This relates to the top answer here: Question on integral, notation and Nikodym derivative
Suppose that $\nu << \mu$. Then we can find a non-negative $f$ s.t.
$$\nu(E) = \int_{E} d\nu = \int_{E} fd\mu$$.
So far, things seem clear to me. My question is the following: Though it makes intuitive sense, how can we be sure that
$$\int_{E} d\nu = \int_{E} fd\mu \Rightarrow \int_{E} gd\nu = \int_{E} gfd\mu$$ for all integrable functions $g$? In the top answer to the question I linked to above, the following claim is made:
For every integrable $g$, the following formula holds:
$$\int_{E} gd(\int_{E} fd\mu) = \int_{E} gfd\mu$$.
It therefore seems that a justification/proof of this claim would answer my question.
I have been exposed to some measure theory and integration theory a few years back, and as I was revising some material recently, this claim was not clear to me. Perhaps this claim is obvious, and my confusion simply arises from a poor understanding of important definitions. Either way, any help in understanding this claim is much appreciated.
For reference to future readers and to mark this question as answered. The comment from Prahlad Vaidyanathan under the original post answers my question. The result is given as a theorem in the book Real Analysis: Measures, Integrals and Applications, by Makarov and Podkorytov (p. 146). The proof in that book is essentially the same as the argument given by Prahlad Vaidyanathan.