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 :(

  • $\begingroup$ Just curious: what book are you reading? $\endgroup$ – Rudy the Reindeer Aug 17 '12 at 14:21
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    $\begingroup$ It is Deitmar, automorphic forms (in german). $\endgroup$ – Fabian Werner Aug 17 '12 at 14:22
  • $\begingroup$ There's a result in Rudin's Real and Complex Analysis which states something like "every Borel measure on an LCH space in which every open set is $\sigma$-compact is regular." Unfortunately I do not have access to my copy of the book right now, so I can't confirm that I'm remember correctly. Perhaps somebody else can. $\endgroup$ – Keenan Kidwell Aug 17 '12 at 14:30
  • $\begingroup$ I am pretty sure this is done in Valenza's Fourier Analysis on Number Fields $\endgroup$ – M Turgeon Aug 17 '12 at 14:54
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    $\begingroup$ @tomasz: what exactly do you mean when you say Radon measure? There are about as many notions of Radon measures as there are books on measure theory. $\endgroup$ – t.b. Aug 17 '12 at 15:37

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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  • $\begingroup$ Or use the statement of the Riesz representation theorem 2.14 directly -- apply it to the positive linear functional on $C_c(\mathbb{Q}_{p}^\times)$ given by $f \mapsto \int f(x) \cdot \frac{1}{|x|_p}\,dx$ $\endgroup$ – t.b. Aug 17 '12 at 15:14
  • $\begingroup$ Ok, thanks!! I will check. $\endgroup$ – Fabian Werner Aug 17 '12 at 15:45
  • $\begingroup$ Thanks for the precise statement, @Makoto. $\endgroup$ – Keenan Kidwell Aug 17 '12 at 17:54
  • $\begingroup$ Hmm, this proof confuses me... After one knows that the measure in question coincides with a (by construction) regular measure on open sets there is nothing left to prove, still he does go on... Does somebody know why this is the case? $\endgroup$ – Fabian Werner Aug 21 '12 at 6:20
  • $\begingroup$ @FabianWerner "After one knows that the measure in question coincides with a (by construction) regular measure on open sets there is nothing left to prove," Could you explain this? $\endgroup$ – Makoto Kato Aug 21 '12 at 18:53

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