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….
Best Answer
$$\cdots=\int\left(\sum_k a_k\chi_{A_k}\right)fd\mu=\int\phi fd\mu$$