Interpretation of dP in Radon-Nikodym Theorem [Radon-Nikodym Theorem]
Let $(\Omega, \Sigma, P)$  be a probability space. Suppose that $(\Omega, \Sigma, \mu)$ be a measure space with $\mu(A)=0$ implies $P(A)=0$, then there exist a function $f:X \rightarrow \mathbb{R}$ such that 
\begin{equation}
P(A)=\int_A f d\mu.
\end{equation}
Further, function $f$ is called Radon-Nikodym derivative of $P$ with respect to $\mu$.
(So far I know, why this definition makes sense, because it's similar with derivative in calculus, that is, $g'$ is derivative of $g$ because $g(x)=\int g'(x) dx$).
In calculus, we also know differential, that is, if $y=f(x)$ be a differentiable function, then differential is defined by $dy=f'(x)dx$. Could we interpret $dP$ as a differential like in calculus? How could we interpret $dP$ in Radon-Nikodym context?
 A: The $dP$ symbol appears first in probability in the context of integration, not the Radon-Nikodym theorem. In that setting it is analogous to the $dx$ in Riemann integration: $dx$ is a notational artifact that remains when passing from a Riemann sum to integration and refers to the Lebesgue measure of an interval of length $\Delta x$ as $\Delta x \to 0$. So, 
$$
  \sum_{i=1}^n f(x_i)\Delta x_i \ \mbox{ is to } \ \sum_{i=1}^n \sup_{x_i \in A_i}f(x_i) P(A_i)
$$
as
$$
  \int_a^b f(x) dx \ \mbox{ is to } \ \int_{\cup A_i} f dP.
$$
Put another way, a differential like $dx$ can be interpreted as an infinitesimally small length, and the $dP$ can be interpreted as infinitesimally small measure.
In the Radon-Nikodym context, if $\Omega$ has a topological structure and if you have 
$$
  P(A) 
= \int_A f(x) d\mu(x)
$$ 
for some $f:\Omega \to \mathbb{R}$ satisfying: $\forall \epsilon > 0$, $\exists$ $\delta > 0$ such that $x,y \in U$ and $P(U) < \delta$ implies $|f(x) - f(y)| < \epsilon$, then for a set $\Delta A$ of small $P$-measure
$$
        P(\Delta A) 
=       \int_{\Delta A} f(x) d\mu(x)
\approx \sup_{x \in \Delta A} f(x) \mu(\Delta A)
$$
which, if you take the limit as $P(\Delta A) \to 0$, you can think of as being analogous to $dy = f'(x) dx$.
A: The Radon-Nikodym theorem is very general. To tone down the generality lets consider $f$ a continuous function and $g$ a differentiable function. This will allow us to understand the problem better and make interpretations from there. 
In this nice Riemann-Stieltjes context we arrive at the following calculation: 
$F(x) = \int_a^x f\ dg = \int_a^x f\cdot g'\ dt$ 
(Page 255 - Proposition 2.10 - Haaser & Sullivan - Real Analysis)
We see that $F'(x)=f(x)\cdot g'(x)$. $F$ could be called the antiderivative of $f$ with respect to $g$. 
In the case that $g=x$ then this new definition reduces to the old definition of antidifferentiation. 
In some sense antidifferentiating $f$ with respect to $g$, means to find a function $h$ such that $f=h'\circ g$. Then $h\circ g$ will be the antiderivative of $f$ with respect to $g$ as in the following calculation. 
$(h\circ g)'=(h'\circ g) \cdot g' = f\cdot g'$. 
So, $P$ can be interpreted as the antiderivative of $f$ with respect to $\mu$, meaning that there is some function $h$ such that $P=h\circ \mu$, $f=h'\circ \mu$ and $P'=f\cdot \mu'$. Of course, the symbols I just wrote may not make much sense in this different context.  
