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On extension of submeasures. (English) Zbl 0596.28002
Let $${\mathcal R}$$ be a ring of subsets of a nonempty set T. According to Definition 1 in an earlier paper by the author [Diss. Math. 112, 35 p. (1974; Zbl 0292.28001)] a set function $$\mu:{\mathcal R}\to [0,+\infty)$$ is a submeasure if it is monotone, continuous $$(A_ n\searrow {\mathcal R}0\Rightarrow \mu (A_ n)\to 0),$$ and subadditively continuous (for all $$A\in {\mathcal R}$$ and all $$\epsilon >0$$ there exists $$\delta >0$$ such that $$B\in {\mathcal R},\mu (B)<\delta \Rightarrow \mu (A)-\epsilon \leq \mu (A- B)\leq \mu (A)\leq \mu (A\cup B)\leq \mu (A)+\epsilon).\quad$$ The property of being subadditively continuous is equivalent to the following property: $$A,A_ n\in {\mathcal R},n=1,2,...$$ and $$\mu (A_ n\Delta A)\to 0\Rightarrow \mu (A_ n)\to \mu (A).$$ A set function $$\mu:{\mathcal R}\to [0,+\infty]$$ is exhaustive if $$\mu (A_ n)\to 0$$ for each infinite sequence $$A_ n\in {\mathcal R},n=1,2,...,$$ of pairwise disjoint sets. In Theorem 18 in the above-cited paper [op. cit.], the author has proved that a uniform, subadditive or additive submeasure $$\mu:{\mathcal R}\to [0,+\infty)$$ has a unique extension of the same type to $$\sigma$$ ($${\mathcal R})$$- the $$\sigma$$-ring generated by $${\mathcal R}$$- if and only if it is exhaustive. Two additional, rather clumsy, conditions were needed to obtain the extension theorem for nonuniform submeasures. In this note, using a more transparent approach he shows that these conditions may be replaced by the following: (i) For each $$\epsilon >0$$ there is a $$\delta >0$$ such that $$A,B\in {\mathcal R},$$ $$\mu (A),\mu (B)\leq \delta$$ implies $$\mu (A\cup B)<\epsilon$$ (pseudometric generating property) and $$(ii)\quad A_ n\in {\mathcal R},n=1,2,...,$$ and $$\mu (A_ n\Delta A_ m)\to 0$$ as $$n,m\to \infty$$ implies that $$\mu (A_ n)-\mu (A_ m)\to 0$$ as $$n,m\to \infty.$$

##### MSC:
 28A12 Contents, measures, outer measures, capacities 28A10 Real- or complex-valued set functions