Let $\mathbb{A}$ denote the ring of adeles of $\mathbb{Q}$, let $\mu$ be the Haar measure of $\mathbb{A}$, and let $\|\cdot\|_{\infty}$ denote the sup-norm of the components in the Archimedean component of $\mathbb{A}^m$ (that is $\mathbb{R}^m$), $m\ge 1$.

For any subset $E\subseteq \mathbb{A}^m$ and any $B>0$, we denote $E(B):=E\cap \{\mathbf{x}\in\mathbb{A}^m\,:\,\|\mathbf{x}\|_{\infty}\le B\}$.

Consider also $\mathbb{Z}^m$ embedded diagonally in $\mathbb{A}^m$, and let $\Lambda(B):=\mathbb{Z}^m\cap E(B)$.

In analogy to what happens in Euclidean spaces, since we still have a natural (uniquely) ergodic action, I would expect that if $E$ is somehow "nicely shaped" then one has $$|\#\Lambda(B)-\mu^m(E(B))|=o(\mu^m(E(B)))$$ as $B\to \infty$.

Are you aware of any sufficient (not obvious) condition on $E$ to ask to make this happen?

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    $\begingroup$ From some viewpoints, the thing like a "lattice" in $\mathbb A^n$ would be $\mathbb Q^n$ here... ??? And for a finite-covolume discrete subgroup $\Gamma$ of a unimodular topological group $G$, a measure-theoretic analogue of Minkowski's lemma holds. Of course, $\mathbb Z^n$ is not finite covolume... $\endgroup$ – paul garrett Jun 29 '19 at 17:15

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