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?


  • $\begingroup$ I have been told of, though have not been shown, a counterexample for a set of measure zero in $\mathbb{R^2}$ which is not contained in a compact set. So maybe this holds only if $A$ is contained in a compact set, or if $A$ is in $\mathbb{R}$. Any stronger or weaker result would be wonderful as well, or just a hint. $\endgroup$ Apr 14, 2011 at 18:40
  • $\begingroup$ @JBeards: I think it would help to clarify your post along the lines of Arturo's comment on user9176's answer. You currently have $\varepsilon$ ambiguously quantified. $\endgroup$ Apr 14, 2011 at 19:22
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    $\begingroup$ @Jonas: I did what I could, but I'm struggling to see how this could be misinterpreted. $\endgroup$ Apr 14, 2011 at 19:42
  • $\begingroup$ One idea I have is to choose $\{V_i^1\}$ such that $\mu(\cup_iV_i^1)<\frac{\varepsilon}{2}$ then choose one element from there. Now, choose $\{V_i^2\}$ such that $\mu(\cup_iV_i^2)<\frac{\varepsilon}{4}$ and choose an element from there, repeating for every natural number. However, I'm not sure the set I end up with will cover $A$. $\endgroup$ Apr 14, 2011 at 20:07
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    $\begingroup$ @JBeardz: This will not work either, though a variant might. Look for example at the rationals in the unit interval. The first choice of interval might be $(1/4,3/4)$. That leaves a bunch of stuff uncovered. Take a cover of this of small total length, and choose the second interval to be in the top half, and keep on doing this. The procedure will never include anything $\le 1/4$. $\endgroup$ Apr 14, 2011 at 20:31

1 Answer 1


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.

  • $\begingroup$ Thanks, I should have looked more carefully at the Cantor Set. $\endgroup$ Apr 14, 2011 at 22:02

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