A question about finitely additive integration

Question

Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space ($\mathbb P$ is countably additive). Let $\{p_\omega: \omega \in \Omega\}$ be a family of (countably additive) probability measures on $(\Omega, \mathcal F)$, and assume that the mapping $\omega \mapsto p_\omega$ is $\mathcal F$-measurable. Let $\mu$ be a finitely additive probability measure on $(\Omega, \mathcal F)$. Suppose the following holds for all $A \in \mathcal F$: $$\mathbb P(A) = \int p_\omega(A) \mu(d\omega) \tag{1}$$

Can (1) be extended to $$\int f d\mathbb P = \int\int f(\omega')p_{\omega}(d\omega')\mu(d\omega) \tag{2}$$ for all bounded $\mathcal F$-measurable $f$?

The definition of the finitely additive integral I'm working with is exactly like the usual definition of the Lebesgue integral. In particular, the integral is linear on the space of boudned $\mathcal F$-measurable functions and continuous in the sup-norm. Thus, by linearity, (2) holds for simple functions.

I'm a bit confused about the general case, though. If $f$ is bounded, it can be uniformly approximated by a sequence $f_n$ of $\mathcal F$-simple functions. Then, $$\int f d\mathbb P = \lim_n \int f_n d\mathbb P = \lim_n \int\int f_n(\omega') p_\omega(d\omega')\mu(d\omega).$$ In order conclude, I think I need that $\int f_n dp_\omega$ converges to $\int f dp_\omega$ uniformly in $\omega$. But I am unable to see that this is the case.

Answer 1

A measurable function $f$ taking values in $[-M,M]$ can be approximated above and below by simple functions: $f_n \le f \le f_n+1/n$ where $f_n(x)=\lfloor nf(x) \rfloor/n$. Note that $f_n$ takes on less than $3Mn$ values. Now from the case of simple functions it follows that the LHS and RHS of (2) differ by at most $1/n$.