[proofplan] The proof has three parts: (1) show $\overline{\mathcal{F}}$ is a $\sigma$-algebra, (2) show $\overline{\mu}$ is well-defined, and (3) show $\overline{\mu}$ is a complete measure extending $\mu$. For (1), we verify closure under complements and countable unions by manipulating the representation $A \cup N$ with $A \in \mathcal{F}$ and $N \subset E \in \mathcal{F}$, $\mu(E) = 0$. For (2), if $A_1 \cup N_1 = A_2 \cup N_2$ with two different representations, we show $\mu(A_1) = \mu(A_2)$ by bounding the symmetric difference $A_1 \triangle A_2$ inside a null set. For (3), countable additivity of $\overline{\mu}$ follows from countable additivity of $\mu$, and completeness follows from the construction: any subset of a $\overline{\mu}$-null set is again in $\overline{\mathcal{F}}$. [/proofplan]

[step:Show $\overline{\mathcal{F}}$ is a $\sigma$-algebra]We verify the three $\sigma$-algebra axioms. **Contains $X$:** Take $A = X \in \mathcal{F}$, $N = \varnothing \subset \varnothing \in \mathcal{F}$ with $\mu(\varnothing) = 0$. Then $X = A \cup N \in \overline{\mathcal{F}}$. **Closure under complements:** Let $S = A \cup N \in \overline{\mathcal{F}}$ with $A \in \mathcal{F}$, $N \subset E \in \mathcal{F}$, $\mu(E) = 0$. Then \begin{align*} S^c = (A \cup N)^c = A^c \cap N^c = A^c \setminus N. \end{align*} Write $S^c = (A^c \setminus E) \cup ((A^c \cap E) \setminus N)$. Set $A' := A^c \setminus E = A^c \cap E^c \in \mathcal{F}$ (since $\mathcal{F}$ is a $\sigma$-algebra), and $N' := (A^c \cap E) \setminus N$. Then $N' \subset A^c \cap E \subset E$, and $E \in \mathcal{F}$ with $\mu(E) = 0$, so $N'$ is a subset of a null set. Therefore $S^c = A' \cup N' \in \overline{\mathcal{F}}$. **Closure under countable unions:** Let $S_k = A_k \cup N_k$ for $k \in \mathbb{N}$, with $A_k \in \mathcal{F}$, $N_k \subset E_k \in \mathcal{F}$, $\mu(E_k) = 0$. Then \begin{align*} \bigcup_{k=1}^\infty S_k = \bigcup_{k=1}^\infty (A_k \cup N_k) = \left(\bigcup_{k=1}^\infty A_k\right) \cup \left(\bigcup_{k=1}^\infty N_k\right). \end{align*} Set $A' := \bigcup_{k=1}^\infty A_k \in \mathcal{F}$ and $N' := \bigcup_{k=1}^\infty N_k$. Then $N' \subset \bigcup_{k=1}^\infty E_k =: E' \in \mathcal{F}$, and by [countable subadditivity](/theorems/1081), $\mu(E') \le \sum_{k=1}^\infty \mu(E_k) = 0$. So $\bigcup_k S_k = A' \cup N' \in \overline{\mathcal{F}}$.[/step]

[guided]The complement step is the most delicate. We need to express $S^c = (A \cup N)^c$ in the form $A' \cup N'$ with $A' \in \mathcal{F}$ and $N'$ a subset of a null set. The decomposition $S^c = (A^c \setminus E) \cup ((A^c \cap E) \setminus N)$ works because $A^c \setminus E \in \mathcal{F}$ (a set operation on $\mathcal{F}$-sets) and the remainder $(A^c \cap E) \setminus N$ is contained in the null set $E$. For countable unions, the key is that a countable union of subsets of null sets is again a subset of a null set (by [countable subadditivity](/theorems/1081), the union of countably many null sets is null).[/guided]

[step:Show $\overline{\mu}$ is well-defined]We must verify: if $A_1 \cup N_1 = A_2 \cup N_2$ with $A_i \in \mathcal{F}$, $N_i \subset E_i \in \mathcal{F}$, $\mu(E_i) = 0$, then $\mu(A_1) = \mu(A_2)$. Since $A_1 \cup N_1 = A_2 \cup N_2$, we have $A_1 \subset A_2 \cup N_2 \subset A_2 \cup E_2$. Therefore \begin{align*} A_1 \setminus A_2 \subset (A_2 \cup E_2) \setminus A_2 = E_2 \setminus A_2 \subset E_2. \end{align*} Since $A_1 \setminus A_2 \in \mathcal{F}$ and $A_1 \setminus A_2 \subset E_2$ with $\mu(E_2) = 0$, [monotonicity](/theorems/1081) gives $\mu(A_1 \setminus A_2) = 0$. By symmetry (exchanging indices $1$ and $2$), $\mu(A_2 \setminus A_1) = 0$. Now \begin{align*} A_1 = (A_1 \cap A_2) \sqcup (A_1 \setminus A_2), \quad A_2 = (A_1 \cap A_2) \sqcup (A_2 \setminus A_1). \end{align*} By countable additivity: $\mu(A_1) = \mu(A_1 \cap A_2) + \mu(A_1 \setminus A_2) = \mu(A_1 \cap A_2) + 0 = \mu(A_1 \cap A_2)$, and similarly $\mu(A_2) = \mu(A_1 \cap A_2)$. Therefore $\mu(A_1) = \mu(A_2)$.[/step]

