# Radon–Nikodym derivative and “normal” derivative

The Radon–Nikodym theorem states that,

given a measurable space $(X,\Sigma)$, if a $\sigma$-finite measure $\nu$ on $(X,\Sigma)$ is absolutely continuous with respect to a $\sigma$-finite measure $\mu$ on $(X,\Sigma)$, then there is a measurable function $f$ on $X$ and taking values in $[0,\infty)$, such that

$$\nu(A) = \int_A f \, d\mu$$

for any measurable set $A$.

$f$ is called the Radon–Nikodym derivative of $\nu$ wrt $\mu$.

I was wondering in what cases the concept of Radon–Nikodym derivative and the concept of derivative in real analysis can coincide and how?

Thanks and regards!

• Of course it isn't a pure coincidence. The main motivation is the chain rule and the change of variable formula. The Radon-Nikodym derivative is often written as $f = \frac{d\mu}{d\nu}$ if $\mu \ll \nu$. That should suffice as a first hint. (I removed my comment and trivially edited in order to remove my downvote). I'm tired and a bit unnerved. Please accept my apologies. – t.b. May 10 '11 at 23:28
• @Theo: Thanks for your hint! No need to apologize. :) – Tim May 12 '11 at 0:33
• @Tim: Thanks! Well, I was very harsh and you were an innocent victim. I'll try my best to answer your queries whenever I find them interesting. – t.b. May 12 '11 at 0:35
• @Tim: You seem to have deleted the question you posted some minutes ago. Here's a comment to it: Doesn't Rudin addresses 1. in his book? I thought so. 2. The Radon-Nokodym derivative makes sense on a general measure space, while the derivative requires some metric structure, which leads me to 3: Yes this is possible, but you need a metric structure on the measure space and some condition that the metric and the measure interact nicely, for example a doubling condition, i.e., if you double the radius of the ball the measure only grows in some controlled way. – t.b. May 12 '11 at 12:03
• @Theo: Thanks! (1) I just found Rudin mentioned it in the corollary of Theorem 8.6, so I deleted it. I have reposted my question, and you might want to repost your reply there and I might later accept it. (2) Do you have some thought about why you related R-N derivative to change of variables as in my comments under ncmathsadist's reply? Thanks! – Tim May 12 '11 at 12:09

## 1 Answer

Consider absolute continuity of functions as defined in Royden's Real Analysis. These functions are integrals of their derivatives and they are, in fact Radon-Nikodym derivatives.

• In particular, if the function is continuously differentiable, this reduces to the classical change of variables formula and shows that the Radon-Nikodym derivative can be seen as generalized Jacobian (see also Rudin for the higher dimensional case). – t.b. May 11 '11 at 0:10
• @Theo: I wonder why is that "if the function is continuously differentiable, this reduces to the classical change of variables formula"? In change of variables, the variable in the integrand and integrator are both changed, while in change of measure/cdf by R-N derivative, the variable of the integrand does not change and it is more of changing from Lebesgue-stieltjes integral to Lebesgue integral than change of variables. – Tim May 12 '11 at 0:58
• My two previous questions are related to this: math.stackexchange.com/questions/31905/… and math.stackexchange.com/questions/38359/… . – Tim May 12 '11 at 0:59
• @ncmathsadist: Thanks! I was wondering where in Royden's Real Analysis "these functions are integrals of their derivatives and they are, in fact Radon-Nikodym derivatives" is mentioned? I searched to no avail. – Tim Mar 9 '12 at 7:00