Haar measure on the p-adic numbers

Example

Let $p$ be a prime, and let $\Q_p$ be the field of p-adic numbers with the p-adic absolute value $\abs{\cdot}_p$. Let $\Z_p = \set{x : \abs{x}_p \le 1}$ be the ring of p-adic integers. The additive group $(\Q_p, +)$ is locally compact, and $\Z_p$ is a compact open subgroup. By the existence and uniqueness of Haar measure, there is a unique Haar measure $\mu$ on $(\Q_p, +)$ with $\mu(\Z_p) = 1$.

Its values on balls follow from invariance alone. For $k \ge 0$, the subgroup $p^k \Z_p$ has index $p^k$ in $\Z_p$, so $\Z_p$ is a disjoint union of $p^k$ translates of $p^k \Z_p$. These translates have equal measure, hence $\mu(a + p^k \Z_p) = p^{-k}$ for every $a \in \Q_p$. For $k < 0$, the ball $p^k \Z_p$ is a disjoint union of $p^{-k}$ translates of $\Z_p$, so again $\mu(a + p^k \Z_p) = p^{-k}$. Since $\abs{p^k}_p = p^{-k}$, the ball $\set{x : \abs{x - a}_p \le p^{-k}}$ has measure equal to its radius $p^{-k}$.

More generally, $\mu(cA) = \abs{c}_p \, \mu(A)$ for every $c \in \Q_p$ with $c \ne 0$ and every Borel set $A$.