The Unapologetic Mathematician

Mathematics for the interested outsider

The Radon-Nikodym Derivative

Okay, so the Radon-Nikodym theorem and its analogue for signed measures tell us that if we have two \sigma-finite signed measures \mu and \nu with \nu\ll\mu, then there’s some function f so that


But we also know that by definition


If both of these integrals were taken with respect to the same measure, we would know that the equality


for all measurable E implies that f=g \mu-almost everywhere. The same thing can’t quite be said here, but it motivates us to say that in some sense we have equality of “differential measures” d\nu=f\,d\mu. In and of itself this doesn’t really make sense, but we define the symbol


and call it the “Radon-Nikodym derivative” of \nu by \mu. Now we can write


The left equality is the Radon-Nikodym theorem, and the right equality is just the substitution of the new symbol for f. Of course, this function — and the symbol \frac{d\nu}{d\mu} — is only defined uniquely \mu-almost everywhere.

The notation and name is obviously suggestive of differentiation, and indeed the usual laws of derivatives hold. We’ll start today by the easy property of linearity.

That is, if \nu_1 and \nu_2 are both \sigma-finite signed measures, and if a_1 and a_2, then a_1\nu_1+a_2\nu_2 is clearly another \sigma-finite signed measure. Further, it’s not hard to see if \nu_i\ll\mu then a_1\nu_1+a_2\nu_2\ll\mu as well. By the Radon-Nikodym theorem we have functions f_1 and f_2 so that


for all measurable sets E. Then it’s clear that


That is, a_1f_1+a_2f_2 can serve as the Radon-Nikodym derivative of a_1\nu_1+a_2\nu_2 with respect to \mu. We can also write this in our suggestive notation as


which equation holds \mu-almost everywhere.


July 9, 2010 - Posted by | Analysis, Measure Theory


  1. […] Radon-Nikodym Chain Rule Today we take the Radon-Nikodym derivative and prove that it satisfies an analogue of the chain […]

    Pingback by The Radon-Nikodym Chain Rule « The Unapologetic Mathematician | July 12, 2010 | Reply

  2. already is w fine…

    Comment by juanmarqz | July 15, 2010 | Reply

  3. all the way at the top, I think you mean v<<u. You have this reversed.

    Comment by Bobby Brown | March 30, 2011 | Reply

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