Steinhaus theorem

Template:Short description 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.

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

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

contains an open neighbourhood of the origin.

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

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

contains an open neighbourhood of unity.

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

For any $\epsilon >0$, there exists an open set $𝒰$, so that $A\subset 𝒰$ and $\mu (𝒰)<\mu (A)+\epsilon$. Since $𝒰$ is a union of open intervals, for a given $\alpha \in (1/2,1)$, we can find an interval $(a,b)$ such that $\mu (B)>\alpha (b-a)$, where $B=(a,b)\cap A$.

Let $\delta =(2\alpha -1)(b-a)$. Suppose for contradiction that there exists $|x|<\delta$ such that $(x+A)\cap A=\varnothing$. Then, $(x+B)\cap B=\varnothing$, and thus

$\mu ((x+B)\cup B)=2\mu (B)$

But, we also have

$\mu ((x+B)\cup B)\le \mu ((a,b+|x|))<b-a+\delta$,

so $2\mu (B)<b-a+\delta =2\alpha (b-a)$, which contradicts $\mu (B)>\alpha (b-a)$.

Hence, $(x+A)\cap A\ne \varnothing$ for all $|x|<\delta$, and it follows immediately that $\{x:|x|<\delta \}\subset A-A$, as desired.

Corollary

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

See also

• Falconer's conjecture

References

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