Hi,
 Consider the set $\lambda P\_n \subset \mathbb{Q}\_p$, where $\lambda \in \mathbb{Q}\_p^{\times}$ and $P\_n$ is the set $\lbrace x \in \mathbb{Q}\_p \mid \exists y \in \mathbb{Q}\_p x= y^n\rbrace$. Is there a way to measure $\lambda P\_n \cap B(r)$ for $r$ in $\mathbb{R}\_{>0}$ the ball of radius $r$ (with Haar measure)?


Also, is $\mathbb{Q}\_p^{\times}/ P\_n^{\times}$ finite? what is its value?
 Thanks.



The set of $p$-adic numbers of some fixed $n^ {\rm th}$-power residue 
What you are denoting $P\_n$ is more usually denoted $\mathbf{Q}\_p^{\times n}$, the image of the endomorphism $(\ )^n$ of $\mathbf{Q}\_p^\times$. 


To compute the quotient $\mathbf{Q}\_p^\times/\mathbf{Q}\_p^{\times n}$, recall that there is an exact sequence 
$$
1\to\mathbf{Z}\_p^\times\to\mathbf{Q}\_p^\times\to\mathbf{Z}\to1
$$  

which can be split by sending $1\in\mathbf{Z}$ to any uniformiser $\pi$ (such as $\pi=p$) of $\mathbf{Q}\_p$. Further, there is an exact sequence
$$
1\to U\_1\to\mathbf{Z}\_p^\times\to\mathbf{F}\_p^\times\to1
$$
which has a canonical splitting $\tau:\mathbf{F}\_p^\times\to\mathbf{Z}\_p^\times$ (the Teichmüller lift) and $U\_r=1+p^r\mathbf{Z}\_p$ is the group of units $\equiv1\pmod{p^r}$. Note that $U\_1$ is a $\mathbf{Z}\_p$-module which is free of rank $1$ if $p\neq2$; if $p=2$, then $U\_1=\mu\_2\times U\_2$, where $\mu\_2$ is the order-$2$ group consisting of $1$ and $-1$, and the $\mathbf{Z}\_2$-module $U\_2$ is free of rank $1$.


As a result, the choice of a uniformiser $\pi$ leads to an isomorphism
$$
\mathbf{Q}\_p^\times=U\_1\times\mathbf{F}\_p^\times\times\pi^{\mathbf{Z}}
$$
which allows you to determine $\mathbf{Q}\_p^\times/\mathbf{Q}\_p^{\times n}$. For example, if $n$ is prime to $p$, then 
$$
\mathbf{Q}\_p^\times/\mathbf{Q}\_p^{\times n}
=\mathbf{F}\_p^\times/\mathbf{F}\_p^{\times n}\times\pi^{\mathbf{Z}/n\mathbf{Z}}
$$
and if $n=p$, then 
$$
\mathbf{Q}\_p^\times/\mathbf{Q}\_p^{\times p}
=U\_1/U\_1^p\times\pi^{\mathbf{Z}/p\mathbf{Z}}\qquad p\neq2
$$
and 
$$
\mathbf{Q}\_2^\times/\mathbf{Q}\_2^{\times 2}
=U\_2/U\_2^2\times\mu\_2\times\pi^{\mathbf{Z}/2\mathbf{Z}}.
$$
Note finally that $U\_1^p=U\_2$ if $p\neq2$ and $U\_2^2=U\_3$ if $p=2$. Similar considerations allow you to prove that $\mathbf{Q}\_p^\times/\mathbf{Q}\_p^{\times n}$ is finite for every $p$ and every $n>0$, and to compute its cardinality.


See for example Serre's *Course in arithmetic*, or any book on local fields.


