Is it known if an open set of $\mathbb{R}^n$ can be covered (up to a set of Lebesgue measure zero) with disjoint open balls?

It is obviously true for $n=1$ where every open set is a union of intervals which are the same as balls, and it is also valid for the p-adics as in here.

It is also known that it can be covered with open-closed boxes of the type $(0,1]^n$ as done in measure theory.

  • $\begingroup$ Lebesgue measure? $\endgroup$
    – Mnifldz
    Nov 25, 2019 at 19:10
  • $\begingroup$ Yes, Lebesgue measure. $\endgroup$
    – Paulo Ney
    Nov 25, 2019 at 19:12

1 Answer 1


Take a look Donald Cohn's Measure theory(second edition),at page 164.

It is about the Vitali Covering theorem.

You can adjust it to work for balls also. The proof is almost the same.

In case of infinite measure it is a known fact that every open set in $\Bbb{R^n}$ for $n>1$ is a countable union of disjoint dyadic cubes (with finite measure).

The family $\mathcal{V}$ of all balls satisfies the regularity condition mentioned in the definition of the lecture(I leave it to you as an exercise to verify it)

So you can apply the theorem to every dyadic cube and you will end up with a countable union of countable disjoint families of disjoint balls which cover each rectangle up to a set of measure zero. and the countably many sets of measure zero whose union will again have measure zero.

  • $\begingroup$ How can I access the lecture notes ? $\endgroup$
    – Pii_jhi
    Apr 5, 2020 at 11:03
  • $\begingroup$ @Pii_jhi oh they removed them....take a look at the book of Donald Cohn,Measure theory ,second edition at page 164 $\endgroup$ Apr 5, 2020 at 19:46
  • $\begingroup$ You can adjust the proof and theorem to work for balls also $\endgroup$ Apr 5, 2020 at 19:55
  • $\begingroup$ Do you think the theorem is true on more general metric spaces ? For instance on riemannian manifolds ? (see also my question on MO : link ) $\endgroup$
    – Pii_jhi
    Apr 5, 2020 at 23:27

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