Baire measures on [O,\(\Omega\) ) and [O,\(\Omega\) ]. II (Q810176)

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scientific article; zbMATH DE number 4212371
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Baire measures on [O,\(\Omega\) ) and [O,\(\Omega\) ]. II
scientific article; zbMATH DE number 4212371

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    Baire measures on [O,\(\Omega\) ) and [O,\(\Omega\) ]. II (English)
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    1991
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    Let \(X=[0,\Omega)\) and \(\bar X=[0,\Omega]\), where \(\Omega\) denotes the first uncountable ordinal. A finite measure \(\mu\) on a measurable space (E,\({\mathcal E})\) is called perfect if for each \({\mathcal E}\)-measurable real valued function f on E and for each set \(A\subset R\) (R - the real line) with \(f^ 1(A)\in {\mathcal E},\) there exists a Borel subset \(B\subset R,\) i.e. \(B\in {\mathcal B}(R)\) such that \(B\subset A\) and \(\mu (f^{-1}(B))=\mu (f^{-1}(A)).\) The author proves the following: Every finite Baire or Borel measure on X or \(\bar X\) is perfect and is complete if and only if it is not purely discontinuous. A \(\sigma\)-algebra is called strongly Blackwell if any two countably generated sub-\(\sigma\)-algebras of \({\mathcal E}\) with the same atoms coincide. Then in the second section of the paper the following theorems have been proved: Theorem. The Baire \(\sigma\)-algebras \({\mathcal B}_ 0(X)\) and \({\mathcal B}_ 0(\bar X)\) are strongly Blackwell. Theorem. The Borel \(\sigma\)-algebras \({\mathcal B}(X)\) and \({\mathcal B}(\bar X)\) are not strongly Blackwell. [For Part I cf. same journal 14, No.2, 454-463 (1989; Zbl 0684.28005)].
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    Baire measure
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    Borel measure
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    discontinuous measure
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    Blackwell \(\sigma \) -algebra
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