The Radon-Nikodym Theorem says:
If a measureable space $(S,\,\Sigma)$ admits $\sigma$-finite measures $\mu$ and $\nu \ll \mu $, there is a unique (up to sets of $\mu$-measure zero) $\Sigma$-measurable function $ f:S\rightarrow \left[\right.0,\,\infty \left.\right)$, such that, for any measurable set $ A\in \Sigma$, $$ \nu (A)\,=\,\int_{A}\,f\,d\mu\,\;. $$ Named the Radon-Nikodym derivative of $\nu$ with respect to $\mu$, the function $f$ is commonly written as $d\nu/d\mu$.
Question.
A function $\,S\longrightarrow\mathbb{R}\,$ is called Lebesgue-measurable iff the preimage of any open set from $\mathbb R$ is measurable in $(S,\,\Sigma)$.
Would it be right to say that the Radon-Nikodym derivative $f$ is Lebesgue-measurable?