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I have already asked this question in this MSE thread, but some people suggested me to ask to the MO community also.

Preliminaries

An algebra of sets in a set $X$ is an $\mathcal{X}\subseteq\mathcal{P}(X)$ such that:

  1. $\emptyset\in\mathcal{X}$.

  2. For $A,B\in\mathcal{X}$ then $A\cup B\in\mathcal{X}$.

  3. For $A,B\in\mathcal{X}$ then $A\setminus B\in\mathcal{X}$.

If $G$ is a topological abelian group and if $f:X\rightarrow G$ is a function and $g\in G$, the sum $\sum_{x\in X}f(x)$ is said to converge to $g$ iff for every neighborhood $U$ of $g$ there is a finite subset $A\subseteq X$ such that for every finite subset $B\subseteq X$ such that $A\subseteq B$ we have $\sum_{x\in B}f(x)\in U$.

If $\kappa$ is a regular infinite cardinal, then a $\kappa$-premeasure is a function $\mu:\mathcal{X}\rightarrow H$ where $H$ is a topological abelian group and:

  1. For $\mathcal{A}\subseteq\mathcal{X}$, if its elements are disjoint, $|\mathcal{A}|<\kappa$ and $\bigcup\mathcal{A}\in\mathcal{X}$, then $\sum_{A\in\mathcal{A}}\mu(A)$ converges to $\mu(\bigcup\mathcal{A})$.

Question

If $G$, $H$ and $K$ are topological abelian groups and $\cdot:G\times H\rightarrow K$ is a continuous biadditive function, $f:X\rightarrow G$ is a function and $\mu:\mathcal{X}\rightarrow H$ is a $\kappa$-premeasure, how we may define the integrability of $f$ so that this definition is equivalent to the usual definitions of Lebesgue integrability of a measurable real/complex valued function in the usual case where we have a measure space $(X,\mathcal{Y},\nu)$ in the usual sense, $\mathcal{X}=\{A\in\mathcal{Y}:\mu(A)\text{ is finite}\}$, $\kappa=\aleph_1$ and $\mu=\nu\upharpoonright\mathcal{X}$?

I have read the definitions of Bochner Integral and Pettis Integral. They try to be quite nice generalizations, but they depend on the integration of real/complex valued functions and I do not know how to generalize it to arbitrary topological fields.

Attempt

I tried to follow the idea of Riemann Integration, because it is easier to generalize the definition of Riemann Integration to topological vector spaces over arbitrary topolofical fields, but I have made some modifications.

Let $\Omega$ be the set of $\mathcal{A}\subseteq\mathcal{X}\setminus\{\emptyset\}$ whose elements are disjoint and such that $|\mathcal{A}|<\kappa$ and $\bigcup\mathcal{A}\in\mathcal{X}$.

For $\mathcal{A},\mathcal{B}\in\Omega$, we say $\mathcal{A}\leq\mathcal{B}$ iff for every $A\in\mathcal{A}$ there is $\mathcal{B}_A\subseteq\mathcal{B}$ such that $A=\bigcup\mathcal{B}_A$. In other words, $\mathcal{A}\leq\mathcal{B}$ iff every $A\in\mathcal{A}$ is a union of members of $\mathcal{B}$. It looks like refinement, but we may have $\bigcup\mathcal{A}\subset\bigcup\mathcal{B}$ (proper inclusion).

It is easy to see that $\leq$ is a preorder. It is also directed. Indeed, for every $\mathcal{A}\in\Omega$ and $E\in\mathcal{X}$ such that $\bigcup\mathcal{A}\subset E$, then consider $\mathcal{B}=\mathcal{A}\cup\{E\setminus\bigcup\mathcal{A}\}$, then $\mathcal{B}\in\Omega$, $\bigcup\mathcal{B}=E$ and $\mathcal{A}\leq\mathcal{B}$. Now, if $\mathcal{A},\mathcal{B}\in\Omega$, then considering $E=\bigcup\mathcal{A}\cup\bigcup\mathcal{B}$ we have $E\in\mathcal{X}$ and there are $\mathcal{A}',\mathcal{B}'\in\Omega$ such that $\mathcal{A}\leq\mathcal{A'}$, $\mathcal{B}\leq\mathcal{B'}$ and $\bigcup\mathcal{A'}=\bigcup\mathcal{B'}=E$, so let:

