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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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    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.2011-04-14
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    @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.2011-04-14
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    @Jonas: I did what I could, but I'm struggling to see how this could be misinterpreted.2011-04-14
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    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 $A$.2011-04-14
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    @JBeardz: It does not have to, even if $A$ is a two-element set.2011-04-14
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    Whoops, that's obvious. I forgot to add that at each step, you take a cover of $A$ minus the part you just covered, since that will still be of measure zero. In the case of a two-element set, this resolves this issue. Sorry, important point that I forgot to mention.2011-04-14
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    It feels to me that if there is a counter-example, it would be the Cantor set on [0,1]. Not sure why that feels right, but it does.2011-04-14
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    @Thomas: Sure. The Cantor set ends up being a counter example to everything. I haven't spent much time looking at it in this case, because I feel like my statement should be true, and I don't really like the Cantor set. But maybe I need to spend some time with it.2011-04-14
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    @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 $\le 1/4$.2011-04-14
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    @user6312: Thanks, good point.2011-04-14
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    I'm wondering if the rationals may provide a counterexample.2011-04-14
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    @JBeardz: A counterexample will have to be uncountable. If the set were $\{a_1,a_2,a_3,\ldots\}$, you could let $V_i$ be an interval of length $\epsilon2^{-i}$ containing $a_i$.2011-04-14
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    Oh lord, what a vast oversight. I remember realizing this earlier. Please excuse me, haha. I am sitting at home sick, coughing up my lungs.2011-04-14
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    Well, I guess a counterexample would be required to be uncountably infinite (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