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Functions which preserve Lebesgue spaces. (English) Zbl 1059.26005

A function \(f:\mathbb R^+\to\mathbb R^+\), where \(\mathbb R^+\) is the set of nonnegative real numbers, is called {metric preserving} if for every metric space \((M,\rho)\), the function \(f\circ\rho\) is also a metric on \(M\). A {Lebesgue number} for an open cover \(\mathcal U\) of a metric space \((M,d)\) is an \(\varepsilon>0\) such that for each point \(p\in X\), \(\{x\in X:d(p,x)<\varepsilon\}\) is contained in at least one member of \(\mathcal U\). A Lebesgue space is a metric space such that every open cover of the space has a Lebesgue number.
The authors tie together these concepts by proving that for every metric space \((M,d)\) and function \(f:\mathbb R^+\to\mathbb R^+\), \((X,f\circ d)\) is a Lebesgue space if and only if the function \(f\) is metric preserving. The authors also consider the question of the image of a Lebesgue number under a metric preserving \(f\). A proof is given that if \((X,d)\) has an open cover \(\mathcal U\) with Lebesgue number \(\epsilon\), then, for any such \(f\), any value between 0 and \(f(\varepsilon)/2\) may be used as a Lebesgue number for \(\mathcal U\) under the metric \(f\circ d\). They provide examples showing that, in general, the preceding question and the following one in the opposite direction both have answers of “no”.
If \(\widehat\epsilon=f(\varepsilon)\) is a Lebesgue number for \(\mathcal U\) under \(f\circ d\), then must \(\varepsilon\) be a Lebesgue number for \(\mathcal U\) under \(d\)?

MSC:

26A30 Singular functions, Cantor functions, functions with other special properties
54E40 Special maps on metric spaces
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