Steinhaus theorem

In the mathematical field of real analysis, the Steinhaus theorem states that the difference set of a set of positive measure contains an open neighbourhood of zero. It was first proved by Hugo Steinhaus. ^[1]

Statement

Let A be a Lebesgue-measurable set on the real line such that the Lebesgue measure of A is not zero. Then the difference set

${\displaystyle A-A=\{a-b\mid a,b\in A\}}$

contains an open neighbourhood of the origin.

The general version of the theorem, first proved by André Weil, ^[2] states that if G is a locally compact group, and A ⊂ G a subset of positive (left) Haar measure, then

${\displaystyle AA^{-1}=\{ab^{-1}\mid a,b\in A\}}$

contains an open neighbourhood of unity.

The theorem can also be extended to nonmeagre sets with the Baire property.

Corollary

A corollary of this theorem is that any measurable proper subgroup of ${\displaystyle (\mathbb {R} ,+)}$ is of measure zero.

Applications

A special case of the Steinhaus Theorem (and the Lebesgue density theorem) deals with the existence of arithmetic progressions in a set of positive Lebesgue measure. In particular, let ${\displaystyle E\subset \mathbb {R} ^{n}}$, for some positive integer ${\displaystyle n}$, be a set of positive Lebesgue measure. Then for any integer ${\displaystyle N>0}$, ${\displaystyle E}$ contains a finite arithmetic progression of length ${\displaystyle N+1}$.

Proof ^[3]

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.

See also

• Falconer's conjecture

Notes

1. Steinhaus (1920); Väth (2002)
2. Weil (1940) p. 50
3. "Show that a Set ${\displaystyle A\subset \mathbb {R} ^{2}}$ of positive Lebesgue measure contains the vertices of an equilateral triangle". Mathematics StackExchange. 10 December 2021. Retrieved 10 January 2026.

References

• Steinhaus, Hugo (1920). "Sur les distances des points dans les ensembles de mesure positive" (PDF). Fund. Math. (in French). 1: 93–104. doi: 10.4064/fm-1-1-93-104 ..
• Weil, André (1940). L'intégration dans les groupes topologiques et ses applications. Hermann.
• Stromberg, K. (1972). "An Elementary Proof of Steinhaus's Theorem". Proceedings of the American Mathematical Society. 36 (1): 308. doi:10.2307/2039082. JSTOR   2039082.
• Sadhukhan, Arpan (2020). "An Alternative Proof of Steinhaus's Theorem". American Mathematical Monthly. 127 (4): 330. arXiv: 1903.07139 . doi:10.1080/00029890.2020.1711693. S2CID   84845966.

• Väth, Martin (2002). Integration theory: a second course. World Scientific. ISBN   981-238-115-5.
• Yueh-Shin, Lee,(1994). Counting Bipartite Steinhaus Graphs. National Chiao Tung University . https://hdl.handle.net/11296/afmq86