Cantor function: Difference between revisions

Revision as of 14:15, 12 May 2025

Revision as of 00:15, 14 May 2025

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===Lack of absolute continuity===

Because theThe [[Lebesgue measure]] of the [[Uncountable set|uncountably infinite]] [[Cantor set]] is 0. Therefore, for any positive ''ε''&nbsp;<&nbsp;1 and any ''δ'' > 0, there exists a finite sequence of [[pairwise disjoint]] sub-intervals with total length <&nbsp;''δ'' over which the Cantor function cumulatively rises more than&nbsp;''ε''.

In fact, for every ''δ''&nbsp;>&nbsp;0 there are finitely many pairwise disjoint intervals (''x<SUB>k</SUB>'',''y<SUB>k</SUB>'') (1&nbsp;≤&nbsp;''k''&nbsp;≤&nbsp;''M'') with <math>\sum\limits_{k=1}^M (y_k-x_k)<\delta</math> and <math>\sum\limits_{k=1}^M (c(y_k)-c(x_k))=1</math>.