Is the mapping from non-negative functions to their finitely additive integrals semi-continuous? This is a follow up to a question I asked at MO, which I think ended up being too easy for that site.
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 $\liminf_\alpha\int f d\mu_\alpha \geq \int f d\mu$ for all non-negative $\mathcal X$-measurable $f: X \to [0,\infty]$?

In the previous question, I asked whether  $\mu_\alpha \to \mu$ implies $\int f d\mu_\alpha \to \int f d\mu$, and this was shown to be false by a simple example in which $\int f d\mu_\alpha = 1$ for all $\alpha$ and $\int f d\mu = 0$, which is consistent with the answer to the present question being affirmative.
 A: 
Is it the case that if $\mu_\alpha \to \mu$, then $\liminf_\alpha\int f d\mu_\alpha \geq \int f d\mu$ for all non-negative $\mathcal X$-measurable $f: X \to [0,\infty]$?

Yes. Let a net $(\mu_\alpha:\alpha\in A)$ of finitely additive probability measures converges to a finitely additive probability measure $\mu$. Let $M<\int f d\mu$ be any real number. There exist a natural number $n$, numbers $a_1,\dots a_n\in\Bbb R\cup\{\infty\}$, and sets $A_1,\dots,  A_n\in\mathcal X$ such that $0\le g = \sum_{j=1}^n a_j 1_{A_j}\le f$ and $\int g d\mu>M$. For each $\varepsilon=(\varepsilon_1,\dots, \varepsilon_n)\in\{-1,1\}^n$ put $A_\varepsilon=\bigcap A_j^{\varepsilon_j}$, where $A_j^{\varepsilon_j}$ equals $A_j$, if $ \varepsilon_j=1$, and equals $X\setminus A_j$, otherwise. Then a family $\mathcal A=\{A_\varepsilon: \varepsilon\in \{-1,1\}^n\}$ consists of pairwise disjoint sets. So $g=\sum_{i=1}^m b_i 1_{B_i}$ for some distinct $B_i\in\mathcal A$ and numbers $b_i\in\Bbb R\cup\{\infty\}$. Since $g\ge 0$, we have $b_i\ge 0$ for each $i$. It is easy to check that we can relax condition $b_i\in \Bbb R\cup\{\infty\}$ to $b_i\in\Bbb R$, keeping $0\le g\le f$ and $\int g d\mu>M$.
Let $\varepsilon>0$ be any real number. Pick $\delta>0$ such that $\delta\sum_{i=1}^m b_i<\varepsilon$. There exists $\beta\in A$ such that $|\mu_\alpha(B)- \mu(B)|< \delta$ for each $\alpha\ge\beta$ and each $B\in\mathcal A$. It follows that
$$\int f d\mu_\alpha\ge \int g d\mu_\alpha=\sum_{i=1}^m b_i\mu_\alpha(B_i)\ge \sum_{i=1}^m b_i(\mu(B_i)- \delta)\ge \int g d\mu-\delta\sum_{i=1}^m b_i\ge \int g d\mu-\varepsilon>M-\varepsilon.$$
A: Alex Ravsky's answer spends a lot of time messing around with sets, but I'm not sure that such argumentation is necessary.
First, suppose $g$ simple.  Then $$\lim_{\alpha}{\int{g\,d\mu_{\alpha}}}=\int{g\,d\mu}$$ since limits interchange with finite linear combinations.  In particular, this common value must also be the liminf.
Now let $f$ be any (nonnegative) function and simple $g\leq f$.  Then, for any $\alpha$, $$\int{f\,d\mu_{\alpha}}\geq\int{g\,d\mu_{\alpha}}$$  Taking the liminf, we have $$\liminf_{\alpha}{\int{f\,d\mu_{\alpha}}}\geq\int{g\,d\mu}$$ by the above.  But since $g$ was arbitrary, we can take the supremum on the RHS to obtain the desired inequality: $$\liminf_{\alpha}{\int{f\,d\mu_{\alpha}}}\geq\int{f\,d\mu}$$
