Let $(\Omega, \mathcal{F})$ be a measurable space and $P$ be a random, $\mathcal{G}$-measurable finite measure on $(\Omega, \mathcal{F})$, with $\mathcal{G} \subseteq \mathcal{F}$.
Is the following proposition true? Is my proof correct?
Proposition. The function $f(\omega) = \int g dP(\omega)$ is $\mathcal{G}$-measurable for all bounded, $\mathcal{F}$-measurable functions $g$.
Proof. The proposition holds if $g$ is the indicator function of $A \in \mathcal{F}$ simply because $P(A)(\omega)$ is $\mathcal{G}$-measurable in $\omega$ by definition. Since the bounded $\mathcal{G}$-measurable functions are closed under linear combinations and uniform limits, the entire proposition follows.
Is that sufficient? I'm somehow not fully confident about the last sentence of the proof. Should I add that the bounded convergence theorem is used for the limit step?
No, your argument is not complete. Given $g$ there exist simple functions $g_n,n=1,2,...$ such that $g_n \to g$ uniformly. $\int g_ndP(\omega)$ is measurable for each $n$. To show that $\int gdP(\omega)=\lim \int g_ndP(\omega)$ you will have to assume that the finite measures $dP(\omega)$ have a uniform bound, i.e. $sup_{\omega}P(\omega,\Omega))<\infty$. I hope this is what you meant when you said the random measure $P$ was finite.