# Does the finitely additive integral preserve convergence for non-negative measurable functions?

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 $$\int f d\mu = \sup\Big\{ \int gd \mu: g \ \text{simple}, \ 0 \leq g \leq f \Big\}.$$

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 $$f: X \to [0,\infty]$$?

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 it is shown that the finitely additive integral remains linear on non-negative functions.

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 $$\int f\,d\mu = 0$$.