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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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