No mid points implies measure $0$ Let $E$ be a Lebesgue measurable set in $\mathbb R$ such that $x \in E, y\in E, x\neq y$ implies $\frac {x+y} 2 \notin E$. Show that $m(E)=0$ where $m$ is the Lebesgue measure on $\mathbb R$ I believe this result is interesting enough to the MSE community so I am posting the question as well as the answer. The proof may also benefit those who are beginning to learn measure theory.
 A: By replacing $E$ by $E \cap (-n,n)$ with $n$ sufficiently large we may suppose $0<m(E)<\infty$. 
Let $0<\epsilon < m(E)/6$. Since $I_E \in L^{1} (\mathbb R)$ there is a continuous function $f$ with compact support such that $0 \leq f \leq 1$ and $\int_{\mathbb R} |I_E -f|dm <\epsilon$. 
The hypothesis implies that $I_E(x) I_E (x+t) I_E (x-t)=0$ for all $x$ and $t \neq 0$. Let us estimate $$\int |I_E(x) I_E (x+t) I_E (x-t) -f(x)f(x+t)f(x-t)|dx.$$ By adding and subtracting $I_E(x) I_E(x+t) f(x-t) $ and $I_E(x) I_E(x) f(x+t) f(x-t)$ and using the triangle inequality, it follows that $$\int |I_E(x) I_E (x+t) I_E (x-t) -f(x)f(x+t)f(x-t)|dx < 3\epsilon.$$ Hence $\int |f(x)f(x+t)f(x-t)|dx < 3\epsilon$. 
Letting $t \to 0$ and using Dominated Convergence Theorem we get $\int |f(x)|^{3}dx \leq 3\epsilon$. 
By putting $t = 0$ in the inequality $\int |I_E(x) I_E (x+t) I_E (x-t) -f(x)f(x+t)f(x-t)|dx < 3\epsilon$ we get $\int |I_E(x)-f^{3}(x)|dx < 3\epsilon$. It now follows that $\int I_E <6\epsilon$. Hence $m(E)< 6\epsilon <m(E)$ which is a contradiction.
