most recent 30 from http://mathoverflow.net 2013-06-18T08:02:16Z http://mathoverflow.net/feeds/question/48558 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/48558/a-collection-of-intervals-that-can-cover-any-measure-zero-set Sergei Ivanov 2010-12-07T11:43:38Z 2010-12-07T13:11:07Z

<p>This is a follow-up to <a href="http://mathoverflow.net/questions/48453" rel="nofollow">this question</a> (in fact, this is what originally motivated me to ask that one.)</p> <p>Let's say that a sequence <code>$\{s_i\}$</code> of positive reals <em>covers</em> a set $X\subset\mathbb R$ if there is a collection if intervals <code>$\{I_i\}$</code> such that $X\subset\bigcup I_i$ and the length of each $I_i$ equals $s_i$.</p> <p>Does there exist a sequence <code>$\{s_i\}$</code> such that $\sum s_i&lt;\infty$ and <code>$\{s_i\}$</code> covers any set of Lebesgue measure zero?</p> <p>For example, simple things like geometric progressions do not work: they cannot cover a union of infinitely many copies of a compact set of positive Hausdorff dimension, separated by a distance at least <code>$\max_i \{s_i\}$</code> from one another.</p> <p>(Sorry for the strange collection of tags. It is hard to see in advance which area this question really belongs to.)</p>

http://mathoverflow.net/questions/48558/a-collection-of-intervals-that-can-cover-any-measure-zero-set/48560#48560 fedja 2010-12-07T13:11:07Z 2010-12-07T13:11:07Z

<p>No. If you can cover every set of measure $0$ by your sequence of intervals, you can certainly scale (shrink all intervals some number of times) and still have covering (just cover the expanded set by the original sequence) . If $\sum s_j&lt;+\infty$, then $\sum H(s_j)&lt;+\infty$ for some measuring function $H$ with $H(x)/x\to+\infty$ as $x\to 0$. Thus, every set of measure $0$ would have the Hausdorff measure associated with $H$ zero, which can be ruled out by the standard Cantor construction.</p>