I have read in a book that if one takes $\mu$ to be the additive Haar measure on $\mathbb{Q}_p$, the p-adic rationals, then
$$\nu(A) := \int_{A} 1/|x|_p dx$$
is a multiplicative Haar measure on $\mathbb{Q}_p^\times$. My question is: why is this the case? I can see three things:
0) $\nu$ is a measure.
1) $\nu(K) < \infty$ for $K$ compact in $\mathbb{Q}_p^\times$.
2) $\nu$ is left multiplicative invariant, i.e. $\nu(xA) = \nu(A)$.
what remains to be shown is that it is regular. In the book i am reading this means that it satisfies
3) For every measurable set $A$, $$\nu(A) = \inf_{U \supset A} \nu(U)$$ where $U$ runs through the open sets containing $A$.
4) For every set $A$ that is either open or has finite measrue, $$\nu(A) = \sup_{K \subset A} \nu(K)$$ where $K$ runs through the compact sets contained in $A$.
Can somebody tell me how one can do that? I tried to play around with monotone convergence, etc but i have the feeling that i am missing something simple :(
Rudin, Real and complex analysis, Theorem 2.18
Let $X$ be a locally compact Hausdorff space in which every open set is $\sigma$-compact. Let $\lambda$ be any positive Borel measure on $X$ such that $\lambda(K) < \infty$ for every compact set $K$. Then $\lambda$ is regular.