Lebesgue Measure - Applications

TheoremSteinhaus Theorem

Let $E \subseteq \mathbb{R}$ be a measurable set with $\lambda(E) > 0$ . Then the difference set $E - E = \{x - y : x, y \in E\}$ contains an open interval around $0$ .

The Steinhaus Theorem is a remarkable result showing that sets of positive measure have algebraic structure. It reveals that even highly irregular sets must contain certain regular configurations when they have positive measure.

ExampleApplication to Sums of Sets

Consider the set $E = [0, 1/4] \cup [3/4, 1]$ . This set has measure $\lambda(E) = 1/2 > 0$ .

The difference set is: $E - E = [-1, -1/2] \cup [-1/4, 1/4] \cup [1/2, 1]$

Notice that $(-1/4, 1/4) \subseteq E - E$ , an open interval containing $0$ , confirming the theorem.

The theorem also applies to the sum set $E + E = \{x + y : x, y \in E\}$ . For any set with positive measure, $E + E$ contains an open interval.

Remark

Important consequences of the Steinhaus Theorem:

Non-existence of certain sets: If $H$ is a subgroup of $\mathbb{R}$ with $\lambda(H \cap [0,1]) > 0$ , then $H = \mathbb{R}$ . This follows because $H - H = H$ must contain an interval, but a proper subgroup cannot contain an interval.
Density of sumsets: If $A, B \subseteq \mathbb{R}$ both have positive measure, then $A + B$ contains an interval, hence has non-empty interior.
Impossibility results: Combined with translation invariance, the theorem implies constraints on the structure of measurable sets that are also algebraic objects.

The proof uses the continuity of translation in $L^1$ : for $E$ with $\lambda(E) > 0$ , the function $h \mapsto \lambda(E \cap (E + h))$ is continuous and positive at $h = 0$ . Therefore, it remains positive on some interval $(-\delta, \delta)$ . But if $\lambda(E \cap (E + h)) > 0$ , then there exist $x, y \in E$ with $x = y + h$ , so $h = x - y \in E - E$ . This elegant argument demonstrates the power of combining measure theory with algebraic operations.