An example illustrating \(P^g(K)\neq P_0^g(K)\) for compact sets of finite premeasure (Q1852443)

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scientific article; zbMATH DE number 1848902
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An example illustrating \(P^g(K)\neq P_0^g(K)\) for compact sets of finite premeasure
scientific article; zbMATH DE number 1848902

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    An example illustrating \(P^g(K)\neq P_0^g(K)\) for compact sets of finite premeasure (English)
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    5 January 2003
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    Let \(K\) be a compact set of \({\mathbb{R}^n}\), and \(P^s(K)\), \(P_0(K)\) be the \(s\)-dimensional packing and prepacking measure of \(K\), respectively. \textit{D. J. Feng, S. Hua} and \textit{Z.-Y. Wen} [``Some relations between packing premeasure and packing measure'', Bull. Lond. Math. Soc. 31, 665-670 (1999; Zbl 1018.28004)] proved \[ P^s(K)=P^s_0(K)\tag{*} \] provided \(P^s_0(K)<\infty\). Now let \(g\) be a gauge function, and \(P^g(K)\) and \(P^g_0(K)\) the corresponding packing measure and prepacking measure, respectively. The author gives a condition such that the equality (*) holds. On the other hand, he constructs a doubling gauge function \(g\) and a compact set \(L\subset {\mathbb R}\) for which \(P^g(L)<P_0^g(L)<\infty\). In conclusion, the equality \((*)\) cannot be generalized to the general gauge functions.
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    packing measure
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    premeasure
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    doubling gauge function
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    prepacking measure
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