Proof that all primitive cells have the same size A primitive cell of a crystal lattice is a set $A$ such that two copies of $A$ which are translated by a lattice vector do not overlap and such that $A$ tiles the entire crystal.

I have read (for example in the german “Festkörperphysik” by Gross, Marx), that all primitive cells have the same size/volume.
Intuitively, this seems plausible, but is there a proof?
My precise, measure theoretic, interpretation of this statement is:
If $a_1, \ldots, a_n$ is a basis of $\mathbf{R}^n$ and $A, B \subset \mathbf{R}^n$ are sets such that $(\cup (A+\alpha_1 a_1+\ldots+\alpha_n a_n))C$ and $(\mathbf{R}^n \cup (B+\alpha_1 a_1+\ldots+\alpha_n a_n))^C$ (where the union is over all $\alpha_1, \ldots,\alpha_n ∈ \mathbf{Z}$) are Lebesgue null sets and such that for all $\alpha_1,\ldots,\alpha_n∈\mathbf{Z}$: $(A+\alpha_1 a_1 + \ldots \alpha_n a_n) \cap A$ and $(B+\alpha_1 a_1 + \ldots \alpha_n a_n) \cap B$ are Lebesgue null sets, then $A$ and $B$ have the same Lebesgue measure.
 A: Let $\mathcal{L}\subseteq\mathbb{R}^n$ be a lattice with a basis $B\in\mathcal{R}^{n\times n}$ and $F\subseteq\text{span}(\mathcal{L})$ be measurable. $F$ tiles $\mathcal{L}$ iff


*

*$(x+F)\cap(y+F)=\emptyset\,\forall x\neq y\in\mathcal{L}$, and

*$\mathcal{L}+F=\text{span}(\mathcal{L})$


It is trivial to show (I'll leave it as an exercise) that 1. implies:
$$ |(\mathcal{L}+x)\cap F|\leq 1 $$
while 2. implies
$$ |(\mathcal{L}+x)\cap F|\geq 1 $$
and therefore
$$ |(\mathcal{L}+x)\cap F|= 1 $$
for $x\in\text{span}(\mathcal{L})$. Then we have
\begin{align} \text{vol}(F) &= \int_{\mathbb{R}^n} 1_F(x)dx \\
&= \int_{B[0,1)^n}\sum_{y\in\mathcal{L}} 1_F(x+y)dx\\
&=\int_{B[0,1)^n}|(\mathcal{L}+x)\cap F|dx=\text{vol}(B[0,1)^n) \end{align}
A: The quotient forming map $\Bbb R^n\to\Bbb R^n/\Lambda$ is a local isometry (as translations by elements of the lattice $\Lambda$ are isometries without fixed points) onto a torus, whose volume is equal to the (absolute value of the) determinant of the basis of $\Lambda$. 
The preimage of each point of the torus has exactly one point in the primitive cell, by definition of the primitive cell, so (as soon as the primitive cell has a well-defined volume) its volume must be equal to that of the torus.
