Utente:Paolo Boldrini/Sandbox

Introduction

These pages are a brief written report on the construction of ${\textstyle p}$-adic integration. Our final goal is to give sense of integrating continuous function defined from ${\textstyle \mathbb {Q} _{p}}$, or at least some particular subsets like ${\textstyle \mathbb {Z} _{p}}$, with value in ${\textstyle \mathbb {Q} _{p}}$. We develop a theory similar to Riemann integration for real valued functions: we define the integral as the limit of Riemann sums. The main result is Theorem 4.2. which states that for continuous functions the sequence of Riemann sums actually converges to a limit in ${\textstyle \mathbb {Q} _{p}}$. We will give particular attention to Bernoulli distribution (in chapter 3) and measures (in chapter 4), because they are the most common and useful. For example they can be used to prove Kummer’s congruences. This work is mostly based on "Koblitz" .

Preliminaries.

Some topology on ${\textstyle \mathbb {Q} _{p}}$ Errore del parser (errore di sintassi): {\textstyle \\} We will consider the topology induced by the metric on ${\textstyle \mathbb {Q} _{p}}$ . The standard basis of this topology consists of all sets of the form Errore del parser (funzione sconosciuta '\begin{split}'): {\displaystyle \begin{split} a+(p^N)&=a+p^N\mathbb{Z}_p\\&=\{x\in\mathbb{Q}_p|~x=a+tp^N, t\in\mathbb{Z}_p \}\\&=\{x\in\mathbb{Q}_p|~|x-a|_p\leq p^{-n}\}\\&=B_{p^{-n-1}}(a). \end{split}}

These sets are called open balls. They are, open, but also closed.

There is also a really unexpected property:

Proposition 1. 'Any (open) ball is compact in ${\textstyle \mathbb {Q} _{p}}$.'

Proof. Consider the open ball ${\textstyle B_{p^{z}}(b)}$ for some ${\textstyle z\in \mathbb {Z} }$ and ${\textstyle b\in \mathbb {Q} _{p}}$. We will prove that ${\textstyle B_{p^{z}}(b)}$ is sequentially compact and therefore (since it is a metric space) compact. We first observe that for any ${\textstyle x\in B_{p^{z}}(b)}$ , ${\textstyle |x|_{p}\leq max(|b|_{p},p^{z})}$. In fact, if ${\textstyle |x|_{p}>max(|b|_{p},p^{z})}$, then ${\textstyle |x-b|_{p}=|x|_{p}>p^{z}}$. We can fix ${\textstyle \alpha }$ such that ${\textstyle max(|b|_{p},p^{z})=p^{\alpha }}$, so if we expand ${\textstyle x}$ we get Errore del parser (funzione sconosciuta '\label'): {\displaystyle \label{starequation} \tag{$\star$} x=\sum_{i=-\alpha}^{\infty}{a_ip^i}}

Now we want to prove that any sequence in ${\textstyle B_{p^{z}}(b)}$ has a convergent subsequence. To do so let ${\textstyle \{x_{n}\}}$ to be a sequence, any of the ${\textstyle x_{n}}$ can be written in the form [starequation]. We observe that for all ${\textstyle i}$, ${\textstyle a_{i}}$ is between ${\textstyle 0}$ and ${\textstyle p-1}$, so it can take only ${\textstyle p}$ possible values. Thus there is a subsequence ${\textstyle \{x_{k}^{0}\}}$ of elements with ${\textstyle a_{-\alpha }=A_{0}}$ for some ${\textstyle A_{0}\in \{0,...,p-1\}}$. Iterating the process, we build a subsequence ${\textstyle \{x_{k}^{m}\}}$ with ${\textstyle a_{-\alpha }=A_{-\alpha }}$, ${\textstyle a_{-\alpha +1}=A_{-\alpha +1}}$ and so on until ${\textstyle a_{-\alpha +m}=A_{-\alpha +m}}$. Then ${\textstyle \{x_{j}^{j}\}}$ is clearly a convergent subsequence of the original sequence ${\textstyle \{x_{n}\}}$. Thus ${\textstyle B_{p^{z}}(b)}$ is sequencially compact. ◻

We say that a subset of ${\textstyle \mathbb {Q} _{p}}$ is compact open if it is compact and it is open.

Lemma 1. 'Let ${\textstyle B_{p^{n}}(a)}$ and ${\textstyle B_{p^{m}}(b)}$ be two balls, suppose ${\textstyle n\leq m}$. If ${\textstyle B_{p^{n}}(a)\cap B_{p^{m}}(b)\neq \emptyset }$, then ${\textstyle B_{p^{n}}(a)\subseteq B_{p^{m}}(b)}$.Errore del parser (errore di sintassi): {\textstyle \\} If two balls have the same radius either they do not intersect or they actually are the same ball.'

Proof. Suppose ${\textstyle x\in B_{p^{n}}(a)\cap B_{p^{m}}(b)}$ and ${\textstyle y\in B_{p^{n}}(a)}$. We have: $$|y-b|_{p}=|(y-a)+(a-x)+(x-b)|_{p}\leq max(|y-a|_{p},|a-x|_{p},|x-b|_{p})\leq p^{m},$$ thus ${\textstyle y\in B_{p^{m}}(b)}$.Errore del parser (errore di sintassi): {\textstyle \\} If ${\textstyle n=m}$ we can apply this result in both directions, thus ${\textstyle B_{p^{n}}(a)=B_{p^{m}}(b)}$. ◻

Proposition 2. 'An open subset of ${\textstyle \mathbb {Q} _{p}}$ is compact if and only if it is a finite union of disjoint open balls.'

Proof. Finite union of disjoint open balls are both open and compact (because balls are compact!). Conversely, if a subset is open, it is the union of open balls, and because of the compactness we can take just a finite number of them. Now Lemma Lemma 1 is sufficient to conclude. ◻

Recall 1.1. For all ${\textstyle x\in \mathbb {Q} _{p}}$, ${\textstyle x=\sum _{i=m_{0}}^{\infty }x_{i}p^{i}}$, for some ${\textstyle m_{0}\in \mathbb {Z} }$ and ${\textstyle x_{i}\in \{0,...,p-1\}}$.

