The Unapologetic Mathematician

Mathematics for the interested outsider

Some Sets of Measure Zero

Here’s a useful little tool:

Let F=\{F_1,F_2,\dots\} be a countable collection of sets of measure zero in \mathrm{R}^n. That is, \overline{m}(F_k)=0 for all k. We define S to be the union

\displaystyle S=\bigcup\limits_{k=1}^\infty F_k

Then it turns out that \overline{m}(S)=0 as well.

To see this, pick some \epsilon>0. For each set F_k we can pick a Lebesgue covering C_k of F_k so that \mathrm{vol}(C_k)<\frac{\epsilon}{2^k}. We can throw all the intervals in each of the C_k together into one big collection C, which will be a Lebesgue covering of all of S. Indeed, the union of a countable collection of countable sets is still countable. We calculate the volume of this covering:


where the final summation converges because it’s a geometric series with initial term \frac{\epsilon}{2} and ratio \frac{1}{2}. This implies that \overline{m}(S)=0.

As an example, a set consisting of a single point has measure zero because we can put it into an arbitrarily small open box. The result above then says that any countable collection of points in \mathbb{R}^n also has measure zero. For instance, the collection of rational numbers in \mathbb{R}^1 is countable (as Kate mentioned in passing recently), and so it has measure zero. The result is useful because otherwise it might be difficult to imagine how to come up with a Lebesgue covering of all the rationals with arbitrarily small volume.

December 14, 2009 - Posted by | Analysis, Measure Theory


  1. […] , we also have , and therefore have as […]

    Pingback by Lebesgue’s Condition « The Unapologetic Mathematician | December 15, 2009 | Reply

  2. […] the other hand, the double integral does not exist. Yes, is countable, and so it has measure zero. However, it’s also dense, which means is discontinuous everywhere in the unit […]

    Pingback by Iterated Integrals III « The Unapologetic Mathematician | December 18, 2009 | Reply

  3. […] have measure zero, and we assume that the discontinuities in each are of measure zero, their (countable) union will also have measure zero. And then so must the set of discontinuities of in have measure zero […]

    Pingback by Integrals are Additive Over Regions « The Unapologetic Mathematician | December 30, 2009 | Reply

  4. […] of all, we observe that any hyperplane has measure zero, and so any finite collection of them will too. Thus the collection of all the hyperplanes perpendicular to vectors cannot fill up all of . We […]

    Pingback by The Existence of Bases for Root Systems « The Unapologetic Mathematician | February 2, 2010 | Reply

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: