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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?

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    You mean, in general?2012-11-06
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    @did I hope so.. But I also thanks for any comment or advice to compute it!2012-11-06
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    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.2012-11-06
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    @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 $?2012-11-06
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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.2012-11-06
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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)2012-11-06
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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.2012-11-07

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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.

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    Excuse me, but have you forgotten the integral sign? Should it be $d\mu=\int f dm$?2017-02-19
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    That would not be standard notation as far as I'm aware.2017-03-01