Recall 1.2.For all p, ${\textstyle \mathbb {Z} _{p}=\{x\in \mathbb {Q} _{p}||x|_{p}\leq 1\}}$. Therefore, if ${\textstyle x\in \mathbb {Z} _{p}}$, then ${\textstyle m_{0}=0}$.

p-adic distributions

Definition 1. Let X and Y be two topological spaces. A map ${\textstyle f:X\rightarrow Y}$ is called locally constant if it is constant in the neighborhood of each point.

Observe that in a connected space, locally constant function are not that interesting, because they have to be globally constant.

Lemma 2. 'Let ${\textstyle a+(p^{N})}$ be a ball, ${\textstyle a+(p^{N})=\bigcup \limits _{b=0}^{p-1}a+bp^{N}+(p^{N+1})}$.'

Proof. Suppose ${\textstyle x\in a+(p^{N})}$, then ${\textstyle x=a+tp^{N}}$ for some ${\textstyle t=\sum _{i=0}^{\infty }t_{i}p^{i}}$. Thus $$x=a+t_{0}p^{N}+p^{N+1}\sum _{i=1}^{\infty }t_{i}p^{i-1}\in a+t_{0}p^{N}+(p^{N+1}),$$ for ${\textstyle 0\leq t_{0}\leq p-1}$. We have proven $$a+(p^{N})\subseteq \bigcup \limits _{b=0}^{p-1}a+bp^{N}+(p^{N+1}).$$ Conversely if ${\textstyle y\in \bigcup \limits _{b=0}^{p-1}a+bp^{N}+(p^{N+1})}$, then $$y=a+b_{0}p^{N}+tp^{N+1}=a+p^{N}(b_{0}+tp).$$ Thus $$a+(p^{N})\supseteq \bigcup \limits _{b=0}^{p-1}a+bp^{N}+(p^{N+1}),$$ and we have the desired equality. ◻

Lemma Lemma 2 means that we can divide a ball of length ${\textstyle p^{-N}}$ into ${\textstyle p-1}$ (disjoint) balls of length ${\textstyle p^{-N-1}}$. Repeating this process, Errore del parser (funzione sconosciuta '\begin{split}'): {\displaystyle \begin{split} a+(p^N)&=\bigcup \limits_{b_0=0}^{p-1}\bigcup \limits _{b_1=0}^{p-1} a+b_0p^N+b_1p^{N+1}+(p^{N+2})\\&=\bigcup \limits _{b_0,...,b_k=0}^{p-1} a+b_0p^N+...+b_kp^{N+k}+(p^{N+k+1}). \end{split}} So we can split any ball into smaller balls of arbitrary length. This construction is also unique: suppose ${\textstyle U}$ is an ball and ${\textstyle U=\bigcup \limits _{i=1}^{k}I_{i}=\bigcup \limits _{j=1}^{h}J_{j}}$ , with each ${\textstyle I_{i}}$ and ${\textstyle J_{j}}$ of length ${\textstyle p^{-l}}$. If ${\textstyle x\in I_{1}}$, ${\textstyle x\in J_{j_{1}}}$ for some ${\textstyle j_{1}}$ and by Lemma Lemma 1 ${\textstyle I_{1}=J_{j_{1}}}$, without loss of generality we can set ${\textstyle j_{1}=1}$. Thus if we proceed we get: ${\textstyle k=h}$ and ${\textstyle I_{i}=J_{i}}$ for all ${\textstyle i}$. It is easy to see that the same property still holds if ${\textstyle U}$ is a finite union of balls. Therefore we have the following:

Corollary 1. If ${\textstyle U}$ is compact open, it can be written in a unique way as the disjoint union of a finite number of balls of length ${\textstyle p^{-N}}$, for any N big enough.

We notice that this corollary implies that any ball is totally bounded. Since balls are closed they are also complete, because ${\textstyle \mathbb {Q} _{p}}$ is complete. Therefore this is another proof of Proposition Proposition 1.

Proposition 3. 'Let X be a compact open subset of ${\textstyle \mathbb {Q} _{p}}$ and ${\textstyle f:X\rightarrow \mathbb {Q} _{p}}$. ${\textstyle f}$ is locally constant if and only if it is a finite linear combinations of characteristic functions associated with compact open subsets of X.'

Proof. Suppose ${\textstyle f=\sum _{i=1}^{n}\phi _{i}\chi _{U_{i}}}$, with ${\textstyle U_{i}}$ compact open. Notice that ${\textstyle X=\bigcup \limits _{i=1}^{n}U_{i}\cup (X\setminus \bigcup \limits _{i=1}^{n}U_{i})}$, this is a disjoint union of open sets and on each of these sets ${\textstyle f}$ is constant. If we take any ${\textstyle x\in X}$, it is in one and only of these sets, thus it has a neighborhood where ${\textstyle f}$ is constant. Conversely, suppose ${\textstyle f}$ is locally constant, which means for all ${\textstyle x\in X}$, there exists a neighborhood of ${\textstyle x}$ (we call it ${\textstyle U_{x}}$), where ${\textstyle f}$ is constant. Since ${\textstyle X\subseteq \bigcup \limits _{x\in X}U_{x}}$, there exist ${\textstyle \{x_{1},...,x_{n}\}}$ such that ${\textstyle X\subseteq \bigcup \limits _{i=1}^{n}U_{x_{i}}}$. By Lemma Lemma 2 we can split each ${\textstyle U_{x_{i}}}$ into balls of length ${\textstyle p^{-N}}$, for some N big enough, and this partition is unique. Thus we have split ${\textstyle X}$ into a finite number of disjointed balls, and on each of these balls ${\textstyle f}$ is constant. Therefore, we can write ${\textstyle f}$ as a linear combination of the characteristic functions associated with these balls. ◻

Definition 2. Let ${\textstyle X}$ be a compact-open, we denote by ${\textstyle {\mathcal {K}}_{loc}(X)}$ the vector space of the locally constant function from X to ${\textstyle \mathbb {Q} _{p}}$ (i.e. the set of linear combinations of characteristic function associated with compact open subsets of X: in fact these characteristic function form a basis).

