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This is a follow-up to this question (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.)

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up vote 19 down vote accepted

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.

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