A subset $E$ of $\mathbb R$ is said to contain arbitrarily long arithmetic progressions, if for every natural $n$, there exists $a, d \in R, d$ nonzero, such that $a + kd$ is in $E$ for all natural $k \leq n$.


Does every measurable subset of $\mathbb R$ of non zero Lebesgue measure contain arbitrarily long arithmetic progressions?



Does every measurable subset of $\mathbb R$ of non zero Lebesgue measure contain arbitrarily long arithmetic progressions?
Yes. For fixed $n$, we approximate our set $E$ from above by an open set $U=\sqcup \Delta\_i$ ($\Delta\_i$ are disjoint intervals) with such accuracy that one of intervals $\Delta\_i$ satisfies $|E\cap \Delta\_i|>(1-\frac1{n+1})|\Delta\_i|$, where $|\cdot|$ denotes Lebesgue measure. Now if $\Delta\_i=(a,a+(n+1)t)$, we consider $n+1$ sets $E\_i:=(E-it)\cap (a,a+t), i=0,1,\ldots,n$. The sum of there measures equals $|E\cap \Delta\_i|>nt$, thus there exists a point covered by them all. It corresponds to an arithmetic progression inside $E$ with difference $t$ and $n+1$ terms.


