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.



Connection between Volkenborn integral and Haar measure on $\mathbb{Q}_p$
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).


