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aduh
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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 $f \in \Phi$ that aren't necessarily bounded?

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 on the right-hand side exist and the right-hand side is not of the form $\infty - \infty$ 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.

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 $\int f d\mu$ for $f \in \Phi$ that aren't necessarily bounded?

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 the integrals on the right-hand side exist and the right-hand side is not of the form $\infty - \infty$.

(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.

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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aduh
  • 869
  • 6
  • 15

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 $\int f d\mu$ for $f \in \Phi$ that aren't necessarily bounded?

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

(1) $\int (f+g) d\mu = \int f d\mu + \int g \mu$ whenever$\int (af+bg) d\mu = a\int f d\mu + b\int g \mu$ for $f,g \in \Phi$ and $a,b \in \mathbb R$, provided the integrals on the right-hand side exist and the right-hand side is not of the form $\infty - \infty$.

(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.

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 $\int f d\mu$ for $f \in \Phi$ that aren't necessarily bounded?

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

(1) $\int (f+g) d\mu = \int f d\mu + \int g \mu$ whenever the right-hand side is not of the form $\infty - \infty$.

(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.

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 $\int f d\mu$ for $f \in \Phi$ that aren't necessarily bounded?

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 the integrals on the right-hand side exist and the right-hand side is not of the form $\infty - \infty$.

(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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aduh
  • 869
  • 6
  • 15

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 $\int f d\mu$ for all $f \in \Phi$ that aren't necessarily bounded?

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

(1) $\int (f+g) d\mu = \int f d\mu + \int g \mu$ whenever the right-hand side is not of the form $\infty - \infty$.

(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.

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 $\int f d\mu$ for all $f \in \Phi$?

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

(1) $\int (f+g) d\mu = \int f d\mu + \int g \mu$ whenever the right-hand side is not of the form $\infty - \infty$.

(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.

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 $\int f d\mu$ for $f \in \Phi$ that aren't necessarily bounded?

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

(1) $\int (f+g) d\mu = \int f d\mu + \int g \mu$ whenever the right-hand side is not of the form $\infty - \infty$.

(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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aduh
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