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Let $\mathbb{Q}_p$ be the p-adic number field, $\mathbb{Z}_p$ its ring of integers. Let $\mathcal B$ be the smallest $\sigma$-algebra containing all the open subsets of $\mathbb{Q}_p$. Can we prove the following assertion without using the knowledge of Haar measures?

There exists a unique measure $\mu$ on $\mathcal B$ with the following properties.

1) $\mu(\mathbb{Z}_p)$ = 1.

2) For $a \in \mathbb{Q}_p$, $M \in \mathcal B$, $a + M \in \mathcal B$(we need to prove this) and $\mu(a + M) = \mu(M)$

The motivation is that we usually prove the existence of the Lebesgue measure on $\mathbb{R}$ without using the knowledge of Haar measures.

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  • $\begingroup$ Are there problems in mimicking the procedure for $\Bbb R$? $\endgroup$
    – Lubin
    Commented Mar 27, 2015 at 15:58

1 Answer 1

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Notation Let $A_1,A_2\cdots$ be a sequence of subsets of a set $X$. Suppose $A_1,A_2\cdots$ is mutually disjoint. We denote $\cup_n A_n$ by $\oplus A_n$ or by $A_1\oplus A_2 \cdots$.

Definition Let $X$ be a set. A non-empty family $\Pi$ of subsets of $X$ is called a semiring on $X$ if it satisfies the following conditions.

(1) $P\cap Q\in \Pi$ whenever $P\in \Pi$ and $Q\in \Pi$.

(2) $P - Q$ is a finite disjoint union of members of $\Pi$ whenever $P\in \Pi$ and $Q\in \Pi$.

Definition Let $\Pi$ be a semiring on a set $X$. A premeasure on $\Pi$ is a function $\lambda: \Pi \rightarrow [0, \infty]$ with the following properties.

(1) $\lambda(\emptyset) = 0$.

(2) If $P$ and $Q_1, Q_2, \cdots$ are members of $\Pi$ such that $P = \oplus_{n=1}^{\infty} Q_n$, then $\lambda(P) = \sum_{n=1}^{\infty} \lambda(P_n)$.

The following theorem is crucial in explicitly constructing a Haar measure on $\mathbb Q_p$. The theorem is slightly more general than the usual extension theorem of a measure defined on an algebra or a ring of sets. The proof is not so much different from the usual ones but I prove it anyway.

Theorem 1 Let $\Pi$ be a semiring on a set $X$ and let $\lambda$ is a premeasure on $\Pi$. Let $\mathcal S(\Pi)$ be the $\sigma$-algebra generated by $\Pi$. Suppose there exists a sequence $R_n, n = 1, 2, \cdots$ of members of $\Pi$ such that $X = \cup_{n=1}^{\infty} R_n$ and $\lambda(R_n) \lt \infty$ for all $n$. Then there exists a unique measure $\mu$ defined on $\mathcal S(\Pi)$ such that $\mu = \lambda$ on $\Pi$.

We need several lemmas to prove this.

Lemma 1 Let $\Pi$ be a semiring on a set $X$ and let $\lambda$ be a premeasure on $\Pi$. Suppose $P \subset Q$ where $P, Q\in \Pi$. Then $\lambda(P) \le \lambda(Q)$.

Proof: Since $\Pi$ is a semiring, $Q - P$ is a finite disjoint union of members of $\Pi$. Suppose $Q - P$ is a disjoint union of $P_1, \cdots, P_n$ where $P_i\in \Pi$ for all $i$. Then $Q$ is a disjoint union of $P, P_1,\cdots,P_n$. Hence $\lambda(Q) = \lambda(P) + \sum_{i=1}^{n} \lambda(P_i) \ge \lambda(P)$.

Lemma 2 Let $\Pi$ be a semiring on a set $X$ and let $\lambda$ be a premeasure on $\Pi$. Suppose there exists a sequence $R_n, n = 1, 2, \cdots$ of members of $\Pi$ such that $X = \cup_n R_n$. Let $A$ be a subset of $X$. Define $\mu^*(A) = \text{inf}\{\sum_n \lambda(P_n): A \subset \cup_n P_n, P_n\in \Pi, n = 1,2,\cdots\}$. Then $\mu^*$ is a Caratheodory outer measure.