Definition 3. Let X be a compact open, a ${\textstyle p}$-adic distribution ${\textstyle \mu }$ on X is a vector space homomorphism from ${\textstyle {\mathcal {K}}_{loc}(X)}$ to ${\textstyle \mathbb {Q} _{p}}$. If ${\textstyle f\in {\mathcal {K}}_{loc}(X)}$ we denote ${\textstyle \mu (f)}$ as ${\textstyle \int fd\mu }$.

In these notes we will often omit the "${\textstyle p}$-adic" in front of "distributions". We can see ${\textstyle p}$-adic distributions in another interesting and useful way. If ${\textstyle \mu }$ is a ${\textstyle p}$-adic distibution, it restricts to a map from the set of compact open (we will call this set ${\textstyle {\mathcal {CO}}(X)}$) to ${\textstyle \mathbb {Q} _{p}}$: ${\textstyle {\widetilde {\mu }}(U)=\mu (\chi _{U})}$ which is (finitely) additive. Conversely if we have such an additive map ${\textstyle {\widetilde {\mu }}}$ we can extends it to a ${\textstyle p}$-adic distribution ${\textstyle \mu (\sum _{i=1}^{n}\phi _{i}\chi _{U_{i}})=\sum _{i=1}^{n}\phi _{i}{\widetilde {\mu }}(U_{i})}$ and the extension is clearly unique. From now on, we will identify ${\textstyle {\widetilde {\mu }}}$ and ${\textstyle \mu }$ in the sense that we will see a ${\textstyle p}$-adic distribution on ${\textstyle X}$ also as a map from ${\textstyle {\mathcal {CO}}(X)}$ to ${\textstyle \mathbb {Q} _{p}}$.

Proposition 4. 'Let ${\textstyle \mu }$ be a map from the set of balls contained in X to ${\textstyle \mathbb {Q} _{p}}$ such that for any ${\textstyle a+(p^{N})\subseteq X}$ we have $$\mu (a+(p^{N}))=\sum _{b=0}^{p-1}\mu (a+bp^{N}+(p^{N+1})),$$ then ${\textstyle \mu }$ extends to a unique ${\textstyle p}$-adic distribution.'

Proof. By proposition Proposition 2, any compact-open is the finite disjoint union of open balls. So if ${\textstyle U\in {\mathcal {CO}}(X)}$, ${\textstyle U=\bigcup \limits _{i=1}^{n}I_{i}}$, thus the only possible definition to be additive is ${\textstyle \mu (U)=\sum _{i=1}^{n}\mu (I_{i})}$. We now have to prove that this is a good definition, i.e. it is independent of the partition of U. First we notice that iterating ${\textstyle (1)}$ we have: Errore del parser (funzione sconosciuta '\begin{split}'): {\displaystyle \begin{split} \mu(a+(p^N))&=\sum_{b_0=0}^{p-1}\mu(a+b_0p^N+(p^{N+1}))\\&=\sum_{b_0=0}^{p-1}\sum_{b_1=0}^{p-1}\mu(a+b_0p^N+b_1p^{N+1}+(p^{N+2}))\\&=\sum \limits _{b_0,...,b_k=0}^{p-1}\mu(a+b_0p^N+...+b_kp^{N+k}+(p^{N+k+1})). \end{split}} Suppose ${\textstyle U=\bigcup \limits _{i=1}^{n}a_{i}+(p^{N_{i}})=\bigcup \limits _{j=1}^{n'}a'_{j}+(p^{N'_{j}})}$. Now we set ${\textstyle M=max(N_{i},N'_{j})}$ , and we split each ${\textstyle a_{i}+(p^{N_{i}})}$ into smaller balls of length ${\textstyle p^{-M}}$. By ${\textstyle (2)}$ (for ${\textstyle k=M-N_{i}}$) we have ${\textstyle \sum _{i=1}^{n}\mu (a_{i}+(p^{N_{i}}))=\sum _{i=1}^{n}\sum \limits _{b_{0},...,b_{k}=0}^{p-1}\mu (a+b_{0}p^{N_{i}}+...+b_{k}p^{N_{i}+k}+(p^{N_{i}+k+1}))}$. This means that in the end the value of ${\textstyle \mu (U)}$ depends only on his value on smaller intervals. By the previous observation we know the decomposition is unique, thus the definition of ${\textstyle \mu }$ is independent of the partition. At this point ${\textstyle \mu }$ is a map from ${\textstyle {\mathcal {CO}}(X)}$ to ${\textstyle \mathbb {Q} _{p}}$ which is clearly additive. Thus ${\textstyle \mu }$ extends to a p-adic distribution. ◻

Examples 1.  

