Theorem: Minkowski's Theorem

Let $L$ be a lattice in $\R^n$.
Let $d$ be the covolume of $L$.
Let $\mu$ be a translation invariant measure on $\R^n$
Let $S$ be a convex subset of $\R^n$ that is symmetric about the origin, i.e. such that:
:$\forall p \in S : -p \in S$ 
Let the volume of $S$ be greater than $2^n d$.
Then $S$ contains a non-zero point of $L$.

Proof:
Let $D$ be any fundamental parallelepiped.
Then by definition:
:$\ds \R^n = \coprod \limits_{\vec x \mathop \in L} \paren {D + \vec x}$
where:
: $A + \vec x := \set {\vec a + \vec x : \vec a \in A}$
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:
:$(1): \quad \ds \frac 1 2 S = \coprod \limits_{\vec x \mathop \in L} \paren {\frac 1 2 S \cap \paren {D + \vec x} }$
where:
: $\dfrac 1 2 S := \set {\dfrac 1 2 \vec s: \vec s \in S}$
{{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:
:$\ds \frac 1 2 S \cap \paren {D + \vec x} = \paren {\paren {\frac 1 2 S - \vec x} \cap D} + \vec x$
Since, {{hypothesis}}, $\mu$ is translation invariant:
{{begin-eqn}}
{{eqn | n = 2
      | l = \map \mu {\frac 1 2 S \cap \paren {D + \vec x} }
      | r = \map \mu {\paren {\paren {\frac 1 2 S - \vec x} \cap D} + \vec x}
}}
{{eqn | r = \map \mu {\paren {\frac 1 2 S - \vec x} \cap D}
}}
{{end-eqn}}
{{AimForCont}} the sets $\ds \set {\frac 1 2 S - \vec x: \vec x \in L}$ are pairwise disjoint.
That is:
:$\ds \forall \vec x, \vec y \in L: \paren {\frac 1 2 S - \vec x} \cap \paren {\frac 1 2 S - \vec y} \ne \O \iff \vec x \ne \vec y$
Then:
{{begin-eqn}}
{{eqn | l = \map \mu S
      | r = 2^n \map \mu {\frac 1 2 S}
      | c = Dilation of Lebesgue-Measurable Set is Lebesgue-Measurable
}}
{{eqn | r = 2^n \map \mu {\coprod \limits_{\vec x \mathop \in L} \paren {\frac 1 2 S \cap \paren {D + \vec x} } }
      | c = from $(1)$ above
}}
{{eqn | r = 2^n \map \mu {\coprod \limits_{\vec x \mathop \in L} \paren {\paren {\frac 1 2 S - \vec x} \cap D} }
      | c = from $(2)$ above
}}
{{eqn | r = 2^n \sum \limits_{\vec x \mathop \in L} \map \mu {\paren {\frac 1 2 S - \vec x} \cap D}
      | c = {{Defof|Measure (Measure Theory)|Measure}}
}}
{{eqn | r = 2^n \map \mu D
      | o = \le
      | c = Measure is Countably Subadditive
}}
{{eqn | r = 2^n d
      | c =
}}
{{end-eqn}}
which is a contradiction.
So $\ds \set {\frac 1 2 S - \vec x: \vec x \in L}$ are not pairwise disjoint.
This means:
:$\ds \exists \vec x, \vec y \in L: \vec x \ne \vec y, \paren {\frac 1 2 S - \vec x} \cap \paren {\frac 1 2 S - \vec y} \ne \O$
Therefore there exist $\vec {p_1}, \vec {p_2} \in L$ such that $\vec {p_1} \ne \vec {p_2}$ and:
:$\ds \frac 1 2 \vec {p_1} - \vec x = \frac 1 2 \vec {p_2} -  \vec y$
and therefore:
:$\ds \frac 1 2 \paren {\vec {p_1} - \vec {p_2} } = \vec x - \vec y \in L$
Since $S$ is convex, we have that:
:$\dfrac 1 2 \paren {\vec {p_1} - \vec {p_2} } \in S$
As $\vec {p_1} \ne \vec {p_2}$ by definition:
:$\dfrac 1 2 \paren {\vec {p_1} - \vec {p_2} } \ne \vec 0$
Hence the result.
{{qed}}
{{Namedfor|Hermann Minkowski|cat = Minkowski}}