Consider the following exercise from [1] (p. 44):
30 If $E$ and $F$ are measurable, and $m(E) > 0$, $M(F) > 0$, prove that
$$
E + F = \{x + y : x \in E, x \in F\}
$$
contains an interval.
If $E = F$, this exercise reduces to exercise 29 (see below), which I am able to prove according to the hint. But I cant figure out how to modify exercise 29's proof to prove exercise 30's more general claim. Any help will be appreciated.
29 Suppose $E$ is a measurable subset of $\mathbb{R}$ with $m(E) > 0$. Prove that the difference set of $E$, which is defined by
$$
\{z \in \mathbb{R} : z = x – y \mbox{ for some } x, y \in E\}
$$
contains an open interval centered at the origin.If $E$ contains an interval, then the conclusion is straightforward. In general, one may rely on Exercise 28.
[Hint: Indeed, by Exercise 28, there exists an open interval $I$ so that $m(E \cap I) \geq (9/10)m(I)$. If we denote $E \cap I$ by $E_0$, and suppose that the difference set of $E_0$ does not contain an open interval around the origin, then for arbitrarily small $a$ the sets $E_0$ and $E_0 + a$ are disjoint. From the fact that $(E_0 \cup (E_0 + a)) \subseteq (I \cup (I + a))$ we get a contradiction, since the lift-hand side has measure $2m(E_0)$, while the right-hand side has measure only slightly larger than $m(I)$.]
For completeness, here's exercise 28 mentioned in exercise 29:
28 Let $E$ be a subset of $\mathbb{R}$ with $m_*(E) > 0$. Prove that for each $0 < \alpha < 1$, there exists an open interval $I$ so that
$$
m_*(E\cap I) \geq \alpha m_*(I).
$$
Loosely speaking, this estimate shows that $E$ contains almost a whole interval.[Hint: Choose an open set $\mathcal{O}$ that contains $E$, and such that $m_*(E) \geq \alpha m_*(\mathcal{O})$. Write $\mathcal{O}$ as the countable union of disjoint open intervals, and show that one of these intervals must satisfy the desired property.]
References
[1] Stein, Elias M. and Shakarchi, Rami. Real Analysis: Measure Theory, Integration, and Hilbert Spaces. Princeton University Press (2005)
Best Answer
I also don't know how to deduce Exercise 30 from the proof of Exercise 29 which you have shown, but I know a direct proof of Exercise 30 (I don't recall where I learned it).
Assume wlog. that $m(E)<\infty$, $m(F)<\infty$ (otherwise work with subsets of finite measure). Consider
$$\mathbf{1}_E*\mathbf{1}_F(x)=\int_{\mathbb{R}} \mathbf{1}_{E}(x-y)\mathbf{1}_F(y)dy$$
and notice that if $\mathbf{1}_E*\mathbf{1}_F(x)>0$ for some $x\in\mathbb{R}$, then this implies $x\in E+F$ (even more: the set of $y$ such that $x-y\in E$ and $y\in F$ must have positive measure). Therefore we have
$$A:=\{x\in\mathbb{R}\,:\,\mathbf{1}_E*\mathbf{1}_F(x)>0\}\subset E+F$$
The point is now that $\mathbf{1}_E*\mathbf{1}_F$ is a continuous function (see comment).
Thus the set $A$ is open. Also, $m(A)>0$ since by Fubini we have
$$\int_{\mathbb{R}} \mathbf{1}_E*\mathbf{1}_F(x) dx = \left(\int_{\mathbb{R}} \mathbf{1}_E(x) dx\right) \left(\int_{\mathbb{R}} \mathbf{1}_{F}(y) dy\right)= m(E)m(F)>0$$
So $A$ and therefore also $E+F$ contains an open interval.