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<n. 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<i\leq n\right\}}

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.

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