# Integration theory for functions and values with values in topological rings

I am curious whether somebody ever tried to generalize the classical theory of Lebesgue integral to functions and measures with values in Hausdorff topological rings.

The generalization of a measure is straightforward: given a topological ring $$R$$ and a $$\sigma$$-algebra $$\mathcal A$$ on a set $$\Omega$$, define an $$R$$-valued measure as a function $$\mu:\mathcal A\to R$$ such that

$$\bullet$$ $$\mu(A\cup B)=\mu(A)+\mu(B)$$ for any disjoint sets $$A,B\in\mathcal A$$;

$$\bullet$$ $$\mu(\bigcup_{n\in\omega}A_n)=\sum_{n\in\omega}\mu(A_n)$$ for any sequence $$(A_n)_{n\in\omega}$$ consisting of pairwise disjoint sets in the algebra $$\mathcal A$$.

Given a simple $$\mathcal A$$-measurable function $$f:\Omega\to R$$ and an $$R$$-valued measure $$\mu$$, define the integral $$\int f d\mu$$ as the (finite) sum $$\sum_{y\in R}y\cdot\mu(f^{-1}(y))$$.

So, the question:

Is anything known about topological rings $$R$$ for which the $$R$$-valued integral can be defined for some reasonably wide class of functions and so-generalized integral has all basic properties of the usual Lebesgue integral?

• You mean Hausdorff topological ring for assumptions to be meaningful (still topological abelian group is enough to define the measure). Also when you write $\sum_\omega \mu(A_n)$, you implicitly mean that this sum exists, which probably means that $\sum_{k=0}^n\mu(A_k)$ converges... or you assume anything stronger? anyway this bare convergence assumption forces $\sum_{k=0}^n\mu(A_{f(k)})$ to converge to the same limit for every permutation $f$ of $\omega$, which is quite close to "absolute convergence" in spirit.
– YCor
Sep 22, 2020 at 11:33
• Yes, I assume that $R$ is a Hausdorff topological ring, if necessary, commutative. And writing $\sum_{n\in\omega}\mu(A_n)=\mu(\bigcup_{n\in\omega}A_n)$ I understand that the series converges to $\mu(\bigcup_{n\in\omega}A_n)$. And the convergence is unconditional (since $\mu(\bigcup_{n\in\omega}A_n)$ does not depend on the order of the summands). In infinite-dimensional Banach spaces the unconditional convergence is strictly weaker than the absolute convergence. But in general topological rings the absolute convergence is undefined (only unconditional can be defined). Sep 22, 2020 at 12:39
• I would start with Kaplansky, Irving. "Topological rings." Bulletin of the American Mathematical Society 54.9 (1948): 809-826. and refer to later works that cite it. Sep 22, 2020 at 12:45
• @rschwieb Thank you. This paper of Kaplansky has only 46 citations. None of them involves measure. There is however another paper of Kaplansky with the same title in Amer. J.Math. with 289 citations. Maybe this will lead to something interesting. Sep 22, 2020 at 12:53
• In general one could consider three Hausdorff topological abelian groups $A,B,C$ with a continuous $\mathbf{Z}$-bilinear map $A\times B\to C$, such a measure valued in a $A$, consider functions $f$ valued in $B$, and define the integral for simple $f$ as $\sum_{y\in B}f(\mu(f^{-1}(\{y\})),y)$.
– YCor
Sep 22, 2020 at 13:13

• Thank you for your answer. Could you give me a reference where the locally convex case is studied. Observe that duality theory cannot be applied since for a linear functional $x^*$ and elements $x,y$ of the ring the value $x^*(xy)$ cannnot be expressed via $x^*(x)$ and $x^*(y)$. So, the vector-valued integral does not project to the field of scalars. This is the case if either you integrate vector-valued functions by scalar measures or integrage scalar functions by vector-valued measures, but not vector-valued functions by vector-valued measures. Sep 23, 2020 at 15:10