1. ${\textstyle \mu _{\mathrm {Dirac} }}$: Suppose X is any compact open in ${\textstyle \mathbb {Q} _{p}}$ and ${\textstyle \alpha \in X}$, we can define the Dirac distribution concentrated at ${\textstyle \alpha }$ as Errore del parser (funzione sconosciuta '\begin{cases}'): {\textstyle \mu_{\alpha}(U)=\begin{cases} 1, & \mbox{if }\alpha\in U \\ 0, & \mbox{otherwise }\ \end{cases}} for any compact-open ${\textstyle U\subseteq X}$
2. ${\textstyle \mu _{\mathrm {Haar} }}$: We define the Haar distribution on the compact open ${\textstyle \mathbb {Z} _{p}}$ (it is a compact open since it is the ball of radius 1 centered at 0) as ${\textstyle \mu _{\mathrm {Haar} }(a+(p^{N}))={\frac {1}{p^{N}}}}$ for all ${\textstyle a+(p^{N})\subseteq \mathbb {Z} _{p}}$ (i.e ${\textstyle a\in \mathbb {Z} _{p}}$ and ${\textstyle N\geq 0}$). We observe that we have defined the distribution only on balls, but it is sufficient to get a distribution by Prop Proposition 4. In fact ${\textstyle \sum _{b=0}^{p-1}\mu _{\mathrm {Haar} }(a+bp^{N}+(p^{N+1}))=\sum _{b=0}^{p-1}{\frac {1}{p^{N+1}}}={\frac {1}{p^{N}}}=\mu _{\mathrm {Haar} }(a+(p^{N}))}$.
3. ${\textstyle \mu _{\mathrm {Mazur} }}$: We define the Mazur distribution on the compact open ${\textstyle \mathbb {Z} _{p}}$ as ${\textstyle \mu _{\mathrm {Mazur} }(a+(p^{N}))={\frac {a}{p^{N}}}-{\frac {1}{2}}}$ for all ${\textstyle a+(p^{N})\subseteq \mathbb {Z} _{p}}$ (i.e ${\textstyle a\in \mathbb {Z} _{p}}$ and ${\textstyle N\geq 0}$), where ${\textstyle a}$ is a positive integer such that ${\textstyle 0\leq \ a\leq p^{N}-1}$. It is easy to show that in any ball ${\textstyle \alpha +(p^{N})\subseteq \mathbb {Z} _{p}}$ exists one and only ${\textstyle a}$ with that property, thus by Lemma Lemma 1 we have that ${\textstyle \alpha +(p^{N})=a+(p^{N})}$, so this is a good definition. As before we can verify that ${\textstyle \mu _{\mathrm {Mazur} }(a+(p^{N}))=\sum _{b=0}^{p-1}\mu _{\mathrm {Mazur} }(a+bp^{N}+(p^{N+1}))}$, thus it extends to a distribution. For example ${\textstyle \mu _{\mathrm {Mazur} }((p))=-{\frac {1}{2}}}$, this may be apparently strange, because in real analysis we usually have distributions with positive value, but here ${\textstyle -{\frac {1}{2}}\in \mathbb {Q} _{p}}$, thus it does not make sense to think it as a "negative number". Someone could ask why this ${\textstyle -{\frac {1}{2}}}$ appears in the definition. There are actually two answer to this question. The first one is the fact that if we don’t do it, we don’t get a distribution, since the condition of Proposition Proposition 4 is not true anymore. The second and more interesting answer is related to Bernoulli numbers and it is the purpose of the next paragraph.
4. Integrating locally constant functions: Let ${\textstyle f:\mathbb {Z} _{p}\rightarrow \mathbb {Q} _{p}}$ be the function which associate to any ${\textstyle x\in \mathbb {Z} _{p}}$ his first digit in p-adic expansion (i.e the coefficient of ${\textstyle p^{0}}$ in its expansion). This is clearly locally constant and naturally splits ${\textstyle \mathbb {Z} _{p}}$ into p subsets: ${\textstyle \mathbb {Z} _{p}=\bigcup \limits _{b=0}^{p-1}b+(p)}$. ${\textstyle f}$ is constant on each ${\textstyle b+(p)}$ and it has value b, thus ${\textstyle f=\sum _{b=0}^{p-1}b\chi _{b+(p)}}$. If we want to integrate ${\textstyle f}$ respect to a distribution ${\textstyle \mu }$ we get ${\textstyle \int fd\mu =\sum _{b=0}^{p-1}b\mu (b+(p))}$. For example if ${\textstyle \mu =\mu _{\mathrm {Haar} }}$ we have ${\textstyle \mu (b+(p))={\frac {1}{p}}}$, thus ${\textstyle \int fd\mu _{\mathrm {Haar} }={\frac {p(p-1)}{2p}}={\frac {p-1}{2}}}$. Instead, if ${\textstyle \mu =\mu _{\mathrm {Mazur} }}$ ${\textstyle \mu (b+(p))={\frac {b}{p}}-{\frac {1}{2}}}$, thus ${\textstyle \int fd\mu _{\mathrm {Mazur} }={\frac {(p-1)(p-2)}{12}}}$.

We will see that distribution are not enough to extends integration to continuous functions, and then we need a more particular tool.

Bernoulli Distributions

Definition 4. We define ${\textstyle k}$-th Bernoulli number as ${\textstyle k!}$ the ${\textstyle k}$-th coefficient of the Taylor expansion of ${\textstyle {\frac {t}{e^{t}-1}}}$.

By definition we can write ${\textstyle {\frac {t}{e^{t}-1}}=\sum \limits _{k=0}^{\infty }{\frac {B_{k}}{k!}}t^{k}}$.Errore del parser (errore di sintassi): {\textstyle \\} We write the first few Bernoulli numbers: Errore del parser (errore di sintassi): {\displaystyle B_0=0,\ B_1=-\frac{1}{2},\ B_3=0,\ B_4=-\frac{1}{30}\} It is not difficult to show that for every odd integer greater than 1, ${\textstyle B_{2m+1}=0}$.Errore del parser (errore di sintassi): {\textstyle \\} Now we define in a similar way the Bernoulli polynomials.

Definition 5. Consider $${\frac {te^{xt}}{e^{t}-1}}=(\sum \limits _{j=0}^{\infty }B_{j}{\frac {t^{j}}{j!}})(\sum \limits _{j=0}^{\infty }{\frac {(xt)^{j}}{j!}})=\sum \limits _{k=0}^{\infty }c_{k},$$ where $$c_{k}=\sum \limits _{j=0}^{k}(B_{j}{\frac {t^{j}}{j!}})({\frac {(xt)^{k-j}}{(k-j)!}})={\frac {t^{k}}{k!}}\sum \limits _{j=0}^{k}{\binom {k}{j}}B_{j}x^{k-j}.$$ We call ${\textstyle k}$-th Bernoulli polynomial: $$B_{k}(x)=\sum \limits _{j=0}^{k}{\binom {k}{j}}B_{j}x^{k-j}.$$

In analogy with Bernoulli numbers, we notice that for a fixed ${\textstyle x}$, the ${\textstyle k}$-th Bernoulli polynomial evalued at ${\textstyle x}$ is ${\textstyle k}$! the ${\textstyle k}$-th coefficient of the Taylor expansion of ${\textstyle {\frac {te^{xt}}{e^{t}-1}}}$.

