Define a measure $\nu$ given by $\nu(E)=\int_E gd\mu$ for all $E\subseteq\Omega$ measurable and for some measurable non-negative $g$. Prove $\int_\Omega fd\nu=\int_\Omega fgd\mu$.
First step I proved this for simple functions, which was pretty straight-forward. Also, proving $\int_\Omega fgd\mu\geq\int_\Omega fd\nu$ for any measurable non-negative $f$ was easy too. It's the other inequality I'm having trouble with.
Following definition, $\int_\Omega fgd\mu=\operatorname{Sup}\{\int_\Omega hd\nu:0\leq h\leq fg, \text{h simple}\}$. Now, if I could say that $fg\geq h$ implies $f\geq \frac{h}{g}$ then solving it would be easy, but I can because it can be undefined. So I'm stuck. Any hint would be helpful.
Once the statement is proven for indicator functions and thus, for nonnegative simple functions, the results follows for all nonnegative measurable functions through the monotone convergence theorem. Indeed, for $f\geq0$ define the sequence of simple functions $$ s_n=\sum^{n2^n-1}_{j=0} 2^{-n}j\mathbb{1}_{2^{-n}j\leq f<(j+1)2^{-n}} + n\mathbb{1}_{\{n\leq f\}}$$
It is not difficult to check that $0\leq s_n\leq s_{n+1}$ and that $\lim_ns_n=f$ pointwise (I leave the details to the OP). Another way to define $s_n$ is by setting $$s_n(x) =\max(n,2^{-n}\lfloor 2^n f(x)\rfloor)$$ where $y\mapsto \lfloor y\rfloor$ is the integer part function.
The result follows then by monotone convergence theorem: $$\int f\,d\nu=\lim_n\int s_n\,d\nu=\lim_n\int s_n g\,d\mu=\int fg\,d\mu$$
By linearity, the result extends to any real measurable function for which $\int|f|\,d\nu=\int|f|g\,d\mu<\infty$.