Steinhaus’s Theorem

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 $A$ is a set of positive, finite Lebesgue measure, then $A-A$ contains an open interval; and the “Strong Steinhaus” theorem says that if $A,B$ are sets of positive, finite Lebesgue measure, then $A+B$ 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 $(X,d)$ be a metric space, and $K \subset X$ a compact subset and $U \supset K$ be an open set. Then there exists $r > 0$ such that

$\displaystyle B(x;r) \subset U, \indent \forall x \in K$

Proof. Assume that for each $r > 0$, there exists $x_{r} \in K$ such that $B(x_{r};r) \cap U^{c} \neq \emptyset$. Thus, there exists a sequence $(x_{n})_{n=1}^{\infty}$ such that $B(x_{n};\frac{1}{n}) \cap U^{c} \neq \emptyset$. Since compactness implies sequential compactness in a metric space, we see that there is some $x \in K$ such that $x_{n} \rightarrow x$. I claim that $x$ is also a limit point of $U^{c}$, hence $x \in U^{c}$. Indeed, $B(x;\frac{2}{n}) \cap U^{c} \neq \emptyset$ for every $n \in \mathbb{Z}^{\geq 1}$. Since $K \cap U^{c} = \emptyset$, we arrive at a contradiction. $\Box$

Now, we prove the weak theorem

Theorem 2. (Weak Steinhaus theorem)

Let $A \subset \mathbb{R}$ be a Lebesgue measurable subset of positive, finite measure. Then the sets

$\displaystyle A - A := \left\{c \in \mathbb{R} : c = a-a', a \in A, a' \in A\right\}, \indent A + A := \left\{c \in \mathbb{R} : c = a+a', a \in A, a' \in A\right\}$

contain an open interval.

Proof. By the regularity of the Lebesgue measure, for $\epsilon > 0$ given, there exists a compact subset $K \subset A$ and an open subset $A \subset U$ such that

$\displaystyle \mu(K) + \epsilon > \mu(A) > \mu(U) - \epsilon$

Choosing $0 < \epsilon < \frac{1}{4}\mu(A)$, we obtain that

$\displaystyle \mu(K) > \mu(U) - 2\epsilon > \mu(U) -2\frac{\mu(A)}{4} > \mu(U_{1}) - \frac{\mu(U)}{2} = \frac{\mu(U)}{2}$

By the preceding lemma, there exist $r > 0$ such that

$\displaystyle K + B(0;r) \subset U$

Since $B(0;r)$ is symmetric (i.e. $B(0;r) = -B(0;r)$), we see that $K - B(0;r) \subset U$. Let $x \in B(0;r)$, and suppose that $(K + x) \cap K = \emptyset$. Then by the translation-invariance of the Lebesgue measure and the result that $(K+x) \cup K \subset U$,

$\displaystyle 2\mu(K) = \mu(K) + \mu(K)= \mu(K + x) + \mu(K) \leq \mu(U),$

which contradicts that $\mu(K) > \frac{\mu(U)}{2}$. $\Box$

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 $A,B \subset \mathbb{R}$ be Lebesgue measurable subsets of positive, finite measure. Then the set $A+B$ contains an open interval.

Proof. By the inner regularity of the Lebesgue measure, for any $\epsilon > 0$, there exists a compact subset $K \subset A$ such that $\mu(A \setminus K) < \epsilon$. Hence, we may assume without loss of generality that $A$ is compact. Since $x \mapsto -x$ is continuous, $C := -A$ is also compact. We can define a continuous function $f: \mathbb{R} \rightarrow \mathbb{R}$ by

$\displaystyle f(t) := \mu(C_{t} \cap B), \indent t \in \mathbb{R}$

where $C_{t} := C + t := \left\{x \in \mathbb{R} : x = c +t\right\}$. Since $f$ is continuous and nonnegative, we have by Fubini’s theorem and the translation invariance of the Lebesgue measure that

$\displaystyle \begin{array}{lcl}\int_{\mathbb{R}}f(t)dt = \int_{\mathbb{R}}\mu(C_{t} \cap B)dt = \int_{\mathbb{R}}\left(\int_{\mathbb{R}}\mathbf{1}_{A}(t-s)\mathbf{1}_{B}(s)ds\right)dt &=& \left(\int_{\mathbb{R}}\mathbf{1}_{A}(t)dt\right)\left(\int_{\mathbb{R}}\mathbf{1}_{B}(s)ds\right)\\&=&\mu(A)\mu(B) > 0\end{array}$

Hence, since $f$ is continuous, there exists $t_{0} \in \mathbb{R}$ and $\delta > 0$ such that $f(t) > 0$ for $t \in (t_{0}-\delta, t_{0}+\delta)$. Hence, $C_{t} \cap B \neq \emptyset$ for all $t \in (t_{0}-\delta, t_{0} + \delta)$. So for any $t \in (t_{0}-\delta,t_{0} + \delta)$, there exist $a \in A, b \in B$ such that

$\displaystyle -a + t = b \Longleftrightarrow t = b+a \in A + B,$

which implies that $(t_{0}-\delta,t_{0}+\delta) \subset A+B$. $\Box$