Proof: Let $A_n, n = 1, 2, \cdots$ be a sequence of subsets of $X$ and let $A = \cup_n A_n$. It suffices to prove that $\mu^*(A) \le \sum_n \mu^*(A_n)$. If $\mu^*(A_n) = \infty$ for some $n$, then the inequality trivially holds. Hence we may assume $\mu^*(A_n) \lt \infty$ for all $n$. Choose $\epsilon\gt 0$. For each $n$, there exists a sequence $P_{nm}, m = 1,2,\cdots$ of members of $\Pi$ such that $A_n\subset \cup_m P_{nm}$ and $\sum_m \lambda(P_{nm}) \lt \mu^*(A_n) + \epsilon/2^n$. Then $\mu^*(A) \le \sum_{n,m} \lambda(P_{nm}) \le \sum_n \mu^*(A_n) + \epsilon$. Letting $\epsilon \rightarrow 0$, we have the desired inequality.

Lemma 3 Let $\Pi$ be a semiring on a set $X$ and let $\lambda$ be a premeasure on $\Pi$. Suppose there exists a sequence $R_n, n = 1, 2, \cdots$ of members of $\Pi$ such that $X = \cup_n R_n$. Let $A$ be a subset of $X$. Define $\mu^*(A) = \text{inf}\{\sum_n P_n: A \subset \cup_n P_n, P_n\in \Pi, n = 1,2,\cdots\}$. A subset $E$ of $X$ is said to be $\mu^*$-measurable if it satisfies the following condition.

For every subset $S$ of $X$, $\mu^*(S) \ge \mu^*(S\cap E) + \mu^*(S-E)$.

Let $\mathcal M$ be the family of $\mu^*$-measurable sets. Let $\mu$ be the restriction of $\mu^*$ on $\mathcal M$. Then $\mathcal M$ is a $\sigma$-algebra and $\mu$ is a measure on $\mathcal M$.

Proof: By Lemma 2, $\mu^*$ is a Caratheodory outer measure. The assertion is well-known and the proof is omitted.

Lemma 4 Let $\Pi$ be a semiring on a set $X$. Let $P_n, n = 1, 2, \cdots$ be a sequence of members of $\Pi$. Let $Q_1 = P_1$. Let $Q_n = P_n - (P_1\cup\cdots\cup P_{n-1})$ for $n\gt 1$. Then $Q_n, n = 1, 2, \cdots$ is a disjoint sequence and $\cup_n P_n = \cup_n Q_n$. Furthermore each $Q_n$ is a finite disjoint union of members of $\Pi$.

Proof: Clearly $Q_n, n = 1, 2, \cdots$ is a disjoint sequence and $\cup_n P_n = \cup_n Q_n$. Since $Q_n = \cap_{i=1}^{n-1} (P_n - P_i)$ for $n \gt 1$, $Q_n$ is a finite disjoint union of members of $\Pi$.

Lemma 5 Let $\Pi$ be a semiring on a set $X$ and let $\lambda$ be a premeasure on $\Pi$. Supose $P\subset \cup_n P_n$ where $P, P_1, P_2,\cdots$ are members of $\Pi$. Then $\lambda(P) \le \sum_n\lambda(P_n)$.

Proof: By Lemma 4, there exists a disjoint sequence $Q_1, Q_2,\cdots$ of subsets of $X$ such that $Q_n\subset P_n$ for all $n$ and $\cup_n P_n = \cup_n Q_n$ and $Q_n$ is a finite disjoint union of members of $\Pi$. Suppose $Q_n$ is a finite disjoint union of $R_{n1},\cdots,R_{nk_n}$. $\lambda(P) = \sum_{n,k} \lambda(P\cap R_{nk}) \le \sum_{n,k} \lambda(R_{nk}) \le \sum_n \lambda(P_n)$ and we are done.

Lemma 6 Let $\Pi$ be a semiring on a set $X$ and let $\lambda$ be a premeasure on $\Pi$. Suppose there exists a sequence $R_n, n = 1, 2, \cdots$ of members of $\Pi$ such that $X = \cup_n R_n$. Let $\mu^*$ is the Caratheodory outer measure as defined in Lemma 2. Then $\mu^*(P) = \lambda(P)$ for all $P\in \Pi$.

