Radon–Nikodym derivative Relation to derivative in calculus?

I am reading chapter 3 of Folland's Real Analysis Book, where it introduces the Radon-Nikodym Derivative of a sigma finite signed measure with respect to another sigma finite positive measure.

Let $$\nu$$ be a $$\sigma$$-finite signed measure and $$\mu$$ be a $$\sigma$$-finite positive measure on $$(X,M)$$. There exists unique $$\sigma$$-finite signed measure $$\lambda,\rho$$ on the space such that $$\lambda \perp \mu \text{ } \rho << \mu \text{ and } \nu=\lambda +\rho$$ and there is an $$\mu$$-integrable function $$f:X\to \mathbb{R}$$ such that $$d\rho=f d\mu$$ where $$f$$ is the Radon-Nikodym derivative.

I am just wondering if this is somehow related to the derivative in calculus in anyway, since Riemann integral of a continuous fct on bounded interval is equal to Lebesgue integral and Lebesgue measure is $$\sigma$$-finite? Or maybe this question is kind of stupid and they are not related in any way.

1 Answer

The two are very much related.

You may be familiar with Lebesgue's Differentation Theorem. In the special case where $$\lambda$$ is the Lebesgue measure in $$\mathbb{R}$$, and for a $$\sigma$$-finite measure $$\nu$$ with $$\nu << \lambda$$ we have $$\frac{d\nu}{d\lambda}(x)=\lim\limits_{h\to 0}\frac{\int\limits_{x-h}^{x+h}{\frac{d\nu}{d\lambda}(y)}d\lambda(y)}{2h}= \lim\limits_{h\to 0}\frac{\nu([x-h,x+h])}{2h} \quad \text{\lambda-a.e.}$$

And this in turn is the differential quotient of the function $$f(x)=\nu((-\infty,x])$$

A more general version of the theorem holds for Radon measures. It can be shown that if $$\nu << \mu$$, then $$\frac{d\nu}{d\mu}(x)=\lim\limits_{r \to 0} \frac{\nu({B_{r}(x)})}{\mu(B_{r}(x))} \quad \text{\mu-a.e.}$$

see Theorem 1.30 in

Evans, Lawrence Craig; Gariepy, Ronald F., Measure theory and fine properties of functions, Textbooks in Mathematics. Boca Raton, FL: CRC Press (ISBN 978-1-4822-4238-6/hbk). p.50 (2015). ZBL1310.28001.