# Radon–Nikodym derivative and “normal” derivative

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!

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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
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