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This may be a rather elementary question, but I haven't been able to figure it out on my own, and the literature appears to be eerily silent on the topic.

Since $\mathbb{Q}_p$ is a locally compact group with $\mathbb{Z}_p$ as a compact subset, there exists a unique Haar measure $\mu$ on $\mathbb{Q}_p$ with $\mu(\mathbb{Z}_p)=1$.

The Volkenborn integral for functions $\mathbb{Z}_p\to\mathbb{Q}_p$ is defined by $$ \int_{\mathbb{Z}_p} f = \lim_{n\to\infty}\frac{1}{p^n}\sum_{x=0}^{p^n-1}f(x), $$ takes values in $\mathbb{Q}_p$, and is not translation invariant.

But now suppose I define an integral over $\mathbb{Z}_p$ for functions $\mathbb{Z}_p\to\mathbb{R}$ using the same expression as the Volkenborn integral. Then it seems that I can integrate the functions $\mathbb{1}_{a+p^n\mathbb{Z}_p}$ to get $\frac{1}{p^n}=\mu(a+p^n\mathbb{Z}_p)$, and because the integral so defined is clearly ($\mathbb{R}$-)linear in $f$, it would appear that it agrees with the integral over the Haar measure (at least when both exist).

Since such a Volkenborn-style integral for $\mathbb{R}$-valued functions is never discussed in the literature that I could find, I wonder if I am missing some obvious problem with the construction here, or if this is just a well-known rarely-mentioned fact.

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    $\begingroup$ This is a well knowm fact, and is not rarely mentioned. In articles and books on $p$-adic distributions, this is usually the first example given, and authors usually call it the ($p$-adic valued) Haar measure (which is not a $p$-adic measure). It is the same as the Volkenborn integral. You may look at Koblitz's book, Mazur's Bourbaki report, Colmez's Asterisque survey, etc. $\endgroup$
    – efs
    Mar 13, 2020 at 17:51
  • $\begingroup$ Thanks for the pointers, but to be quite clear: my question is about functions $\mathbb{Q}_p\to\mathbb{R}$, so I'm not sure how this is a $p$-adic valued measure. Is this just nomenclature? $\endgroup$
    – gmvh
    Mar 13, 2020 at 20:10
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    $\begingroup$ Probably I'm not understanding you correctly. The Haar measure is always $\mathbb{R}$ valued, by definition. Now, in this case it is $\mathbb{Q}$ valued, hence may be interpreted as $\mathbb{Q}_p$ valued. Hence the name "$p$-adic Haar measure" is just nomenclature (as far as I know). Actually, there is no translation invariant, $p$-adic valued, bounded measure on $\mathbb{Z}_p$. $\endgroup$
    – efs
    Mar 13, 2020 at 23:11
  • $\begingroup$ OK, now I understand you: the Haar measure is interpreted as $\mathbb{Q}_p$ valued, which leads to statements such as it being unbounded (cf. Jon's answer), i.e. relative to $|\cdot|_p$. When integrating $\mathbb{Q}_p$-valued functions, that makes sense, but when integrating $\mathbb{R}$-valued functions, I think the real value is the only one that fits; I never saw the latter case being discussed, but that may be because number theorists just aren't terribly interested in it. $\endgroup$
    – gmvh
    Mar 14, 2020 at 7:09
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    $\begingroup$ Perhaps you can find some discussion on this in Schikhof's book on ultrametric calculus and/or in Van Rooij's book on ultrametric functional analysis. I don't have my copies at hand, but I'm pretty sure they are in libgen. $\endgroup$
    – efs
    Mar 14, 2020 at 13:19

2 Answers 2

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This does not just vary with translation, it's non-canonical. The idea is that knowing the size of the compact open subsets $p^n\mathbb{Z}_p$ should give you a measure (i.e., an integral) by the following, if it exists:

$$ \int_{\mathbb{Z}_p} f d\mu = \lim_{n \to \infty} \sum_{a=0}^{p^n-1} f(\text{any representative of }a+p^n\mathbb{Z}_p) \mu(a+p^n\mathbb{Z}_p). $$

You should compare this with the usual Riemann integral on $\mathbb{R}$ to see that it makes sense. It seems that your Volkenborn integral just chooses a specific representative, $a$, for $a+p^n\mathbb{Z}_p$.

To see the issue, try to integrate $f(x)=x$ where you choose $\{0,1,\dots,p^n-1\}$ as your representatives, and then do it where you choose $\{p^n,1,\dots,p^n-1\}$. You should get different answers! (I believe they are $\pm\frac{1}{2}$.)

In general, algebraic number theorists say that the "Haar measure" is actually only a $\textit{distribution}$, since $|\mu(p^n\mathbb{Z}_p)|_p = p^n$ is unbounded. If $\mu$ tells us the size of each compact open $|\mu(p^n\mathbb{Z}_p)|_p$ is bounded, we call it a measure. Standard analysis techniques can show that integrals of continuous functions are well-defined with respect to such a measure. EDITED TO ADD: The same proof should work regardless of the metric on your target space, as long as you have that the sizes are bounded. So since $|\mu_{\text{Haar}}(a+p^n\mathbb{Z}_p)|_\infty = \frac{1}{p^n} \leq 1$ is bounded when the target is $\mathbb{R}$, this shows that your Volkenborn integral agrees with the more canonical definition for functions $f \colon \mathbb{Z}_p \to \mathbb{R}$.

(On the other hand, I'm told that analytic number theorists call distributions measures, and call measures "strict measures" -- with these names, integration continuous functions is well defined with respect to strict measures, but not necessarily over measures).

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  • $\begingroup$ Thanks, but I'm not sure how your example relates to the question: I specifically asked about $\mathbb{R}$-valued functions, which $f(x)=x$ isn't. EDITED TO ADD: With your addition in edit, it sounds like you agree that this works in the case where the target is $\mathbb{R}$? $\endgroup$
    – gmvh
    Mar 13, 2020 at 20:01
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This is an addendum to Jon's answer, just to clarify some points on the Volkenborn integral. This was studied by Volkenborn, in a slightly more general form, because of its relation to Bernoulli numbers and polyomials, hence to zeta functions. Namely, $$B_n(x)=\int_{\mathbb{Z}_p}(x+t)^n dt$$ is the $n$-th Bernoulli polynomial (I think this identity is due to Witt). This was probably inspired by the sums that appeared in the work of Kubota and Leopoldt on the $p$-adic zeta function. Hence, in practice it is an extremely useful way to handle Bernoulli polynomials, and actually a lot of their properties have trivial proofs using the Volkenborn integral.

But, the definition of this "integral" by means of "Riemann sums" is kind of artificial once you notice that this is part of a more general theory, and because this integral depends on the choice of the representatives as Joel point out. I think that the theoretically correct way of thinking the Volkenborn integral of a (say, strictly differentiable) function $f$ is as a "Riemann sum" which happens to give the correct value of the $p$-adic distribution $\mu(a+p^n\mathbb{Z}_p)=p^{-n}$ (which modulo normalization, happens to "be" the Haar measure) applied to $f$. This is done in detail in Colmez's Fonctions d'une variable $p$-adic, section II.3.3 in page 32.

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