Prove combination of two sets contains an open ball

[aegis90]

So this was an exam question that our professor handed out ( In class. I didn't get the question right)

Let E be a subset of R^n, n>= 2. Suppose that E measurable and m(E)>0. Prove that:

E+E = {x+y: x in E, y in E } contains an open ball.

(The text Zygmund that we used showed an example that E-E defined in similar sense contains an open interval centered at the origin, where E is a subset of R. Stein had another problem that asked to show that E+E contains an open interval.

I'm assuming that's where he got the problem, but I'm not sure that the same method works, since he gave a hint to prove that the convolution: chi(e)*chi(e) is continuous at the origin. )

 

[fresh_42]

I'd say that measurability guarantees an open ball $B$ in $E$. Now try to construct an open ball in $B+B$.

 

[Infrared]

@fresh_42 That is not the case. The irrationals are subset of a full measure in, say, $[0,1]$, but do not contain any interval.

Following the hint: consider the convolution $f=\chi_E*\chi_E$, where $\chi_E$ is the indicator function for $E$. By the definition of convolution, $f(x)=m(E\cap (x-E))$. From this formula, we have that $f$ is supported on $E+E$. Since $f$ is continuous and not identically zero (as it integrates to $m(E)^2$), this means that there is an open set in $E+E$ on which $f$ is positive.

 

[WWGD]

Or try a fat Cantor set, i.e., remove less than 1/3 at each stage. Of course, if E itself contains an interval, we are done. EDIT: This reminds me of convolutions, exhustions by open and compact sets, maybe additive Lie groups and neighborhoods of unity.