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Let $A \subset \mathbb{R}$ be a Lebesgue measurable set with positive Lebesgue measure. Show that for any $k\in\mathbb{N}$, there exists $a,t \in\mathbb{R}$ ($t\ne0$) such that \begin{align*} a,a+t,a+2t,\cdots,a+kt\in A\end{align*}

I knew there exists $\delta>0$ s.t $(-\delta,\delta)\subset A-A$ because of $\mu(A)>0$.

The problem seems to be similar with that. However, I hard to think how to prove that..

Any help is appreciated..

Thank you!

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  • $\begingroup$ What do you mean by $A - A$? Is that the empty set? Anyways, note that a set of positive measure doesn't have to contain an open interval, for instance, the set of irrational numbers. $\endgroup$
    – silvascientist
    Apr 9, 2018 at 18:30
  • $\begingroup$ @MonstrousMoonshiner $$A\setminus B=\{x:x\in A,\ x\notin B\}$$ $$A-B=\{a-b:a\in A,\ b\in B\}$$ $\endgroup$
    – bof
    Apr 9, 2018 at 18:50
  • $\begingroup$ @bof, Right, I'll modify. Thank you! $\endgroup$
    – user453447
    Apr 9, 2018 at 19:12
  • $\begingroup$ @bof Thanks, I was confused as I have in fact seen the subtraction symbol used to indicate actual set difference - it wasn't clear what the notation meant. Anyways, is it still necessarily true that $A - A$ contains an open interval? It certainly doesn't seem clear to me. $\endgroup$
    – silvascientist
    Apr 10, 2018 at 7:10
  • $\begingroup$ @Monstrous Moonshiner Firstly, we can show that $\lim _{t \rightarrow 0} \mu(A\cap (A+t)) =\mu(A)$ by using Tonelli's theorem. Then, we can choose $\delta >0$ such that $A\cap (A+t) \neq \emptyset$ for $\vert t \vert \leq \delta$ if $\mu(A)>0$. Finally, we can get $A-A$ contains an open interval. $\endgroup$
    – user453447
    Apr 10, 2018 at 7:15

1 Answer 1

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Choose an interval $I$ such that $$\mu(A\cap I)\gt\frac k{k+1}\mu(I),$$ in other words, $$\mu(I\setminus A)\lt\frac{\mu(I)}{k+1}.$$ Choose $t\gt0$ so that $$\mu(I\setminus A)+kt\lt\frac{\mu(I)}{k+1}.$$ Then, for $0\le j\le k,$ we have $$\mu(I\setminus(A-jt))\le\mu(I\setminus A)+jt\le\mu(I\setminus A)+kt\lt\frac{\mu(I)}{k+1},$$ and so $$\mu\left(\bigcup_{j=0}^k(I\setminus(A-jt)\right)\le\sum_{j=0}^k\mu(I\setminus(A-jt))\lt\mu(I),$$ whence $$I\cap\bigcap_{j=0}^k(A-jt)\ne\emptyset.$$ Choose $$a\in\bigcap_{j=0}^k(A-jt);$$ then $a+jt\in A$ for $j=0,1,\dots,k.$


P.S. I have been asked to explain the inequality $$\mu(I\cap(A-jt)^c)\leq\mu(I\cap A^c)+jt.\tag1$$
Lemma. If $I$ is an interval and $s$ a real number, $$\mu(I\cap(X+s))\le\mu(I\cap X)+|s|.$$
Proof. Since $$I\cap(X+s)\subseteq((I\cap X)+s)\cup(I\setminus(I+s)),$$ we have $$\mu(I\cap(X+s))\le\mu((I\cap X)+s)+\mu(I\setminus(I+s))\le\mu(I\cap X)+|s|.$$
Now let $X=A^c$ and $s=-jt,$ so that $X+s=A^c-jt=(A-jt)^c.$ By the lemma we have $$\mu(I\cap(A-jt)^c)=\mu(I\cap(X+s))\le\mu(I\cap X)+|s|=\mu(I\cap A^c)+jt.$$

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  • $\begingroup$ Could you explain why $\mu(I \cap (A-jt)^c) \leq \mu (I \cap A^c) +jt $?? Thank you for your detailed solution! $\endgroup$
    – user453447
    Apr 9, 2018 at 21:08
  • $\begingroup$ Thank you! It's clear to me! $\endgroup$
    – user453447
    Apr 10, 2018 at 6:23

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