Induced Outer Measure Restricted to Semiring is Pre-Measure

Theorem
Let $\mathcal S$ be a semiring over a set $X$.

Let $\mu: \mathcal S \to \overline{\R}_{\ge0}$ be a pre-measure on $\mathcal S$, where $\overline{\R}_{\ge 0}$ denotes the set of positive extended real numbers.

Let $\mu^*: \mathcal P \left({X}\right) \to \overline{\R}_{\ge0}$ be the outer measure induced by $\mu$.

Then:
 * $\displaystyle \mu^*\restriction_{\mathcal S} \, = \mu$

where $\restriction$ denotes restriction.

Proof
Let $S \in \mathcal S$.

It follows immediately from the definition of the induced outer measure that $\mu^* \left({S}\right) \le \mu \left({S}\right)$.

Therefore, it suffices to show that if $\left({A_n}\right)_{n=1}^{\infty}$ is a countable cover for $S$, then:
 * $\displaystyle \mu \left({S}\right) \le \sum_{n=1}^\infty \mu \left({A_n}\right)$

If the above statement is true, then it follows directly from the definition of infimum that $\mu \left({S}\right) \le \mu^* \left({S}\right)$, thus proving the theorem.

Define, for all $n \in \N$:
 * $\displaystyle B_n = A_n \setminus A_{n-1} \setminus \cdots \setminus A_1$

When $n = 1$, the above expression is taken to be $A_1$.

Using the mathematical induction, we will prove that for all natural numbers $m < n$, $B_{n, m} = A_n \setminus A_{n-1} \setminus \cdots \setminus A_{n-m}$ is the union of finitely many pairwise disjoint elements of $\mathcal S$.

The base case $m = 1$ follows directly from the definition of a semiring.

Now assume that the above statement is true for some natural number $m < n - 1$, and let $D_1, D_2, \ldots, D_{n_m}$ be pairwise disjoint elements of $\mathcal S$ such that:
 * $\displaystyle B_{n, m} = \bigcup_{k=1}^{n_m} D_k$

Then:

Now, for all natural numbers $k \le n_m$, $D_k \setminus A_{n-m-1}$ is the union of finitely many pairwise disjoint elements of $\mathcal S$.

Hence $B_{n, m+1}$ is the union of finitely many pairwise disjoint elements of $\mathcal S$.

Therefore, $B_{n, n - 1} = B_n$ is the union of finitely many pairwise disjoint elements of $\mathcal S$, as desired.

For all $n \in \N$, let $\mathcal F_n$ be the set of all finite collections $\left\{{D_1, D_2, \ldots, D_m}\right\}$ of pairwise disjoint elements of $\mathcal S$ such that $\displaystyle B_n = \bigcup_{k=1}^m D_k$.

Applying the axiom of countable choice, there exists a sequence $\left\langle{f_n}\right\rangle_{n \in \N} = \left\langle{\left\{{D_{n, 1}, D_{n, 2}, \ldots, D_{n, m_n} }\right\}}\right\rangle_{n \in \N}$ such that $f_n \in \mathcal F_n$ for all $n \in \N$.

Now, $x \in S$ if and only if there exists an $n \in \N$ such that $x \in S \cap A_n$.

Taking the least possible $n$, it follows that $x \notin A_1, A_2, \ldots, A_{n-1}$, and so $x \in S \cap B_n$.

Therefore:
 * $\displaystyle S = \bigcup_{n=1}^\infty \, \left({S \cap B_n}\right)$

Hence: