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.



A question about finitely additive integration
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 $1/n$.


