Let's look at the usual distribution function of a discrete probability distribution, say Poisson$(\lambda)$; call this function $F_X(x\vert\lambda)$. However, let's modify that function slightly so that it is left-continuous instead of right-continuous; call this function $\bar{F}_X(x\vert\lambda)$.
Now let $\mu$ be the counting measure. With the usual CDF, $\frac{dF(X\vert\lambda)}{d\mu} = \frac{\lambda^xe^{-\lambda}}{x!}$. What happens when we take the Radon-Nikodyn derivative of the left-contnuous function, $\frac{d\bar{F}}{d\mu}$?
AFAIK we can only speak of a Radon-Nikodym derivative if there are two measures $\nu,\mu$ with: $$\nu(E)=\int_E f\;d\mu$$for every measurable $E$.
In that case $f$ is a Radon-Nikodym derivative of $\nu$ with respect to $\mu$ and can be denoted as $\frac{d\nu}{d\mu}$.
I understand that for $\mu$ you want to take the counting measure but what about $\nu$?
Just some affected (lost right-continuity) CDF at most indicates that - in order to determine a measure $\nu$ - we should go back to the original CDF and do it with the probability measure that is determined by this CDF.
That is an uninteresting situation because nothing changes then.