Induced Outer Measure Restricted to Semiring is Pre-Measure

Theorem
Let $\mu^*$ be the outer measure induced by a pre-measure $\mu$ defined on a semiring $\mathcal S$ over a set $X$.

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:
 * $\displaystyle \mu \left({S}\right) \le \sum_{n=1}^\infty \mu \left({A_n}\right)$

Define:
 * $\displaystyle B_n = A_n \setminus A_{n-1} \setminus \cdots \setminus A_1$

for all $n \in \N$. When $n = 1$, the above expression is taken to be $A_1$.

By the definition of a semiring, $A_n \setminus A_{n-1}$ is the union of finitely many pairwise disjoint elements of $\mathcal S$.

Thus by Set Difference is Right Distributive over Union, $A_n \setminus A_{n-1} \setminus A_{n-2}$ is the union of finitely many pairwise disjoint sets of the form $A \setminus B$ where $A, B \in \mathcal S$, and is therefore also the union of finitely many pairwise disjoint elements of $\mathcal S$.

Repeating the process until one arrives at $B_n$, it follows that $B_n$ is the union of finitely many pairwise disjoint $D_{n, 1}, D_{n, 2}, \ldots, D_{n, m_n} \in \mathcal S$.

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: