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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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    Just curious: what book are you reading?2012-08-17
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    It is Deitmar, automorphic forms (in german).2012-08-17
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    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.2012-08-17
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    I am pretty sure this is done in Valenza's *Fourier Analysis on Number Fields*2012-08-17
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    @KeenanKidwell: I think that it is enough for every open set to be $F_\sigma$, but I think you also need some finiteness assumptions.2012-08-17
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    @KeenanKidwell: In fact (at least for finite measures), every Borel measure on a Polish space is Radon. $\mathbf Q_p$ is Polish, so...2012-08-17
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    @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.2012-08-17
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    @t.b. locally finite (trivial for finite measures), inner regular with respect to compact sets, outer regular with respect to open sets (follows from inner regular for finite measures).2012-08-17
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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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    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$2012-08-17
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    Ok, thanks!! I will check.2012-08-17
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    Thanks for the precise statement, @Makoto.2012-08-17
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    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?2012-08-21
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    @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?2012-08-21
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    In this proof, one shows that for every open set $V$ one has $\mu(V) = \lambda(V)$ where $\lambda$ is the Borel-measure in question and $\mu$ is a !!regular!! Borel measure. Since $\lambda$ and $\mu$ coincide on open sets, they coincide on the whole Borel $\sigma$-algebra, hence, in particular, $\lambda$ is regular. Now Rudin does not stop after equation (3), he goes on and i do not understand what/why he does it. Can someone explain? Thx!!2012-09-03
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    @FabianWerner "Since $\lambda$ and $\mu$ coincide on open sets, they coincide on the whole Borel $\sigma$-algebra," Could you explain why this is so?2012-09-03
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    Hmm... isnt this the usual argument? Both are measures at least on the Borel $\sigma$-algebra, hence, the collection of sets $\mathcal{A} := \{ M \in \mathcal{B}(X) : \mu(M) = \lambda(M)\}$ forms a (sub)sigma algebra which contains the open sets. Since the Borel sigma algebra $\mathcal{B}(X)$ is the minimal sigma algebra containing all the open sets, $\mathcal{A} \subseteq \mathcal{B}(X)$2012-09-03
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    @FabianWerner "hence, the collection of sets $\mathcal{A}:=\{M\in\mathcal{B}(X)\colon\mu(M)=\lambda(M)\}$ forms a (sub)sigma algebra" Could you explain why this is so?2012-09-03
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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