Let $D$ be any fundamental parallelepiped. Then by definition: :$\displaystyle \R^n = \coprod \limits_{\vec x \mathop \in L} \left({D + \vec x}\right)$ where: : $A + \vec x := \left\{{\vec a + \vec x : \vec a \in A}\right\}$ By Intersection with Subset is Subset: :$\dfrac 1 2 S \cap \R^n = \dfrac 1 2 S \iff \dfrac 1 2 S \subseteq \R^n$ Hence by Intersection Distributes over Union: :$(1): \quad \displaystyle \frac 1 2 S = \coprod \limits_{\vec x \mathop \in L} \left({\frac 1 2 S \cap \left({D + \vec x}\right)}\right)$ where: : $\dfrac 1 2 S := \left\{{\dfrac 1 2 \vec s: \vec s \in S}\right\}$ {{explain|The translation and scaling operations need to link to their definitions on ProofWiki}} Consider the intersection of $D + \vec x$ and $\dfrac 1 2 S$. This is obtained by adding $\vec x$ to every point in $D$ then taking those points that are also in $S$. However, this is the same as subtracting $\vec x$ from all elements of $\dfrac 1 2 S$, taking the elements that are also in $D$ and adding $\vec x$ to restore them to their original position. Thus: :$\displaystyle \frac 1 2 S \cap \left({D + \vec x}\right) = \left({\left({\frac 1 2 S - \vec x}\right) \cap D}\right) + \vec x$ Since, by hypothesis, $\mu$ is translation invariant: {{begin-eqn}} {{eqn | n = 2       | l = \mu \left({\frac 1 2 S \cap \left({D + \vec x}\right)}\right)       | r = \mu \left({\left({\left({\frac 1 2 S - \vec x}\right) \cap D}\right) + \vec x}\right) }} {{eqn | r = \mu \left({\left({\frac 1 2 S - \vec x}\right) \cap D}\right) }} {{end-eqn}} Aiming for a contradiction, suppose the sets $\displaystyle \left\{{\frac 1 2 S - \vec x: \vec x \in L}\right\}$ are pairwise disjoint. That is: :$\displaystyle \forall \vec x, \vec y \in L: \left({\frac 1 2 S - \vec x}\right) \cap \left({\frac 1 2 S - \vec y}\right) \ne \varnothing \iff \vec x \ne \vec y$ Then: {{begin-eqn}} {{eqn | l = \mu \left({S}\right)       | r = 2^n \mu \left({\frac 1 2 S}\right)       | c = Dilation of Lebesgue-Measurable Set is Lebesgue-Measurable }} {{eqn | r = 2^n \mu \left({\coprod \limits_{\vec x \mathop \in L} \left({\frac 1 2 S \cap \left({D + \vec x}\right)}\right)}\right)       | c = from $(1)$ above }} {{eqn | r = 2^n \mu \left({\coprod \limits_{\vec x \mathop \in L} \left({\left({\frac 1 2 S - \vec x}\right) \cap D}\right)}\right)       | c = from $(2)$ above }} {{eqn | r = 2^n \sum \limits_{\vec x \mathop \in L} \mu \left({\left({\frac 1 2 S - \vec x}\right) \cap D}\right)       | c = {{Defof|Measure}} }} {{eqn | r = 2^n \mu \left({D}\right)       | o = \le       | c = Measure is Countably Subadditive }} {{eqn | r = 2^n d       | c = }} {{end-eqn}} which is a contradiction. So $\displaystyle \left\{{\frac 1 2 S - \vec x: \vec x \in L}\right\}$ are not pairwise disjoint. This means: :$\displaystyle \exists \vec x, \vec y \in L: \vec x \ne \vec y, \left({\frac 1 2 S - \vec x}\right) \cap \left({\frac 1 2 S - \vec y}\right) \ne \varnothing$ Therefore there exist $\vec {p_1}, \vec {p_2} \in L$ such that $\vec {p_1} \ne \vec {p_2}$ and: :$\displaystyle \frac 1 2 \vec {p_1} - \vec x = \frac 1 2 \vec {p_2} -  \vec y$ and therefore: :$\displaystyle \frac 1 2 \left({\vec {p_1} - \vec {p_2} }\right) = \vec x - \vec y \in L$ Since $S$ is convex, we have that: : $\dfrac 1 2 \left({\vec {p_1} - \vec {p_2} }\right) \in S$ As $\vec {p_1} \ne \vec {p_2}$ by definition: : $\dfrac 1 2 \left({\vec {p_1} - \vec {p_2} }\right) \ne \vec 0$ Hence the result. {{qed}} {{namedfor|Hermann Minkowski|cat = Minkowski}} Category:Number Theory Category:Geometry of Numbers 2ke2lgagmchy78lj04fjxykx8b1c7rc