Steinhaus theorem

Let ${\displaystyle E\subset \mathbb {R} ^{n}}$ be a set of positive Lebesgue measure, ${\displaystyle \{a_{1},a_{2},\ldots ,a_{N}\}}$ be an arbitrary collection of unit vectors in ${\displaystyle \mathbb {R} ^{n}}$, and ${\displaystyle \epsilon \in (0,(2^{N}-1)^{-1})}$. Also denote the ${\displaystyle n}$-dimensional Lebesgue measure by ${\displaystyle m^{n}}$. By inner regularity of the Lebesgue measure, we obtain a compact set ${\displaystyle K_{1}\subset E}$ such that ${\displaystyle m^{n}(K_{1})>0}$, and by outer regularity an open set ${\displaystyle U\supset K_{1}}$ such that ${\displaystyle m^{n}(U)\leq (1+\epsilon )m^{n}(K_{1}).}$

Because ${\displaystyle K_{1}}$ is compact, the distance ${\displaystyle R=d(K_{1},U^{c})}$ is strictly positive. Let ${\displaystyle \delta \in (0,R)}$ be arbitrary, and consider the set ${\displaystyle K_{1}+\delta a_{1}}$. If this subset is not contained in ${\displaystyle U}$, then we would have

$${\displaystyle {\begin{aligned}d(K_{1},U^{c})<\Vert \delta a_{1}\Vert =\delta <R,\end{aligned}}}$$ which is a contradiction. Therefore, ${\displaystyle K_{1}\cap (K_{1}+\delta a_{1})\subset U}$. This means that

$${\displaystyle {\begin{aligned}m^{n}(U)\geq m^{n}(K_{1}\cup (K_{1}+\delta a_{1}))=m^{n}(K_{1})+m^{n}(K_{1}+\delta a_{1})-m^{n}(K_{1}\cap (K_{1}+\delta a_{1})).\end{aligned}}}$$

By translation invariance of the Lebesgue measure, we note that ${\displaystyle m^{n}(K_{1}+\delta a_{1})=m^{n}(K_{1})}$, and so

$${\displaystyle {\begin{aligned}m^{n}(K_{1}\cap (K_{1}+\delta a_{1}))\geq 2m^{n}(K_{1})-m^{n}(U)\geq (1-\epsilon )m^{n}(K_{1}).\end{aligned}}}$$

Since ${\displaystyle \epsilon <1}$, we see that the measure on the left side is strictly positive, which means ${\displaystyle K_{1}+\delta a_{1}\neq \emptyset .}$ Now for each ${\displaystyle i=1,\ldots ,N}$, define the sets ${\displaystyle K_{i+1}=K_{i}\cap (K_{i}+\delta a_{i})}$. By a generalization of the argument above, each ${\displaystyle K_{i}}$ is contained in ${\displaystyle U}$. Moreover, for each ${\displaystyle i}$, ${\displaystyle m^{n}(K_{i})\geq (1-(2^{i}-1))m^{n}(K_{1})}$ (a simple application of induction immediately yields this result) so that each ${\displaystyle K_{i}}$ is nonempty. This yields a nested sequence of sets ${\displaystyle \emptyset \neq K_{N+1}\subset K_{N}\subset \dotsm \subset K_{1}\subset E}$. Let ${\displaystyle q\in K_{N+1}}$. Since ${\displaystyle K_{N+1}=K_{N}+\delta a_{N}}$, ${\displaystyle q-\delta a_{N}\in K_{N}}$. Likewise, since ${\displaystyle K_{N}=K_{N-1}+\delta a_{N-1}}$, ${\displaystyle q-\delta a_{N}-\delta a_{N-1}\in K_{N-1}}$. Repeating this procedure iteratively and eventually denoting ${\displaystyle p=q-\delta \sum _{i=1}^{N}a_{i}}$, we recover the finite arithmetic progression ${\displaystyle \{p,p+\delta a_{1},p+\delta a_{1}+\delta a_{2},\ldots ,p+\delta a_{1}+\dotsm +\delta a_{N}\}\subset E}$ consisting of ${\displaystyle N+1}$ points. Hence, the proof concludes.