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Suppose we have a subset $A\subset\mathbb{R}$ of Lebesgue measure zero contained in a compact interval, say $[0,1]$. We know that since $A$ has measure zero we can cover $A$ with a countable set of open intervals, say $\{U_i\}$, such that $\mu(\cup_iU_i)\leq \varepsilon$ for any $\varepsilon$. Now, if we fix some $\varepsilon>0$, can we cover $A$ with a countable set of open intervals, say now $\{V_i\}$, such that $\mu(V_i)\leq\frac{\varepsilon}{2^i}$ for each $i$? That is, can we control the size of each individual set in some way? I have been thinking about this for a bit, and keep running into having to do things an infinite number of times, or having to choose the wrong indices first. Any ideas?

Thanks!

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    Well, I guess a counterexample would be required to be uncountably in$f$inite (because a countable set always has such a cover) and the compactness of the Cantor set means that any such cover must have a finite sub-cover.2011-04-14

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If the property holds, then $A$ has Hausdorff dimension $0$, because $\sum_{n=1}^\infty \left(\frac{\varepsilon}{2^n}\right)^d=\frac{\varepsilon^d}{2^d-1}$ can be made arbitrarily small for each fixed $d>0$ by choosing $\varepsilon$ small enough. The Cantor set has Lebesgue measure $0$ and Hausdorff dimension $\log_3(2)$, so it is a counterexample.

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    Thanks, I should have looked more carefully at the Cantor Set.2011-04-14