Steinhaus theorem

Let A be a Lebesgue-measurable set in \mathbb{R}^{n} 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. {{Math proof|title=Proof|proof= Let E \subset \mathbb{R}^{n} be a subset of positive Lebesgue measure. First, we consider the case where m(A) . In this case, it follows that the characteristic functions \chi_{A}(x) and \chi_{-A}(x) are contained in L^{p}(\mathbb{R}^{n}) for all 1 \leq p . Then h(x) = \chi_{A} \ast \chi_{-A}(x) is continuous on \mathbb{R}^{n} (where \ast denotes convolution) and h(0) = \int_{\mathbb{R}^{n}}\chi_{A}(-y)\chi_{A}(y)\;dy = \int_{\mathbb{R}^{n}}\chi_{A}(s)\;ds = m(A) > 0. Then since h is continuous and h(0) > 0, there exists an open neighborhood U of 0 so that h(x) > 0 for all x \in U. But by definition of h(x), h(x) > 0 if and only if x \in A - A. Hence, 0 \in U \subset A - A. Now suppose m(A) = \infty. We can write A as the following union: A = \bigcup_{R = 0}^{\infty}(A \cap B_{R}(0)), where B_{R}(0) is the ball of radius R centered at 0. By countable subadditivity, there exists at least one R_{0} so that m(A \cap B_{R_{0}}(0)) > 0. Hence, since A \cap B_{R_{0}}(0) has finite Lebesgue measure, by the first part of the proof, there exists a neighborhood U contained in (A \cap B_{R_0}(0)) - (A \cap B_{R_0}(0)) \subset A - A. Hence, the proof concludes. }} The general version of the theorem, first proved by André Weil, states that if G is a locally compact group, and A ⊂ G 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. == Corollary ==

A corollary of this theorem is that any measurable proper subgroup of (\R,+) is of measure zero. == Applications ==

A special case of the Steinhaus Theorem (and the Lebesgue density theorem) deals with the existence of arithmetic progressions in a set of positive Lebesgue measure. In particular, let E \subset \mathbb{R}^{n}, for some positive integer n, be a set of positive Lebesgue measure. Then for any integer N > 0, E contains a finite arithmetic progression of length N + 1. {{Math proof|title=Proof|proof= Let E \subset \mathbb{R}^{n} be a set of positive Lebesgue measure, \{a_{1}, a_{2}, \ldots, a_{N}\} be an arbitrary collection of unit vectors in \mathbb{R}^{n}, and \epsilon \in (0, (2^{N} - 1)^{-1}). Also denote the n-dimensional Lebesgue measure by m^{n}. By inner regularity of the Lebesgue measure, we obtain a compact set K_{1} \subset E such that m^{n}(K_{1}) > 0, and by outer regularity an open set U \supset K_{1} such that m^{n}(U) \leq (1 + \epsilon)m^{n}(K_{1}). Because K_{1} is compact, the distance R = d(K_{1}, U^{c}) is strictly positive. Let \delta \in (0, R) be arbitrary, and consider the set K_{1} + \delta a_{1}. If this subset is not contained in U, then we would have \begin{align} d(K_{1}, U^{c}) which is a contradiction. Therefore, K_{1} \cap (K_{1} + \delta a_{1}) \subset U. This means that \begin{align} m^{n}(U) \geq m^{n}(K_{1} \cup (K_{1} + \delta a_{1})) = m^{n}(K_{1}) + m^{n}(K_{1} + \delta a_{1}) - m^{n}(K_{1} \cap (K_{1} + \delta a_{1})). \end{align} By translation invariance of the Lebesgue measure, we note that m^{n}(K_{1} + \delta a_{1}) = m^{n}(K_{1}), and so \begin{align} m^{n}(K_{1} \cap (K_{1} + \delta a_{1})) \geq 2m^{n}(K_{1}) - m^{n}(U) \geq (1 - \epsilon)m^{n}(K_{1}). \end{align} Since \epsilon , we see that the measure on the left side is strictly positive, which means K_{1} + \delta a_{1} \neq \emptyset. Now for each i = 1, \ldots, N, define the sets K_{i + 1} = K_{i} \cap (K_{i} + \delta a_{i}). By a generalization of the argument above, each K_{i} is contained in U. Moreover, for each i, m^{n}(K_{i}) \geq (1 - (2^{i} - 1))m^{n}(K_{1}) (a simple application of induction immediately yields this result) so that each K_{i} is nonempty. This yields a nested sequence of sets \emptyset \neq K_{N + 1} \subset K_{N} \subset \dotsm \subset K_{1} \subset E. Let q \in K_{N + 1}. Since K_{N + 1} = K_{N} + \delta a_{N}, q - \delta a_{N} \in K_{N}. Likewise, since K_{N} = K_{N - 1} + \delta a_{N - 1}, q - \delta a_{N} - \delta a_{N - 1} \in K_{N - 1}. Repeating this procedure iteratively and eventually denoting p = q - \delta\sum_{i = 1}^{N}a_{i}, we recover the finite arithmetic progression \{p, p + \delta a_{1}, p + \delta a_{1} +\delta a_{2}, \ldots, p + \delta a_{1} + \dotsm + \delta a_{N}\} \subset E consisting of N + 1 points. Hence, the proof concludes. }} ==See also==