Introduction

Let $p$ be a prime number. Then the space of $p$-adic numbers $\mathbb{Q}_p$ is a locally compact abelian group. This can be observed through the local basis

where $|\cdot|_p$ is the $p$-adic norm such that, whenever we write $r=p^mq$ such that $q$ is prime to $p$, we have $|r|_p=p^{-m}$.

We remind the reader that every locally compact abelian group $G$ admits a Haar measure, which is unique up to a scalar multiplication (proof). In this post, we try to find the Haar measure on $\mathbb{Q}_p$, which makes it possible to do harmonic analysis on it. For this reason, in future posts, we also find the dual group of $\mathbb{Q}_p$ as well as the dual measure.

Haar Measure

Let us first recall the basic structure of $\mathbb{Q}_p$. Every element is in the form of Laurent series

where $m \in \mathbb{Z}$ and $c_j \in \{0,\dots,p-1\}$. The ring of integers $\mathbb{Z}_p$ is exactly the closed disc of radius $1$ at the origin. That is, $\mathbb{Z}_p=\overline{B}(0,1)$ is a compact set. Let $\mu$ be an arbitrary Haar measure on $\mathbb{Q}_p$. Then $\mu(\mathbb{Z}_p)$ is non-zero and finite. We can therefore put

Then in particular $m_p(\mathbb{Z}_p)=1$. This is the canonical Haar measure we are looking for. But it would be hilarious to end the post here. We will give a closer look at it, at least on a $p$-adic level.

Recall that when studying the Lebesgue measure on $\mathbb{R}$ we have encountered some definition in the form of

where the infimum is taken over all countable collections of open intervals $\{I_j\}$ such that $\bigcup_j I_j \supset E$, and $\ell(I_j)$ is the length of $I_j$. In fact, we can actually write

On $\mathbb{Q}_p$, we write

The point here is how to express $V$. For this reason we need to recall some topology of $\mathbb{Q}_p$.

Some p-adic topology

$\mathbb{Q}_p$ is a separable metric space. Therefore every open set $V$ is a union of open balls.

There is nothing special about this statement. The space has already been equipped with a norm. Besides, as $\mathbb{Q}$ is dense in $\mathbb{Q}_p$, we have nothing to worry about second countability.

Every closed ball of $\mathbb{Q}_p$ is open (hence we call them “balls” thereafter). Every point in the ball is a “centre”. If two balls intersect then one is contained in the other.

This is dramatically different from our understanding of $\mathbb{R}$ or $\mathbb{C}$. Notice that the $p$-adic norm $|\cdot|_p$ only takes the values from $p^k$ with $k \in \mathbb{Z}$ or $0$. For any $r>0$, there exists some $\varepsilon>0$ such that

The clopenness of balls in $\mathbb{Q}_p$ follows.

Next, recall that $|\cdot|_p$ is non-Archimedean. Consider $y \in \overline{B}(x,r)$. It follows that $|x-y|_p=|y-x|_p \le r$. On the other hand, for any $z \in \overline{B}(x,r)$, we have $|x-z|_p \le r$. Therefore $|y-z|_p \le r$. Hence $\overline{B}(x,r)\subset \overline{B}(y,r)$. Symmetrically we see $\overline{B}(y,r) \subset \overline{B}(x,r)$. Hence they are equal.

Let $\overline{B}(x,r)$ and $\overline{B}(x’,r’)$ be two balls that intersect, and without loss of generality we assume that $r \le r’$. Let $y$ be a point in the intersection, then we see

So far so good. We next try to compute the Haar measure of every ball.

Measure of a ball

Every ball of radius $p^k$ has measure $p^k$ ($k \in \mathbb{Z}$).

First of all notice that $\overline{B}(0,1)=\mathbb{Z}_p$, and we defined $m_p$ so that $m_p(\mathbb{Z}_p)=1$. Therefore every ball of the form $\overline{B}(x,1)$ has measure $1$. Next, notice that $\overline{B}(0,p^k)=p^{-k}\mathbb{Z}_p$ for all $k \in \mathbb{Z}$, it is necessary to unwind $\mathbb{Z}_p$ a little bit more.

We have

Therefore $\mathbb{Z}_p$ is a disjoint union of $p^k$ balls of radius $p^{-k}$ when $k>0$. Hence in this case,

as expected. In other words, for $k<0$, the ball $\overline{B}(0,p^k)$ has measure $p^k$.

For the counterpart, we notice that

which is to say $\overline{B}(0,p^k)=p^{-k}\mathbb{Z}_p$ is a disjoint union of $p^k$ balls of radius $1$. Hence its measure is $p^k$. This concludes our computation of balls in $\mathbb{Q}_p$.

Back to the remaining problem

Now we come back to the definition of $m_p$. Now every open set $V$ can be written in the form

The union is countable because $\mathbb{Q}_p$ is second countable. By combining intersecting balls, we can assume that the union is also disjoint. It follows that

Note: this should be understood in the sense of real series, instead of $p$-adic number, because $m_p$ takes the values in $\mathbb{R}$. So for an arbitrary measurable set, we have