I have posted this theorem before but I am re-posting it again because I have a different question.
Theorem 1.9 – Suppose that $(X,M,\mu)$ is a measure space. Let $\mathcal{N} = \{N\in M:\mu(N) = 0\}$ and $\overline{M} = \{E\cup F: E\in M, F\subset N, N\in\mathcal{N}\}$. Then $\overline{M}$ is a $\sigma$-algebra and there is a unique extension $\overline{\mu}$ of $\mu$ to a complete measure on $\overline{M}$.
Claim 1 – $\overline{M}$ is a $\sigma$-algebra
proof:
i.) Since $M$ is a $\sigma$-algebra, $\emptyset\in M\subset \overline{M}$ so $\emptyset\in \overline{M}$.
ii.) Suppose $B\in\overline{M}$ then there is an $E\in M$ such that $B = E\cup F$ where $F\subset N$ and $N\in\mathcal{N}$. Then
\begin{align*}
X\setminus B &= X\cap B^c\\
&= X\cap (E\cup F)^c\\
&= X\cap ((E^c\cap N^c\cap F^c)\cup(F^c\setminus N^c))\\
&= X\cap ((E^c\cap N^c)\cup (F^c\cap N))\\
&= X\cap ((E\cup N)^c\cup (N\setminus F))
\end{align*}
Since $(E\cup N)^c\in M$ and $(N\setminus F)\subset N\in M$ with $\mu(N) = 0$, then $B^c\in\overline{M}$.
iii.) Let $\{B_j\}_{1}^{\infty}\in\overline{M}$ then for each $j$ there is an $E\in M$ such that $B_j = E_j\cup F_j$ where $F_j\subset N_j$ and $\mu(N_j) = 0$. So, $$\bigcup_{1}^{\infty}B_j = \bigcup_{1}^{\infty}(E_j\cup F_j) = \bigcup_{1}^{\infty}E_j \cup \bigcup_{1}^{\infty}F_j$$ Note that $\bigcup_{1}^{\infty}F_j\subset \bigcup_{1}^{\infty}N_j$ and $\mu\left(\bigcup_{1}^{\infty}N_j\right) = 0$. So we have $\bigcup_{1}^{\infty}B_j\in\overline{M}$. Therefore $\overline{M}$ is a $\sigma$-algebra.
Claim 2 – There is a unique extention $\overline{\mu}$ of $\mu$ to a complete measure on $\overline{M}$
Proof:
We first need to show that $\overline{\mu}$ is well-defined. Suppose $E\cup F\in\overline{M}$, set $\overline{\mu}(E\cup F) = \overline{\mu}(E)$. This is well-defined since if $E_1\cup F_1 = E_2\cup F_2$ where $F_j\subset N_j\in\mathcal{N}$. Then we know that $E_1\subset E_1\cup F_1 = E_2\cup F_2 = E_2\cup N_2$ then $E_1\subset E_2\cup N_2$ and so by monotonicity $\mu(E_1)\leq \mu(E_2) + \mu(N_2) = \mu(E_2)$. Also, since $E_2\subset E_2\cup F_2 = E_1\cup F_1 = E_1\cup N_1$ then again by monotonicity $\mu(E_2)\leq \mu(E_1)+\mu(N_1) = \mu(E_1)$. Thus $\overline{\mu}$ is well-defined.
I now need to show that $\overline{\mu}$ is a complete measure on $\overline{M}$ and that $\overline{\mu}$ is the only measure on $\overline{M}$ that extends $\mu$.
Step 1 – Show $\overline{\mu}$ is a measure.
Proof:
i.) $\overline{\mu}(\emptyset) = \overline{\mu}(\emptyset \cup F) = 0$ since $F\subset N$ with $\mu(N) = 0$.
