Today, we prove a little gem from measure theory known in the literature as Steinhaus’s theorem. There are variations on the statement of the theorem, but we will consider two cases. The “Weak Steinhaus” theorem says that if is a set of positive, finite Lebesgue measure, then contains an open interval; and the “Strong Steinhaus” theorem says that if are sets of positive, finite Lebesgue measure, then contains an open interval. The reader will see below that our method of proof differs in each case, so the distinction is somewhat nontrivial.
First, a lemma about metric spaces.
Lemma 1. Let be a metric space, and a compact subset and be an open set. Then there exists such that
Proof. Assume that for each , there exists such that . Thus, there exists a sequence such that . Since compactness implies sequential compactness in a metric space, we see that there is some such that . I claim that is also a limit point of , hence . Indeed, for every . Since , we arrive at a contradiction.
Now, we prove the weak theorem
Theorem 2. (Weak Steinhaus theorem)
Let be a Lebesgue measurable subset of positive, finite measure. Then the sets
contain an open interval.
Proof. By the regularity of the Lebesgue measure, for given, there exists a compact subset and an open subset such that
Choosing , we obtain that
By the preceding lemma, there exist such that
Since is symmetric (i.e. ), we see that . Let , and suppose that . Then by the translation-invariance of the Lebesgue measure and the result that ,
which contradicts that .
We now prove the strong version of the Steinhaus theorem. The proof we will give below will make use of many of the same arguments above, but we will also make use of Fubini’s theorem.
Theorem 3. (Strong Steinhaus theorem)
Let be Lebesgue measurable subsets of positive, finite measure. Then the set contains an open interval.
Proof. By the inner regularity of the Lebesgue measure, for any , there exists a compact subset such that . Hence, we may assume without loss of generality that is compact. Since is continuous, is also compact. We can define a continuous function by
where . Since is continuous and nonnegative, we have by Fubini’s theorem and the translation invariance of the Lebesgue measure that
Hence, since is continuous, there exists and such that for . Hence, for all . So for any , there exist such that
which implies that .