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.
 A: 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 $\frac{b-a}{2n}$?
A: ** 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.
A: 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 $\mathbb{R}$.