[guided]The idea is to show $A_1$ and $A_2$ differ by a subset of a null set. Since $A_1 \cup N_1 = A_2 \cup N_2$, any point in $A_1 \setminus A_2$ must be in $N_2$ (hence in $E_2$), so $A_1 \setminus A_2 \subset E_2$. Since $E_2$ has measure zero and $A_1 \setminus A_2$ is measurable (both $A_1, A_2 \in \mathcal{F}$), monotonicity gives $\mu(A_1 \setminus A_2) = 0$. Symmetrically, $\mu(A_2 \setminus A_1) = 0$. The decomposition $A_i = (A_1 \cap A_2) \sqcup (A_i \setminus A_{3-i})$ then gives $\mu(A_1) = \mu(A_1 \cap A_2) = \mu(A_2)$.[/guided]

[step:Show $\overline{\mu}$ extends $\mu$] For $A \in \mathcal{F}$, write $A = A \cup \varnothing$ with $\varnothing \subset \varnothing \in \mathcal{F}$ and $\mu(\varnothing) = 0$. Then $A \in \overline{\mathcal{F}}$ and $\overline{\mu}(A) = \mu(A)$. Therefore $\mathcal{F} \subset \overline{\mathcal{F}}$ and $\overline{\mu}|_{\mathcal{F}} = \mu$. [/step]

[step:Show $\overline{\mu}$ is countably additive]Let $(S_k)_{k=1}^\infty$ be pairwise disjoint sets in $\overline{\mathcal{F}}$ with $S_k = A_k \cup N_k$, $A_k \in \mathcal{F}$, $N_k \subset E_k \in \mathcal{F}$, $\mu(E_k) = 0$. Then $S := \bigsqcup_{k=1}^\infty S_k = (\bigcup_k A_k) \cup (\bigcup_k N_k)$ with $\overline{\mu}(S) = \mu(\bigcup_k A_k)$. Since $A_k \subset S_k$ and $A_l \subset S_l$ for $k \ne l$, and $S_k \cap S_l = \varnothing$, we have $A_k \cap A_l \subset S_k \cap S_l = \varnothing$. So the $A_k$ are pairwise disjoint. By countable additivity of $\mu$: \begin{align*} \overline{\mu}(S) = \mu\!\left(\bigcup_{k=1}^\infty A_k\right) = \mu\!\left(\bigsqcup_{k=1}^\infty A_k\right) = \sum_{k=1}^\infty \mu(A_k) = \sum_{k=1}^\infty \overline{\mu}(S_k). \end{align*} Also $\overline{\mu}(\varnothing) = \mu(\varnothing) = 0$. Therefore $\overline{\mu}$ is a measure on $\overline{\mathcal{F}}$.[/step]

[guided]The key observation: when the $S_k$ are disjoint and $A_k \subset S_k$, the $A_k$ are automatically disjoint. This allows us to apply countable additivity of $\mu$ to the disjoint union $\bigsqcup A_k = \bigcup A_k$. The "null debris" $N_k$ does not affect the measure since $\overline{\mu}(S_k) = \mu(A_k)$ by definition.[/guided]

[step:Show $\overline{\mu}$ is complete]A measure $\overline{\mu}$ on $(X, \overline{\mathcal{F}})$ is complete if: whenever $S \in \overline{\mathcal{F}}$ with $\overline{\mu}(S) = 0$ and $T \subset S$, then $T \in \overline{\mathcal{F}}$. Let $S = A \cup N \in \overline{\mathcal{F}}$ with $\overline{\mu}(S) = \mu(A) = 0$, $N \subset E$, $\mu(E) = 0$. Let $T \subset S$. Then $T \subset A \cup N \subset A \cup E$. Set $E' := A \cup E \in \mathcal{F}$. By [countable subadditivity](/theorems/1081), $\mu(E') \le \mu(A) + \mu(E) = 0$. Write $T = \varnothing \cup T$ with $\varnothing \in \mathcal{F}$ and $T \subset E' \in \mathcal{F}$ with $\mu(E') = 0$. Therefore $T \in \overline{\mathcal{F}}$, and $\overline{\mu}(T) = \mu(\varnothing) = 0$.[/step]

[guided]Completeness means every subset of a null set is measurable (and null). Given $T \subset S$ with $\overline{\mu}(S) = 0$: we have $\mu(A) = 0$ and $N \subset E$ with $\mu(E) = 0$. The set $T$ sits inside $A \cup E$, which has measure $0$. Representing $T = \varnothing \cup T$ with $T \subset A \cup E$ shows $T \in \overline{\mathcal{F}}$. This is where the definition of $\overline{\mathcal{F}}$ — allowing unions with arbitrary subsets of null sets — pays off: the construction is tailor-made to capture all subsets of null sets.[/guided]