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.

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    $\begingroup$ Doing it for all functions involves $\infty - \infty$ calculations. Even for the countably additive case of counting measure on $\mathbb N$, we do not evaluate all series $\sum a_n$. (Equivalent to a probability measure by using $\mu(\{n\}) = 2^{-n}$.) $\endgroup$ – Gerald Edgar Sep 29 at 13:28
  • $\begingroup$ @GeraldEdgar Right, I shouldn't have said for all functions. I edited a bit. The question is, Is there a notion of integrability that doesn't require the function to be boudned? $\endgroup$ – aduh Sep 29 at 21:05
  • $\begingroup$ @aduh Do you also require $\int (-f)d\mu = -\int f d\mu$ ? $\endgroup$ – user95282 Sep 30 at 15:13
  • $\begingroup$ @user95282 Yes, I will edit, thanks. $\endgroup$ – aduh Sep 30 at 21:52
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    $\begingroup$ @aduh Do you mean that $\Phi$ is the set of all functions 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 $\infty-\infty$. $\endgroup$ – user95282 Oct 2 at 13:59

Once the integral over measurable bounded functions is defined, a “standard” way is this: For a nonnegative (measurable) function $f$ one takes the supremum of the integral over all bounded measurable functions $g$ with $0\le g\le f$. For general measurable $f$ one splits into integrals over the positive and negative part and takes the difference, requiring that at least one of the integrals (according to the previous definition) is finite. (For functions for which both integrals are infinite, there is no “standard” way.)

For functions with values in a Banach space, one cannot deal with infinite integrals, and one cannot use the supremum-argument. A “standard” procedure in this case is carried out in the “bible” of functional analysis, Dunford N., Schwarz Jacob T., Linear Operators I, Int. Publ./Wiley-Interscience 1966, but be aware that there are some oversights (IIRC, e.g. in the proof of the monotone convergence theorem; not sure whether they were corrected in later editions).

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  • $\begingroup$ "Theory of Charges: A Study of Finitely Additive Measures" by K.P.S. Bhaskara Rao and M. Bhaskara Rao has a chapter on integration against charges (i.e., against signed finitely additive measures). They cover Dunford-Schwarz's notion of integration in that chapter, along with a different notion of integration which they describe as "of Stieltjes type". $\endgroup$ – asahay Oct 3 at 7:01

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