$$\mathcal{C}=\{A\cap B:A\in\mathcal{A'},B\in\mathcal{B'}\}\setminus\{\emptyset\},$$

then $\mathcal{C}\in\Omega$ and $\mathcal{A}\leq\mathcal{C}$ and $\mathcal{B}\leq\mathcal{C}$.

A choice function for $\mathcal{A}\in\Omega$ is a function $a:\mathcal{A}\rightarrow X$ such that $a_A\in A$ for every $A\in\mathcal{A}$.

Let $\Theta$ be the set of pairs $(\mathcal{A},a)$ where $\mathcal{A}\in\Omega$ and $a$ is a choice function for $\mathcal{A}$.

Let us say an $s\in K$ is an integral of $f$ iff for every neighborhood $U$ of $s$ there is a $\mathcal{A}\in\Omega$ such that for every $(\mathcal{B},b)\in\Theta$ satisfying $\mathcal{A}\leq\mathcal{B}$ the sum $\sum_{B\in\mathcal{B}}f(b_B)\mu(B)$ converges to some element of $U$.

Now consider the usual case where we have a measure space $(X,\mathcal{Y},\nu)$ in the usual sense, $\mathcal{X}=\{A\in\mathcal{Y}:\mu(A)\text{ is finite}\}$, $\kappa=\aleph_1$ and $\mu=\nu\upharpoonright\mathcal{X}$, and we have a measurable function $f:X\rightarrow\mathbb{R}_{\geq0}$. Let us say a simple function is a linear combination of characteristic functions of sets with finite measure (elements of $\mathcal{X}$). Then consider the following proprerties:

  1. $f$ is Lebesgue-integrable in the usual sense, that is, the set $\{\int\varphi:\varphi\text{ simple and }0\leq\varphi\leq f\}$ is bounded.

  2. $f$ is integrable in the new sense.

I was able to prove (2)$\Rightarrow$(1) and, if $X$ has finite measure, then (1)$\Rightarrow$(2).

$\bullet$ (2)$\Rightarrow$(1): Suppose $s$ is an integral of $f$ in the new sense. For every simple function $\varphi$ such that $0\leq\varphi\leq f$, let $\varphi=\sum_{A\in\mathcal{A}}s_A\chi_A$ where $\mathcal{A}\in\Omega$, then for every $\varepsilon>0$ there is a $\mathcal{B}\in\Omega$ such that for every $(\mathcal{C},c)\in\Theta$ satisfying $\mathcal{B}\leq\mathcal{C}$ then $s-\varepsilon<\sum_{C\in\mathcal{C}}f(c_C)\mu(C)<s+\varepsilon$; so for every $(\mathcal{C},c)\in\Theta$ satisfying $\mathcal{A},\mathcal{B}\leq\mathcal{C}$, then:

$$\int\varphi=\sum_{A\in\mathcal{A}}s_A\mu(A)=\sum_{A\in\mathcal{A}}s_A\sum_{C\in\mathcal{C}\\C\subseteq A}\mu(C)=\sum_{A\in\mathcal{A}}\sum_{C\in\mathcal{C}\\C\subseteq A}s_A\mu(C)\leq\sum_{A\in\mathcal{A}}\sum_{C\in\mathcal{C}\\C\subseteq A}f(c_C)\mu(C)\leq\sum_{C\in\mathcal{C}}f(c_C)\mu(C)<s+\varepsilon.$$

Then $\int\varphi\leq s$. Therefore $\sup\{\int\varphi:\varphi\text{ simple and }0\leq\varphi\leq f\}\leq s$.

