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.

  • 2
    $\begingroup$ I read from here that my E contains an interval. How to show that it is itself an interval? $\endgroup$
    – MikeG
    Sep 22, 2021 at 4:38
  • $\begingroup$ The result is true if $E$ is closed math.stackexchange.com/questions/1371147/… $\endgroup$ Sep 22, 2021 at 9:05
  • 1
    $\begingroup$ @KaviRamaMurthy If we use Steinhaus to $E/2$ then from hypothesis $E$ contains an interval. $\endgroup$ Sep 22, 2021 at 9:17

2 Answers 2


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?

  • $\begingroup$ It has a positive measure. Or am I wrong? $\endgroup$
    – Youem
    Sep 22, 2021 at 4:12

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