This thread is Q&A.
Problem
Given Borel spaces $X$ and $Y$.
Consider a Borel measure: $$\mu:\mathcal{B}(X)\to\mathbb{C}:\quad\mu\geq0$$
Regard a pushforward: $$h\in\mathcal{B}(X,Y):\quad\nu:=\mu\circ h^{-1}$$
Then for integrability: $$g\in\mathcal{L}(Y;\nu)\iff g\circ h\in\mathcal{L}(X;\mu)$$
And one obtains: $$\int_Y g\,\mathrm{d}\nu=\int_X(g\circ h)\,\mathrm{d}\mu$$
How can I prove this?
Reference
This is a lemma for: Pushforward (SM)
For simple functions: $$s(h(x))=\sum_kb_k\delta_{h(x)\in B_k}=\sum_kb_k\delta_{x\in h^{-1}B_k}$$
Note that it remains: $$\nu(B_k)<\infty\implies\mu(h^{-1}B_k)<\infty$$
So one obtains: $$\int_Y s\mathrm{d}\nu=\sum_kb_k\nu(B_k)=\sum_kb_k\mu(h^{-1}B_k)=\int_X h\circ h\mathrm{d}\mu$$
By Beppo-Levi one gets: $$\int_Y|g|\,\mathrm{d}\nu=\lim_n\int_Ys_n\,\mathrm{d}\nu=\lim_n\int_X(s_n\circ h)\,\mathrm{d}\mu=\int_X|g\circ h|\,\mathrm{d}\mu$$
And equality follows from: $$g=\Re_+g-\Re_-g+i\Im_+g-i\Im_-g$$
Concluding the assertion.
