[proofplan]
We construct an outer measure $\mu^*$ from $\mu_0$ by covering arbitrary sets with countable unions from $\mathcal{A}$, then identify a $\sigma$-algebra of $\mu^*$-measurable sets (via the Caratheodory criterion) on which $\mu^*$ is countably additive. The key ideas are: (1) $\mu^*$ agrees with $\mu_0$ on $\mathcal{A}$; (2) the Caratheodory measurability criterion carves out a $\sigma$-algebra; (3) every set in $\mathcal{A}$ satisfies this criterion, so $\sigma(\mathcal{A})$ is contained in the measurable sets.
[/proofplan]
[step:Define the outer measure $\mu^*$ and verify it agrees with $\mu_0$ on $\mathcal{A}$]
Define $\mu^*: \mathcal{P}(E) \to [0, \infty]$ by
\begin{align*}
\mu^*(B) = \inf\!\left\{ \sum_{n=1}^\infty \mu_0(A_n) : A_n \in \mathcal{A},\; B \subseteq \bigcup_{n=1}^\infty A_n \right\}.
\end{align*}
[claim:Mu Star Is An Outer Measure]
$\mu^*$ satisfies: (i) $\mu^*(\varnothing) = 0$; (ii) monotonicity; (iii) countable subadditivity.
[/claim]
[proof]
For (i): cover $\varnothing$ by $A_n = \varnothing$ for all $n$, giving $\mu^*(\varnothing) \leq 0$.
For (ii): every covering of a larger set covers the smaller set.
For (iii): fix $\varepsilon > 0$; for each $n$, choose $(A_{n,k})_{k=1}^\infty$ from $\mathcal{A}$ with $B_n \subseteq \bigcup_k A_{n,k}$ and $\sum_k \mu_0(A_{n,k}) \leq \mu^*(B_n) + \varepsilon/2^n$.
Then $(A_{n,k})_{n,k}$ covers $\bigcup_n B_n$, so
\begin{align*}
\mu^*\!\left(\bigcup_n B_n\right) \leq \sum_{n,k} \mu_0(A_{n,k}) \leq \sum_{n=1}^\infty \mu^*(B_n) + \varepsilon.
\end{align*}
Since $\varepsilon$ was arbitrary, (iii) follows.
[/proof]
[claim:Mu Star Agrees With Mu Zero On The Ring]
For every $A \in \mathcal{A}$, $\mu^*(A) = \mu_0(A)$.
[/claim]
[proof]
The inequality $\mu^*(A) \leq \mu_0(A)$ is immediate: take $A_1 = A$, $A_n = \varnothing$ for $n \geq 2$.
For the reverse, let $(A_n)$ cover $A$.
Define $B_n = A \cap A_n \cap (\bigcup_{k=1}^{n-1} A_k)^c$, which are pairwise disjoint, lie in $\mathcal{A}$, satisfy $B_n \subseteq A_n$, and $\bigcup_n B_n = A$.
By countable additivity of $\mu_0$,
\begin{align*}
\mu_0(A) = \sum_{n=1}^\infty \mu_0(B_n) \leq \sum_{n=1}^\infty \mu_0(A_n).
\end{align*}
Taking the infimum gives $\mu_0(A) \leq \mu^*(A)$.
[/proof]
[/step]
[step:Define the Caratheodory measurability criterion and show measurable sets form a $\sigma$-algebra]
Call $B \subseteq E$ $\mu^*$-measurable if for every test set $T \subseteq E$,
\begin{align*}
\mu^*(T) = \mu^*(T \cap B) + \mu^*(T \cap B^c).
\end{align*}
Let $\mathcal{M}$ denote the collection of all $\mu^*$-measurable sets.
[claim:Measurable Sets Form A Sigma Algebra]
$\mathcal{M}$ is a $\sigma$-algebra on $E$.
[/claim]
[proof]
The definition is symmetric in $B$ and $B^c$, so $\mathcal{M}$ is closed under complements.
For finite unions: if $B_1, B_2 \in \mathcal{M}$, applying measurability of $B_2$ to test set $T \cap B_1^c$ gives
\begin{align*}
\mu^*(T) = \mu^*(T \cap B_1) + \mu^*(T \cap B_1^c \cap B_2) + \mu^*(T \cap B_1^c \cap B_2^c).
\end{align*}
By subadditivity, the first two terms bound $\mu^*(T \cap (B_1 \cup B_2))$ from above, giving $B_1 \cup B_2 \in \mathcal{M}$.
For countable unions of pairwise disjoint sets $B_n \in \mathcal{M}$ with $B = \bigcup_n B_n$: iterating measurability gives $\mu^*(T \cap C_N) = \sum_{n=1}^N \mu^*(T \cap B_n)$ where $C_N = \bigcup_{n=1}^N B_n$.
Since $B^c \subseteq C_N^c$,
\begin{align*}
\mu^*(T) \geq \sum_{n=1}^N \mu^*(T \cap B_n) + \mu^*(T \cap B^c).
\end{align*}
Taking $N \to \infty$ and applying subadditivity gives $B \in \mathcal{M}$.
General sequences are handled by disjointification.
[/proof]
[/step]
[step:Show every set in $\mathcal{A}$ is $\mu^*$-measurable]
[claim:Ring Is Contained In M]
$\mathcal{A} \subseteq \mathcal{M}$.
[/claim]
[proof]
Fix $A \in \mathcal{A}$ and test set $T$ with $\mu^*(T) < \infty$.
Fix $\varepsilon > 0$ and choose $(A_n)$ from $\mathcal{A}$ with $T \subseteq \bigcup_n A_n$ and $\sum_n \mu_0(A_n) \leq \mu^*(T) + \varepsilon$.
Since $\mathcal{A}$ is a ring, $A_n \cap A \in \mathcal{A}$ and $A_n \setminus A \in \mathcal{A}$, with $\mu_0(A_n) = \mu_0(A_n \cap A) + \mu_0(A_n \setminus A)$.
Moreover, $T \cap A \subseteq \bigcup_n (A_n \cap A)$ and $T \cap A^c \subseteq \bigcup_n (A_n \setminus A)$, so
\begin{align*}
\mu^*(T \cap A) + \mu^*(T \cap A^c) \leq \sum_n \mu_0(A_n \cap A) + \sum_n \mu_0(A_n \setminus A) = \sum_n \mu_0(A_n) \leq \mu^*(T) + \varepsilon.
\end{align*}
Since $\varepsilon$ was arbitrary, $A \in \mathcal{M}$.
[/proof]
[/step]
[step:Restrict $\mu^*$ to $\sigma(\mathcal{A})$ to obtain the extension]
Since $\mathcal{A} \subseteq \mathcal{M}$ and $\mathcal{M}$ is a $\sigma$-algebra, $\sigma(\mathcal{A}) \subseteq \mathcal{M}$.
Define $\mu = \mu^*|_{\sigma(\mathcal{A})}$.
The countable additivity of $\mu$ on $\sigma(\mathcal{A})$ follows from the $\sigma$-algebra structure of $\mathcal{M}$ and the additive property established in the union step.
By the agreement claim, $\mu(A) = \mu^*(A) = \mu_0(A)$ for all $A \in \mathcal{A}$, so $\mu$ extends $\mu_0$.
[/step]
