Question:
Given measure space $(X,\mathcal(A),\mu)$ and non-negative measurable function $f:X\to[0,\infty]$, consider the measure $\nu(A) = \int_{A}fd\mu$ defined for all $A\in\mathcal{A}$. Prove that for any $\mu$-measurable function $g$, $$ \int_{X}gd\nu = \int_{X}fgd\mu $$
Attempted Solution:
I am thinking we can use (1) the fact that any measurable function $g$ is a pointwise limit of simple functions, and (2) monotone convergence theorem, to extend the result for simple functions
So let $\varphi$ be a simple function with canonical form $\sum_{k}a_k \chi_{a_k}$. Then $$ \int_{X}\varphi d\nu = \sum_{k}a_{k}\nu(A_{k}) = \sum_{k}a_{k}\int_{A_{k}}fd\mu = \sum_{k}\int_{A_{k}}a_{k}fd\mu $$
And hereI am stuck....
Let $\varphi$ be a non-negative simple measurable function of the cannonical form $\sum_{k}a_k \chi_{A_k}$ where $a_k\in [0,\infty)$ and $A_k$ are measurable sets. Then we have $$ \int_{X}\varphi\ \mathrm d\nu = \sum_{k}a_{k}\nu(A_{k}) = \sum_{k}a_{k}\int_Xf\chi_{A_k}\ \mathrm d\mu = \int_X\sum_ka_{k}\chi_{A_k}f\ \mathrm d\mu=\int_X\varphi f\ \mathrm d\mu$$
Now whenever $g$ is a non-negative measurable function we have a sequence $\{\varphi_n\}$ simple functions such that $0≤\varphi_1≤\varphi_2≤\cdots≤g$ and $\varphi_n \rightarrow g$ pointwise. Then we can apply Monotone Convergence Theorem, $$\int_X g\ \mathrm d\nu=\lim\int_X\varphi_n \ \mathrm d\nu=\lim\int_X\varphi_nf\ \mathrm d\mu=\int_X\lim(\varphi_nf)\ \mathrm d\mu=\int_Xfg\ \mathrm d\mu$$ (Since $\{f\varphi_n\}$ is an increasing sequence of non-negative measurable functions converging point-wise to $fg$)
Now whenever $g$ is arbitrary measurable function write $g$ as $g=g^+-g^-$ where both $g^+=\max(g,0),\ g^-=\max(-g,0)$ both are non-negative measurable. Then we have $$\int_X fg^+\ \mathrm d\mu=\int_X g^+\ \mathrm d\nu$$ and $$\int_X fg^-\ \mathrm d\mu=\int_X g^-\ \mathrm d\nu$$ Hence $g$ is $\nu$ integrable if $g$ is $\mu$ integrable, and in this case we can write $$\int_Xg\ \mathrm d\nu=\int_X g^+\ \mathrm d\nu-\int_Xg^-\ \mathrm d\nu=\int_Xfg^+\ \mathrm d\mu-\int_Xfg^-\ \mathrm d\mu=\int_Xfg\ \mathrm d\mu$$