Steinhaus theorem

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Short description: Mathematical theorem in real analysis

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

[math]\displaystyle{ A-A=\{a-b\mid a,b\in A\} }[/math]

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

[math]\displaystyle{ AA^{-1} = \{ ab^{-1} \mid a,b \in A \} }[/math]

contains an open neighbourhood of unity.

The theorem can also be extended to nonmeagre sets with the Baire property. The proof of these extensions, sometimes also called Steinhaus theorem, is almost identical to the one below.

Proof

The following simple proof can be found in a collection of problems by late professor H.M. Martirosian from the Yerevan State University, Armenia (Russian).

Let's keep in mind that for any [math]\displaystyle{ \varepsilon\gt 0 }[/math], there exists an open set [math]\displaystyle{ \, {\cal U} }[/math], so that [math]\displaystyle{ A\subset{\cal U} }[/math] and [math]\displaystyle{ \mu ({\cal U})\lt \mu (A)+\varepsilon }[/math]. As a consequence, for a given [math]\displaystyle{ \alpha \in (1/2,1) }[/math], we can find an appropriate interval [math]\displaystyle{ \Delta=(a,b) }[/math] so that taking just an appropriate part of positive measure of the set [math]\displaystyle{ A }[/math] we can assume that [math]\displaystyle{ A\subset\Delta }[/math], and that [math]\displaystyle{ \mu(A)\gt \alpha(b-a) }[/math].

Now assume that [math]\displaystyle{ |x|\lt \delta }[/math], where [math]\displaystyle{ \delta=(2\alpha-1)(b-a) }[/math]. We'll show that there are common points in the sets [math]\displaystyle{ x+A }[/math] and [math]\displaystyle{ A }[/math]. Otherwise [math]\displaystyle{ 2\mu(A)=\mu \{(x+A)\cup A\}\leq \mu \{(x+\Delta)\cup \Delta\} }[/math]. But since [math]\displaystyle{ \delta\lt b-a }[/math], and

[math]\displaystyle{ \mu \{(x+\Delta)\cup \Delta\}=b-a+|x|\lt b-a+\delta }[/math],

we would get [math]\displaystyle{ 2\mu(A)\lt b-a+\delta=2\alpha(b-a) }[/math], which contradicts the initial property of the set. Hence, since [math]\displaystyle{ (x+A)\cap A\neq\varnothing }[/math], when [math]\displaystyle{ |x|\lt \delta }[/math], it follows immediately that [math]\displaystyle{ \{x; |x|\lt \delta\}\subset A-A }[/math], what we needed to establish.

Corollary

A corollary of this theorem is that any measurable proper subgroup of [math]\displaystyle{ (\R,+) }[/math] is of measure zero.

See also

Notes

  1. (Steinhaus 1920); (Väth 2002)
  2. (Weil 1940) p. 50

References

  • Väth, Martin (2002). Integration theory: a second course. World Scientific. ISBN 981-238-115-5.