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.

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 $\mathbb{R}$.
