Question on integral, notation and Nikodym derivative I have a very general question for those measure theoric, real analysis guy out there .
I am very confused by the concept of Nikodym derivative. If $v << \mu$, we can find a non negative function $f$ s.t:
\begin{align}
v(E) = \int_E fd\mu
\end{align}
We call $f$ the Radon-Nikodym derivative. So far I have no problem it's just notation.
My book says that $f$ is sometimes written $\frac{dv}{d\mu}$ because it is the derivative of v w.r to $\mu$. Well this is again notation. What I don't get is why in the book they start to use this notation in calculus. Like simplifying fractions and even write : $fd\mu = dv$.
Isn't $\frac{dv}{d\mu}$ suppose to be just a notation ?
If it's more than a notation what does it represent ? I understand the concept of function, measure, and integral but i'm lost when it comes to this notation. It has to represent something otherwise we won't be using it. And the fact that this fraction can be nicely used to derive the product rule for example doesn't comes for free. 
So my question is clear, what does $\frac{dv}{d\mu}$ represent. The notion of derivative hasn't been introduced (yet) in the book so I am not supposed to conclude that if I derive v w.r to $\mu$ I get f.
Thanks for any help !!!

When we're integrating on the real line, and if $\nu$ and $\mu$ are identified with sufficiently smooth functions giving rise to Stieltjes measures, then in the case that $f$ is differentiable, it is literally true that $f = d\nu/d\mu$ in the sense of strong derivatives.
For example, suppose that $d\mu = dx$, the regular Lebesgue measure, and $d\nu$ is given the Stieltjes measure arising from $x^2$, so $d\nu([a,b]) = b^2 - a^2$. Then the Radon-Nikodym derivative is $f = d\nu/d\mu = d(x^2)/dx = 2x$. 
