# The Unapologetic Mathematician

## Mathematics for the interested outsider

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

$\displaystyle\frac{d\nu}{d\lambda}=\frac{d\nu}{d\mu}\frac{d\mu}{d\lambda}$

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

\displaystyle\begin{aligned}f&=\frac{d\nu}{d\mu}\\g&=\frac{d\mu}{d\lambda}\end{aligned}

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

\displaystyle\begin{aligned}\lim\limits_{n\to\infty}\int\limits_Ef_n\,d\mu&=\int\limits_Ef\,d\mu\\\lim\limits_{n\to\infty}\int\limits_Ef_ng\,d\lambda&=\int\limits_Efg\,d\lambda\end{aligned}

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

$\displaystyle\int\limits_Ef_n\,d\mu=\int\limits_Ef_ng\,d\lambda$

Passing to the limit, we find that

$\displaystyle\nu(E)=\int\limits_E\,d\nu=\int\limits_Ef\,d\mu=\int\limits_Efg\,d\lambda$

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 | 10 Comments