We say a set $A$ contains an arithmetic progression of length $N$ if it contains a subset of the form $\{a, a + b, \ldots, a + (N − 1)b\}$ where $b \neq 0$. Show that if $A \subset [0, 1]$ with $\lambda(A) > 0$, then $A$ contains an arithmetic progression of length $N$ for every $N$. Here, $\lambda$ denotes the Lebesgue measure on $\mathbb{R}$.
I somehow would like to use density points, but I'm not really sure how. Any suggestions?
Let $K \subset A \subset V$ with $K$ compact, $V$ open, and $\lambda(K) > \frac{1}{1+\delta} \lambda(V)$ for some small $\delta$. Let $U = (0,\epsilon)$ with $\epsilon$ small enough that $K+U \subset V$. This is possible because by compactness we can find a finite cover of $K$ contained in $V$ by open balls and take the minimum radius.
Let $u\in U$ and suppose that $\lambda( ( K+u ) \cap K ) < (1 - \delta)\lambda(K)$. Then, since $K \cup (K+u) \subset V$, $$ \lambda(V) > \lambda( K ) + \lambda(K+u ) - \lambda( (K+u) \cap K ) > (1 + \delta )\cdot \lambda(K) $$ but we specifically chose $K$ and $V$ so that $(1 + \delta)\lambda(K) > \lambda(V)$, so this is a contradiction. Therefore $\lambda( ( K+u ) \cap K ) \geq ( 1 - \delta ) \lambda(K)$.
Therefore, since $\delta$ did not depend on $u$, we get that $\lambda( K \cap ( K + \frac{m}{N}u ) ) \geq (1-\delta)\lambda(K) $. From this, we get
$$ \lambda( K \cap ( K + \frac{1}{N}u )\cap \ldots \cap ( K + \frac{N-1}{N}u ) ) \geq ( 1 - N\delta ) \lambda( K ) > 0 $$
In particular, since we could have chosen $\delta$ as small as we wanted, this intersection is non-empty, which is what we wanted to show.