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 | 8 Comments



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