Steinhaus theorem

Proof

Let E ⊂ R n {\displaystyle E\subset \mathbb {R} ^{n}} be a subset of positive Lebesgue measure. First, we consider the case where m ( A ) < ∞ {\displaystyle m(A)<\infty } . In this case, it follows that the characteristic functions χ A ( x ) {\displaystyle \chi _{A}(x)} and χ − A ( x ) {\displaystyle \chi _{-A}(x)} are contained in L p ( R n ) {\displaystyle L^{p}(\mathbb {R} ^{n})} for all 1 ≤ p < ∞ {\displaystyle 1\leq p<\infty } . Then h ( x ) = χ A ∗ χ − A ( x ) {\displaystyle h(x)=\chi _{A}\ast \chi _{-A}(x)} is continuous on R n {\displaystyle \mathbb {R} ^{n}} (where ∗ {\displaystyle \ast } denotes convolution) and

h ( 0 ) = ∫ R n χ A ( − y ) χ A ( y ) d y = ∫ R n χ A ( s ) d s = m ( A ) > 0. {\displaystyle 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 {\displaystyle h} is continuous and h ( 0 ) > 0 {\displaystyle h(0)>0} , there exists an open neighborhood U {\displaystyle U} of 0 so that h ( x ) > 0 {\displaystyle h(x)>0} for all x ∈ U {\displaystyle x\in U} . But by definition of h ( x ) {\displaystyle h(x)} , h ( x ) > 0 {\displaystyle h(x)>0} if and only if x ∈ A − A {\displaystyle x\in A-A} . Hence, 0 ∈ U ⊂ A − A {\displaystyle 0\in U\subset A-A} .

Now suppose m ( A ) = ∞ {\displaystyle m(A)=\infty } . We can write A {\displaystyle A} as the following union:

A = ⋃ R = 0 ∞ ( A ∩ B R ( 0 ) ) , {\displaystyle A=\bigcup _{R=0}^{\infty }(A\cap B_{R}(0)),} where B R ( 0 ) {\displaystyle B_{R}(0)} is the ball of radius R {\displaystyle R} centered at 0. By countable subadditivity, there exists at least one R 0 {\displaystyle R_{0}} so that m ( A ∩ B R 0 ( 0 ) ) > 0 {\displaystyle m(A\cap B_{R_{0}}(0))>0} . Hence, since A ∩ B R 0 ( 0 ) {\displaystyle A\cap B_{R_{0}}(0)} has finite Lebesgue measure, by the first part of the proof, there exists a neighborhood U {\displaystyle U} contained in ( A ∩ B R 0 ( 0 ) ) − ( A ∩ B R 0 ( 0 ) ) ⊂ A − A {\displaystyle (A\cap B_{R_{0}}(0))-(A\cap B_{R_{0}}(0))\subset A-A} . Hence, the proof concludes.

Proof^[3]

Let E ⊂ R n {\displaystyle E\subset \mathbb {R} ^{n}} be a set of positive Lebesgue measure, { a 1 , a 2 , … , a N } {\displaystyle \{a_{1},a_{2},\ldots ,a_{N}\}} be an arbitrary collection of unit vectors in R n {\displaystyle \mathbb {R} ^{n}} , and ϵ ∈ ( 0 , ( 2 N − 1 ) − 1 ) {\displaystyle \epsilon \in (0,(2^{N}-1)^{-1})} . Also denote the n {\displaystyle n} -dimensional Lebesgue measure by m n {\displaystyle m^{n}} . By inner regularity of the Lebesgue measure, we obtain a compact set K 1 ⊂ E {\displaystyle K_{1}\subset E} such that m n ( K 1 ) > 0 {\displaystyle m^{n}(K_{1})>0} , and by outer regularity an open set U ⊃ K 1 {\displaystyle U\supset K_{1}} such that m n ( U ) ≤ ( 1 + ϵ ) m n ( K 1 ) . {\displaystyle m^{n}(U)\leq (1+\epsilon )m^{n}(K_{1}).}

