# Explicit construction of Haar mesure on the p-adic number field

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.

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

## 1 Answer

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.