Excercise 2.1 from book Probability for Statistians: Completion of measure space Below is the whole set up the question, I am decomposing it into several parts and tried to tackle them one by one:
Let $(\Omega, \mathcal{A},\mu)$ denote a measure space. Show that
\begin{equation*} 
\begin{split}
\hat{ \mathcal{A}_\mu} & \equiv \{A:A_1 \subset A\subset A_2 \text{ with } A_1, A_2\in \mathcal{A} \text{ and } \mu(A_2\backslash A_1)=0\} \\
 & = \{A \cup N: A\in \mathcal{A}, \text{ and } N\subset (\text{some }B)\in \mathcal{A} \text{ having } \mu(B)=0\} \\
 & = \{A \triangle N: A\in \mathcal{A}, \text{ and } N\subset (\text{some }B)\in \mathcal{A} \text{ having } \mu(B)=0\},
\end{split}
\end{equation*}
and that $\hat{ \mathcal{A}_\mu}$ is a $\sigma\text{-field}$. Define $\hat{\mu}$ on $\hat{ \mathcal{A}_\mu}$ by $\hat{\mu}(A \cup N)=\mu(A)$ for all $N\subset (\text{some }B)\in \mathcal{A}$ having $\mu(B)=0$. Show that $(\Omega,\hat{ \mathcal{A}_\mu},\hat{\mu})$ is a complete measure space for which $\hat{\mu}| \mathcal{A}=\mu$. [Note: A proof must include a demonstration that definition leads to a well-defined $\hat{\mu}$. That is, whenever $A_1 \cup N_1 = A_2 \cup N_2$ we must have $\mu(A_1)=\mu(A_2)$, so that $\hat{\mu}(A_1 \cup N_1)=\hat{\mu}(A_2 \cup N_2).$]
First, I showed that $ \hat{ \mathcal{A}_\mu}$ is a $\sigma-$field, then I wanted to show that the first two set expressions are the same. Here's my proof:

*

*Let $A\in \hat{ \mathcal{A}_\mu}\Rightarrow A\in \mathcal{A}$. Notice that $A_2 \backslash A_1=A_2\backslash A + A\backslash A_1$. Let $N=A_2\backslash A,B=A_2\backslash A_1\in \mathcal{A}$, we have $N\subset B, \mu(B)=0$. We have $A\in\{A \cup N: A\in \mathcal{A}, \text{ and } N\subset (\text{some }B)\in \mathcal{A} \text{ having } \mu(B)=0\}$. Hence $ \hat{ \mathcal{A}_\mu}\subset \{A \cup N: A\in \mathcal{A}, \text{ and } N\subset (\text{some }B)\in \mathcal{A} \text{ having } \mu(B)=0\}$.


*We only need to consider when $N=B_{n_0}$ for some $B$ since any $ A \in \{A \cup N'\} \subsetneqq \{A \cup N \}$ will be in $ \hat{ \mathcal{A}_\mu}$ provided that if $\{A \cup N \} \subset \hat{ \mathcal{A}_\mu}$. Let $A_2=A \cup B_{n_0}\in \mathcal{A}, A_1= A \in \mathcal{A} $, then we have $A_1\subset A \subset A_2$, $A_2\backslash A_1=B_{n_0}\backslash A \subset B_{n_0}.$ Therefore $\mu(A_2\backslash A_1)\leq \mu(B_{n_0})=0$. In so doing any $A\in \{A \cup N: A\in \mathcal{A}, \text{ and } N\subset B_{n_0}\in \mathcal{A} \text{ having } \mu(B_{n_0})=0\} \Rightarrow A\in \hat{ \mathcal{A}_\mu}$.
I am not sure if my proof makes sense as it is all too confusing. For example, I thought about letting $N= \cap B$ but I thought the set defined in the question means as long as $N \subset $ at least one $B$ such that $\mu(B)=0$ would be enough. Also, what I am thinking is given $A\in \hat{ \mathcal{A}_\mu}$, as long as I can find expression for $N$ and $B$ which satisfies the condition, and similarly, given $A \in \{A \cup B\}$ I only need to find $A_1, A_2$ which satisfies the condition. I am not quite sure what I am doing actually makes sense or not. I wish someone could point out to me whether I am on the right track. I have noticed that this seems to be a well-known extension or something that many have asked on the site. I am not very familiar with the set language and I am not a well-trained maths major students. I wished by showing the steps clear I could have someone points out my mistakes so I can spot flaws in my mind and have better understanding of the course I am learning.

I didn't quite understand the meaning of the competition theorem when I was doing the homework... With the help of many sites on Mathematics Stackexchange, I managed to finish the proof:
Prove that $\hat{ \mathcal{A}_\mu}$ is a $\sigma$-field:

*

*Show $\emptyset,\Omega\in \hat{ \mathcal{A}_\mu}$:

Since $(\Omega, \mathcal{A},\mu)$ is a measure space, $ \mathcal{A}$ is a $ \sigma\text{-field}$. Let $A_1=A_2\in \mathcal{A}$, we have $\mu(\emptyset)=0$. Therefore $\emptyset,\Omega\in \hat{ \mathcal{A}_\mu}$ as they are in $ \mathcal{A}$.

