Browsing my favorite measure theory reference text, the encyclopedic two-volume set Measure Theory by Vladimir I. Bogachev, I found an application of Steinhaus’ theorem among the Chapter 3 exercises that I want to share with the readers of Math by Matt.
Recall the statement of Steinhaus theorem:
Theorem 1. Let be Lebesgue measurable subsets of positive, finite measure. Then the set contains an open interval.
Exercise 3.10.40 of Volume 1 of Bogachev’s text defines a subset of the closed unit square by
and asks the reader to show that has planar measure zero but intersects every set , where have positive measure. Proving that has measure zero does not require Steinhaus theorem; really, we just need to write as a countable union of null sets. But the second part, as Bogachev hints, follows readily from showing that contains an open interval.
Let be an enumeration of . We can write as the countable union of line segments,
I claim that each set has planar measure zero. To establish this claim, we prove a more general lemma on the measure of affine subspaces of a given Euclidean space.
Lemma 2. Suppose is -dimensional affine subspace of , where . If denotes the -dimensional Lebesgue measure, then .
Proof. Since is translation-invariant, without loss of generality, we may assume that is a linear subspace. If , then it is obvious that , so assume otherwise. Let be an orthonormal basis for , and let denote the standard basis for . Since can be extended to an orthonormal basis for the entire space and since is invariant under orthogonal transformations, we may assume that
Relabeling if necessary, we may furthermore assume that for . Observe that, for any , we have that
since by hypothesis. Since , we conclude that .
For the second assertion, fix two such sets and , and consider the set
Steinhaus’ theorem tells us that contains an open interval , for some and . If be a rational number in , then there exist numbers and such that , which implies that .