Proof: Suppose $P \subset \cup_n P_n$ where $P_n\in\Pi$ for $n = 1, 2, \cdots$. Then $\lambda(P) \le \sum_n\lambda(P_n)$ by Lemma 5. Hence $\lambda(P) \le \mu^*(P)$. Since $P\subset P$, $\mu^*(P)\le \lambda(P)$ and we are done.

Lemma 7 Let $\Pi$ be a semiring on a set $X$ and let $\lambda$ be a premeasure on $\Pi$. Suppose there exists a sequence $R_n, n = 1, 2, \cdots$ of members of $\Pi$ such that $X = \cup_n R_n$. Let $\mathcal S(\Pi)$ be the $\sigma$-algebra generated by $\Pi$. Let $\mu^*$ be the Caratheodory outer measure as defined in Lemma 2. A subset $E$ of $X$ is said to be $\mu^*$-measurable if it satisfies the following condition.

For every subset $S$ of $X$, $\mu^*(S) \ge \mu^*(S\cap E) + \mu^*(S-E)$.

Then every set in $\mathcal S(\Pi)$ is $\mu^*$-measurable.

Proof: Let $\mathcal M$ be the family of $\mu^*$-measurable sets. By Lemma 3, $\mathcal M$ is a $\sigma$-algebra. Hence it suffices to prove that $\Pi\subset \mathcal M$. Let $P$ be a member of $\Pi$. Let $S$ be any subset of $X$. We need to prove that $\mu^*(S) \ge \mu^*(S\cap P) + \mu^*(S-P)$. If $\mu^*(S) = \infty$, this is trivially true. So we may assume $\mu^*(S) \lt \infty$. Choose $\epsilon \gt 0$. There exists a sequence $Q_n, n = 1,2,\cdots$ of members of $\Pi$ such that $S\subset \cup_n Q_n$ and $\sum_n \lambda(Q_n) \lt \mu^*(S) + \epsilon$. Then $S\cap P \subset \cup_n(Q_n\cap P)$,$S - P \subset \cup_n(Q_n - P)$. Hence $\mu^*(S\cap P) + \mu^*(S - P) \le \sum_n \mu^*(Q_n\cap P) + \sum_n \mu^*(Q_n - P)$ $= \sum_n(\mu^*(Q_n\cap P) + \mu^*(Q_n - P))$. Since $\Pi$ is a semiring, for each $n$ there exists a finite disjoint sequence $Q_{n1},\cdots, Q_{nk_n}$ of members of $\Pi$ such that $Q_n - P = \oplus_{i=1}^{k_n} Q_{ni}$ Hence $\mu^*(Q_n - P) \le \sum_{i=1}^{k_n} \mu^*(Q_{ni})$. Hence $\mu^*(S\cap P) + \mu^*(S - P) \le \sum_n(\mu^*(Q_n\cap P) + \mu^*(Q_{ni}))$ $= \sum_n(\lambda(Q_n\cap P) + \lambda(Q_{ni})) = \sum_n \lambda(Q_n) \lt \mu^*(S) + \epsilon$. Letting $\epsilon \rightarrow 0$, we have $\mu^*(S\cap P) + \mu^*(S - P) \le \mu^*(S)$ as desired.

The proof of Theorem 1 Let $\mu^*$ be the Caratheodory outer measure as defined in Lemma 2. Let $\mathcal M$ be the family of $\mu^*$-measurable sets. Let $\mu$ be the restriction of $\mu^*$ on $\mathcal M$. By Lemma 3, $\mathcal M$ is a $\sigma$-algebra and $\mu$ is a measure on $\mathcal M$. By Lemma 7, $\mathcal S(\Pi) \subset \mathcal M$. By Lemma 6, $\mu = \lambda$ on $\Pi$. Hence $\mu$ satisfies the condition of the theorem.

