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Common extension of group-valued charges. (English) Zbl 0741.28008

Let \(G\) be an Abelian group, let \(X\) be a non-empty set and let \({\mathcal A}\) be a field of subsets of \(X\). A function \(\mu:{\mathcal A}\to G\) is called a \(G\)-valued charge if \(\mu(\emptyset)=0\) and \(\mu(A_ 1\cup A_ 2)=\mu(A_ 1)+\mu(A_ 2)\) whenever \(A_ 1\) and \(A_ 2\) are disjoint sets in \({\mathcal A}\).
Let \({\mathcal A}\) and \({\mathcal B}\) be two fields of subsets of \(X\) and let \(\mu:{\mathcal A}\to G\) and \(\nu:{\mathcal B}\to G\) be two \(G\)-valued charges. Then
(a) We say that \(\mu\) and \(\nu\) are consistent if \(\mu|{\mathcal A}\cap{\mathcal B}=\nu|{\mathcal A}\cap{\mathcal B}\).
(b) If \(\mu\) and \(\nu\) are consistent, we say that \(\mu\) and \(\nu\) have a common extension if there exists a \(G\)-valued charge \(\rho\) defined on the field generated by \({\mathcal A}\cup{\mathcal B}\) such that \(\rho\mid{\mathcal A}=\mu\) and \(\rho\mid{\mathcal B}=\nu\).
The main result of the paper under review can be stated now as follows:
Theorem. Let \(G\) be an Abelian compact Hausdorff topological group. Then any two consistent \(G\)-valued charges have a common extension.

MSC:

28B10 Group- or semigroup-valued set functions, measures and integrals
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