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

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?

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.