Measure theoretic proof of $|\Bbb{Z}^d/A\Bbb{Z}^d| = |\det(A)|$ Let $A \in \Bbb{Z}^{d\times d}$ be an invertible matrix with entries in $\Bbb{Z}$.
It is well-known (and can be proved using algebraic properties of matrices) that the index of the group $A \Bbb{Z}^d \leq \Bbb{Z}^d$ in $\Bbb{Z}^d$ is given by 
$$|\Bbb{Z}^d/A\Bbb{Z}^d| = |\det(A)|. \qquad (\dagger)$$
Looking at the right hand side of this question, I was immediately reminded of the change of variables formula for the Lebesgue integral, e.g.
$$
\lambda_d (AM) = |\det(A)| \cdot \lambda_d(M),
$$
where $\lambda_d$ is the $d$-dimensional Lebesgue measure.
Thus, I want to know if there is a proof of $(\dagger)$ using this formula.
One thing I tried is to use that
$$
|\Bbb{Z}^d / A\Bbb{Z}^d| \equiv \sum_{k \in \Bbb{Z}^d} \chi_{A[0,1)^d + k} (x) \qquad \forall x \in \Bbb{R}^d,
$$
but since the corresponding integral diverges, the only hope would be to multiply this identity with an arbitrary function $f \in L^1$.
If I do this, I seem to be unable to show that
$$
\int  \sum_{k \in \Bbb{Z}^d} f(x) \cdot  \chi_{A[0,1)^d + k} (x) \, dx = |\det(A)| \cdot \int f \, dx.
$$
Any help would be appreciated.
 A: With a bit of help from a colleague of mine (i.e.: this is his proof :) ),
I arrived at the following solution:
Let $G:=A\mathbb{Z}^{d}$. The first fact that we have to use is that
any two measurable systems of representatives of $\mathbb{R}^{d}/G$
have equal measure. To see this, let $C,D\subset\mathbb{R}^{d}$ be
measurable with
$$
\biguplus_{n\in G}D+n=\mathbb{R}^{d}=\biguplus_{n\in G}C+n.
$$
This implies
$$
D=\biguplus_{n\in G}\left[D\cap\left(C+n\right)\right]\qquad\text{ and }\qquad C=\biguplus_{n\in G}\left[C\cap\left(D+n\right)\right].
$$
Using the translation invariance of the Lebesgue measure $\lambda$
on $\mathbb{R}^{d}$, we derive
\begin{eqnarray*}
\lambda\left(D\right) & = & \sum_{n\in G}\lambda\left(D\cap\left(C+n\right)\right)\\
 & = & \sum_{n\in G}\lambda\left(\left[D\cap\left(C+n\right)\right]-n\right)\\
 & = & \sum_{n\in G}\lambda\left(\left[\left(D-n\right)\cap C\right]\right)\\
 & \overset{k=-n}{=} & \sum_{k\in G}\lambda\left(\left[\left(D+k\right)\cap C\right]\right)\\
 & = & \lambda\left(C\right).
\end{eqnarray*}
Here, we made crucial use of the fact that $G=A\mathbb{Z}^{d}$ is
countable.
We now need to fine two measurable systems of representatives of $\mathbb{R}^{d}/G$
to which we can apply the preceding claim. We first note that
$$
\mathbb{R}^{d}=\biguplus_{k\in\mathbb{Z}^{d}}\left(\left[0,1\right)^{d}+k\right)\text{ and hence }\mathbb{R}^{d}=A\mathbb{R}^{d}=\biguplus_{k\in\mathbb{Z}^{d}}A\left(\left[0,1\right)^{d}+k\right)=\biguplus_{n\in G}\left(A\left[0,1\right)^{d}+n\right).
$$
In the last step we used the fact that $A$ is invertible. Hence,
$C:=A\left[0,1\right)^{d}$ is a measurable system of representatives
of $\mathbb{R}^{d}/G$.
Since $\mathbb{Z}^{d}/G$ is countable, there is a countable system
of representatives $\left(y_{i}\right)_{i\in I}$ of $\mathbb{Z}^{d}/G$,
i.e. $I$ is countable with
$$
\mathbb{Z}^{d}=\biguplus_{i\in I}\left(G+y_{i}\right).\qquad\left(\dagger\right)
$$
This implies that $D:=\biguplus_{i\in I}\left(\left[0,1\right)^{d}+y_{i}\right)\subset\mathbb{R}^{d}$
is measurable (because $I$ is countable). Furthermore,
$$
\mathbb{R}^{d}=\biguplus_{k\in\mathbb{Z}^{d}}\left(\left[0,1\right)^{d}+k\right)\overset{\left(\dagger\right)}{=}\biguplus_{n\in G}\biguplus_{i\in I}\left(\left[0,1\right)^{d}+n+y_{i}\right)=\biguplus_{n\in G}\left(D+n\right).
$$
All in all, we finally conclude
$$
\left|\mathbb{Z}^{d}/A\mathbb{Z}^{d}\right|=\left|\mathbb{Z}^{d}/G\right|=\left|I\right|=\sum_{i\in I}\lambda\left(\left[0,1\right)^{d}+y_{i}\right)=\lambda\left(D\right)=\lambda\left(C\right)=\left|\det\left(A\right)\right|.
$$
