Since the adeles $\mathbb{A}$ (with addition) are a locally compact Hausdorff topological group there exists a Haar measure $\mu$. Now people claim that it can be normalized such that for every function of the form $f = \prod_p f_p$ such that $f_p = \mathbb{1}_{Z_p}$ for almost all $p$ and $f_p$ integrable for all the rest, one has
$\int_{\mathbb{A}} f = \prod_p \int_{Q_p} f_p(x_p) dx_p$
where $dx_p$ is the additive Haar measure on $Q_p$, normalized such that the measure of $Z_p$ is one (the lebesgue measure up to some factor for $p=\infty$ respectively).
My question is: why is this the case?
In one of the books i tried to find the answer in, the author proceeds as follows: he defines simple sets to be sets of the form $M = \prod_p M_p$ where $M_p = Z_p$ for almost all $p$ and $M_p = U_p$ is open in $Q_p$ for all the rest. Then he defines another measure $\nu(M) := \prod_p \mu_p(M_p)$ where $\mu_p$ is the additive Haar measure on $Q_p$. Since simple sets are stable under finite intersections, one can continue this to a measure on the whole Borel-$\sigma$-algebra. The question here is: why is it a Radon measure, i.e. one has to show that $\nu(K) < \infty$ for compact $K$ and that it is outer regular for all borel sets and inner regular for open sets and sets of finite measure. How to do that?
I guess that there is a better way to achieve this by starting with the abstract nice measure $\mu$ and then show that the relation above holds up to a factor. Does somebody know where to find that or can somebody point out a basic reference for the construction of the Haar-measure on the adeles?
Thanks,
Fabian Werner