I've struggling with this problem for days now :(, I managed to solve part a) by simple functions but I havent been able to do anything in b).Any help would be really appreciated.
Problem Let $(\Omega,F,\mu)$ a measure space, $(X,G)$ a measurable space and $f:\Omega \rightarrow X$ measurable function. For $A \in G$, we define $\upsilon(A)=\mu(f^{-1}(A))$.
a) Prove that if $g:X\rightarrow [0,\infty]$ measurable, then $\int_{X}gd\upsilon = \int_{\Omega} g \circ fd\mu.$
b) $g:X\rightarrow \mathbb{R}$. Show that g is $\upsilon$-integrable iff $g\circ f$ is $\mu$-integrable and in this case show the equality of a).
Thanks so much in advance <3
The identity (a) holds obviously if $g$ is an indicator function (and therefore if it is a simple function).
Then, let $g\ge 0$ and consider a sequence of non-negative indicator functions $h_n\nearrow g$. By Beppo-Levi, $$\lim_{n\to\infty} \int_X h_n\,d\nu=\int_X f\,d\nu$$
On the other hand, $\psi_n=h_n\circ f$ is itself a sequence of non-negative indicator functions such that $\psi_h\nearrow g\circ f\ge 0$, therefore $$\lim_{n\to\infty}\int_\Omega \psi_n\,d\mu=\int_\Omega g\circ f\,d\mu$$ Since $\int_\Omega\psi_n\,d\mu=\int_X h_n\,d\nu$, the identity is proved for non-negative functions.
Finally, a function $g$ is $\nu$-integrable if and only if $\int_X\lvert\bullet\rvert\circ g\,d\nu<\infty$, in which case $\int_X g\,d\nu=\int_X(\bullet)^+\circ g\,d\nu-\int_X(\bullet)^-\circ g\,d\nu$.
Point (b) now follows by the identity in (a) and the fact that $$(\lvert\bullet\rvert\circ g)\circ f=\lvert\bullet\rvert\circ (g\circ f)\\ ((\bullet)^+\circ g)\circ f=(\bullet)^+\circ( g\circ f)\\ ((\bullet)^-\circ g)\circ f=(\bullet)^-\circ( g\circ f)$$