Haar measure of $p$-adic integers with valuation 0 modulo 3

Question

I am reading (trying to read) this paper, when I came across some statement regarding the Haar measure of $p$-adic integers (it appears in the proof of Lemma 2.5). The statement is the following:

$$ \mu(\{ x \in \mathbb{Z}_p : 3|v_p(x) \}) = \frac{1-1/p}{1-(1/p)^3}$$

where $\mu$ is the (normalised) Haar measure in $\mathbb{Z}_p$.

I have tried proving it myself, but I don't seem to get anywhere. I am aware of the basic facts of p-adic measures, such as: $$ \mu (p\mathbb{Z}_p) = \frac{1}{p} \mu(\mathbb{Z}_p),$$ but not much more than that. I was wondering if there is any clever way of partitioning the space that would yield the desired measure, or if this result requires more complicated machinery. Would anyone mind sharing some helpful tips?

Thanks in advance

Answer 1

Let's first describe $\{ x \in \mathbb{Z}_p \ : \ 3 | v_p(x)\}$. We have \begin{align} \{ x \in \mathbb{Z}_p \ : \ 3 | v_p(x)\} &= \coprod_{n \in \mathbb{N}}\{ x \in \mathbb{Z}_p \ : \ v_p(x)=3n\} \\ &= \coprod_{n \in \mathbb{N}}(p^{3n}\mathbb{Z}_p \backslash p^{3n+1}\mathbb{Z}_p ) \end{align} Note that $\mu(p^{3n}\mathbb{Z}_p \backslash p^{3n+1}\mathbb{Z}_p)=\mu(p^{3n}\mathbb{Z}_p)-\mu(p^{3n+1}\mathbb{Z}_p)=\frac{1}{p^{3n}}-\frac{1}{p^{3n+1}} $. Thus \begin{align} \mu(\{ x \in \mathbb{Z}_p \ : \ 3 | v_p(x)\}) &= \sum_{n=0}^{\infty}\left(\frac{1}{p^{3n}}-\frac{1}{p^{3n+1}} \right) \\ &=\left(1-\frac{1}{p}\right)\sum_{n=0}^{\infty}\frac{1}{p^{3n}} \\ &= \frac{1-1/p}{1-(1/p)^3} \end{align}

Giving the desired result.

Answer 2

A similar approach is to let the self-similarity of the p-adics do the work for you, think about the geometric series.

$$\mu(\{x \in \mathbb{Z}_p : 3 |v_p(x) \}) = \mu(\mathbb{Z}_p\setminus p\mathbb{Z}_p) + \mu(\{x \in \mathbb{Z}_p : 3 |v_p(x), v_p(x) > 0 \}) $$

So here I've pulled out the first set with $v_p(x)=0$ and left everything with a higher power. I will now simplify that first bit as just $1-p^{-1}$ and the second bit is just $p^{-3}$ times the original measure. We just factor it out of all our terms using that elementary fact that we know, $\mu(p^{3+n}\mathbb{Z}_p) = p^{-3} \mu(p^n \mathbb{Z}_p)$ since we know it must be just some linear combination of terms of that form.

$$\mu(\{x \in \mathbb{Z}_p : 3 |v_p(x) \}) = 1- p^{-1} + p^{-3} \mu(\{x \in \mathbb{Z}_p : 3 |v_p(x) \})$$

At this point it's pretty trivial to rearrange but I'll do it anyways cause it's fun.

$$\mu(\{x \in \mathbb{Z}_p : 3 |v_p(x) \}) - p^{-3} \mu(\{x \in \mathbb{Z}_p : 3 |v_p(x) \}) = 1- p^{-1} $$

$$\mu(\{x \in \mathbb{Z}_p : 3 |v_p(x) \}) (1- p^{-3}) = 1- p^{-1} $$

$$\mu(\{x \in \mathbb{Z}_p : 3 |v_p(x) \}) = \frac{1- p^{-1}}{1- p^{-3}} $$