# The Unapologetic Mathematician

## 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.

December 14, 2009 Posted by | Analysis, Measure Theory | 4 Comments