Let $A \subset \mathbb{R}$ be a Lebesgue measurable set with positive Lebesgue measure.

Question

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!

Answer 1

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 $$\mu(I\cap(A-jt)^c)=\mu(I\cap(X+s))\le\mu(I\cap X)+|s|=\mu(I\cap A^c)+jt.$$