Definition 6. We define the ${\textstyle k}$-th Bernoulli distribution as $$\mu _{B,k}(a+(p^{N}))=p^{(k-1)N}B_{k}({\frac {a}{p^{N}}}),$$ with ${\textstyle a\in \mathbb {Z} }$ and ${\textstyle 0\leq a\leq p^{N}-1}$.

It is a distribution in the sense of Proposition Proposition 4, by the next:

Proposition 5. Errore del parser (funzione sconosciuta '\label'): {\displaystyle \label{1} \tag{1} \mu_{B,k}(a+(p^N))=\sum_{b=0}^{p-1}\mu_{B,k}(a+bp^N+(p^{N+1})).} Thus ${\textstyle \mu _{B,k}}$ extends to a p-adic distribution.

Proof. By definition the left hand side is ${\textstyle p^{(k-1)N}B_{k}({\frac {a}{p^{N}}})}$ and the right hand side is the sum of terms of the form ${\textstyle p^{(k-1)(N+1)}B_{k}({\frac {a+bp^{N}}{p^{N+1}}})}$. Setting ${\textstyle \alpha ={\frac {a}{p^{N+1}}}}$, we have to prove ${\textstyle B_{k}(\alpha p)=p^{k-1}\sum _{b=0}^{p-1}B_{k}(\alpha +{\frac {b}{p}})}$. To do so, consider Errore del parser (funzione sconosciuta '\label'): {\displaystyle \label{2} \tag{2} p^{k-1}\sum_{b=0}^{p-1}\frac{te^{\alpha t+\frac{b}{p}t}}{e^t-1}=p^k\sum_{b=0}^{p-1}\frac{\frac{t}{p}e^{(\alpha p)\frac{t}{p}}}{e^{\frac{t}{p}}-1}=p^k\sum\limits_{j=0}^{\infty}\frac{B_j(\alpha p)}{j!}\frac{t^j}{p^j}.} Thus the right hand side in [1], which is ${\textstyle k}$! times the ${\textstyle k-th}$ coefficient of [2], is exactly ${\textstyle B_{k}(\alpha p)}$ . ◻

Examples 2.

1. ${\textstyle k=0}$: ${\textstyle \mu _{B,0}(a+(p^{N}))=p^{-N}B_{0}({\frac {a}{p^{N}}})}$, since ${\textstyle B_{0}(x)=1}$ we simply have ${\textstyle \mu _{B,0}=\mu _{\mathrm {Haar} }}$.
2. ${\textstyle k=1}$: ${\textstyle \mu _{B,1}(a+(p^{N})=B_{k}({\frac {a}{p^{N}}})}$, since ${\textstyle B_{1}(x)=x-{\frac {1}{2}}}$ we simply have Errore del parser (errore di sintassi): {\textstyle \\} ${\textstyle \mu _{B,1}=\mu _{\mathrm {Mazur} }}$. This is the interesting reason why we need the ${\textstyle -{\frac {1}{2}}}$ in the definition of ${\textstyle \mu _{\mathrm {Mazur} }}$: we want it to be a Bernoulli distribution.

Measures and integration

Definition 7. A distribution ${\textstyle \mu }$ on X is called ${\textstyle p}$-adic measure if for all ${\textstyle U}$ compact open subsets of X we have ${\textstyle |\mu (U)|_{p}\leq C<\infty }$, for some ${\textstyle C\in \mathbb {R} _{\geq 0}}$.

In these notes, we will often omit the "${\textstyle p}$-adic" in front of "measures".

Examples 3.

1. The Dirac distribution is bounded above by the constant 1, thus it is a measure.
2. Bernoulli distributions are not measures. In fact for all k we can consider ${\textstyle |\mu _{B,k}(1+(p^{N}))|_{p}=|p^{(k-1)N}B_{k}({\frac {1}{p^{N}}})|_{p}=p^{-Nk+N}|B_{k}({\frac {1}{p^{N}}})|_{p}}$, since ${\textstyle B_{k}}$ is a polynomial of degree k and the coefficient of ${\textstyle x^{k}}$ is ${\textstyle B_{0}=1}$, ${\textstyle |B_{k}({\frac {1}{p^{N}}})|_{p}=p^{Nk}}$. Thus ${\textstyle |\mu _{B,k}(1+(p^{N}))|_{p}=p^{N}}$ which is not bounded above.

Definition 8. If ${\textstyle \mu }$ and ${\textstyle \nu }$ are two distribution (resp. measures) on X and ${\textstyle \alpha \in \mathbb {Q} _{p}}$ we define:

${\textstyle (\mu +\nu )(U)=\mu (U)+\nu (U)}$

${\textstyle (\alpha \mu )(U)=\alpha \mu (U)}$

${\textstyle \mu _{\alpha }(U)=\mu (\alpha U)}$ , where ${\textstyle \alpha \neq 0}$ and ${\textstyle \alpha U=\{x\in \mathbb {Q} _{p}|{\frac {x}{\alpha }}\in U\}}$

for all U compact open subsets of X.

Proposition 6. '${\textstyle \mu +\nu }$, ${\textstyle \alpha \mu }$, ${\textstyle \mu _{\alpha }}$ are distribution (resp. measures).'