For every $\varepsilon>0$ there is a $\mathcal{A}\in\Omega$ such that for every $(\mathcal{B},b)\in\Theta$ satisfying $\mathcal{A}\leq\mathcal{B}$ then $s-\frac{\varepsilon}{2}<\sum_{B\in\mathcal{B}}f(b_B)\mu(B)<s+\frac{\varepsilon}{2}$, then for every choice function $a$ for $\mathcal{A}$ we have:

$$s-\frac{\epsilon}{2}<\sum_{A\in\mathcal{A}}f(a_A)\mu(A),$$

so, if $s_A=\inf_{x\in A}f(x)$, and $\varphi=\sum_{A\in\mathcal{A}}s_A\chi_A$, then $\varphi$ is simple, $0\leq\varphi\leq f$ and:

$$s-\varepsilon<s-\frac{\varepsilon}{2}\leq\sum_{A\in\mathcal{A}}s_A\mu(A)=\int\varphi.$$

Therefore $\sup\{\int\varphi:\varphi\text{ simple and }0\leq\varphi\leq f\}=s$.

$\bullet$ (1)$\Rightarrow$(2) assuming $\mu(X)<\infty$: Suppose $\sup\{\int\varphi:\varphi\text{ simple and }0\leq\varphi\leq f\}=s$. For every $\varepsilon>0$ there is a simple function $\varphi$ such that $0\leq\varphi\leq f$ and $\int\varphi>s-\varepsilon$. Let $\varphi=\sum_{A\in\mathcal{A}}s_A\chi_A$, where $A\in\Omega$. There is an $N\geq 1$ such that $\frac{1}{N}\mu(X)<\varepsilon$, and consider $\mathcal{B}=\{X_k:k\geq 0\}$, where:

$$X_k=\{x\in X:\frac{k}{N}\leq f(x)<\frac{k+1}{N}\},$$

then, because $f$ is measurable and $\mu(X)<\infty$, we have $\mathcal{B}\in\Omega$ and $\bigcup\mathcal{B}=X$. For every $(\mathcal{C},c)\in\Theta$ such that $\mathcal{A},\mathcal{B}\leq\mathcal{C}$, then for every $\mathcal{C}'\subseteq\mathcal{C}$ finite there is an $n\geq 0$ such that $\forall C\in\mathcal{C}':\exists k\in\{0,\dots,n\}:C\subseteq X_k$, so that:

$$\sum_{C\in\mathcal{C}'}f(c_C)\mu(C)=\sum_{k=0}^n\sum_{C\in\mathcal{C}'\\C\subseteq X_k}f(c_C)\mu(C)\leq\sum_{k=0}^n\frac{k+1}{N}\sum_{C\in\mathcal{C}'\\C\subseteq X_k}\mu(C)\leq\sum_{k=0}^n\frac{k}{N}\mu(X_k)+\frac{1}{N}\mu(X)\leq s+\frac{1}{N}\mu(X)<s+\varepsilon,$$

because $\psi=\sum_{k=0}^n\frac{k}{N}\chi_{X_k}$ is a simple function satisfying $0\leq\psi\leq f$, moreover:

$$\sum_{C\in\mathcal{C}}f(c_C)\mu(C)\geq\sum_{A\in\mathcal{A}}\sum_{C\in\mathcal{C}\\C\subseteq A}f(c_C)\mu(C)\geq\sum_{A\in\mathcal{A}}\sum_{C\in\mathcal{C}\\C\subseteq A}s_A\mu(C)=\sum_{A\in\mathcal{A}}s_A\mu(A)=\int\varphi>s-\varepsilon,$$

therefore:

$$s-\varepsilon<\sum_{C\in\mathcal{C}}f(c_C)\mu(C)<s+\varepsilon.$$

Thus $s$ is an integral of $f$ in the new sense.

However, I do not know how to prove (1)$\Rightarrow$(2) in the general case. I do not even know if the new definition of integration is a good definition or I have to think of something else.

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