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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.

tfjaervik
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    The formula holds when $g$ is a characteristic function (by definition). By linearity, it holds whenever $g$ is a simple function. Now use the monotone convergence theorem to conclude that it holds for any non-negative measurable function. Finally, for an integrable function, express it as a linear combination of non-negative functions. – Prahlad Vaidyanathan Dec 11 '21 at 12:40
  • @PrahladVaidyanathan Thanks for the comment! Would it be possible for you (or anyone else that sees this) to add some more details and post it a a reply? That would make it more clear to me (and hopefully others later), and I could also mark the question as answered. – tfjaervik Dec 11 '21 at 12:54
  • Tip: use \ll and \gg to render $\ll$ and $\gg$ properly – FShrike Dec 11 '21 at 13:53
  • @PrahladVaidyanathan 's argument is the standard one used ubiquitously in measure theory; if you are advanced to the point where you have studied Radon-Nikodym, you have almost surely seen their point before! – FShrike Dec 11 '21 at 13:55
  • @FShrike Thanks for the comment (appreciate the pun). Yes, after I read his argument I realized this result was presented as a theorem in the book we used (Real Analysis - Measures, Integrals and Applications, by Makarov and Podkorytov (p. 146). The argument given there is more or less the same. – tfjaervik Dec 11 '21 at 15:23

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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.

tfjaervik
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