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?

  • 1
    $\begingroup$ 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. $\endgroup$
    – YCor
    Sep 22, 2020 at 11:33
  • $\begingroup$ 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). $\endgroup$ Sep 22, 2020 at 12:39
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    $\begingroup$ 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. $\endgroup$
    – rschwieb
    Sep 22, 2020 at 12:45
  • $\begingroup$ @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. $\endgroup$ Sep 22, 2020 at 12:53
  • $\begingroup$ 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 $\sum_{y\in B}f(\mu(f^{-1}(\{y\})),y)$. $\endgroup$
    – YCor
    Sep 22, 2020 at 13:13

1 Answer 1


This is a comment but too long. The leap from the classical case to your general one is, of course, huge but if one takes a more modest one, namely to functions with values in locally convex algebras and their non locally convex analogues, one sees quite clearly what can happen. To be concrete, we consider the rings of continuous, resp. (equivalent classes of) measurable functions say on the reals, functions from the interval with values in these spaces and finally their integrals with respect to Lebesgue measure). (This fits into your scheme since we can consider the reals as the subring of the constant functions). The first case is well-studied and well-behaved, in particular, continuous or even bounded measurable functions are integrable, but this is no longer true in the non-locally convex case. The kindergarten reason for this is that while convex combinations of small things are small in the first situation, this can fail in the second one—the standard way of defining an integral (e.g. the Riemann one) is to take convex combinations of function values and then proceed to take a limit.

The two things you lose in going from the l.c.s. case to the t.v.s. are duality and convexity arguments which play a vital role in vector or algebra valued integration. I suppose that what I am trying to say is that without some substitute for these, problems could arise in the very general situation you are envisaging.

  • $\begingroup$ 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. $\endgroup$ Sep 23, 2020 at 15:10
  • $\begingroup$ I only considered the case of integration with respect to a scalar measure since I wanted to show how thing can go wrong even in this special case and to indicate some of the pitfalls that would have to be avoided in the general situation you described. I have no knowledge about work on vector valued functions being integrated with respect to vector valued measures. $\endgroup$
    – user131781
    Sep 23, 2020 at 20:41

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