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.

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July 13, 2010 - Posted by John Armstrong | Analysis, Measure Theory

6 Comments »

Why is that initial change of variables thing justifiable? You hint that it follows from the chain rule or was closely related, but I don’t see how.

Comment by Bobby Brown | March 30, 2011 | Reply
The first equation is basically just notation, when we’re talking about measures and the Radon-Nikodym derivative. It’s the original change of variables formula that I meant is tied to the chain rule. Follow that link and you should find more information about that connection.

Comment by John Armstrong | March 30, 2011 | Reply
Actually, I was confusing myself in a far different way. I have figured it out now. THanks John

Comment by Bobby Brown | March 31, 2011 | Reply
I think it is not very clear from the first paragraph that you are actually going to prove (or at least make very acceptable) this change of variables formula in the second paragraph. This only became clear to me when I was at “and the substitution formula follows by putting X in for E.” and I realized that “substitution formula” = “change of variables formula”.

Comment by Evert De Nolf | June 21, 2011 | Reply
- Great series of posts though!
  
  Comment by Evert De Nolf | June 21, 2011 | Reply
Thanks, Evert. Glad you find them useful.

Comment by John Armstrong | June 21, 2011 | Reply

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