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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

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1 Answer 1

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The answer goes like this:

Take any index set $\mathcal{I}$ and for each $i \in \mathcal{I}$, an LCH group $G_i$ with a compact, open subgroup $K_i$. Let $\mathcal{B}(\cdot)$ denote the Borel-Sigma-Algebra, i.e. the Sigma-Algebra generated by the open sets. A Haar-measure on an LCH Group $G$ is a measure $\mu : \mathcal{B}(G) \to [0, \infty]$ such that

1) $\mu(gM) = \mu(M)$ for all $g \in G, M \in \mathcal{B}(G)$

2) $\mu(M) = \inf\{ \mu(O) : M \subset O, O \text{open}\}$ for all $M \in \mathcal{B}(G)$

3) $\mu(M) = \sup\{ \mu(K) : K \subset M, K \text{compact}\}$ not for all $M$ but for those $M$ which are open or have finite measure.

There is different notions for a Haar-measure. This one is used by [F] = "Folland, Real Analysis, Modern Techniques and their Applications" for example. Take nontrivial Haar-measures $\mu_i$ on every $G_i$. Assume further that $\mu_i(K_i) \neq 0$ so we can normalize and assume $\mu_i(K_i) = 1$. By [F], Thm 7.28 we get a measure $\mu$ on the product $K = \prod K_i$ that satisfies

$$\mu(\prod_{i \in \mathcal{I}} M_i) = \prod_{i \in \mathcal{I}} \mu_i(M_i)$$

for all sets $M$ such that $M_i = K_i$ for almost all $i$. Let us further assume that all $G_i$ are $\sigma$-finite with respect to their measure. Then we can form finite product measures, see [F], Thm 7.27. For a finite set $F \subset \mathcal{I}$ put

$$G_F := \prod_{i \in F} G_i \times \prod_{i \notin F} K_i$$

The groups $G_F$ are open subgroups of the LCH-group $G$, so by [F], Thm 7.27 the product measure is the nontrivial Haar-measures $\mu_F$ on $\mathcal{B}(G_F)$. We also have the choice to get another Haar-measure $\nu_F := \mu|_{\mathcal{B}(G_F)}$. As generally, Haar measures on LCH groups are unique up to normalization, for each $F$ there is a $c_F$ such that $\nu_F = c_F \mu_F$. Selecting for every $i \in F$ a set $A_i \in \mathcal{B}(G_i)$ such that $0 < \mu_i(A_i) < \infty$, and measuring the set $M = \prod_{i \in F} A_i \times \prod_{i \notin F} K_i$ in the equation

$$ c_F \mu_F = \nu_F = \mu|_{\mathcal{B}(G_F)} = \mu|_{\mathcal{B}(G_{E \cup F})} |_{\mathcal{B}(G_F)} = c_{E \cup F} \mu_{E \cup F} |_{\mathcal{B}(G_F)}$$

one gets $c_F = c_{E \cup F}$ for any other finite set $E \subset \mathcal{I}$. Hence, the $c_F$ does actually not depend on the $F$ and hence we can put $\tilde{\mu} = \frac{1}{c} \mu$ and this measure has (by construction) the desired property.

The product-integral-formula $$\int_G \prod f_i = \prod \int_{G_i} f_i $$ for measurable $f_i$ with $f_i = 1_{K_i}$ for almost all $i$, now is a direct consequence:

The property of the measure shows that it holds for characteristic functions. By a straightforward calculation, one shows that the product formula holds for "step functions", i.e. functions $f = \prod f_i$ such that $f_i = 1_{K_i}$ for almost all $i$ and the rest of the $f_i$ being step functions. For a general nonnegative function $f = \prod f_i$ one chooses monotone step functions $f_{i, v} \to f_i$ for those finitely many $f_i \neq 1_{K_i}$, then $f_v = \prod f_{i, v} \cdot \prod_{i \notin F} f_i$ converges monotonously to $f$, so by interchanging the limit (Thm of monotone convergence) and applying the above for "step functions", one gets the desired result.

Fabian Werner

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