I have read Radon–Nikodym derivative and "normal" derivative but I am still quite confused. Here are the two kinds of derivative and I am looking for the analog:
The derivative of a function of real variables measures the change of function value w.r.t. its input.
The $\mu$-integrable function $f=d\nu/d\mu$ is called the Radon-Nikodym derivative whenever $\nu\ll\mu$.
My idea is that the absolute continuity gives us a hint of the change of value $\nu$ w.r.t. $\mu$. Is it correct?
It is called a derivative as a tribute to the (Riemann) fundamental theorem of calculus. It is not a classic derivative, since $\nu$ is most likely not a function of $\mu$.
The Radon-Nikodym theorem just states that whenever some (possibly signed) measure $\nu$ is absolutely continuous with respect to a measure $\mu$, over the same sigma algebra in a sigma finite space, there exists a measurable function $f$ such that:
$$\nu(A)=\int_Af\,\mathrm{d}\mu$$
If we treated this like a more familiar integral we might be tempted to differentiate both sides “with respect to $\mu$” because we are used to derivatives being the opposite of integrals, and we would “get” $d\nu=f\,d\mu$ as a convenient heuristic notation.
There do exist other types of derivative definition for more abstract cases (e.g. Fréchet’s derivative) but Radon Nikodym is not that. However, saying $f$ is the R-N derivative of $\nu$ with respect to $\mu$ does tell us “the density” of $\nu$, or how $\nu$ measure changes with $\mu$ measure, so intuitively it is a good notation/terminology.
Just remember that the following:
$$\lim_{h\to0}\frac{\nu(A+h)-\nu(A)}{\mu(h)}$$
Would perhaps be a way to express your classical derivative, and here that expression does not make any sense unless we make many more definitions. It is the same idea as derivative, but formally different to, say, $d/dx\,x^2=2x$.
You may be interested in the Lebesgue differentiation theorem. It is again not talking about a classical derivative, but is a much closer analogue.
