Let $G$ be a profinite group. It is known that in $G$, every neighborhood of the identity element contains an open compact subgroup. I would like to explicitly construct the Haar measure on $G$. The Haar measure $\mu$ on $G$ is a measure defined on the $\sigma$-algebra $\mathcal B(G)$ of all Borel sets with the following properties.
1) $\mu(G) = 1$.
2) $\mu(E) = \text{sup} \{\mu(K) : K \subset E$, where $K$ is compact$\}$ for every $E \in \mathcal B(G)$.
3) $\mu(E) = \text{inf} \{\mu(U): E \subset U$, where $U$ is open $\}$ for every $E\in \mathcal B(G)$.
4) $\mu(aE) = \mu(E)$ whenever $a\in G$ and $E\in \mathcal B(G)$.
Let $\Pi$ be the set of subsets of $G$ that are either the empty set or of the form $aH$, where $a\in G$ and $H$ is an open compact subgroup. It can be easily proved that $\Pi$ is a semiring of sets: See Intersection of cosets from possibly distinct subgroups is either empty or a coset of the intersection between the two subgroups
Define $\lambda(\emptyset) = 0$ and $\lambda(aH) = 1/(G : H)$ where $a\in G$ and $H$ is an open compact subgroup of $G$.
Here are my questions:
1) How do you prove that $\lambda$ is finitely additive on $\Pi$. In other words, if $U, U_1, \cdots, U_n \in \Pi$ and $U = \bigcup_{i=1}^n U_i$ is a disjoint union, how do you prove that $\lambda(U) = \sum_{i=1}^n \lambda(U_i)$?
2) How do you construct the Haar measure $\mu$ by using $\lambda$?
