The Unapologetic Mathematician

Mathematics for the interested outsider

The Radon-Nikodym Chain Rule

Today we take the Radon-Nikodym derivative and prove that it satisfies an analogue of the chain rule.

If \lambda, \mu, and \nu are totally \sigma-finite signed measures so that \nu\ll\mu and \mu\ll\lambda, then \lambda-a.e. we have


By the linearity we showed last time, if this holds for the upper and lower variations of \nu then it holds for \nu itself, and so we may assume that \nu is also a measure. We can further simplify by using Hahn decompositions with respect to both \lambda and \mu, passing to subspaces on which each of our signed measures has a constant sign. We will from here on assume that \lambda and \mu are (positive) measures, and the case where one (or the other, or both) has a constant negative sign has a similar proof.

Let’s also simplify things by writing


Since \mu and \nu are both non-negative there is also no loss of generality in assuming that f and g are everywhere non-negative.

So, let \{f_n\} be an increasing sequence of non-negative simple functions converging pointwise to f. Then monotone convergence tells us that


for every measurable E. For every measurable set F we find that

\displaystyle\int\limits_E\chi_F\,d\mu=\mu(E\cap F)=\int\limits_{E\cap F}\,d\mu=\int\limits_{E\cap F}g\,d\lambda=\int\limits_E\chi_Fg\,d\lambda

and so for all the simple f_n we conclude that


Passing to the limit, we find that


and so the product fg serves as the Radon-Nikodym derivative of \nu in terms of \lambda, and it’s uniquely defined \lambda-almost everywhere.

July 12, 2010 - Posted by | Analysis, Measure Theory


  1. […] Corollaries of the Chain Rule Today we’ll look at a couple corollaries of the Radon-Nikodym chain rule. […]

    Pingback by Corollaries of the Chain Rule « The Unapologetic Mathematician | July 13, 2010 | Reply

  2. Could you quickly clarify why there is no loss of generality in assuming that f and g are everywhere non-negative? Sorry for the frequent recent posts, and thank you for the great resource too!

    Comment by Bobby Brown | March 30, 2011 | Reply

  3. Basically, if they’re not you can always decompose into positive and negative parts, use the result there, and put everything back together.

    Comment by John Armstrong | March 30, 2011 | Reply

  4. Could anyone tell me, were we exactly use the increasing sequences and why? I dont really see it 🙂

    Comment by Wiebs91 | November 16, 2012 | Reply

  5. You mean where we use the fact that the sequence \{f_n\} is increasing? That’s a requirement of the monotone convergence theorem.

    Comment by John Armstrong | November 16, 2012 | Reply

  6. I know that theorem. I just wanted to know why we need it here, where it is used in the last part V(E)=… . I’m sorry, im really not used to this notations at all and don’t get the point of it by now

    Comment by Wiebs91 | November 16, 2012 | Reply

  7. Okay, well the point is that any measurable function f can be approximated as the limit of an increasing sequence of measurable functions f_n, and the simple functions are basically constants times characteristic functions. So what we do is prove our result for the characteristic functions \chi(F). Then the fact that everything in sight is linear means that it holds for simple functions. And finally we can pass to the limit (using monotone convergence) and get the result for all measurable functions.

    Comment by John Armstrong | November 16, 2012 | Reply

  8. Ah, now I see it, thanks for your time an help 🙂

    Comment by Wiebs91 | November 16, 2012 | Reply

  9. how do we know fg is lamda-integrable?

    Comment by egfdgfg | April 2, 2015 | Reply

  10. Because fg is the Radon-Nikodym derivative of \nu with respect to \lambda. To put it another way, you know that there has to be one, since \nu\ll\lambda (because \ll is transitive), so we can call it h, and it’s uniquely defined (\lambda-a.e.). But the product fg satisfies the same condition as h: you can get the \nu-measure of any set E by integrating either fg or h with respect to \lambda over E. So they’re equal \lambda-a.e. and fg must be \lambda integrable, and so it’s as good a representative for the Radon-Nikodym derivative as any.

    Comment by John Armstrong | April 2, 2015 | Reply

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