TITLE: Steinhaus theorem (sums version)
QUESTION [37 upvotes]: 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.

REPLY [12 votes]: 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 $\epsilon$.