So how I see it $\nu <<\mu $ on $\mathcal{N}$. Thus there exists a Borel measurable function such that $\nu(A)=\int_Ahd\mu|\mathcal{N}=\int_Ahd\mu$ where the equality holds by the observations here integral computed with respect to a sub-$\sigma$-algebra and $h$ is the Nikodym derivative. Now if $f$ is $\mathcal{N}$ measurable we can say that $g=\frac{f}{h}$ ($g$ is $\mathcal{N}$ measurable and $h>0$ a,e) since $\int_A\frac{f}{h}d\nu=\int_Afd\mu$. However, this does not work when $f$ is not $\mathcal{N}$ measurable. What do I do in this case?
On $(X,\mathcal N, \nu)$ define a new measure $\lambda $ by $\lambda (E)=\int_E fd\mu$. This is a well defined (real) measure and it is absolutely continuous w.r.t. $\nu$. By RNT there exists $g\in L^{1}(\nu)$ such that $\lambda (E)=\int_E gd\nu$ for all $E \in \mathcal N$. Uniqueness is also clear from RNT.
If you are not familiar with RNT for real measures just apply the argument to $\lambda_1(E)=\int_E f^{+}d\mu$ and $\lambda_2(E)=\int_E f^{-}d\mu$.