The first step is to extend the set function $\mu_0$ to the power set $P(X)$ . For any subset $S\subseteq X$ the value of $\mu^*(S)$ is defined by taking sequences $S_i$ in $A$ which cover $S$ , \begin{equation}\label{eq:1} \mu^*(S)\equiv\inf\left\{\sum_{i=1}^\infty \mu_0(S_i): S_i\in A,\ S\subseteq\bigcup_{i=1}^\infty S_i\right\}. \end{equation}We show that this is an outer measure. First, it is clearly non-negative. Secondly, if $S=\emptyset$ then we can take $S_i=\emptyset$ in () to obtain $\mu^*(S)\le\sum_i\mu_0(\emptyset)=0$ , giving $\mu^*(\emptyset)$ =0. It is also clear that $\mu^*$ is increasing, so that if $S\subseteq T$ then $\mu^*(S)\le\mu^*(T)$ . The only remaining property to be proven is subadditivity. That is, if $S_i$ is a sequence in $P(X)$ then \begin{equation}\label{eq:2} \mu^*\left(\bigcup_i S_i\right)\le\sum_i\mu^*(S_i). \end{equation}To prove this inequality, choose any $\epsilon>0$ and, by the definition () of $\mu^*$ , for each $i$ there exists a sequence $S_{i,j}\in A$ such that $S_i\subseteq\bigcup_j S_{i,j}$ and, \begin{equation*} \sum_{j=1}^\infty\mu_0(S_{i,j})\le\mu^*(S_i)+2^{-i}\epsilon. \end{equation*}As $\bigcup_iS_i\subseteq\bigcup_{i,j}S_{i,j}$ , equation () defining $\mu^*$ gives \begin{equation*} \mu^*\left(\bigcup_iS_i\right)\le\sum_{i,j}\mu_0(S_{i,j})=\sum_i\sum_j\mu_0(S_{i,j})\le\sum_i(\mu^*(S_i)+2^{-i}\epsilon)=\sum_i\mu^*(S_i)+\epsilon. \end{equation*}As $\epsilon>0$ is arbitrary, this proves subadditivity (). So, $\mu^*$ is indeed an outer measure.

The next step is to show that $\mu^*$ agrees with $\mu_0$ on $A$ . So, choose any $S\in A$ . The inequality $\mu^*(S)\le\mu_0(S)$ follows from taking $S_1=S$ and $S_i=\emptyset$ in (), and it remains to prove the reverse inequality. So, let $S_i$ be a sequence in $A$ covering $S$ , and set \begin{equation*} S^\prime_i=(S\cap S_i)\setminus\bigcup_{j=1}^{i-1}S_j\in A. \end{equation*}Then, $S^\prime_i$ are disjoint sets satisfying $\bigcup_{j=1}^iS^\prime_j=S\cap \bigcup_{j=1}^iS_j$ and, therefore, $\bigcup_iS^\prime_i=S$ . By the countable additivity of $\mu_0$ , \begin{equation*} \sum_i\mu_0(S_i)=\sum_i(\mu_0(S^\prime_i)+\mu_0(S_i\setminus S^\prime_i))\ge\sum_i\mu_0(S^\prime_i)=\mu_0(S). \end{equation*}As this inequality hold for any sequence $S_i\in A$ covering $S$ , equation () gives $\mu^*(S)\ge\mu_0(S)$ and, by combining with the reverse inequality, shows that $\mu^*$ does indeed agree with $\mu_0$ on $A$ .

We have shown that $\mu_0$ extends to an outer measure $\mu^*$ on the power set of $X$ . The final step is to apply Carathéodory's lemma on the restriction of outer measures. A set $S\subseteq X$ is said to be $\mu^*$ -measurable if the inequality \begin{equation}\label{eq:3} \mu^*(E)\ge\mu^*(E\cap S)+\mu^*(E\cap S^c) \end{equation}is satisfied for all subsets $E$ of $X$ . Carathéodory's lemma then states that the collection $\mathcal{F}$ of $\mu^*$ -measurable sets is a $\sigma$ -algebra and that the restriction of $\mu^*$ to $\mathcal{F}$ is a measure. To complete the proof of the theorem it only remains to be shown that every set in $A$ is $\mu^*$ -measurable, as it will then follow that $\mathcal{F}$ contains $\mathcal{A}=\sigma(A)$ and the restriction of $\mu^*$ to $\mathcal{A}$ is a measure.

So, choosing any $S\in A$ and $E\subseteq X$ , the proof will be complete once it is shown that () is satisfied. Given any $\epsilon>0$ , equation () says that there is a sequence $E_i$ in $A$ such that $E\subseteq\bigcup_iE_i$ and \begin{equation*} \sum_i\mu_0(E_i)\le\mu^*(E)+\epsilon. \end{equation*}As $E\cap S\subseteq\bigcup_i(E_i\cap S)$ and $E\cap S^c\subseteq\bigcup_i(E_i\cap S^c)$ , \begin{equation*} \mu^*(E\cap S)+\mu^*(E\cap S^c) \le\sum_i\mu_0(E_i\cap S)+\sum_i\mu_0(E_i\cap S^c)=\sum_i\mu_0(E_i)\le\mu^*(E)+\epsilon. \end{equation*}Since $\epsilon$ is arbitrary, this shows that () is satisfied and $S$ is $\mu^*$ -measurable.