# Arithmetic sequence in a Lebesgue measurable set

Let $A\subseteq[a,b]$ be Lebesgue measurable, such that: $m(A)>\frac{2n-1}{2n}(b-a)$. I need to show that $A$ contains an arithmetic sequence with n numbers ($a_1,a_1+d,...,a_1+(n-1)*d$ for some d).

I thought about dividing [a,b] into n equal parts, and show that if I put one part on top of the other, there must be at least one lapping point, that will occur in every part. but I haven't succeeded in showing that.

Thank you.

• I think this may follow from Steinhaus' Theorem: The difference set $A-A \triangleq \{a-b:a,b\in A\}$ of a Lebesgue measurable set $A$ of positive measure contains an open neighborhood of the origin. Commented Oct 29, 2013 at 22:07
• Actually there is a second part of the question, to prove just that (just they didn't call it Steinhaus therorem). I can see how this follows from Steinhaus theorem but I think what they meant is that there is a way to prove the theorem using this exercise. Commented Oct 30, 2013 at 7:23

Hint: You are on the right track. Have you noticed that the length of each of your sub-intervals is $\frac{b-a}{n}$, while the total length of all the missing pieces is only $\frac{b-a}{2n}$?

** Fact 1.** Let $p$ be a linear normalized Lebesgue measure on a circle $C$ centered at the origin of the plane $R^2$. Suppose that $p(A)> (n-1)/n$. Then there exist $n$ points in $A$ which are vertices of a regular $n$-sided polygon.

Proof. Let us consider a rotation defined by $f(x,y)=e^{\frac{2\pi i}{n}}(x+iy)$. Let consider sets $f^{0}(A),f^{1}(A), \cdots, f^{n-1}(A)$. From rotation invariance of $p$ we get $p(f^{0}(A))=p(f^{1}(A))= \cdots= p(f^{n-1}(A))>(n-1)/n$ and $p(C \setminus f^{0}(A))=p(C \setminus f^{1}(A))= \cdots= p(C \setminus f^{n-1}(A))<1/n$. We claim that $p(f^{0}(A) \cap f^{1}(A) \cap \cdots \cap f^{n-1}(A))>0$. Assume the contrary. Then by our assumption and De'Morgan rule we get $$1=p(C \setminus \cap_{k=0}^{n-1}f^{k}(A))=p(\cup_{k=0}^{n-1} (C \setminus f^{k}(A)))\le \sum_{k=0}^{n-1}p(C \setminus f^{k}(A))<n \times 1/n=1,$$ which is a contradiction.

Hence there is $x \in \cap_{k=0}^{n-1}f^{k}(A)$, equivalently, $x \in f^{0}(A),x \in f^{1}(A), \cdots, x \in f^{n-1}(A)$. The latter relations imply that $x \in A, f^{-1}(x) \in A, \cdots, f^{-(n-1)}(x) \in A$. Notice that the points $M_0=x, M_1=f^{-1}(x), \cdots, M_{n-1}=f^{-(n-1)}(x)$ are vertices of a regular $n$-sided polygon.

Fact 2. Let $B$ be a subset of the real axis whose linear Lebesgue measure is positive. Then for an arbitrary $n >1$ there are $n$ points in $B$ which constitute an arithmetic progression.

Proof. By Lebesgue theorem about density points, there is a density point $x_0 \in B$. Let $\epsilon$ be such a positive number that $\frac{m([x_0 -\epsilon,x_0+\epsilon[ \cap B)}{2\epsilon }>(n-1)/n$, where $m$ denotes the standard linear Lebesgue measure in the real axis $\mathbb{R}$. Let consider a circle $C$ of the length $2 \epsilon$ and centered at the origin of real plane. Let $p$ be a linear normalized Lebesgue measure on $C$. Let reel up the set $[x_0 -\epsilon,x_0+\epsilon[$ on the circle $C$ by the unique transformation $\phi$ such that $\phi(x_0 -\epsilon)=(0, \frac{2\epsilon}{2\pi})$ and $\phi(x_0 -\frac{\epsilon}{2})=(- \frac{2\epsilon}{2\pi},0)$. Obviously, $$p(\phi(B \cap [x_0 -\epsilon,x_0+\epsilon[))=\frac{m([x_0 -\epsilon,x_0+\epsilon[ \cap B)}{2\epsilon }>(n-1)/n$$ and by Fact 1, there exist $n$ points $M_0, M_1, \cdots, M_{n-1}$ in $\phi(B \cap [x_0 -\epsilon,x_0+\epsilon[)$ which are vertices of a regular $n$-sided polygon. From these points we choose a point $M_{k_0}$ which is a nearest point for the point $(0, \frac{2\epsilon}{2\pi})$ from the left (along the circle $C$). Then the points $\phi^{-1}(M_{k_0}), \phi^{-1}(f^{1}(M_{k_0})),\cdots,\phi^{-1}(f^{n-1}(M_{k_0})$ are in $B \cap [x_0 -\epsilon,x_0+\epsilon[$ and they constitute an arithmetic progression.

Fact 2 implies that we need no a restriction that $m(A)>\frac{2n-1}{2n}(b-a)$. It is sufficient to require that $m(A)>0$.

It is subject of interest that there is a Lebesgue null set $X$ in $\mathbb{R}$ which satisfies the following conditions:

(i) $\mbox{card}(X)=2^{\aleph_0}$;

(ii) for an arbitrary $n >2$ there are no $n$ points in $X$ which constitute an arithmetic progression.

The proof of this fact essentially implies an existence of Lebesgue measurable Hamel bases(over of all rationals $\mathbb{Q}$) in $\mathbb{R}$.