[proofplan]
We verify directly that the collection $\mathcal{M}$ of $\mu^*$-measurable sets satisfies the three axioms of a sigma-algebra: it contains $\varnothing$, is closed under complements, and is closed under countable unions. Closure under finite unions is established first by an inductive splitting argument, then extended to countable unions of disjoint sets via a telescoping identity and passage to the limit using countable subadditivity. Countable additivity of $\mu$ on $\mathcal{M}$ follows from the same limiting argument. Completeness is proved by showing that every $\mu^*$-null set satisfies the Carathéodory splitting condition.
[/proofplan]
[step:Verify that $\varnothing$ and $X$ belong to $\mathcal{M}$]
Recall that $A \in \mathcal{M}$ means: for every $E \subset X$,
\begin{align*}
\mu^*(E) = \mu^*(E \cap A) + \mu^*(E \setminus A).
\end{align*}
For $A = \varnothing$: given any $E \subset X$, we have $E \cap \varnothing = \varnothing$ and $E \setminus \varnothing = E$, so $\mu^*(E \cap \varnothing) + \mu^*(E \setminus \varnothing) = 0 + \mu^*(E) = \mu^*(E)$. Hence $\varnothing \in \mathcal{M}$.
For $A = X$: $E \cap X = E$ and $E \setminus X = \varnothing$, so $\mu^*(E \cap X) + \mu^*(E \setminus X) = \mu^*(E) + 0 = \mu^*(E)$. Hence $X \in \mathcal{M}$.
[/step]
[step:Verify that $\mathcal{M}$ is closed under complements]
Let $A \in \mathcal{M}$. For any test set $E \subset X$,
\begin{align*}
\mu^*(E) = \mu^*(E \cap A) + \mu^*(E \setminus A).
\end{align*}
Since $E \cap A = E \setminus A^c$ and $E \setminus A = E \cap A^c$, this reads
\begin{align*}
\mu^*(E) = \mu^*(E \setminus A^c) + \mu^*(E \cap A^c),
\end{align*}
which is the splitting condition for $A^c$. Hence $A^c \in \mathcal{M}$.
[/step]
[step:Verify that $\mathcal{M}$ is closed under finite unions]
Let $A, B \in \mathcal{M}$. We show $A \cup B \in \mathcal{M}$. Fix an arbitrary test set $E \subset X$. Apply the splitting condition for $A$ with test set $E$:
\begin{align*}
\mu^*(E) = \mu^*(E \cap A) + \mu^*(E \setminus A).
\end{align*}
Apply the splitting condition for $B$ with test set $E \setminus A$:
\begin{align*}
\mu^*(E \setminus A) = \mu^*((E \setminus A) \cap B) + \mu^*((E \setminus A) \setminus B).
\end{align*}
Substituting into the first equation:
\begin{align*}
\mu^*(E) = \mu^*(E \cap A) + \mu^*((E \setminus A) \cap B) + \mu^*(E \setminus (A \cup B)).
\end{align*}
Now observe that $E \cap (A \cup B) = (E \cap A) \cup ((E \setminus A) \cap B)$, and these two sets are disjoint. By countable subadditivity,
\begin{align*}
\mu^*(E \cap (A \cup B)) \leq \mu^*(E \cap A) + \mu^*((E \setminus A) \cap B).
\end{align*}
Combining:
\begin{align*}
\mu^*(E) \geq \mu^*(E \cap (A \cup B)) + \mu^*(E \setminus (A \cup B)).
\end{align*}
The reverse inequality $\mu^*(E) \leq \mu^*(E \cap (A \cup B)) + \mu^*(E \setminus (A \cup B))$ holds by countable subadditivity. Therefore equality holds, and $A \cup B \in \mathcal{M}$.
By induction, $\mathcal{M}$ is closed under finite unions.
[/step]
[step:Establish finite additivity of $\mu^*$ on disjoint sets in $\mathcal{M}$]
Let $A_1, \ldots, A_n \in \mathcal{M}$ be pairwise disjoint. Define $B_n = \bigcup_{k=1}^n A_k \in \mathcal{M}$ (by closure under finite unions). We show by induction that for every $E \subset X$,
\begin{align*}
\mu^*(E \cap B_n) = \sum_{k=1}^n \mu^*(E \cap A_k).
\end{align*}
**Base case** ($n = 1$): $\mu^*(E \cap B_1) = \mu^*(E \cap A_1)$.
**Inductive step**: Suppose the identity holds for $n$. Apply the splitting condition for $A_{n+1}$ with test set $E \cap B_{n+1}$:
\begin{align*}
\mu^*(E \cap B_{n+1}) = \mu^*((E \cap B_{n+1}) \cap A_{n+1}) + \mu^*((E \cap B_{n+1}) \setminus A_{n+1}).
\end{align*}
Since $A_{n+1}$ is disjoint from $A_1, \ldots, A_n$, we have $(E \cap B_{n+1}) \cap A_{n+1} = E \cap A_{n+1}$ and $(E \cap B_{n+1}) \setminus A_{n+1} = E \cap B_n$. By the inductive hypothesis:
\begin{align*}
\mu^*(E \cap B_{n+1}) = \mu^*(E \cap A_{n+1}) + \mu^*(E \cap B_n) = \mu^*(E \cap A_{n+1}) + \sum_{k=1}^n \mu^*(E \cap A_k) = \sum_{k=1}^{n+1} \mu^*(E \cap A_k).
