The $p$-adic field $\mathbb{Q}_p$ has topological basis of open sets of the form $a+p^N \mathbb{Z}_p$ for $0 \leq a \leq p^N-1$ and $N \in \mathbb{Z}$. These are indeed compact open sets. One can define Bernoulli distributions by $$\mu_{B,k}(a+p^N \mathbb{Z}_p)=p^{N(k-1)}B_k \left(\frac{a}{p^N}\right), $$ where $B_k(x)$ are Bernoulli polynomials and $B_k=B_k(0)$ are Bernoulli numbers.
These $\mu_{B,k}$ extends to a distribution on $\mathbb{Z}_p$.
But these Bernoulli distributions $\mu_{B,k}$ do not define measure on $\mathbb{Z}_p$.
To integrate $p$-adic continuous functions over the compact open subsets $\mathbb{Z}_p$ or $\mathbb{Z}_p^{*}$.of $\mathbb{Q}_p$, one defines the regularized Bernoulli distribution by $$\mu_{k,\alpha}(U)=\mu_{B,k}(U)-\alpha^{-k}\mu_{B,k}(\alpha U)$$ for $\alpha \in \mathbb{Z}_p^{*}$, $k \in \{0\} \cup \mathbb{N}$ and $U=a+p^N \mathbb{Z}_p$.
For $k=1,2, \cdots$, we have the measures $ \mu_{1, \alpha}, \mu_{2,\alpha}, \cdots, \mu_{k, \alpha}, \cdots$. These measure or regularized Bernoulli distributions are related by the following relation $$\int_{\mathbb{Z}_p} f(x) d \mu_{k,\alpha}=\int_{\mathbb{Z}_p} f(x) \cdot kx^{k-1} d \mu_{1, \alpha},$$ for any continuous function $f: \mathbb{Z}_p \to \mathbb{Z}_p$.
This is all the story on $\mathbb{Q}_p$.
Now consider a finite extension $K$ of $\mathbb{Q}_p$. It has similar
compact-open subsets $\mathcal{O}_K$, the ring of integers and topological
basis consisting of open sets $a+\pi O_K$, where $\pi$ is the
uniformizer in the ring $\mathcal{O}_K$.
My question-
Can we extend $\mu_{B,k}$ and $\mu_{k,\alpha}$ from $\mathbb{Z}_p$ to $\mathcal{O}_K$?
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Clearly, we can define Haar measure normalized with $\mu(O_K)=1$ and
so $\mu(m_K)=1/q$, where $m_K$ is the maximal ideal and $q=\lvert\mathcal{O}_K/m_K\rvert$.
Also we know that Haar measure $\mu$ coincide with the Bernoulli distribution $\mu_{B,k}$ for $k=0$.
So it appears trivially that we can define the Bernoulli distribution $\mu_{B,0}$ on the finite extension $K$ as it equals to Haar measure and which can be defined on any locally compact Hausdorff space $K$.
I appreciate any help. Thanks
Best Answer
Here is why I doubt that the question has been looked at and why it is not well-formulated, but I am really not confident in saying this. If it has been considered, I'd hope someone corrects me here.
$\newcommand{\ZZ}{\mathbb{Z}}\newcommand{\QQ}{\mathbb{Q}}$ Let $\mu$ be a measure taking open subsets of a group $\Gamma$ to a ring $A$. In your case $\Gamma$ is isomorphic to $\ZZ_p^\times$ and $A$ is $\ZZ_p$ or $\QQ_p$ or $\mathbb{C}_p$. Conceptually for the Bernoulli distribution, $\Gamma$ is the Galois group of the global extension of $\QQ$ defined by adjoining all $p$-power roots of unity. This $\mu$ corresponds to an element of the Iwasawa Algebra $A[\![\Gamma]\!]$. In your case the measures using Bernoulli numbers link to the $p$-adic $L$-functions and zeta-functions, interpolating values of the classical complex $L$-functions and the Riemann zeta function.
A natural change of the group would be a surjection $\Gamma'\to \Gamma$ corresponding to a further extension of the number field. You could then ask to interpolate $L$-values for more characters of that group. That has been done to some extend and conjecturally we have a good picture of what the right extensions of $\mu$ should be.
In your situation, however, you have a larger group $\Gamma'=\mathcal{O}_K^\times$ that contains $\Gamma=\ZZ^\times_p$ where $K/\QQ_p$ is an extension. I changed the notation to emphasis that the maps $\mu$ really only see the group $\Gamma'$ rather than the ring of integers $\mathcal{O}_K$. As a group $\Gamma'$ is a direct product of a finite group and $\ZZ_p^{[K:\QQ_p]}$; up to the finite bit it won't depend much on $K$, but only on its degree. Since there is no subextension of $\QQ$, this $\Gamma\to\Gamma'$ cannot correspond to a global Galois extension and to $L$-values.
When you are asking for an extension, the most natural thing would be to take the surjection $\mu\circ N_{K/\QQ_p}$. This would now take $N=N_{K/\QQ_p}:\Gamma'\to \Gamma$ and the natural map $A[\![\Gamma]\!]\to A[\![\Gamma']\!]$. It still could not correspond to a larger extension of $\QQ$ as there are no such extensions. We would have to start with a larger number field $L$, but then the Bernoulli measures even for $\Gamma$ have to be changed depending on $L$. Up to the finite bit, your group $\Gamma'$ is the direct sum of $\Gamma$ and the group $\ker (N_{K/\QQ_p})$. That natural extension is then rather boring on this second group. Still ignoring the finite bits, you could more or less randomly choose a different measure on this second group as an extension, but I would not know what meaning to give to it.
(I said ignoring finite subgroups. There is a little interesting question as there are cases when $\Gamma'$ is not the direct product of $\Gamma$ and $\ker(N)$, which I have not thought about.)