Counterexamples in Analysis by Bernard R. Gelbaum

Author:Bernard R. Gelbaum
Language: eng
Format: epub
Publisher: Dover Publications
Published: 2012-06-26T16:00:00+00:00

for every and , then μ(A) = 0 for every . We shall now prove this fact.

We start with an equivalence relation ∼ defined on (0, 1] × (0, 1] as follows: x ∼ y iff . By means of ∼ the half-open interval (0, 1] is partitioned into disjoint equivalence classes C. The axiom of choice, applied to this family of equivalence classes, produces a set A having the two properties: (1) no two distinct points of A belong to the same equivalence class C; (2) every equivalence class C contains a point of A. In terms of the equivalence relation ∼ these two properties take the form: (1) no two distinct members of A are equivalent to each other; (2) every point x of (0, 1] is equivalent to some member of A. We now define, for each , an operation on the set A, called translation modulo 1, as follows:

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