Because K 1 {\displaystyle K_{1}} is compact, the distance R = d ( K 1 , U c ) {\displaystyle R=d(K_{1},U^{c})} is strictly positive. Let δ ∈ ( 0 , R ) {\displaystyle \delta \in (0,R)} be arbitrary, and consider the set K 1 + δ a 1 {\displaystyle K_{1}+\delta a_{1}} . If this subset is not contained in U {\displaystyle U} , then we would have

d ( K 1 , U c ) < ‖ δ a 1 ‖ = δ < R , {\displaystyle {\begin{aligned}d(K_{1},U^{c})<\Vert \delta a_{1}\Vert =\delta <R,\end{aligned}}} which is a contradiction. Therefore, K 1 ∩ ( K 1 + δ a 1 ) ⊂ U {\displaystyle K_{1}\cap (K_{1}+\delta a_{1})\subset U} . This means that

m n ( U ) ≥ m n ( K 1 ∪ ( K 1 + δ a 1 ) ) = m n ( K 1 ) + m n ( K 1 + δ a 1 ) − m n ( K 1 ∩ ( K 1 + δ a 1 ) ) . {\displaystyle {\begin{aligned}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{aligned}}}

By translation invariance of the Lebesgue measure, we note that m n ( K 1 + δ a 1 ) = m n ( K 1 ) {\displaystyle m^{n}(K_{1}+\delta a_{1})=m^{n}(K_{1})} , and so

m n ( K 1 ∩ ( K 1 + δ a 1 ) ) ≥ 2 m n ( K 1 ) − m n ( U ) ≥ ( 1 − ϵ ) m n ( K 1 ) . {\displaystyle {\begin{aligned}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{aligned}}}

Since ϵ < 1 {\displaystyle \epsilon <1} , we see that the measure on the left side is strictly positive, which means K 1 + δ a 1 ≠ ∅ . {\displaystyle K_{1}+\delta a_{1}\neq \emptyset .} Now for each i = 1 , … , N {\displaystyle i=1,\ldots ,N} , define the sets K i + 1 = K i ∩ ( K i + δ a i ) {\displaystyle K_{i+1}=K_{i}\cap (K_{i}+\delta a_{i})} . By a generalization of the argument above, each K i {\displaystyle K_{i}} is contained in U {\displaystyle U} . Moreover, for each i {\displaystyle i} , m n ( K i ) ≥ ( 1 − ( 2 i − 1 ) ) m n ( K 1 ) {\displaystyle 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 {\displaystyle K_{i}} is nonempty. This yields a nested sequence of sets ∅ ≠ K N + 1 ⊂ K N ⊂ ⋯ ⊂ K 1 ⊂ E {\displaystyle \emptyset \neq K_{N+1}\subset K_{N}\subset \dotsm \subset K_{1}\subset E} . Let q ∈ K N + 1 {\displaystyle q\in K_{N+1}} . Since K N + 1 = K N + δ a N {\displaystyle K_{N+1}=K_{N}+\delta a_{N}} , q − δ a N ∈ K N {\displaystyle q-\delta a_{N}\in K_{N}} . Likewise, since K N = K N − 1 + δ a N − 1 {\displaystyle K_{N}=K_{N-1}+\delta a_{N-1}} , q − δ a N − δ a N − 1 ∈ K N − 1 {\displaystyle q-\delta a_{N}-\delta a_{N-1}\in K_{N-1}} . Repeating this procedure iteratively and eventually denoting p = q − δ ∑ i = 1 N a i {\displaystyle p=q-\delta \sum _{i=1}^{N}a_{i}} , we recover the finite arithmetic progression { p , p + δ a 1 , p + δ a 1 + δ a 2 , … , p + δ a 1 + ⋯ + δ a N } ⊂ E {\displaystyle \{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 {\displaystyle N+1} points. Hence, the proof concludes.