This is a follow-up to [this question](https://mathoverflow.net/questions/48453) (in fact, this is what originally motivated me to ask that one.)


Let's say that a sequence $\{s\_i\}$ of positive reals *covers* a set $X\subset\mathbb R$ if there is a collection if intervals $\{I\_i\}$ such that $X\subset\bigcup I\_i$ and the length of each $I\_i$ equals $s\_i$.


Does there exist a sequence $\{s\_i\}$ such that $\sum s\_i<\infty$ and $\{s\_i\}$ covers any set of Lebesgue measure zero?


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 $\max\_i \{s\_i\}$ from one another.


(Sorry for the strange collection of tags. It is hard to see in advance which area this question really belongs to.)



A collection of intervals that can cover any measure zero set
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 $\sum s\_j<+\infty$, then $\sum H(s\_j)<+\infty$ 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.


