Blichfeldt-Minkowski Lemma I'm trying to understand a proof of the following result

Theorem: Let $K$ be a number field, and $|| \cdot ||$ the idelic norm (product of the normalized absolute values at each place).  There exists a constant $C > 0$, depending only on $K$, such that if $\alpha \in \mathbb{I}_K$ with $||\alpha|| > C$, there exists an $x \in K^{\ast}$ such that $||x||_v \leq ||\alpha||_v$ at each place $v$.

found in Caessels and Frohlich (page 66).  The proof uses the existence of a Haar measure $\mu$ on $\mathbb{A}_K$ such that $\mu(\alpha E) = ||\alpha|| \mu(E)$.  That's fine, but it also resorts to computations on the measure of the compact group $\mathbb A_K/K$.  The problem?  They haven't said anything about what the measure on this quotient is or how it's normalized.  So I have no idea what they're talking about.  The only thing I know that might be of use is the following result I found on some notes on topological groups:

Lemma: Let $G$ be a locally compact abelian group with Haar measure $\mu$, and $H$ a closed subgroup with Haar measure $\lambda$.  If $f: G \rightarrow \mathbb{C}$ is continuous of compact support, then so is the function $\bar{f}: G/H \rightarrow \mathbb{C}$ given by $$\bar{f}(gH) = \int\limits_H f(gh) d \lambda(h)$$  Moreover, there exists a Haar measure $\bar{\mu}$ on $G/H$ such that $$\int\limits_{G/H} \bar{f} d \bar{\mu} = \int\limits_G f d \mu$$

 A: Note that $K$ is discrete in $\mathbb A$, so that $\mathbb A \to \mathbb A/K$ is a covering map.  (You should compare it with the map $\mathbb R \to \mathbb R/\mathbb Z$.)  
So a very small open set in $\mathbb A$ will map bijectively onto its image in $\mathbb A/K$, and give us an open subset of the latter space.
The measure on $\mathbb A/K$ is defined so that, in the context of the preceding sentence, the open set in $\mathbb A$ and its image in $\mathbb A/K$ have the same measure.   (Just as measure on $\mathbb R/\mathbb Z$ is defined so that a small interval in $\mathbb R$ and its image in $\mathbb R/\mathbb Z$ have the same measure.)
Another way to think about it is that we can find a fundamental domain for the translation action of $K$ on $\mathbb A$, which is say locally closed as a subset of $\mathbb A$, which maps bijectively onto $\mathbb A/K$, and restricting Haar measure on $\mathbb A$ to this fundamental domain gives a measure on this fundamental domain which we may then identify with a measure on $\mathbb A/K$.  (Just as the interval $[0,1)$ maps bijectively onto $\mathbb R/\mathbb Z$, and the measure on $\mathbb R/\mathbb Z$ may be identified with the usual Lebesgue measure on the interval $[0,1)$.) 
Note that the identification of $[0,1)$ with $\mathbb R/\mathbb Z$ is not a homeomorphism --- there are problems at the end-points --- but is a homeomorphism away from a subset of measure zero --- i.e. the end-points --- and this is good enough for studying the induced measure on the quotient.  The fundamental domain for $K$ in $\mathbb A$ behaves more-or-less the same way.  E.g. in the case $K = \mathbb Q$, we may identify $\mathbb A/\mathbb Q$ with $\mathbb R/\mathbb Z \times \widehat{\mathbb Z},$ and we can take a fundamental domain in $\mathbb A$ to be $[0,1) \times \widehat{\mathbb Z}$; so the analogy with the $\mathbb R/\mathbb Z$ situation becomes very tight.  The picture for general $K$ (say of degree $d$ over $\mathbb Q$) is then just a product of $d$ copies of this picture.  (I mention this to emphasize that you don't need to think purely abstractly about measures, locally compact groups, etc.;  the situation is actually very concrete.)
