I was trying to understand the proof given in this question: Radon-Nikodym derivative composed with function
Especially the last equality: $$ \int_{F(A)} \frac{d\mu}{d\nu} \, d\nu = \int_A (\frac{d\mu}{d\nu} \circ F) \, d\nu\circ F $$
I can see that: $$ \int_{F(A)} \frac{d\mu}{d\nu} \, d\nu = \int_{F(A)} (\frac{d\mu}{d\nu} \circ F) \, d\nu\circ F $$
Since $F_*(\nu\circ F)=\nu$. However, I can’t see why we could exchange to an integral over A, as opposed to this integral over F(A).
Can someone explain to me this part? Is this result true at all? If not, is there another similar result for a composition of a Radon-Nikodym derivative with a measurable function?
Thanks!