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Let $f:\mathbb Z^d \rightarrow \mathbb Z^d$ be a linear map having determinant 1. Is there an obvious way to see that if $U\subseteq \mathbb Z_p^d$ is a measurable set, then the p-adic measure of $U$ is equal to the p-adic measure of $f(U)$? I mean, without involving change of coordinates and jacobians.

Many thanks!

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$\def\Gl{\mathop{\rm Gl}\nolimits}$ $\newcommand{\Zp}{{\mathbb{Z}_p}}$ I assume that you meant $f:\Zp^d\rightarrow \Zp^d$ with determinant of norm 1.

This is only a sketch of a proof.

I endow $\Zp^d$ with the unique Haar measure, calibrated such that $\mu(\Zp^d)=1$. Then as every $f\in\Gl_d(\Zp)$ is continuous, the measure defined by $\mu_f:=\mu\circ f$ is also a Haar measure, thus only differs to $\mu$ by a constant. By bijectivity $f(\Zp^d)=\Zp^d$ thus $\mu = \mu_f$. The elements $f\in\Gl_d(\Zp)$ are exactly those with $|\det f|_p = 1$.

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