Proof. It is a straightforward application of the definitions, so we omit the tedious details. ◻

Definition 9. Let ${\textstyle \alpha \in \mathbb {Z} }$ such that ${\textstyle \alpha \neq 1}$ and ${\textstyle p\not |\alpha }$ , we define the ${\textstyle k}$-th Bernoulli measure regularized by ${\textstyle \alpha }$ as $$\mu _{k,\alpha }(U)=\mu _{B,k}(U)-\alpha ^{-k}\mu _{B,k}(\alpha U).$$

By Proposition Proposition 6 we know that ${\textstyle \mu _{k,\alpha }}$ is a distribution, we will prove that it is also a measure. To do so we are going to focus on ${\textstyle \mu _{1,\alpha }}$ which is the basic case and then we will extend to ${\textstyle \mu _{k,\alpha }}$.

Examples 4.

1. ${\textstyle k=0}$: $$\mu _{0,\alpha }(a+(p^{N}))=\mu _{B,0}(a+(p^{N}))-\mu _{B,0}(\alpha (a+(p^{N})))=0,$$ since ${\textstyle \mu _{B,0}}$ is invariant for translation and ${\textstyle \alpha (a+(p^{N}))=\alpha a+(p^{N})}$.
2. ${\textstyle k=1}$: Errore del parser (funzione sconosciuta '\begin{split}'): {\displaystyle \begin{split} &\mu_{1,\alpha}(a+(p^N))=\mu_{B,1}(a+(p^N))-\alpha^{-1}\mu_{B,1}(\alpha(a+(p^N)))\\=&\frac{a}{p^N}-\frac{1}{2} -\alpha^{-1}(\frac{<\alpha a>_N}{p^N}-\frac{1}{2})=\frac{a}{p^N}+\frac{1-\alpha}{2\alpha}-\frac{1}{\alpha}(\frac{\alpha a}{p^N}-\left[\frac{\alpha a}{p^N}\right])\\=&\frac{1-\alpha}{2\alpha}+\frac{1}{\alpha}\left[\frac{\alpha a}{p^N}\right], \end{split}} where ${\textstyle <\alpha a>_{N}}$ is the only integer between ${\textstyle 0}$ and ${\textstyle p^{N}-1}$ which is congruent to ${\textstyle \alpha a}$ modulo ${\textstyle p^{N}}$ and ${\textstyle [\cdot ]}$ is the greatest integer function.

Lemma 3. 'A distribution ${\textstyle \mu }$ on X is a measure if and only if ${\textstyle |\mu (a+(p^{N}))|_{p}\leq C_{0}}$ for all ${\textstyle a+(p^{N})\subseteq X}$.'

Proof. If ${\textstyle \mu }$ is a measure ${\textstyle |\mu (a+(p^{N}))|_{p}}$ is bounded above since ${\textstyle a+(p^{N})}$ is a compact open. Conversely, if ${\textstyle U}$ is a compact open, by Proposition Proposition 2, Errore del parser (errore di sintassi): {\textstyle \\U=\bigcup\limits_{i=1}^{n}a_j+(p^{N_j})} , and this is a disjointed union. Thus, Errore del parser (errore di sintassi): {\textstyle \\|\mu(U)|_p=|\sum_{i=1}^n\mu(a_j+(p^{N_j}))|_p\leq \underset{i=1,...,n}{max}\{\mu(a_j+(p^{N_j}))\}\leq C_0} . ◻

Proposition 7. '${\textstyle |\mu _{1,\alpha }(U)|_{p}\leq 1}$ for all ${\textstyle U}$ compact-open of ${\textstyle \mathbb {Z} _{p}}$.'

Proof. By Lemma Lemma 3 it is sufficient to prove it for ${\textstyle a+(p^{N})}$ with ${\textstyle N\geq 0}$. By the example (2) ${\textstyle \mu _{1,\alpha }(a+(p^{N}))={\frac {1-\alpha }{2\alpha }}+{\frac {1}{\alpha }}\left[{\frac {\alpha a}{p^{N}}}\right]}$. Let’s first consider ${\textstyle p\neq 2}$, ${\textstyle \alpha }$ is an integer and ${\textstyle |\alpha |_{p}=1}$, thus ${\textstyle {\frac {1-\alpha }{2\alpha }}\in \mathbb {Z} _{p}}$ and ${\textstyle {\frac {1}{\alpha }}\left[{\frac {\alpha a}{p^{N}}}\right]\in \mathbb {Z} _{p}}$. Therefore ${\textstyle |\mu _{1,\alpha }(a+(p^{N}))|_{p}\leq 1}$. We have now to check the case ${\textstyle p=2}$. By definition ${\textstyle 2\not |p}$, thus ${\textstyle \alpha }$ is odd and ${\textstyle 1-\alpha }$ is even. ${\textstyle {\frac {1-\alpha }{2\alpha }}={\frac {\beta }{\alpha }}}$ for some ${\textstyle \beta \in \mathbb {Z} }$. Therefore ${\textstyle {\frac {1-\alpha }{2\alpha }}\in \mathbb {Z} _{p}}$ and we can conclude as before. ◻

Up to now we have proven ${\textstyle \mu _{1,\alpha }}$ is a measure, we want to extend this result to ${\textstyle \mu _{k,\alpha }}$ for all k.

