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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?

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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.

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  • $\begingroup$ Lemma (pigeonhole principle): if $n+1$ subsets of a set $X$ have total measure greater than $n$ times measure of $X$, then there exists a point covered by them all. For example, you may assume the contrary and look at the sum of integrals of characteristic functions of these sets (which equals to the integral of the sum of these functions). $\endgroup$
    – Fedor Petrov
    Commented Nov 29, 2021 at 6:59

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