It remains to prove the uniqueness of such a measure. Let $\nu$ be a measure on $\mathcal S(\Pi)$ such that $\nu = \lambda$ on $\Pi$. Let $E$ be a member of $\mathcal S(\Pi)$. Let $P_n, n = 1, 2, \cdots$ be a sequence of members of $\Pi$ such that $E\subset \cup_n P_n$. Then $\nu(E) \le \sum_n \nu(P_n) = \sum_n \lambda(P_n)$. Hence $\nu(E) \le \mu(E)$. We may assume that $X = \oplus_n R_n$. Then $E = \oplus_n (R_n\cap E)$. Since $\nu(E) = \sum_n \nu(R_n\cap E)$ and $\mu(E) = \sum_n \mu(R_n\cap E)$, we may suppose that $E \subset R_n$ for some $n$. Then $\lambda(R_n) = \nu(E) + \nu(R_n - E)$. Hence $\lambda(R_n) - \nu(E) = \nu(R_n - E) \le \mu(R_n - E) = \lambda(R_n) - \mu(E)$. Hence $-\nu(E) \le -\mu(E)$ and we are done.

Definition Let $\mathcal B(\mathbb Q_p)$ be the $\sigma$-algebra generated by the family of open subsets of $\mathbb Q_p$. A Haar measure $\mu$ on $\mathbb Q_p$ is a measure defined on $\mathcal B(\mathbb Q_p)$ with the following properties.

(1) $\mu\neq 0$.

(2) $\mu(K) \lt \infty$ for all compact subset $K$ of $\mathbb Q_p$.

(3) $\mu(a + M) = \mu(M)$ for all $a\in \mathbb Q_p$ and all $M\in \mathcal B(\mathbb Q_p)$.

We will explicitly construct a Haar measure $\mu$ on $\mathbb Q_p$ such that $\mu(\mathbb Z_p) = 1$. Furthermore we will prove the following fact: If $\nu$ is a Haar measure on $\mathbb Q_p$, then there exists $c\gt 0$ such that $\nu = c\mu$.

Definition A member of the family $\Gamma = \{a + p^n\mathbb Z_p: a\in \mathbb Q, n \ge 0\} \cup \{\emptyset\}$ is called a fundamental set. A fundamental set is a compact and open subset of $\mathbb Q_p$ and $\Gamma$ is a base of the topology of $\mathbb Q_p$.

Proposition 1 The family $\Gamma$ of fundamental sets of $\mathbb Q_p$ is a semiring on $\mathbb Q_p$.

Proof: Suppose $P = a + p^n\mathbb Z_p$ and $Q = b + p^m\mathbb Z_p$. Without loss of generality we may assume $m \ge n$. Let $Q' = b + p^n\mathbb Z_p$. Since $p^m\mathbb Z_p \subset p^n\mathbb Z_p$, $Q \subset Q'$. Since $P$ and $Q'$ are cosets of $\mathbb Q_p/p^n\mathbb Z_p$, either $P = Q'$ or $P\cap Q' = \emptyset$. If $P = Q'$, then $Q \subset P$. Hence $P \cap Q = Q$ and $P - Q$ is a finite disjoint union of members of $\Gamma$. If $P\cap Q' = \emptyset$, then $P \cap Q = \emptyset$ and we are done.

Definition Let $\Gamma$ be the family of fundamental sets of $\mathbb Q_p$. If $P = a + p^n\mathbb Z_p$ is a member of $\Gamma$, we write $|P| = 1/p^n$. We write $|\emptyset| = 0$.

Lemma 8 Let $n\ge 0$ be an integer and let $P$ be a fundamental set of $\mathbb Q_p$ such that $|P| = 1/p^n$. Suppose $P = Q_1 \oplus \cdots \oplus Q_r$ and $|Q_1| = \cdots = |Q_r|$, where $Q_1,\cdots, Q_r$ are fundamental sets. Then $|P| = |Q_1| + \cdots + |Q_r|$.

Proof: Suppose $P = a + p^n\mathbb Z_p$. Then $p^n\mathbb Z_p = -a + P = (-a + Q_1) \oplus \cdots \oplus (-a + Q_r)$.Since $|-a + P| = |P|$ and $|-a + Q_i| = |Q_i|$ for all $i$, we may assume that $a = 0$. Suppose $|Q_1| = \cdots = |Q_r| = 1/p^m$. Then $Q_1,\cdots, Q_r$ are the full cosets of $p^n\mathbb Z_p/p^m\mathbb Z_p$. Hence $r = p^{m-n}$ and $|P| = r(1/p^m) = |Q_1| + \cdots + |Q_r|$.

