I am studying a function whose Fourier transform is zero on a set of strictly positive Lebesgue measure and I need to know this:

If a set has a strictly positive Lebesgue measure can we prove that it contains an interval?

Help is much appreciated

  • 7
    $\begingroup$ How about the irrationals? $\endgroup$
    – Alex R.
    Commented Jun 10, 2016 at 21:43

2 Answers 2


The irrationals are an easy example of not having this property.

  • $\begingroup$ The irrationals have measure 0, being a countable union of points. $\endgroup$ Commented May 1, 2020 at 7:01
  • 2
    $\begingroup$ @JyotirmoyBhattacharya The irrationals are not countable. $\endgroup$ Commented May 27, 2020 at 21:24
  • 1
    $\begingroup$ @eigenvalues_question You are right. I dont know how I made that comment. $\endgroup$ Commented May 28, 2020 at 5:47

Nope. There is a standard construction of "fat Cantor sets" which have positive measure but are in fact nowhere dense, which is considerably stronger than merely containing no intervals. The construction proceeds much like constructing a Cantor set with zero measure except that the lengths of the deleted intervals decay more rapidly, and so they sum up to less than the total measure of the interval you started with. See https://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set for details.

  • $\begingroup$ Out of curiosity, ist there a null set N and an interval (a,b) such that the Intervall is included in the union of N and the fat Cantor Set? $\endgroup$ Commented Oct 2, 2019 at 17:47
  • $\begingroup$ @JosephDoob No, because a fat Cantor set always undershoots the measure of the interval in which it is contained by some $\epsilon$. $\endgroup$
    – Ian
    Commented Aug 18, 2021 at 12:01

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