Let $E\subset \mathbb{R}$ be a Lebesgue measurable subset of reals such that $\mu(E)>0.$ Consider the set $E+E=\{x+y: x,y\in E\},$ prove that $E+E$ contains an interval of length greater than $0$.
Is this problem even true? Let's say the set $E$ is our set of irrational numbers, then $E+E$ doesn't contain positive length interval. If this is indeed true, how can we go about proving this statement?
On
First, let me give some intuition on the idea to be used: If the set $E$ were finite, an idea to "detect" whether a certain point $x$ belongs to $E+E$ would be to look at $\chi_{E}(x-y)$ for every $y \in E;$ that is, to study
$\sum_{y \in E}\chi_{E}(y)\chi_{E}(x-y)$
In the continuous case, we could try something similar by studying the convolution. Consider thus the function:
$\phi(x)=\chi_{E} \ast \chi_{E}(x).$
Notice that it is continuous, as you can easily check by using the dominated convergence theorem. Then the set $A=\{x \in \mathbb{R} \ : \ \phi(x)>0 \}$ is open, which means that it must contain an open interval.
It would be enough then to check that $A \subset E+E.$ But this actually obvious, so we are already done.
EDIT. Sorry, I didn't see that another answer was posted... it happened while I was writing. Let me know if I should delete mine.
This is true in general, for sets $A,B\subseteq\mathbb R$ with $\mu(A),\mu(B)>0$.
A way to do this is to consider the characteristic functions $\chi_A,\chi_B$, and set $f=\chi_A*\chi_B$ to be their convolution. Then $f$ is nonnegative continuous, and we can compute that $\int_{\mathbb R}f=\mu(A)\mu(B)>0$, so, from continuity, $$\int_{\mathbb R}\chi_A(y)\chi_B(x-y)\,dy\geq\varepsilon$$ for all $x$ in an interval $(a,b)$. This shows that there exists $y\in\mathbb R$ such that $\chi_A(y)\chi_B(x-y)>0$, therefore $y\in A$ and $x-y\in B$. This will show that $x\in A+B$, therefore $(a,b)\subseteq A+B$.