Lemma 9 Let $P_1,\cdots, P_m, Q$ be fundamental sets of $\mathbb Q_p$. Suppose $|P_1| = \cdots = |P_r| = 1/p^n$ and $Q\subset P_1 \cup \cdots \cup P_r$ and suppose $|Q| = 1/p^m$ and $m \ge n$. Then $Q \subset P_i$ for some $i$.

Proof: Suppose $P_i = a_i + p^n\mathbb Z_p$, $i = 1,2,\cdots$ and $Q = b + p^m\mathbb Z_p$. Then $b \in a_i + p^n\mathbb Z_p$ for some $i$. Hence $b - a_i \in p^n\mathbb Z_p$. This implies $b + p^n\mathbb Z_p = a_i + p^n\mathbb Z_p = P_i$. Since $m\ge n$, $Q \subset b + p^n\mathbb Z_p$. Hence $Q \subset P_i$ as desired.

Proposition 2 Let $\Gamma$ be the family of fundamental sets of $\mathbb Q_p$. Let $P, P_1,\cdots,P_m$ be members of $\Gamma$ such that $P = P_1 \oplus \cdots \oplus P_m$. Then $|P| = |P_1| + \cdots + |P_m|$.

Proof: We use indunction on $m$. The assertion trivially holds if $m = 1$. Suppose $m\gt 1$ and $P = a + p^n\mathbb Z_p$ and $P_i = a_i + p^{n_i}\mathbb Z_p$, $i = 1,\cdots, m$. Without loss of generality, we may assume that $n_1 = \text{min}\{n_1,\cdots,n_m\}$. Since $P$ has a partition consisting of cosets of $\mathbb Q_p/p^{n_1}\mathbb Z_p$, there exist $Q_2,\cdots,Q_r \in \Pi$ such that $|P_1| = |Q_j| = 1/p^{n_1}, j = 2,\cdots,r$ and $P = P_1 \oplus Q_2 \oplus \cdots \oplus Q_r$. Since $P_1, Q_2,\cdots,Q_r$ are cosets of $\mathbb Q_p/p^{n_1}\mathbb Z_p$, we have $|P| = |P_1| + |Q_2| + \cdots + |Q_r|$. Since $n_1 = \text{min}\{n_1,\cdots,n_m\}$, by Lemma 9 every $P_i, i\gt 1$ belongs to some $Q_j$. Hence $Q_j$ has a partition consisting of members of $\{P_2,\cdots,P_m\}$ whose cardinality is less than $m$. By using the induction hypothesis, we are done.

Lemma 10 Let $\Gamma$ be the family of fundamental sets of $\mathbb Q_p$. Let $P, P_1,\cdots,P_n$ be members of $\Gamma$ such that $P \subset P_1 \cup \cdots \cup P_n$. Then $|P| \le |P_1| + \cdots + |P_n|$.

Proof: Since $P = (P\cap P_1) \cup\cdots \cup (P\cap P_n)$ and each $P\cap P_i \in \Gamma$ and $|P\cap P_i| \le |P_i|$, we may assume that $P = P_1\cup \cdots \cup P_n$. By the lemma of this question(Finitely additive function on a product of semirings of sets), there exist $Q_1, ..., Q_m \in \Gamma$ with the following properties.

(1) $P_1 \cup \cdots \cup P_n = Q_1 \oplus \cdots \oplus Q_m$

(2) Each $P_i$ is a union of a subset of $\{Q_1, ... ,Q_m\}$.

By Proposition 2, $|P| = |Q_1| + \cdots + |Q_m|$. Then the assertion follows immediately.

Proposition 3 Let $\Gamma$ be the family of fundamental sets of $\mathbb Q_p$. Let $P, P_1,P_2,\cdots$ be members of $\Gamma$ such that $P \subset \cup_{n=1}^{\infty} P_n$. Then $|P| \le \sum_{n=1}^{\infty} |P_n|$.

