Let $(X, \mathcal X)$ be a measurable space. Say that a net $(\mu\_\alpha)$ of finitely additive probability measures converges to a finitely additive probability measure $\mu$ if and only if $\mu\_\alpha(A) \to \mu(A)$ for all $A \in \mathcal X$.


If $f$ is an extended-real-valued simple $\mathcal X$-measurable function of the form $f = \sum\_{j=1}^n a\_j 1\_{A\_j}$, then the integral of $f$ with respect to a finitely additive probability measure is defined in the usual way:
$$\int fd\mu = \sum\_{j=1}^n a\_j \mu(A\_j).$$
If $f: X \to [0,\infty]$ is non-negative, then define
$$\int f d\mu = \sup\Big\{ \int gd \mu: g \ \text{simple}, \ 0 \leq g \leq f \Big\}.$$



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> **Question.** Is it the case that if $\mu\_\alpha \to \mu$, then $\int f d\mu\_\alpha \to \int f d\mu$ for all non-negative $\mathcal X$-measurable $f: X \to [0,\infty]$?
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If $f$ is bounded (and not necessarily non-negative), then the result holds. The motivation for the question is that I'm wondering to what extent the "usual properties" of the finitely additive integral extend from bounded functions to non-negative ones. For example, in [this post](https://mathoverflow.net/questions/79663/does-integrating-with-respect-to-a-finitely-additive-measure-respect-addition) it is shown that the finitely additive integral remains linear on non-negative functions.



Does the finitely additive integral preserve convergence for non-negative measurable functions?
No, not even for sequences of countably additive measures.


Take $X = \mathbb{N} = \{0,1,2,\dots\}$ with its discrete $\sigma$-algebra, and let $\mu\_n$ put mass $1/n$ at the point $n$ and mass $1-1/n$ at $0$. Let $\mu$ put mass $1$ at $0$. Then it is clear that $\mu\_n(A) \to \mu(A)$ for every set $A$ (consider the cases $0 \in A$ and $0 \notin A$).


Set $f(n) = n$. Then we have $\int f\,d\mu\_n = 1$ for all $n$ but $\int f\,d\mu = 0$.


