# The Unapologetic Mathematician

## Corollaries of the Chain Rule

Today we’ll look at a couple corollaries of the Radon-Nikodym chain rule.

First up, we have an analogue of the change of variables formula, which was closely tied to the chain rule in the first place. If $\lambda$ and $\mu$ are totally $\sigma$-finite signed measures with $\mu\ll\lambda$, and if $f$ is a finite-valued $\mu$-integrable function, then

$\displaystyle\int f\,d\mu=\int f\frac{d\mu}{d\lambda}\,d\lambda$

which further justifies the the substitution of one “differential measure” for another.

So, define a signed measure $\nu$ as the indefinite integral of $f$. Immediately we know that $\nu$ is totally $\sigma$-finite and that $\nu\ll\mu$. And, obviously, $f$ is the Radon-Nikodym derivative of $\nu$ with respect to $\mu$. Thus we can invoke the above chain rule to conclude that $\lambda$-a.e. we have

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

We then know that for every measurable $E$

$\displaystyle\nu(E)=\int\limits_Ef\frac{d\mu}{d\lambda}\,d\lambda$

and the substitution formula follows by putting $X$ in for $E$.

Secondly, if $\mu$ and $\nu$ are totally $\sigma$-finite signed measures so that $\mu\equiv\nu$ — that is, $\mu\ll\nu$ and $\nu\ll\mu$ — then $\mu$-a.e. we have

$\displaystyle\frac{d\mu}{d\nu}\frac{d\nu}{d\mu}=1$

Indeed, $\mu\ll\mu$, and by definition we have

$\displaystyle\mu(E)=\int\limits_E1\,d\mu$

so $1$ serves as the Radon-Nikodym derivative of $\mu$ with respect to itself. Putting this into the chain rule immediately gives us the desired result.

July 13, 2010 Posted by | Analysis, Measure Theory | 6 Comments