ii.) Let $\{A_n\}_{1}^{\infty}\in\overline{M}$, disjoint, then there is an $\{E_n\}_{1}^{\infty}\in M$ and a sequence $\{F_n\}_{1}^{\infty}\subset N\in\mathcal{N}$ such that $A_n = E_n\cup F_n$ for all $n$. Note from part (iii.) of claim 1, $\cup_{1}^{\infty}F_n\subset \cup_{1}^{\infty}N_n$. Thus,
\begin{align*}
\overline{\mu}\left(\bigcup_{1}^{\infty}A_n\right) = \overline{\mu}\left(\bigcup_{1}^{\infty}E_n\cup F_n\right) &=
\overline{\mu}\left(\bigcup_{1}^{\infty}E_n\cup \bigcup_{1}^{\infty}F_n\right)\\
&= \overline{\mu}\left(\bigcup_{1}^{\infty}E_n\right)\cup \overline{\mu}\left(\bigcup_{1}^{\infty}F_n\right)\\
&= \sum_{1}^{\infty}\overline{\mu}(E_n) + \sum_{1}^{\infty}\overline{\mu}(F_n)\\
&= \sum_{1}^{\infty}\overline{\mu}(A_n)
\end{align*}
Thus $\overline{\mu}$ is a measure.
Step 2 – Show $\overline{\mu}$ is a complete measure.
Proof:
Let $A\subset X$ and suppose there is an $B\in\overline{M}$ such that $A\subset B$ and $\overline{\mu}(B) = 0$. Set $B = E\cup F$ where $E\in M$ and $F\subset N\in M$ with $\mu(N) = 0$. Since $A\subset B = E\cup F = E\cup N$ since $F\subset N$ then by monotonicity $\mu(A)\leq \mu(E) + \mu(N) = \overline{\mu}(E) + 0 = \overline{\mu}(E)\leq \overline{\mu}(B) = 0$. Thus $A\in \overline{M}$?
Step 3: Show $\overline{\mu}$ is a unique extention of $\mu$
Not exactly sure how to show this or if I already have shown this.
These are the steps I have taken, please let me know if this is the correct sequence or any other additional comments that you have in regards to my proofs.
Best Answer
Your proof is in the right direction. Here I copied it and I changed some points in it to make it a complete proof.
Claim 1 - $\overline{M}$ is a $\sigma$-algebra
proof:
i.) Since $M$ is a $\sigma$-algebra, $\emptyset\in M\subset \overline{M}$ so $\emptyset\in \overline{M}$.
ii.) Suppose $B\in\overline{M}$ then there is an $E\in M$ such that $B = E\cup F$ where $F\subset N$ and $N\in\mathcal{N}$. Then \begin{align*} X\setminus B &= X\cap B^c\\ &= X\cap (E\cup F)^c\\ &= X\cap ((E^c\cap N^c\cap F^c)\cup(F^c\setminus N^c))\\ &= X\cap ((E^c\cap N^c)\cup (F^c\cap N))\\ &= X\cap ((E\cup N)^c\cup (N\setminus F)) \end{align*} Since $(E\cup N)^c\in M$ and $(N\setminus F)\subset N\in M$ with $\mu(N) = 0$, then $B^c\in\overline{M}$.
iii.) Let $\{B_j\}_{1}^{\infty}\in\overline{M}$ then for each $j$ there is an $E\in M$ such that $B_j = E_j\cup F_j$ where $F_j\subset N_j$ and $\mu(N_j) = 0$. So, $$\bigcup_{1}^{\infty}B_j = \bigcup_{1}^{\infty}(E_j\cup F_j) = \bigcup_{1}^{\infty}E_j \cup \bigcup_{1}^{\infty}F_j$$ Note that $\bigcup_{1}^{\infty}F_j\subset \bigcup_{1}^{\infty}N_j$ and $\mu\left(\bigcup_{1}^{\infty}N_j\right) = 0$. So we have $\bigcup_{1}^{\infty}B_j\in\overline{M}$. Therefore $\overline{M}$ is a $\sigma$-algebra.
Claim 2 - There is a unique extention $\overline{\mu}$ of $\mu$ to a complete measure on $\overline{M}$
Proof:
We first need to show that $\overline{\mu}$ is well-defined. Suppose $E\cup F\in\overline{M}$, set $\overline{\mu}(E\cup F) = \mu(E)$. This is well-defined since if $E_1\cup F_1 = E_2\cup F_2$ where $F_j\subset N_j\in\mathcal{N}$. Then we know that $E_1\subset E_1\cup F_1 = E_2\cup F_2 = E_2\cup N_2$ then $E_1\subset E_2\cup N_2$ and so by monotonicity $\mu(E_1)\leq \mu(E_2) + \mu(N_2) = \mu(E_2)$. Also, since $E_2\subset E_2\cup F_2 = E_1\cup F_1 = E_1\cup N_1$ then again by monotonicity $\mu(E_2)\leq \mu(E_1)+\mu(N_1) = \mu(E_1)$. So we have $\mu(E_1) = \mu(E_2)$. Thus $\overline{\mu}$ is well-defined.