Proof: Since $P$ is compact and each $P_i$ is open, there exists $m$ such that $P \subset P_1\cup \cdots \cup P_m$. By Lemma 1, $|P| \le |P_1| + \cdots + |P_m| \le \sum_{n=1}^{\infty} |P_n|$ and we are done. Proposition 4 Let $P, P_1,P_2,\cdots$ be fundamental sets of $\mathbb Q_p$ such that $P = \oplus_{n=1}^{\infty} P_n$. Then $|P| = \sum_{n=1}^{\infty} |P_n|$.

Proof: Choose $m \ge 1$. By Proposition 2, $|P_1 \oplus \cdots \oplus P_m| = |P_1| + \cdots + |P_m|$. Since $P_1 \oplus \cdots \oplus P_m \subset P$, $|P_1| + \cdots + |P_m| \le |P|$. Letting $m \rightarrow \infty$, we have $\sum_{n=1}^{\infty} |P_n| \le |P|$. By Proposition 3, $|P| \le \sum_{n=1}^{\infty} |P_n|$ and we are done.

Theorem 2 There exists a unique Haar measure $\mu$ on $\mathbb Q_p$ such that $\mu(\mathbb Z_p) = 1$. If $\nu$ is a Haar measure on $\mathbb Q_p$, then there exists $c\gt 0$ such that $\nu = c\mu$.

Proof: Let $\Gamma$ be the family of fundamental sets of $\mathbb Q_p$. By Proposition 1, $\Gamma$ is a semiring. By Proposition 4, $|*|$ is a premeasure on $\Gamma$. Let $\mathcal S(\Gamma)$ be the $\sigma$-algebra generated by $\Gamma$. Since $\Gamma$ is a base of the topology of $\mathbb Q_p$, $\mathcal S(\Gamma) = \mathcal B(\mathbb Q_p)$. Since $\Gamma$ is countable, there exists a sequence $R_n, n = 1, 2, \cdots$ of members of $\Gamma$ such that $\mathbb Q_p = \cup_n R_n$. Hence by Theorem 1, there exists a unique measure $\mu$ on $\mathcal B(\mathbb Q_p)$ such that $\mu(P) = |P|$ for all $P\in \Gamma$. Let $K$ be a compact subset of $\mathbb Q_p$. There exists a finite sequence $P_1,\cdots, P_n$ of members of $\Pi$ such that $K \subset \cup_{i=1}^{n} P_i$. Then $\mu(K) \le \sum_{i=1}^{n} |P_i| \lt \infty$. Choose $x_0\in \mathbb Q_p$. Define $\nu(E) = \mu(x_0 + E)$ for all $E \in \mathcal B(\mathbb Q_p)$. Then $\nu$ is a measure on $\mathcal B(\mathbb Q_p)$. Let $P\in\Gamma$. Since $\mathbb Q$ is dense in $\mathbb Q_p$, there exists $a\in \mathbb Q$ such that $x_0 + P = a + P$. Hence $\nu(P) = \mu(a + P) = |P| = \mu(P)$. Hence by Theorem 1, $\nu = \mu$. Hence $\mu(x_0 + E) = \mu(E)$ for all $E\in \mathcal B(\mathbb Q_p)$. Thus $\mu$ is a Haar measure.

Let $\nu$ be a Haar measure on $\mathbb Q_p$ such that $\nu(\mathbb Z_p) = 1$. Let $n\ge 0$ be an integer. Since the order of the group $\mathbb Z_p/p^n\mathbb Z_p$ is $p^n$, $\nu(p^n\mathbb Z_p) = 1/p^n$. Hence $\nu(P) = |P|$ for all $P\in\Gamma$. Then by Theorem 1, $\nu = \mu$.

Let $\gamma$ be a Haar measure on $\mathbb Q_p$. Suppose $\gamma(\mathbb Z_p) = c$. Since $\mathbb Q_p = \cup \{a + \mathbb Z_p: a \in \mathbb Q\}$, If $c = 0$, then $\gamma(\mathbb Q_p) = 0$. This is a contradiction. Hence $c\gt 0$. Then $(1/c)\gamma$ is a Haar measure such that $(1/c)\gamma(\mathbb Z_p) = 1$. By what we have just proved, $(1/c)\gamma = \mu$. Hence $\gamma = c\mu$. This completes the proof.

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