Jordan fields and improper integrals
Let \({\mathcal E}\) be a field of subsets of a set X and let \(\mu:{\mathcal E}\to [0,\infty]\) be additive and satisfy \(\mu (E)=\sup \{\mu (A):A\subset E\quad and\quad \mu (A)<\infty \}\) for all \(E\in {\mathcal E}.\) The author calls a function \(f:X\to [0,\infty)\) integrable, in symbol, \(f\in \hat {\mathcal I},\) if the usual Riemann type integral\(\int_{A}\min (f,c)d\mu\) exists for all \(c>0\) and \(A\in {\mathcal E}\) with \(\mu (A)<\infty.\) The supremum of those integrals is denoted by \(\hat I(f)\). For finite \(\mu\) this integral coincides with that due to Dunford and Schwartz. In general, it can happen that \(f\in \hat {\mathcal I}\) and \(\hat I(f)<\infty\) but f is not integrable in the Dunford-Schwartz sense. Denote by \(\hat {\mathcal J}\) the ``extended Jordan field'' generated by \(\mu\), i.e., \(\hat {\mathcal J}=\{H\subset X:\chi_ H\in \hat {\mathcal I}\}\) and let \({\hat \mu}\) stand for the restriction of \(\hat I\) to \(\hat {\mathcal J}.\) The main result of the paper is as follows. For \(f:X\to [0,\infty)\) such that \(\mu_*f^{-1}[t,\infty)\) is finite for all \(t>0\) the following two conditions are equivalent: \((1)\quad f\in \hat {\mathcal I};\) \((2)\quad f^{-1}[t,\infty)\in \hat {\mathcal J}\) for all but countably many t. These conditions imply \(\hat I(f)=\int^{\infty}_{0}{\hat \mu}f^{- 1}[t,\infty)dt,\) this last integral being a Lebesgue integral. (The formula goes back, in the case where \(\mu (X)<\infty\) and f is bounded, to Choquet.) The proof is quite involved. \{Reviewer's remarks: (1) Reference [2] by \textit{C. Klein} has appeared in Stud. Math. 85, 247-255 (1987; Zbl 0618.28010). (2) For related results see \textit{G. H. Greco}, Ann. Mat. Pura Appl., IV. Ser. 132, 243-255 (1982; Zbl 0553.28005).\}
scientific article; zbMATH DE number 4091791
scientific article