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Lipschitz one sets modulo sets of measure zero
Mathematica Slovaca ( IF 1.6 ) Pub Date : 2020-06-25 , DOI: 10.1515/ms-2017-0372
Zoltán Buczolich 1 , Bruce Hanson 2 , Balázs Maga 1 , Gáspár Vértesy 1
Affiliation  

Abstract We denote the local “little” and “big” Lipschitz functions of a function f : ℝ → ℝ by lip f and Lip f. In this paper we continue our research concerning the following question. Given a set E ⊂ ℝ is it possible to find a continuous function f such that lip f = 1E or Lip f = 1E? In giving some partial answers to this question uniform density type (UDT) and strong uniform density type (SUDT) sets play an important role. In this paper we show that modulo sets of zero Lebesgue measure any measurable set coincides with a Lip 1 set. On the other hand, we prove that there exists a measurable SUDT set E such that for any Gδ set E͠ satisfying ∣EΔE͠∣ = 0 the set E͠ does not have UDT. Combining these two results we obtain that there exist Lip 1 sets not having UDT, that is, the converse of one of our earlier results does not hold.

中文翻译:

Lipschitz 一组模零测量集

摘要 我们用 lip f 和 Lip f 表示函数 f 的局部“小”和“大”Lipschitz 函数:ℝ → ℝ。在本文中,我们继续对以下问题进行研究。给定一个集合 E ⊂ ℝ 是否有可能找到一个连续函数 f 使得 lip f = 1E 或 Lip f = 1E?在对这个问题给出部分答案时,均匀密度类型 (UDT) 和强均匀密度类型 (SUDT) 集发挥了重要作用。在本文中,我们展示了零 Lebesgue 模集测量任何与 Lip 1 集重合的可测集。另一方面,我们证明存在一个可测的 SUDT 集合 E,使得对于任何满足 ∣EΔE͠∣ = 0 的 Gδ 集合 E͠,集合 E͠ 没有 UDT。结合这两个结果,我们得到存在没有 UDT 的 Lip 1 集合,即我们之前的结果之一的逆集不成立。
更新日期:2020-06-25
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