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This is partially motivated by a question I saw earlier here, Does such a finitely additive function exist?

I've been reading about the topology of $\mathbb{Q}_p$ in Knapp's Advanced Algebra in Chapter 6, and I'm wondering if it's possible to impose a finitely additive function on $\mathbb{Q}_p$ in a natural way. By this I mean, suppose I take closed balls of form $$ B(x,r)=\{y\in\mathbb{Q}_p\mid|x-y|\leq p^{-r}\} $$ and let $\mathscr{B}$ be the set ring consisting of finite unions of such balls.

I suspect there must also be a finitely additive function $\mu\colon\mathscr{B}\to\mathbb{R}^{+}$, the nonnegative reals, such that $\mu(B(x,r))=p^{-r}$, as one would expect naturally?

In the case for $\mathbb{Q}$, such a function was constructed by first constructing a function which mapped a member of the set ring to its intersection with $\mathbb{Q}$. However, I view elements of $\mathbb{Q}_p$ as sequences in $\prod_{j=1}^\infty\mathbb{Q}$, so would their be some similar way of constructing such a function by first considering finite unions of such closed balls of sequences in $\prod_{j=1}^\infty\mathbb{R}$?

And unlike the case in $\mathbb{Q}$, would it in fact be possible to extend $\mu$ to a unique measure on the generated $\sigma$-algebra of $\mathscr{B}$?

Edit: Harry Altman's answer shows that the Haar measure is a measure which restricts back to this property. I guess I'm curious on how this would be built up. I mean, what would be the basic finitely additive function on $\mathscr{B}$ that could be uniquely extended to the Haar measure on the generated $\sigma$-algebra?

Thank you for your considerations.

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2 Answers 2

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EDIT: Deleting much of this answer because it was wrong. But not the part that answers your actual question.

OK, here is an answer to the revised question. In fact, this is a better answer, because it's elementary, whereas my previous answer hit it with a sledgehammer. This one also says a bit more about the topology of $\mathbb{Q}_p$.

You want to describe the measure on finite unions of balls in the p-adics. But in fact, any finite union of balls in the p-adics is a finite disjoint union of balls. (Indeed, this works in any ultrametric space.) This immediately shows that the function you want exists, and also what it has to be: the measure of a set is the sum of the measures of the disjoint balls that make it up.

This is because in an ultrametric space, if two balls overlap, one must contain the other. (I'm leaving ambiguous whether the balls are open or closed since it works either way, and since in this case the space we care about is the p-adics, where it makes no difference.) I'll just do the proof for open balls, it's obviously the same either way. I'll use $B_r(x)$ to denote the open ball of radius r about x. So, say $B_r(x)$ and $B_s(y)$ are two balls; suppose $r\le s$. Now suppose $z\in B_r(x) \cap B_s(y)$, and $w\in B_r(x)$. Then $d(w,y)\le \max(d(w,z),d(z,y))<\max(r,s)=s$, so $w\in B_s(y)$; i.e., given the existence of such a z, we have $B_r(x)\subseteq B_s(y)$. (Note therefore that if r=s, if $B_r(x)\cap B_r(y)\ne \emptyset$, then $B_r(x)=B_r(y)$; of course, this also follows from the well-known fact that if $y\in B_r(x)$, then $B_r(y)=B_r(x)$, which I guess is really just another way of stating the same thing.)

Thus it is easy to check that if we have the union of a finite set S of balls, this is the disjoint union of the maximal elements of S, and we get what I described in the second paragraph.

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Many thanks Harry!, I see what you mean now. A last note though, what is the extended measure of this finitely additive function on the generated $\sigma$-algebra? Is it essentially just the same, or do some modifications need to be made? –  Buble Nov 5 '11 at 10:21
    
I don't know that there's any easy way to describe extending it to all the Borel sets; general measure-theoretic stuff gives a description but not, I don't think, a nice one. The now-deleted part of this answer claimed that something like the above worked for all open sets in $\mathbb{Q}_p$ but this was wrong. –  Harry Altman Nov 5 '11 at 23:08
    
Thanks Harry, I appreciate your help. –  Buble Nov 6 '11 at 0:26
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Yes; under addition, $\mathbb{Q}_p$ forms a locally compact group, so it has a Haar measure. I.e. there is a measure on $\mathscr{B}$ which is translation invariant, positive on nonempty open sets, finite on compact sets, and satisfies some regularity conditions (see the page). It's unique up to scaling. If we normalize it so that $\mu(\mathbb{Z}_p)=1$, as you've done (which is standard), then we get what you've described above.

Also, if we take what you've described above and just assume the regularity conditions in addition I think this should be enough to force you to get the Haar measure? In any case, I think it's safe to say the Haar measure is what you're looking for. :)

Note, by the way, that if we consider p-adic integers as p-adic expansions, then Haar measure on $\mathbb{Z}_p$ is just the product measure of the uniform measures on the digits. (I.e. we can reason about randomly chosen p-adic integers in the same way we would reason about randomly chosen infinite strings of digits.)

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Thanks Harry. I'm afraid I don't see how this answers my question, which is probably due to my ignorance. Do you mind explaining more explicitly what exactly the Haar measure is in this case that makes it a finitely additive function and extendable to a measure on the generated $\sigma$-algebra? I see a lot of general theory, but can't decipher how it applies in this case. –  Buble Nov 4 '11 at 7:17
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I'm afraid I'm not exactly an expert on this; it is a measure, though, so it's countably additive, in particular finitely additive. It is a measure on the generated sigma-algebra (the Borel sets); I suppose properly speaking it restricts to the finitely additive function you want. It agrees with your function on balls; since $p^k\mathbb{Z}_p$ consists of p translates of $p^{k+1}\mathbb{Z}_p$, it therefore has p times the measure. So assuming $\mu(\mathbb{Z}_p)=1$, $\mu(B(x,r))=\mu(x+p^{-r}\mathbb{Z}_p)=p^{-r}$. As for what it is explicitly on other sets... well, it's a regular measure. :P –  Harry Altman Nov 4 '11 at 22:03
    
Thanks again Hary. This gives me a good idea of what I'm asking for. –  Buble Nov 4 '11 at 22:12
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