Let $A \subset \mathbb{R}$ be a Lebesgue measurable set with positive Lebesgue measure. 
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!  
 A: 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
$$\mu(I\cap(A-jt)^c)=\mu(I\cap(X+s))\le\mu(I\cap X)+|s|=\mu(I\cap A^c)+jt.$$