Theorem 1. 'Let ${\textstyle d_{k}}$ be the least common denominator of the polynomial ${\textstyle B_{k}(x)}$. $$d_{k}\mu _{k,\alpha }(a+(p^{N}))\equiv d_{k}ka^{k-1}\mu _{1,\alpha }(a+(p^{N}))~(mod~p^{N}).$$'

Proof. For simplicity we omit to write ${\textstyle (mod~p^{N})}$, any congruence in this proof should be thought in this sense. First we notice that $$B_{k}({\frac {a}{p^{N}}})={\frac {a^{k}}{p^{Nk}}}-{\frac {k}{2}}{\frac {a^{k-1}}{p^{N(k-1)}}}+...$$ We can consider just these two first terms because the others have ${\textstyle p^{N(k-h)}}$ in the denominator for ${\textstyle h\geq 2}$, thus when we multiply by ${\textstyle p^{N(k-1)}}$ at least a factor ${\textstyle p^{N}}$ survives (and ${\textstyle d_{k}}$ removes every denominator of the polynomial). Thus,

Errore del parser (funzione sconosciuta '\begin{split}'): {\displaystyle \begin{split} &d_k\mu_{k,\alpha}(a+(p^N))\\\equiv &d_k\left(\mu_{B,k}(a+(p^N))-\alpha^{-k}\mu_{B,k}(\alpha a+(p^N))\right)\\ \equiv & d_kp^{N(k-1)}\left(B_k(\frac{a}{p^N})-\alpha^{-k}B_k(\frac{<\alpha a>_N)}{p^N}\right)\\\equiv & d_kp^{N(k-1)}\left( \frac{a^k}{p^{Nk}}-\frac{k}{2}\frac{a^{k-1}}{p^{N(k-1)}}-\alpha^{-k}\left((\frac{<\alpha a>_N}{p^N})^k-\frac{k}{2}(\frac{<\alpha a>}{p^N})^{k-1}\right)\right), \end{split}} As in the previous example ${\textstyle {\frac {<\alpha a>_{N}}{p^{N}}}={\frac {\alpha a}{p^{N}}}-\left[{\frac {\alpha a}{p^{N}}}\right]}$, in fact $${\frac {\alpha a}{p^{N}}}={\frac {<\alpha a>_{N}+tp^{N}}{p^{N}}}={\frac {<\alpha a>_{N}}{p^{N}}}+t,$$ also we notice that ${\textstyle 0\leq {\frac {<\alpha a>_{N}}{p^{N}}}\leq 1}$, thus ${\textstyle t=\left[{\frac {\alpha a}{p^{N}}}\right]}$. So the extreme right side of the previous congruence is congruent to $$d_{k}p^{N(k-1)}\left(\left({\frac {a^{k}}{p^{Nk}}}-\alpha ^{-k}({\frac {\alpha a}{p^{N}}}-\left[{\frac {\alpha a}{p^{N}}}\right])^{k}\right)-{\frac {k}{2}}\left({\frac {a^{k-1}}{p^{N(k-1)}}}-\alpha ^{-k}({\frac {\alpha a}{p^{N}}}-\left[{\frac {\alpha a}{p^{N}}}\right])^{k-1}\right)\right).$$Now, using the binomial expansion, and omitting every term with ${\textstyle p^{N(k-h)}}$ in the denominator for ${\textstyle h\geq 2}$, this is congruent to Errore del parser (funzione sconosciuta '\begin{split}'): {\displaystyle \begin{split} &d_k p^{N(k-1)}\left(\left( \frac{a^k}{p^{Nk}}-\alpha^{-k}(\frac{\alpha^k a^k}{p^{Nk}}-\frac{k\alpha^{k-1}a^{k-1}}{p^{N(k-1)}}\left[\frac{\alpha a}{p^N}\right])\right) -\frac{k}{2}\left(\frac{a^{k-1}}{p^{N(k-1)}}-\alpha^{-k}(\frac{\alpha^{k-1} a^{k-1}}{p^{N(k-1)}}\right)\right)\\ \equiv &d_kka^{k-1}\left(\frac{1}{\alpha}\left[\frac{\alpha a}{p^N}\right]+\frac{1-\alpha}{2\alpha}\right)\equiv d_kka^{k-1}\mu_{1,\alpha}(a+(p^N)). \end{split}} ◻

Corollary 2. 'for all k and ${\textstyle \alpha }$ (for which we have defined it) ${\textstyle \mu _{k,\alpha }}$ is a measure.'

Proof. By Theorem Theorem 1 $$|d_{k}\mu _{k,\alpha }(a+(p^{N}))-d_{k}ka^{k-1}\mu _{1,\alpha }(a+(p^{N}))|_{p}\leq p^{-N}\leq 1.$$ Thus, Errore del parser (funzione sconosciuta '\begin{split}'): {\displaystyle \begin{split} &|\mu_{k,\alpha}(a+(p^N))|_p \\ &\leq max(|\mu_{k,\alpha}(a+(p^N))-ka^{k-1}\mu_{1,\alpha}(a+(p^N))|_p,~ |ka^{k-1}\mu_{1,\alpha}(a+(p^N))|_p)\\& \leq max(|\frac{1}{d_k}|_p,~ 1), \end{split}} where we used the fact that ${\textstyle ka^{k-1}\in \mathbb {Z} _{p}}$ and Proposition Proposition 7. Since ${\textstyle d_{k}}$ is fixed when we choose k, we have proven that ${\textstyle \mu _{k,\alpha }}$ is bounded on balls, and therefore, by Lemma Lemma 3, on every compact-open. ◻

As we mentioned at the end of the third paragraph distributions are not sufficient to introduce integrals for continuous functions, but we can use measures to give a good and powerful definition of "${\textstyle p}$-adic integrals". We will first give the basic idea of the construction of integrals for real valued function. The definition for ${\textstyle \mathbb {Q} _{p}}$ is the same as we will see.

Definition 10. Let ${\textstyle f:[a,b]\rightarrow \mathbb {R} }$ be a continuous function defined on ${\textstyle [a,b]}$, a compact subset of ${\textstyle \mathbb {R} }$. We define the Riemann sums as: $$S_{N,\{x_{i,N}\}}(f)=\sum \limits _{0\leq i<N}f(x_{i,N}){\frac {b-a}{n}},$$ where ${\textstyle x_{i,N}\in [a+i{\frac {b-a}{n}},a+(i+1){\frac {b-a}{n}}].}$

Here we are basically splitting the interval ${\textstyle [a,b]}$ into smaller intervals Errore del parser (errore di sintassi): {\textstyle \\} ${\textstyle [a+i{\frac {b-a}{n}},a+(i+1){\frac {b-a}{n}}]}$. Then, we take a "random" point in each subinterval and we multiply the value of the function in this point by the length of the subinterval. We want this sums to converge to something as ${\textstyle N}$ grows to ${\textstyle \infty }$, so we can define the integral as the limit. This is actually true for real valued continuous functions. We want to do the same thing but with ${\textstyle \mathbb {Q} _{p}}$ valued functions instead.

