Let $X$ be a set, and let $\mu$ be a finitely additive probability measure defined on $2^X$. Let $\Phi$ be the set of functions from $X$ to $\mathbb R \cup \{-\infty, \infty\}$.

Is there a standard way of defining a integrable function that allows us to define $\int f d\mu$ for some unbounded $f \in \Phi$?

To be clear, the integral should satisfy the following properties:

(1) $\int (af+bg) d\mu = a\int f d\mu + b\int g \mu$ for $f,g \in \Phi$ and $a,b \in \mathbb R$, provided all the integrals exist and $\infty - \infty$ doesn't occur anywhere.

(2) If $f \geq 0$, then $\int f d\mu \geq 0$.

(3) If $1_A$ is the indicator of $A \subset X$, then $\int 1_A d\mu = \mu(A)$.

The standard way of defining the integral for bounded, real-valued functions is the familiar one: first define the integral for simple functions, then uniformly approximate any bounded function by a sequence of simple ones. I'm wondering to what extent this can be pushed beyond bounded functions. Obviously the integral will be badly behaved in the sense that it won't satisfy the convergence theorems that we take for granted when $\mu$ is countably additive, but as (1)-(3) indicate, I don't mind that.

allfunctions involves $\infty - \infty$ calculations. Even for the countably additive case of counting measure on $\mathbb N$, we do not evaluateallseries $\sum a_n$. (Equivalent to a probability measure by using $\mu(\{n\}) = 2^{-n}$.) $\endgroup$allfunctions from $X$ to $\mathbb{R} \cup \{ - \infty, \infty \}$? In that case the expression within the left hand side of (1) is sometimes also of the form $\infty-\infty$. $\endgroup$3more comments