to be the Borel measure? – LudvigH Mar 30 '21 at 09:36
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    @LudvigH Yes, as long as our limit is taken as given above (limit of decreasing open balls), since Lebesgue measure and Borel measure agree on open sets (in standard Euclidean topology). – Christopher A. Wong Mar 30 '21 at 21:01
  • Thank you for the addition. :) – LudvigH Mar 31 '21 at 06:51
  • @PedroGomes See Folland's Real Analysis: Modern Techniques and Their Applications (Second Edition) pp. 89. He defines $d\mu = f, dm$ as meaning $\mu(A) = \int_A f , dm$ for all measurable sets $A$. – ashman Feb 17 '22 at 13:46
  • @ChristopherA.Wong This is really nice, thanks. Do you happen to have a reference for this? And any link to an example computation for some concrete $\mu$? (Would just be interested to see one as a template, as I do not have much experience computing limits with balls.)

    Probably my questions will be answered once I get to Sec 3.4 of Folland. Currently at Sec 3.2. :-)

     – ashman Feb 17 '22 at 13:48