Exercise 1.6.8 (Steinhaus theorem)

The key observation that links the Steinhaus theorem to the convolution is following: if the convolution of indicators $1_E$ and $1_{-E}$ at point $\mathrm{\Delta }\in {ℝ}^d$ is is greater than 0

then $\mathrm{\Delta }$ is definitely contained in $E-E$ (otherwise the integral would be zero, since $\mathrm{\Delta }\notin E-E⟹\nexists x,y\in E:x-y=\mathrm{\Delta }⟹\forall <!--\; FUNCTION\; APPLICATION\; -->y\in E:1_E(y)1_{-E}(\mathrm{\Delta }-y)=0$.) We know that this is definitely the case at the origin $0\in {ℝ}^d$:

Suppose that $m(E)\ne \infty$. Since the convolution is continuous (by the previous exercise) there must be a $\delta >0$ such that for all points in the $\delta$-neighbourhood of $0$ the value of the convolution is sufficiently close to $m(E)$ and thus positive. Using our observation from before we see that for this $\delta >0$ all the points in the neighbourhood of $0$ are contained in $E-E$. In case where $m(E)=\infty$, we can find a ball $B(0,M)\subset {ℝ}^d$ and rather work on $B(0,M)\cap E<\infty$.