## Consequence of Steinhaus’ Theorem

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 $A,B\subset\mathbb{R}$ be Lebesgue measurable subsets of positive, finite measure. Then the set $A+B$ contains an open interval.

Exercise 3.10.40 of Volume 1 of Bogachev’s text defines a subset $T$ of the closed unit square by

$\displaystyle T:=\left\{(x,y)\in[0,1]^{2}:x-y\in\mathbb{Q}\right\}$

and asks the reader to show that $T$ has planar measure zero but intersects every set $A\times B$, where $A,B\subset[0,1]$ have positive measure. Proving that $T$ has measure zero does not require Steinhaus theorem; really, we just need to write $T$ as a countable union of null sets. But the second part, as Bogachev hints, follows readily from showing that $A-B$ contains an open interval.

Let $(q_{n})_{n=1}^{\infty}$ be an enumeration of $\mathbb{Q}\cap[0,1]$. We can write $T$ as the countable union of line segments,

$\displaystyle T=\coprod_{n=1}^{\infty}\underbrace{\left\{(y+q_{n},y):0\leq y\leq1-q_{n}\right\}}_{T_{n}}$

I claim that each set $T_{n}$ 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 $A$ is $k$-dimensional affine subspace of $\mathbb{R}^{n}$, where $0\leq k. If $\lambda_{n}$ denotes the $n$-dimensional Lebesgue measure, then $\lambda_{n}(A)=0$.

Proof. Since $\lambda_{n}$ is translation-invariant, without loss of generality, we may assume that $A$ is a linear subspace. If $A=\left\{0\right\}$, then it is obvious that $\lambda_{n}(A)=0$, so assume otherwise. Let $\left\{f_{1},\ldots,f_{k}\right\}$ be an orthonormal basis for $A$, and let $\left\{e_{1},\ldots,e_{n}\right\}$ denote the standard basis for $\mathbb{R}^{n}$. Since $\left\{f_{1},\ldots,f_{k}\right\}$ can be extended to an orthonormal basis for the entire space $\mathbb{R}^{n}$ and since $\lambda_{n}$ is invariant under orthogonal transformations, we may assume that

$\displaystyle\left\{f_{1},\ldots,f_{k}\right\}\subset\left\{e_{1},\ldots,e_{n}\right\}$

Relabeling if necessary, we may furthermore assume that $f_{i}=e_{i}$ for $i=1,\ldots,k$. Observe that, for any $N,M\in\mathbb{N}$, we have that

$\displaystyle\underbrace{A\cap[-N,N]^{n}\times\mathbb{R}^{n-k}}_{A_{N}}\subset{\left\{a_{1}e_{1}+\cdots+a_{n}e_{n}:\left|a_{i}\right|\leq N,1\leq i\leq k,\left|a_{i}\right|\leq\dfrac{1}{M},k

so that

$\displaystyle\lambda_{n}(A_{N})\leq\dfrac{N^{n}}{M^{n-k}}\rightarrow0,M\rightarrow\infty$

since $n-k>0$ by hypothesis. Since $A=\bigcup_{N=1}^{\infty}A_{N}$, we conclude that $\lambda_{n}(A)=0$. $\Box$

For the second assertion, fix two such sets $A$ and $B$, and consider the set

$\displaystyle A-B=\left\{z\in\mathbb{R}:z=x-y, x\in A,y\in B\right\}$

Steinhaus’ theorem tells us that $A-B$ contains an open interval $(t_{0}-\delta,t_{0}+\delta)$, for some $t_{0}\in\mathbb{R}$ and $\delta>0$. If $q$ be a rational number in $(t_{0}-\delta,t_{0}+\delta)$, then there exist numbers $x\in A$ and $y\in B$ such that $x-y=q$, which implies that $(x,y)\in T$.