In Goldfeld's book, we find the following integrability criterion.
Suppose $f = \prod_v f_v$ an adelic function $\mathbb{A} \to \mathbb{C}$ such that $f_v \equiv 1$ on $\mathbb{Z}_p$ for almost every prime $p$. Suppose $f_\infty$ is integrable on $\mathbb{R}$, $f_p$ locally constant and integrable. Then $$\int_{\mathbb{A}} f(x)dx = \int_{\mathbb{R}} f_\infty(x_\infty) dx_\infty \lim_{N \to \infty} \prod_{p < N} \int_{\mathbb{Q}_p} f_p(x_p)dx_p$$ as soon as the limit of $$\prod_{p<N} \int_{\mathbb{Q}_p} |f_p(x_p)| dx_p$$ exists.
Here, $f_p$ is not assumed to be the characteristic function of $\mathbb{Z}_p$, but just $\equiv 1$ on $\mathbb{Z}_p$. How can this be valid?
My problem is that $\mathbb{A}$ is not $\prod_v \mathbb{Q}_v$, since it requires to be in $\mathbb{Z}_p$ for almost all $p$, but his criterion writes the integral over $\mathbb{A}$ as a limit of products over $\mathbb{Q}_p$, without anything insuring to be in $\mathbb{Z}_p$ for almost all $p$. Could the convergence condition ensure that the difference between both, in volume, is zero?
If $\int_{\mathbb{A}} |f(x)|dx<\infty$ and $f_p=1$ on $\Bbb{Z}_p$ for all but finitely many $p$ then $$\int_{\mathbb{A}} f(x)dx= \lim_{N\to \infty} \int_{\Bbb{R}\times \prod_{p< N} \Bbb{Q}_p \times \prod_{p\ge N} \Bbb{Z}_p} f(x)dx$$ $$= \lim_{N\to \infty}( \int_\Bbb{R}f_\infty(x)dx)( \prod_{p< N} \int_{\Bbb{Q}_p} f_p(x)dx ) (\prod_{p \ge N} \int_{\Bbb{Z}_p} f(x)dx) $$ $$ = \int_{\mathbb{R}} f_\infty(x_\infty) dx_\infty \lim_{N \to \infty} \prod_{p < N} \int_{\mathbb{Q}_p} f_p(x_p)dx_p$$