I now need to show that $\overline{\mu}$ is a complete measure on $\overline{M}$ and that $\overline{\mu}$ is the only measure on $\overline{M}$ that extends $\mu$.
Step 1 - Show $\overline{\mu}$ is a measure.
Proof:
i.) $\overline{\mu}(\emptyset) = \overline{\mu}(\emptyset \cup \emptyset)= \mu(\emptyset) = 0$ (here we took $E=\emptyset$ and $F= N=\emptyset$).
ii.) Let $\{A_n\}_{1}^{\infty}\in\overline{M}$, disjoint, then there is an $\{E_n\}_{1}^{\infty}\in M$ and a sequence $\{F_n\}_{1}^{\infty}\subset N\in\mathcal{N}$ such that $A_n = E_n\cup F_n$ for all $n$. Note that $\{E_n\}_{1}^{\infty}$ is a family of disjoint sets in $M$. Note also from part (iii.) of claim 1, $\cup_{1}^{\infty}F_n\subset \cup_{1}^{\infty}N_n$ and $\mu(\cup_{1}^{\infty}N_n)=0$. Thus, \begin{align*} \overline{\mu}\left(\bigcup_{1}^{\infty}A_n\right) = \overline{\mu}\left(\bigcup_{1}^{\infty}E_n\cup F_n\right) &= \overline{\mu}\left(\bigcup_{1}^{\infty}E_n\cup \bigcup_{1}^{\infty}F_n\right)\\ &= \mu\left(\bigcup_{1}^{\infty}E_n\right)=\\ &= \sum_{1}^{\infty}\mu(E_n)=\\ &= \sum_{1}^{\infty}\overline{\mu}(E_n\cup F_n) \\ &= \sum_{1}^{\infty}\overline{\mu}(A_n) \end{align*}
Thus $\overline{\mu}$ is a measure.
Step 2 - Show $\overline{\mu}$ is a complete measure.
Proof:
Let $A\subset X$ and suppose there is an $B\in\overline{M}$ such that $A\subset B$ and $\overline{\mu}(B) = 0$. Since $B\in\overline{M}$, we know we can write $B = E\cup F$ where $E\in M$ and $F\subset N\in M$ with $\mu(N) = 0$. Since $\overline{\mu}(B) = 0$ and $\overline{\mu}(B) = \mu(E)$, we have $\mu(E)=0$. So $E\cup N \in M$ and $\mu(E\cup N)=0$.
Now note that $A=\emptyset \cup A$, $\emptyset \in M$ and $A\subseteq B = E\cup F \subseteq E\cup N$ and $E\cup N \in M$ and $\mu(E\cup N)=0$. So $A \in \overline{M}$ and $\overline{\mu}(A)=\mu(\emptyset)=0$. So $\overline{\mu}$ is complete.
Step 3: Show $\overline{\mu}$ is a unique extension of $\mu$
3.1. $\overline{\mu}$ is an extension of $\mu$.
Given $A \in M$, we have that $A=A\cup\emptyset$ so $A=A\cup F$ where $F\subset N$, $N\in M$ and $\mu(N)=0$ (just take $F=N=\emptyset$). Thus, $A \in \overline{M}$ and $\overline{\mu}(A) =\overline{\mu}(A\cup \emptyset)=\mu(A)$.
So we proved that $M \subseteq \overline{M}$ and $\overline{\mu}$ is an extension of $\mu$.
3.2. Uniqueness.
Suppose $\nu$ is an extension of $\mu$ to $\overline{M}$. Let $A\in\overline{M}$, then there is $E\in M$, $N\in M$ with $\mu(N)=0$ and $F\subseteq N$ such that $A=E\cup F$. Then, since $E\in M$ and $E\cup N\in M$, $$\mu(E)=\nu(E)\leqslant \nu(A)=\nu(E\cup F)\leqslant \nu(E\cup N)=\mu(E\cup N) \leqslant \mu(E)+\mu(N)=\mu(E)$$ So $\nu(A)=\mu(E)=\overline{\mu}(A)$. So $\nu = \overline{\mu}$.