The Radon-Nikodym derivative $f$ for a measurable space $(X,F)$ and measures $\mu,\nu$ where $\nu$'s support contains $\mu$'s support, is defined as follows:
$$\frac {d\mu}{d\nu}(x)=f(x) \text { s.t. }\quad \forall A\in F: \quad \mu(A)=\int_A fd\nu$$
My question is: Can we also define the Radon-Nikodym derivative "directly"? i.e. as follows:
$$f(x):=...$$
I am looking for something analogous to the classical derivative in real analysis which is defined as
$$\frac {dF(x)}{dx}=f(x)=\lim_{d\to 0}\frac {F(x+d)-F(x)}{d}$$
Rather than: $$\frac {dF(x)}{dx}=f(x) \text { s.t. }\quad \forall a,b\in \mathbb R: \quad F(b)-F(a)=\int_a^b f(x)dx$$
EDIT:
I came up with the following idea, but it only works for some measurable spaces: Take a sequence of sets $A_n$ that are both $\mu$ and $\nu$ measure positve for all $n$, such that $\lim_{n\to\infty}A_n=\{x \}$. Then $$f(x)=\lim_{n\to \infty}\frac {\mu(A_n)}{\nu(A_n)}$$
However, out of the examples I've checked, this only works for discrete measurable spaces (in which case the definition is useless anyway), and one-dimensional continuous measure spaces (not sure if it works for all of them, only the examples I picked). It already fails in the measure space $(\mathbb R^2, B)$ because the value you get depends on how you approach the limit.
If $\nu$ is a positive Radon measure in $\mathbb{R}^n$ and $\mu$ is a $\mathbb{R}^m$-valued Radon measures in $\mathbb{R}^n$, then for $\nu$-a.e. $x$ the Radon-Nikodym derivative of $\mu$ with respect to $\nu$ is given by $$ f(x) = \lim_{r\to 0} \frac{\mu(B_r(x))}{\nu(B_r(x))}. $$ (This result is known as Besicovitch derivation theorem.)