# Mid-point convex measurable subset of $\mathbb{R}$ with positive Lebesgue measure is an interval

The question is right as the title:

Let $$E$$ be a measurable subset of $$\mathbb{R}$$ w.r.t. Lebesgue measure, and has positive measure. For any $$x,y\in E$$, $$\frac{x+y}{2}\in E$$. Prove that $$E$$ is an interval(like $$[a,b],[a,b),(a,b)$$ etc., possibly infinity endpoint).

The hint is to find some function on it, and I tried looking at its characteristic function, but I have no clue.

• I read from here that my E contains an interval. How to show that it is itself an interval? Sep 22, 2021 at 4:38
• The result is true if $E$ is closed math.stackexchange.com/questions/1371147/… Sep 22, 2021 at 9:05
• @KaviRamaMurthy If we use Steinhaus to $E/2$ then from hypothesis $E$ contains an interval. Sep 22, 2021 at 9:17

Since you already know that $$E$$ contains an interval, it will suffice to prove the following.

Lemma. If a subset $$E$$ of $$\mathbb R$$ is "midpoint-convex" and has nonempty interior, then $$E$$ is an interval.

Proof. Let $$I$$ be a maximal interval contained in $$E$$, and assume for a contradiction that $$E\ne I$$. Then there is a point $$a\in E$$ such that $$a\lt\inf I$$ or $$a\gt\sup I$$; without loss of generality we assume that $$a\lt c=\inf I$$. Choose $$d\in I$$ so that $$c\lt d$$; then $$(c,d)\subseteq E$$.

By repeated application of midpoint-convexity, we can find points in $$E\cap(a,c)$$ which are arbitrarily close to $$c$$. Choose a point $$e\in E\cap(a,c)$$ which is closer to $$c$$ than $$c$$ is to $$d$$.

Since $$e\in E$$ and $$(c,d)\subseteq E$$, it follows by midpoint-convexity that $$\left(\frac{e+c}2,\frac{e+d}2\right)\subseteq E$$. Since $$\frac{a+c}2\lt c\lt\frac{a+d}2$$, it follows that the set $$J=\left(\frac{e+c}2,\frac{e+d}2\right)\cup I$$ is an interval. Since $$I\subsetneqq J\subseteq E$$, this contradicts the maximality of $$I$$.

I am assuming that you have proved that $$E$$ contains an interval.

Start by using induction to prove that for every $$x$$ and $$y$$ in $$E$$ $$\left(1-\frac k{2^n}\right)x + \frac{k}{2^n} y\in E, \quad \forall n\in \mathbb N, 0\le k\le 2^{n}.$$ Let $$z$$ be an element of $$E$$ such $$\left]z-\epsilon,z+\epsilon\right[\subset E$$. You can prove easily (by induction) that for every $$x\in E$$ $$\left]\left(1-\frac k{2^n}\right)z + \frac{k}{2^n} z-\epsilon, \left(1-\frac k{2^n}\right)z + \frac{k}{2^n}x +\epsilon\right[\subset E, \quad \forall n\in \mathbb N, 0\le k < 2^{n}.$$

Finally let $$n\in\mathbb N$$ such that $$\frac1{2^n} \le \epsilon$$ so \begin{align} [z,x] \subset \bigcup_{k=0}^{2^{n}-1}\left]\left(1-\frac k{2^n}\right)z + \frac{k}{2^n} z-\epsilon, \left(1-\frac k{2^n}\right)z + \frac{k}{2^n}x +\epsilon\right[\subset E \end{align}

Can you continue from here?

• It has a positive measure. Or am I wrong? Sep 22, 2021 at 4:12