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.

$\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.