# How to define Lebesgue Integrability of functions assuming values in an arbitrary topological vector space over an arbitrary topological field?

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 $$s-\varepsilon<\sum_{C\in\mathcal{C}}f(c_C)\mu(C); so for every $$(\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)

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 $$s-\frac{\varepsilon}{2}<\sum_{B\in\mathcal{B}}f(b_B)\mu(B), then for every choice function $$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

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)

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)

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.