Let $(\Omega_1, \mathcal{A}_1)$, $(\Omega_2, \mathcal{A}_2)$ be measurable spaces and $f\colon \Omega_1 \rightarrow \Omega_2$ be an $(\mathcal{A}_1,\mathcal{A}_2)$-measurable function. Furthermore, let $\mu\colon \mathcal{A}_1 \rightarrow [0,\infty]$ be a measure and define $f_*\mu =\mu\circ f^{-1}$ as the pushforward measure of $\mu$.
For every measurable $g\colon \Omega_2 \rightarrow [0,\infty]$ we then have
$\displaystyle \int_{\Omega_2} g\,\mathrm{d}\left( f_*\mu\right)=\int_{\Omega_1} (g\circ f)\,\mathrm{d}\mu$
My proof is as follows:
Step 1: Show the statement for $g=\chi_{A_i},\ A_i\in\mathcal{A}_2$ and $\chi$ is the indicator function ($1$ on $A_i$ and $0$ elsewhere)
$\displaystyle\int_{\Omega_2}\chi_{A_i}\,\mathrm{d}\left( f_*\mu\right)=\mu\circ f^{-1}(A_i)=\int_{\Omega_1}\chi_{f^{-1}(A_i)}\,\mathrm{d}\mu=\int_{\Omega_1}\chi_{A_i}\circ f\,\mathrm{d}\mu$
Here I used that $\chi_{A_i}\circ f=\chi_{f^{-1}(A_i)}$
Step 2: Show for $g=\sum_{i=1}^n\alpha_i\chi_{A_i},\ \alpha_i\in(0,\infty], A_i\in\mathcal{A}_2$, that is, for a simple function.
$\displaystyle \int_{\Omega_2} \sum_{i=1}^n\alpha_i\chi_{A_i}\,\mathrm{d}\left( f_*\mu\right)=\sum_{i=1}^n\alpha_i\int_{\Omega_2}\chi_{A_i}\,\mathrm{d}\left( f_*\mu\right)=\sum_{i=1}^n\alpha_i\int_{\Omega_1}\left( \chi_{A_i}\circ f\right)\,\mathrm{d}\mu\\ \displaystyle =\int_{\Omega_1}\left ( \sum_{i=1}^n \alpha_i \chi_{A_i}\right)\circ f\,\mathrm{d}\mu=\int_{\Omega_1}\left (g\circ f\right)\,\mathrm{d}\mu$
In the second equality I used Step 1 and in the third I used that $\chi_{A_i}\circ f$ is again a simple function.
Step 3: Show the statement for the $g$ as stated at the top. Since $g$ is measurable, there exists a sequence of simple functions $g_n$ which is monotonically increasing to $g$.
$\displaystyle \int_{\Omega_2}g\,\mathrm{d}\left(f_*\mu\right)=\lim_{n\rightarrow\infty}\int_{\Omega_2}g_n\,\mathrm{d}\left(f_*\mu\right)=\lim_{n\rightarrow\infty}\int_{\Omega_1}g_n\circ f\,\mathrm{d}\mu=\int_{\Omega_1}g\circ f\,\mathrm{d}\mu$
In the first equality I used the monotonicity of the Lebesgue Integral for simple functions (already proved in my book). In the second inequality I am not sure if my argument is correct. I used that $g_n\circ f$ converges monotonically to $g\circ f$. Is this correct? I know that $g_n\circ f$ is again a step function, but I am unsure if it indeed converges to $g\circ f$.