Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


share|cite|improve this question
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. – Jon Beardsley Apr 14 '11 at 18:40
@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. – Jonas Meyer Apr 14 '11 at 19:22
@Jonas: I did what I could, but I'm struggling to see how this could be misinterpreted. – Jon Beardsley Apr 14 '11 at 19:42
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 $A$. – Jon Beardsley Apr 14 '11 at 20:07
@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 $\le 1/4$. – André Nicolas Apr 14 '11 at 20:31
up vote 12 down vote accepted

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.

share|cite|improve this answer
Thanks, I should have looked more carefully at the Cantor Set. – Jon Beardsley Apr 14 '11 at 22:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.