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 $\displaystyle \left({A_n}\right)_{n \mathop = 0}^{\infty}$ is a countable cover for $S$, then: :$\displaystyle \mu \left({S}\right) \le \sum_{n \mathop = 0}^\infty \mu \left({A_n}\right)$ 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. {{refactor|The structure of this proof needs to be clarified. I recommend that the induction part of it be extracted into a separately-paged lemma, structured as our usual house style.}} Define, for all natural numbers $n \in \N$: :$\displaystyle B_n = A_n \setminus A_{n-1} \setminus \cdots \setminus A_0$ 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 $\mathcal S$. We take $B_{n, 0} = A_n$. The base case $m = 0$ is trivial. {{Disambiguate|Definition: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 $\mathcal S$ such that: :$\displaystyle B_{n, m} = \bigcup_{k \mathop = 1}^N D_k$ Then: {{begin-eqn}} {{eqn | l = B_{n, m+1}       | r = B_{n, m} \setminus A_{n-m-1} }} {{eqn | r = \left({\bigcup_{k \mathop = 1}^N D_k}\right) \setminus A_{n-m-1} }} {{eqn | r = \bigcup_{k \mathop = 1}^N \, \left({D_k \setminus A_{n-m-1} }\right)       | c = by Set Difference is Right Distributive over Union }} {{end-eqn}} 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 $\mathcal S$. Hence $B_{n, m+1}$ is the finite union of pairwise disjoint elements of $\mathcal S$, completing the induction step. Therefore, $B_{n, n-1} = B_n$ is the finite union of pairwise disjoint elements of $\mathcal S$, as desired. Using the above result and applying the axiom of countable choice, we can, for all $n \in \N$, choose a finite set $\mathcal F_n$ of pairwise disjoint elements of $\mathcal S$ for which: :$\displaystyle B_n = \bigcup \mathcal F_n$ Now, $x \in S$ if and only if there exists an $n \in \N$ such that $x \in S \cap A_n$. Taking the 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: :$\displaystyle S = \bigcup_{n \mathop = 0}^\infty \, \left({S \cap B_n}\right)$ Hence: {{begin-eqn}} {{eqn | l = \mu \left({S}\right)       | r = \mu \left({\bigcup_{n \mathop = 0}^\infty \, \left({S \cap B_n}\right)}\right) }} {{eqn | r = \mu \left({\bigcup_{n \mathop = 0}^\infty \, \left({S \cap \bigcup_{T \mathop \in \mathcal F_n} T}\right)}\right)       | c = by definition of $\mathcal F_n$ }} {{eqn | r = \mu \left({\bigcup_{n \mathop = 0}^\infty \, \bigcup_{T \mathop \in \mathcal F_n} \, \left({S \cap T}\right)}\right)       | c = by Intersection Distributes over Union }} {{eqn | o = \le       | r = \sum_{n \mathop = 0}^\infty \, \sum_{T \mathop \in \mathcal F_n} \, \mu \left({S \cap T}\right)       | c = because the countable union of finite sets is countable, and by the countable subadditivity of $\mu$ }} {{eqn | o = \le       | r = \sum_{n \mathop = 0}^\infty \, \sum_{T \mathop \in \mathcal F_n} \, \mu \left({T}\right)       | c = by monotonicity }} {{eqn | r = \sum_{n \mathop = 0}^\infty \, \mu \left({\bigcup \mathcal F_n}\right)       | c = by additivity }} {{eqn | r = \sum_{n \mathop = 0}^\infty \, \mu \left({B_n}\right)       | c = by definition of $\mathcal F_n$ }} {{eqn | o = \le       | r = \sum_{n \mathop = 0}^\infty \, \mu \left({A_n}\right)       | c = by Set Difference is Subset and monotonicity }} {{end-eqn}} {{qed}} {{ACC}} Category:Measure Theory bcf6t68sq0w2z9x8ibtxotyyug6dso8