Definition 11. Let ${\textstyle \mu }$ be a measure and X a compact open, ${\textstyle X\subseteq \mathbb {Z} _{p}}$ . Suppose Errore del parser (errore di sintassi): {\textstyle \\f:X\rightarrow\mathbb{Q}_{p}} is a continuous function. We define the Riemann sums as: $$S_{N,\{x_{a,N}\}}(f)=\sum \limits _{\underset {a+(p^{N})\subseteq X}{0\leq a<p^{N}}}f(x_{a,N})\mu (a+(p^{N})),$$ where ${\textstyle x_{a,N}}$ is an element in ${\textstyle a+(p^{N})}$.

Theorem 2. 'For any measure ${\textstyle \mu }$ and any continuous function ${\textstyle f:X\rightarrow \mathbb {Q} _{p}}$ ${\textstyle S_{N,\{x_{a,N}\}}(f)}$ converges to a limit in ${\textstyle \mathbb {Q} _{p}}$ as N goes to ${\textstyle \infty }$.'

Proof. Since ${\textstyle \mu }$ is a measure we can fix a constant C such that ${\textstyle |\mu (a+(p^{N}))|_{p}\leq C}$ for any ball ${\textstyle a+(p^{N})}$. Since ${\textstyle f}$ is continuous on a compact set, it is also uniformly continuous, i.e. for all ${\textstyle \epsilon >0}$ there exist N s.t. for all ${\textstyle x,y\in X}$ $$|x-y|_{p}\leq p^{-N}\Rightarrow |f(x)-f(y)|_{p}\leq \epsilon .$$ We have to prove ${\textstyle S_{N,\{x_{a,N}\}}(f)}$ is a Cauchy sequence. To do so we will consider only positive integers N such that any ${\textstyle a+(p^{N})}$ is either contained or outside ${\textstyle X}$, and we try to estimate ${\textstyle |S_{N,\{x_{a,N}\}}(f)-S_{M,\{x_{a,M}\}}(f)|_{p}}$ for some ${\textstyle M>N}$. Notice that: $$S_{N,\{x_{a,N}\}}(f)=\sum \limits _{\underset {a+(p^{N})\subseteq X}{0\leq a<p^{N}}}f(x_{a,N})\mu (a+(p^{N}))=\sum \limits _{\underset {a+(p^{M})\subseteq X}{0\leq a<p^{M}}}f(x_{{\overline {a}},N})\mu (a+(p^{M})),$$ where ${\textstyle {\overline {a}}}$ is the residue of a ${\textstyle (mod~p^{N})}$ (in this way the ${\textstyle x_{{\overline {a}},N}}$ are the same as the ${\textstyle x_{a,N}}$). Thus, Errore del parser (funzione sconosciuta '\begin{split}'): {\displaystyle \begin{split} |S_{N,\{x_{a,N}\}}(f)-S_{M,\{x_{a,M}\}}(f)|_p&=\left |\sum\limits_{\underset{a+(p^N)\subseteq X}{0\leq a<p^N} }(f(x_{\overline{a},N})-f(x_{a,M}))\mu(a+(p^M))\right|_p \\& \leq Cmax(|f(x_{\overline{a},N})-f(x_{a,M})|_p). \end{split}} But ${\textstyle |x_{{\overline {a}},N}-x_{a,M}|_{p}\leq p^{-N}}$. Therefore if we take N big enough, by the uniform continuity ${\textstyle |S_{N,\{x_{a,N}\}}(f)-S_{M,\{x_{a,M}\}}(f)|_{p}\leq C\epsilon }$. So we have proven the convergence of the Riemann sums. Errore del parser (errore di sintassi): {\textstyle \\} Now we only need to show the limit is independent of the choice of ${\textstyle \{x_{a,N}\}}$. Suppose for example to fix another sequence of elements ${\textstyle \{y_{a,N}\}}$ such that ${\textstyle y_{a,N}\in a+(p^{N})}$ and ${\textstyle \epsilon >0}$. Then Errore del parser (funzione sconosciuta '\begin{split}'): {\displaystyle \begin{split} |S_{N,\{x_{a,N}\}}(f)-S_{N,\{y_{a,N}\}}(f)|_p&=\left |\sum\limits_{\underset{a+(p^N)\subseteq X}{0\leq a<p^N}}(f(x_{a,N})-f(y_{a,N}))\mu(a+(p^N))\right|_p\\&\leq Cmax(|f(x_{a,N})-f(y_{a,N}|_p). \end{split}} Since ${\textstyle x_{a,N}}$ and ${\textstyle y_{a,N}}$ are elements of ${\textstyle a+(p^{N})}$, ${\textstyle |x_{a,N}-y_{a,N}|_{p}\leq p^{-N}}$. Thus if we take N big enough ${\textstyle max(|f(x_{a,N})-f(y_{a,N})|_{p})\leq \epsilon }$, by the uniform continuity of ${\textstyle f}$. Therefore the sequences ${\textstyle \{S_{N,{x_{a,N}}}(f)\}}$ and ${\textstyle \{S_{N,{y_{a,N}}}(f)\}}$ converge to the same limit. ◻

Definition 12. Let ${\textstyle f}$ be a continuous function from ${\textstyle X}$ to ${\textstyle \mathbb {Q} _{p}}$ , where X is a compact-open subset of ${\textstyle \mathbb {Q} _{p}}$. We define ${\textstyle \int \limits _{X}fd\mu }$ to be the limit of the Riemann sums defined respect to the measure ${\textstyle \mu }$. Theorem Theorem 2 guarantees this is a good definition.