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 1


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.


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