Finitely additive probabilities and Integrals My question is the following: if $(\Omega, \mathcal{A}, P)$ is a probabilistic space with $P$ simply additive (not necessarily $\sigma$-additive) and $f,g$ two real valued, positive, bounded, $\mathcal{A}$-measurable function, then $\int f+g\,dP=\int f\,dP+\int g\,dP$?
The definition of $P$ simply additive is the following (I took it from Example for finitely additive but not countably additive probability measure):


*

*For each $E \subset \Omega$, $0 \le P(E) \le 1$

*$P(\Omega) = 1$

*If $E_1$ and $E_2$ are disjoint subsets $P(E_1 \cup E_2) = P(E_1) + P(E_2)$.

 A: You would need to define the meaning of the integral for a finitely additive measure. I am not sure if there is standard and well established definition in that context. And I think the notion will generally not be very well behaved.
For instance, suppose you allow not all sets to have a measure, but allow measures on $\mathbb{N}$ to be defined 
only on finite and cofinite sets (as with normal measures, you only define them on $\sigma$-algebras). Then, you have a nice measure $\mu$ that is $0$ on finite sets and $1$ on cofinite sets. Let $f(n) := n$, or $g(n) = n \pmod{2}$. What values would you assign to $\int f $ and $\int g$ ?
That being said, linearity is a "must" for all notions of integrals I have ever seen. I am far from being an expert, but (variously generalised) integrals are things that take a map, map it to a number, and always (1) preserve positivity and (2) are linear.

Edit: If you are referring to the example of ultrafilter measures mentioned under the link, they do lead to well defined "integrals". That is, you can define $\int f = c$ if and only if the set  $$\{ n \ : \ f(n) \in (c-\epsilon, c+\epsilon) \}$$ has measure $1$, or equivalently belongs to the ultrafilter. This gives you a well defined integral, that has all the usual nice properties (except Fubini's theorem, perhaps).