\end{align*}
[/step]
[step:Extend to countable unions of disjoint measurable sets]
Let $(A_k)_{k=1}^\infty$ be a sequence of pairwise disjoint sets in $\mathcal{M}$. Define $A = \bigcup_{k=1}^\infty A_k$ and $B_n = \bigcup_{k=1}^n A_k$. We show $A \in \mathcal{M}$.
Fix an arbitrary test set $E \subset X$. Since $B_n \in \mathcal{M}$, the splitting condition gives
\begin{align*}
\mu^*(E) = \mu^*(E \cap B_n) + \mu^*(E \setminus B_n).
\end{align*}
By finite additivity (established in the previous step):
\begin{align*}
\mu^*(E) = \sum_{k=1}^n \mu^*(E \cap A_k) + \mu^*(E \setminus B_n).
\end{align*}
Since $A \supset B_n$, monotonicity gives $E \setminus A \subset E \setminus B_n$, so $\mu^*(E \setminus A) \leq \mu^*(E \setminus B_n)$. Therefore:
\begin{align*}
\mu^*(E) \geq \sum_{k=1}^n \mu^*(E \cap A_k) + \mu^*(E \setminus A).
\end{align*}
Letting $n \to \infty$:
\begin{align*}
\mu^*(E) \geq \sum_{k=1}^\infty \mu^*(E \cap A_k) + \mu^*(E \setminus A).
\end{align*}
By countable subadditivity, $\mu^*(E \cap A) \leq \sum_{k=1}^\infty \mu^*(E \cap A_k)$, so:
\begin{align*}
\mu^*(E) \geq \mu^*(E \cap A) + \mu^*(E \setminus A).
\end{align*}
The reverse inequality $\mu^*(E) \leq \mu^*(E \cap A) + \mu^*(E \setminus A)$ holds by countable subadditivity. Therefore $A \in \mathcal{M}$.
[/step]
[step:Reduce general countable unions to disjoint ones and conclude $\mathcal{M}$ is a sigma-algebra]
Let $(C_k)_{k=1}^\infty$ be an arbitrary sequence in $\mathcal{M}$ (not necessarily disjoint). Define $A_1 = C_1$ and $A_k = C_k \setminus \bigcup_{j=1}^{k-1} C_j$ for $k \geq 2$. Since $\mathcal{M}$ is closed under complements and finite unions, each $A_k \in \mathcal{M}$. The sets $A_k$ are pairwise disjoint, and $\bigcup_{k=1}^\infty A_k = \bigcup_{k=1}^\infty C_k$. By the previous step, $\bigcup_{k=1}^\infty C_k \in \mathcal{M}$.
Combining: $\mathcal{M}$ contains $\varnothing$, is closed under complements, and is closed under countable unions. Hence $\mathcal{M}$ is a sigma-algebra.
[/step]
[step:Prove that $\mu = \mu^*|_{\mathcal{M}}$ is countably additive]
Let $(A_k)_{k=1}^\infty$ be pairwise disjoint sets in $\mathcal{M}$, and set $A = \bigcup_{k=1}^\infty A_k$. Taking $E = A$ in the inequality established during the countable union step:
\begin{align*}
\mu^*(A) \geq \sum_{k=1}^\infty \mu^*(A \cap A_k) + \mu^*(A \setminus A) = \sum_{k=1}^\infty \mu^*(A_k) + 0 = \sum_{k=1}^\infty \mu^*(A_k).
\end{align*}
The reverse inequality $\mu^*(A) \leq \sum_{k=1}^\infty \mu^*(A_k)$ is countable subadditivity. Therefore
\begin{align*}
\mu(A) = \mu^*(A) = \sum_{k=1}^\infty \mu^*(A_k) = \sum_{k=1}^\infty \mu(A_k).
\end{align*}
Hence $\mu$ is countably additive on $\mathcal{M}$.
[/step]
[step:Prove completeness of $\mu$]
Let $N \in \mathcal{M}$ with $\mu(N) = \mu^*(N) = 0$, and let $S \subset N$. We show $S \in \mathcal{M}$. By monotonicity of $\mu^*$, $\mu^*(S) \leq \mu^*(N) = 0$, so $\mu^*(S) = 0$.
For any test set $E \subset X$:
\begin{align*}
\mu^*(E \cap S) \leq \mu^*(S) = 0,
\end{align*}
so $\mu^*(E \cap S) = 0$. Then
\begin{align*}
\mu^*(E \cap S) + \mu^*(E \setminus S) = \mu^*(E \setminus S) \leq \mu^*(E).
\end{align*}
For the reverse inequality, monotonicity gives $\mu^*(E \setminus S) \leq \mu^*(E)$ (since $E \setminus S \subset E$), so
\begin{align*}
\mu^*(E) \leq \mu^*(E \cap S) + \mu^*(E \setminus S) = 0 + \mu^*(E \setminus S) \leq \mu^*(E).
\end{align*}
Wait — we need the reverse direction. By subadditivity: $\mu^*(E) \leq \mu^*(E \cap S) + \mu^*(E \setminus S) = 0 + \mu^*(E \setminus S)$, and by monotonicity $\mu^*(E \setminus S) \leq \mu^*(E)$. Combining: $\mu^*(E) \leq \mu^*(E \setminus S) \leq \mu^*(E)$, so $\mu^*(E \setminus S) = \mu^*(E)$. Therefore
\begin{align*}
\mu^*(E \cap S) + \mu^*(E \setminus S) = 0 + \mu^*(E) = \mu^*(E).
\end{align*}
Hence $S \in \mathcal{M}$. Every subset of a $\mu$-null set is measurable, so $\mu$ is complete.
[/step]
