Measurable set contains a sequence

Question

I found this question and didn't manage to extrapolate from the hint, could anyone help?

Here's the question for the sake of completeness:

Let $A\subseteq[a,b]$ be Lebesgue measurable, such that: $m(A)>\frac{2n-1}{2n}(b-a)$. I need to show that $A$ contains an arithmetic sequence with n numbers ($a_1,a_1+d,...,a_1+(n-1)*d$ for some d).

Answer 1

Let $d=\dfrac{b-a}{n}$ and $a_1 \in [a,a+d]$.

Using the stacking argument in the linked question, the measure of the $\{a_1\}$ for which the requirement is true is at least $m([a,a+d]) -\left( m([a,b]) - m(A) \right)$.

Using the hint, this is greater than $\dfrac{b-a}{n} - \dfrac{b-a}{2n}=\dfrac{b-a}{2n} \gt 0$.

A set with positive measure has at least one element and you are done.