Suppose $(\Omega,\mathcal{F},P)$ and $(\tilde{\Omega},\tilde{\mathcal{F}})$ both are measurable space. Define a measurable map $f:\Omega \rightarrow \tilde{\Omega}$.
Let's assume that $g:\tilde{\Omega}\rightarrow \Omega$ is non-negative measurable function. Then
$\int_{\tilde{\Omega}}g\ dP_f = \int_\Omega g \circ f\ dP$
is true, where $P_f(A) = P \circ f^{-1}(A),\ A\in\tilde{\mathcal{F}}$. In the book, the proof of this is somehow omitted. But I was wondering how we can prove it.
Also, let's don't assume that $g$ is non-negative and measurable. Are the two statements equivalent: 1. $g$ is $P_f$ integrable 2. $g \circ f$ is $P$ integrable?
If we show this hold for simple functions $g=\mathbb{1}_B,B\in\mathcal{F}$ then by Monotone Class Lemma everything you mentioned follows.
$$\int_\Omega g\circ fdP=\int_\Omega \mathbb{1}_B\circ f(x)dP(x)=\int_\Omega\mathbb{1}_{\{x:f(x)\in B\}}(x)dP(x)=P(\{x:f(x)\in B\})=P_f(\{y:y\in B\})=\int_{\tilde\Omega}\mathbb{1}_{\{y:y\in B\}}(y)dP_f(y)=\int_{\tilde\Omega}gdP_f$$.