Induced Outer Measure Restricted to Semiring is Pre-Measure

Theorem
Let $\SS$ be a semiring over a set $X$.

Let $\mu: \SS \to \overline \R_{\ge 0}$ be a pre-measure on $\SS$, where $\overline \R_{\ge 0}$ denotes the set of positive |extended real numbers.

Let $\mu^*: \powerset X \to \overline \R_{\ge 0}$ be the outer measure induced by $\mu$.

Then:
 * $\ds \mu^*\restriction_\SS \, = \mu$

where $\restriction$ denotes restriction.

Proof
Let $S \in \SS$.

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

Therefore, it suffices to show that if $\ds \sequence {A_n}_{n \mathop = 0}^\infty$ is a countable cover for $S$, then:
 * $\ds \map \mu S \le \sum_{n \mathop = 0}^\infty \map \mu {A_n}$

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

Define, for all natural numbers $n \in \N$:
 * $\ds B_n = A_n \setminus A_{n - 1} \setminus \cdots \setminus A_0$

where $\setminus$ denotes set difference.

We take $B_0 = A_0$.

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 finite union of pairwise disjoint elements of $\SS$.

We take $B_{n, 0} = A_n$.

The base case $m = 0$ is trivial.

Now assume the induction hypothesis that the above statement is true for some natural number $m < n - 1$, and let $D_1, D_2, \ldots, D_N$ be pairwise disjoint elements of $\SS$ such that:
 * $\ds B_{n, m} = \bigcup_{k \mathop = 1}^N D_k$

Then:

By the definition of a semiring, for all natural numbers $k \le N$, $D_k \setminus A_{n - m - 1}$ is the finite union of pairwise disjoint elements of $\SS$.

Hence $B_{n, m+1}$ is the finite union of pairwise disjoint elements of $\SS$, completing the induction step.

Therefore, $B_{n, n-1} = B_n$ is the finite union of pairwise disjoint elements of $\SS$, as desired.

Using the above result and applying the axiom of countable choice, we can, for all $n \in \N$, choose a finite set $\FF_n$ of pairwise disjoint elements of $\SS$ for which:
 * $\ds B_n = \bigcup \FF_n$

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

Taking the smallest such $n$, which exists because $\N$ is well-ordered, it follows that $x \notin A_0, A_1, \ldots, A_{n - 1}$, and so $x \in S \cap B_n$.

Therefore:
 * $\ds S = \bigcup_{n \mathop = 0}^\infty \paren {S \cap B_n}$

Hence: