Let $G$ be a locally compact abelian group. A *lattice* in $G$ is a discrete subgroup $\Lambda$ such that the quotient $G / \Lambda$ is compact. A *Borel fundamental domain* of a lattice $\Lambda$ in $G$ is a Borel set $F \subseteq G$ that intersects each coset in $G / \Lambda$ in exactly one point, i.e., $F$ is a set of coset representatives.


My question is now the following: 


Given two lattices $\Lambda\_1, \Lambda\_2 \subseteq G$ such that $\Lambda\_1 \subseteq \Lambda 2$, is it always possible to take corresponding fundamental domains $F\_i$ of $\Lambda\_i$, where $i = 1,2$, such that $F\_2 \subseteq F\_1$?


By looking at special cases in, say, $G=\mathbb{R}$ or $G = \mathbb{Z}$, it seems that this is always possible.


Any comment or reference is highly appreciated.



Inclusion of lattices and fundamental domains
The image of $\Lambda\_2$ in $G/\Lambda\_1$ is finite. Take a set of representatives $s\_1, s\_2, \ldots s\_n$ in $\Lambda\_2$. Now consider $\cup s\_i F\_2$. I claim this is a Borel fundamental domain for $F\_1$.


It is Borel as it is a finite union of Borel sets. Let $g\Lambda\_1$ be a coset of $\Lambda\_1$. Then $g\Lambda\_2$ intersects $F\_2$ in a single point $x$. Now, $\Lambda\_2 = \cup s\_i \Lambda\_1$, and so $x$ is in some $s\_i$ translate of $F\_2$, and hence in $F\_1$. It is unique as otherwise we have $s\_i \lambda\_1 = s\_j \lambda\_1'$, which would mean our set of representatives was too big in the first place.


