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:

\displaystyle\mathrm{vol}(C)\leq\sum\limits_{k=1}^\infty\mathrm{vol}(C_k)<\sum\limits_{k=1}^\infty\frac{\epsilon}{2^k}=\epsilon

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.

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

4 Comments »

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