Steinhaus theorem

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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]


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.

More generally, if G is a locally compact group, and A ⊂ G is a subset of positive (left) Haar measure, then

 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.


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


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

 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

2\mu (K)>\mu(U).\,

Since K ⊂ U, for each k\in K, there is a neighborhood V_k of 0 such that k+V_k\sub U. The family \{k+V_k : k\in K\} is an open cover of K. K is compact, hence one can choose a finite subcover \{k_1+V_{k_1},\dots,k_n+V_{k_n}\}. Let V:=V_{k_1}\cap\dots\cap V_{k_n}. Then K + V ⊂ U.

Let v ∈ V, and suppose

 (K+v)\cap K=\varnothing.\,



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

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

such that


which means that V ⊂ A − A. Q.E.D.


A consequence is, that any measurable proper subgroup of (R,+) is of measure zero.

See also[edit]



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