For $E \subseteq [0, 1]$, $m(E) > 0$, show that there are $\alpha$ and $\beta$ such that $\alpha, \alpha + \beta, \alpha + 2\beta \in E$. This was originally a proof verification question, but I have since moved the proof to an answer as discussed on meta. I still welcome comments on the proof as well as any alternative proofs.


Let $E$ be a Lebesgue measurable subset of $[0, 1]$ with positive measure. Show that there are $\alpha$ and $\beta$ such that $\alpha, \alpha + \beta, \alpha + 2\beta \in E$.

The only idea I have had is to use Lebesgue density and the Lebesgue Density Theorem, but so far no luck. 
 A: Let $x \in E$ be a point with Lebesgue density $1$ (by the Lebesgue Density Theorem, such a point exists). Fix $\varepsilon > 0$ such that 
$$\frac{m(E\cap(x - \varepsilon, x + \varepsilon))}{m((x - \varepsilon, x+ \varepsilon))} > \frac{1}{2}.$$
Now suppose there is no $y \in (0, \varepsilon)$ such that $x - y, x + y \in E$. Then $A:= -(E\cap(x - \varepsilon, x) - x)$ and $B := E\cap(x, x + \varepsilon) - x$ are disjoint${}^1$ measurable subsets of $(0, \varepsilon)$. Then 
\begin{align*}
m(E\cap(x - \varepsilon, x + \varepsilon)) &= m((E\cap(x - \varepsilon, x))\cup (E\cap(x, x + \varepsilon)))\\
&= m(E\cap(x - \varepsilon, x)) + m(E\cap(x, x + \varepsilon))\\
&= m(A) + m(B)\\
&= m(A\cup B)\\
&\leq m((0, \varepsilon))\\
&= \varepsilon.
\end{align*}
But then
$$\frac{m(E\cap(x - \varepsilon, x + \varepsilon))}{m(x - \varepsilon, x + \varepsilon))} \leq \frac{\varepsilon}{2\varepsilon} = \frac{1}{2}$$
which is a contradiction. Therefore there is $y \in (0, \varepsilon)$ such that $x - y, x + y \in E$. Then setting $\alpha = x - y$ and $\beta = y$ we have $\{\alpha, \alpha + \beta, \alpha + 2\beta\} = \{x - y, x, x + y\} \subset E$.
${}^1$ Suppose $z \in A\cap B$, then $z \in (0, \varepsilon)$, $x - z \in E$ (as $z \in A$), and $x + z \in E$ (as $z \in B$). Therefore, the hypothesis implies $A\cap B = \emptyset$.
