Consider a fundamental parallelotope $F$ for a $n$-dimensional lattice $\Lambda$ and consider a convex, measurable, centrally symmetric subset of $\mathbb R^{n}$, which we call $E$, such that $m(E)>2^nm(F)$, where $m$ is the $n$-dimensional Lebesgue measure.
Clear that $$ \mathbb R^{n}=\bigsqcup_{x\in\Lambda}(x+F) $$ hence $$ \frac12E=\frac12E\cap\mathbb R^n=\bigsqcup_{x\in\Lambda}\left(\frac12E\;\cap(x+F)\right) $$ which yelds to \begin{align*} m(F)&<\frac1{2^n}m(E)=m(1/2E)\\ &=\sum_{x\in\Lambda}m\left(\frac12E\;\cap(x+F)\right)\\ &=\sum_{x\in\Lambda}m\left[\left(\frac12E-x\right)\cap F\right] \end{align*} Then my book says that from the last one, it follows that the sets $\left(\frac12E-x\right)\cap F$ cannot be pairwise disjoint. Why?
It's clear that if $$ \bigcup_{x\in\Lambda}\left(\frac12E-x\right)=\mathbb R^{n}\;\;\; (*) $$ supposing by contradiction that the $\left(\frac12E-x\right)\cap F$ be pairwise disjoint then we'd have \begin{align*} \sum_{x\in\Lambda}m\left[\left(\frac12E-x\right)\cap F\right] &=m\left(\bigsqcup_{x\in\Lambda}\left(\frac12E-x\right)\cap F\right)\\ &=m\left(F\cap\bigsqcup_{x\in\Lambda}\left(\frac12E-x\right)\right)\\ &=m(F\cap\mathbb R^n)\\ &=m(F) \end{align*} and finally $m(F)<m(F)$, absurd!
The problem is I don't know how to prove (*), permitted it's true!
Can someone help me? Thank you all