This is a question from Stromberg related to Steinhaus' Theorem:

If $A$ is a set of positive Lebesgue measure, show that $A + A$ contains an interval.

I can't quite see how to modify the Steinhaus proof though.


I don't know what proof of Steinhaus theorem is used, but we can show the following result:

If $A$ and $B$ have a positive Lebesgue measure, then $A+B$ contains an interval.

We can assume that $A$ and $B$ have finite measure. Indeed, if $\lambda(A)$ is infinite, $A=\bigcup_{n\in\mathbb N}A\cap\left[-n,n\right]$ and we only have to pick $n_0$ such that $\lambda(A\cap \left[-n_0,n_0\right])>0$. If $n_1$ is such that $\lambda(B\cap \left[-n_1,n_1\right])>0$, and we have shown the result for $A$ and $B$ of finite measure, then $A+B\supset (A\cap \left[-n_0,n_0\right])+(B\cap \left[-n_1,n_1\right])\supset I$ and we are done.

Thank to the fact the indicator functions are in $L^2$ and the density of the continuous functions with compact support in $L^2(\mathbb R)$ $$f\colon x\mapsto \mathbf{1}_A*\mathbf{1}_B(x)=\int_{\mathbb R}\mathbf{1}_A(x-t)\mathbf {1}_B(t)d\lambda(t)$$ is continuous . Hence the set $O:=\left\{x\in\mathbb R,f(x)>0\right\}$ is open. Since $\int_{\mathbb R}f(x)d\lambda(x)=\lambda(A)\cdot\lambda(B)>0$, $O$ is non-empty and therefore contains an open non-empty interval $I$. If $x\notin A+B$, $A\cap(-B+x)=\emptyset$. Indeed, if $y\in A\cap(-B+x)$ then $y=a$ for some $a\in A$, and $y=-b+x$ for some $b\in B$, hence $x=a+b$. So if $x\notin A+B$, $f(x)=0$, and taking the complement, if $f(x)\neq 0$ then $x\in A+B$, hence we got $$I\subset O\subset A+B.$$

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    $\begingroup$ extremely clear. Thanks. $\endgroup$ – Bombyx mori Jul 28 '12 at 7:04
  • $\begingroup$ For $A$ and $B$ having infinite measure, Is the result follows by the case of finite measure and $A_n+B_n \subset A+B$ where$ A_n=A \cap [-n,n]$? $\endgroup$ – mnmn1993 Feb 6 '18 at 12:59
  • $\begingroup$ How do you know $f$ is continuous? $\endgroup$ – h3fr43nd May 21 '20 at 0:42

A proof using metric density is outlined in Exercise 5 of Chapter 7 (Differentiation) of Rudin's Real and Complex Analysis, 3rd edition. I present my version.

We generalize to possibly distinct sets $A$ and $B$ of positive measure. The set $A$ has a point $a$ of metric density where

$$m(A\cap (a-\delta, a +\delta ))/2\delta > 3/4,$$

and it suffices to prove the proposition with $A$ is replaced by this intersection. Similarly, we may replace $B$ by some set concentrated in a length $2\delta$ interval near a point $b$.

Let $a_0=a+b \in A+B$. The point is that for small enough $\epsilon$ (positive or negative!), $a_0 +\epsilon \in A+B$. If not, then $a_0+\epsilon -B$ does not intersect $A$. But $A$ and $a_0+\epsilon -B$ both lie in the interval

$$(a-(\delta+|\epsilon|), a + (\delta +|\epsilon|)),$$

which has measure $2(\delta+\epsilon)$. Together $A$ and $a_0+\epsilon -B$ have measure $3\delta$, so they must intersect for small $\epsilon$.

  • $\begingroup$ Not a very important remark : you wrote that $A$ and $a_0+\varepsilon-B$ both lie in the interval $(a_0-(\delta+\varepsilon),a_0+(\delta+\varepsilon))$ : I think it should be $|\varepsilon|$ instead of $\varepsilon$ (as $\varepsilon$ can be negative) $\endgroup$ – charmd May 17 '17 at 8:13
  • $\begingroup$ @charMD Yup, thanks! $\endgroup$ – Potato May 18 '17 at 20:02

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