Lipschitz one sets modulo sets of measure zero

Abstract

<jats:title>Abstract</jats:title><jats:p>We denote the local “little” and “big” Lipschitz functions of a function <jats:italic>f</jats:italic> : ℝ → ℝ by lip <jats:italic>f</jats:italic> and Lip <jats:italic>f</jats:italic>. In this paper we continue our research concerning the following question. Given a set <jats:italic>E</jats:italic> ⊂ ℝ is it possible to find a continuous function <jats:italic>f</jats:italic> such that lip <jats:italic>f</jats:italic> = <jats:bold>1</jats:bold><jats:sub><jats:italic>E</jats:italic></jats:sub> or Lip <jats:italic>f</jats:italic> = <jats:bold>1</jats:bold><jats:sub><jats:italic>E</jats:italic></jats:sub>?</jats:p><jats:p>In giving some partial answers to this question uniform density type (UDT) and strong uniform density type (SUDT) sets play an important role.</jats:p><jats:p>In this paper we show that modulo sets of zero Lebesgue measure any measurable set coincides with a Lip 1 set.</jats:p><jats:p>On the other hand, we prove that there exists a measurable SUDT set <jats:italic>E</jats:italic> such that for any <jats:italic>G<jats:sub>δ</jats:sub></jats:italic> set <jats:italic>E͠</jats:italic> satisfying ∣<jats:italic>E</jats:italic>Δ<jats:italic>E͠</jats:italic>∣ = 0 the set <jats:italic>E͠</jats:italic> 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.</jats:p>

Big and little Lipschitz one sets

“…5, Theorem 5.6, states that G δ sets which are UDT are Lip1. As we show in [5], the converse of this statement does not hold. There exist Lip1 sets which are not UDT.…”

“…• Characterize Lip1 sets. This paper and [5] provide progress in this direction, but there is still more work to be done. • Characterize the sets E ⊂ R for which there is a continuous function f such that {x : lip f (x) < ∞} = E. See [8] for a partial result on this problem.…”

“…We note that apart from unions of non-degenerate closed intervals, it is not completely trivial to construct closed and strongly one-sided dense sets. However, in [5,Theorem 3.1] we construct a nowhere dense closed set which has SUDT and hence is strongly one-sided dense.…”

Strong one-sided density without uniform density

“…The main result of [2] states that if E is G δ and E has UDT then there exists a continuous function f satisfying Lip f = 1 E , that is, the set E is Lip1. On the other hand, in [4] we showed that the converse of this statement does not hold. There exist Lip1 sets which are not UDT.…”

“…In [4] we proved that there exists a measurable SUDT set E such that for any G δ set E satisfying |E∆ E| = 0 the set E does not have UDT. In the same paper we also showed that modulo sets of zero Lebesgue measure any measurable set coincides with a Lip1 set.…”