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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 :(

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    @tomasz: thanks!2012-08-17

1 Answers 1

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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.

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    Ok, i see... it is not clear that the "complement rule" is satisfied... But this confuses me even more: if there are two maps on a single set generated by something and the maps are compatible with whatever structure on the set is and they coincide on the generators then they should also coincide on the whole set (group homomorphisms, vector space homomorphisms, ...) but in measure theory this seems to be false... If two measures coincide on the generators then they need not coincide on the sigma algebra generated by the generators... strange... Thanks, now i understand!!2012-09-03