# 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.

More generally, if G is a locally compact group, and A ⊂ G is 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. The proof of these extensions, sometimes also called Steinhaus theorem, is almost identical to the one below.

## Proof

The following is a simple proof due to Karl Stromberg.[2] If μ is the Lebesgue measure and A is a measurable set with positive finite measure

${\displaystyle 0<\mu (A)<\infty ,}$

then for every ε > 0 there are a compact set K and an open set U such that

${\displaystyle K\subset A\subset U,\quad \mu (K)+\epsilon >\mu (A)>\mu (U)-\epsilon .}$

For our purpose it is enough to choose K and U such that

${\displaystyle 2\mu (K)>\mu (U).}$

Since K ⊂ U, for each ${\displaystyle k\in K}$, there is a neighborhood ${\displaystyle W_{k}}$ of 0 such that ${\displaystyle k+W_{k}\subset U}$, and, further, there is a neighborhood ${\displaystyle V_{k}}$ of 0 such that ${\displaystyle 2V_{k}\subset W_{k}}$. For example, if ${\displaystyle W_{k}}$ contains ${\displaystyle (-\epsilon ,\epsilon )}$, we can take ${\displaystyle V_{k}=(-\epsilon /2,\epsilon /2)}$. The family ${\displaystyle \{k+V_{k}:k\in K\}}$ is an open cover of K. Since K is compact, one can choose a finite subcover ${\displaystyle \{k_{1}+V_{k_{1}},\dots ,k_{n}+V_{k_{n}}\}}$. Let ${\displaystyle V:=V_{k_{1}}\cap \dots \cap V_{k_{n}}}$. Then,

${\displaystyle K+V\subset ((k_{1}+V_{k_{1}})\cup \dots \cup (k_{n}+V_{k_{n}}))+V\subset ((k_{1}+2V_{k_{1}})\cup \dots \cup (k_{n}+2V_{k_{n}}))\subset ((k_{1}+W_{k_{1}})\cup \dots \cup (k_{n}+W_{k_{n}}))\subset U}$.

Let v ∈ V, and suppose

${\displaystyle (K+v)\cap K=\varnothing .}$

Then,

${\displaystyle 2\mu (K)=\mu (K+v)+\mu (K)<\mu (U)}$

contradicting our choice of K and U. Hence for all v ∈ V there exist

${\displaystyle k_{1},k_{2}\in K\subset A}$

such that

${\displaystyle v+k_{1}=k_{2},}$

which means that V ⊂ A − A. Q.E.D.

## Corollary

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