*

*Show if $A\in \hat{ \mathcal{A}_\mu}\Rightarrow A^c \in \hat{ \mathcal{A}_\mu}$:

Let $A\in \hat{ \mathcal{A}_\mu}$, we have $A_1 \subset A\subset A_2 \text{ with } A_1, A_2\in \mathcal{A} \text{ and } \mu(A_2\backslash A_1)=0$. Therefore $A_2^c\subset A^c \subset A_1^c$ with $A_1^c,A_2^c\in \mathcal{A}$ by definition of a $\sigma-$field and $\mu(A_1^c\backslash A_2^c)=\mu(A_1^c \cap A_2)=\mu(A_2\cap A_1^c)=\mu(A_2\backslash A_1)=0$.

*

*Show if $A_1,A_2,...\in \hat{ \mathcal{A}_\mu} \Rightarrow \cup^\infty A_n \in \hat{ \mathcal{A}_\mu}$:\

Let $A_1,A_2,...\in \hat{ \mathcal{A}_\mu}$, for any given $A_k\in \hat{ \mathcal{A}_\mu}$, we have $A_{1k}\subset A_k \subset A_{2k}$ such that $A_{1k},A_{2k}\in \mathcal{A}$ and that $\mu(A_{2k}\backslash A_{1k})=0,k\geq 1$. Taking the union of sequence $A_k$, we have $ \cup^\infty A_{1k}\subset \cup^\infty A_k \subset \cup^\infty A_{2k} $, and by definition of a $ \sigma\text{-field}$ we know $ \cup^\infty A_{1k}, \cup^\infty A_{2k}\in \mathcal{A}$.
\begin{equation*} 
\begin{split}
\mu(\cup A_{2k}\backslash\cup A_{1k}) & = \mu(\cup A_{2k} \cap (\cup A_{1k})^c)  \\ 
 & = \mu(\cup A_{2k})-\mu(\cup A_{1k}) \\
 & \leq \sum \mu(A_{2k})- (\sum \mu(A_{1k})-\epsilon) \\
 & = \sum (\mu(A_{2k})-\mu(A_{1k}))+\epsilon \\
 & = \epsilon,
\end{split}
\end{equation*}
$\epsilon$ can be any number greater than 0, hence we have $\mu(\cup A_{2k}\backslash\cup A_{1k})\leq 0$, but by definition $\mu \geq 0$, hence $\mu(\cup A_{2k}\backslash\cup A_{1k})=0$.

Claim: Denote three class of subset of $\Omega$ as $ \mathcal{A}_1, \mathcal{A}_2, \mathcal{A}_3. $ Show that $ \mathcal{A}_1 \subset \mathcal{A}_2 \subset \mathcal{A}_3 \subset \mathcal{A}_1.$ So that the three terms are equivalent.
Reference: Definition of completion of a measure space

*

*$ \mathcal{A}_1 \subset \mathcal{A}_2$: let $A_1 \subset A \subset A_2 $ with $A_1, A_2 \in \mathcal{A}, \mu(A_2\backslash A_1)=0$, then $A=A_1 \cup N$, where $N= A \cap (A_2 \backslash A_1) $.


*$\mathcal{A}_2 \subset \mathcal{A}_3$: $A \cup N = A\triangle N' $ where $N'=N\backslash A$ as $A\triangle N'= A\backslash N' + N' \backslash A = A \cup N' - AN' =  A \cup N' = A \cup N$. $N' \subset B\backslash A \subset B $ having $\mu(B \backslash A)=0$.


*$\mathcal{A}_3 \subset \mathcal{A}_1$: let $A_1=A\backslash B, A_2 = A \cup B$. Then $\mu(A_2\backslash A_1) = \mu(B) = 0. A_1 = A \backslash B \subset A \backslash N \subset  A\backslash N + N \backslash A = A\triangle N \subset A \cup B = A_2.$

Reference: Real Analysis, Folland Theorem 1.9, extention of a measure to a complete measure
Claim: $\hat{\mu}(A \cup N)=\mu(A)$ is well-defined.

*

*Suppose $A \cup N \in \hat{ \mathcal{A}_\mu}$, set $\hat{\mu}(A \cup N)= \mu(A)$ by definition. If $A_1 \cup N_1 = A_2 \cup N_2$ where $N_i \subset B_i \in \mathcal{A} \text{ with } \mu(B_i)=0$. $A_1\subset A_1 \cup N_1 = A_2 \cup N_2 \subset A_2 \cup B_2$. By monotonicity $\mu(A_1)\leq \mu(A_2)+\mu(B_2)=\mu(A_2)$. By symmetry $\mu(A_1)=\mu(A_2)$.

Claim: $\hat{\mu}$ is a measure.

*

*$\hat{\mu}(\emptyset)=\hat{\mu}(\emptyset \cup \emptyset)=\mu(\emptyset)=0$.

*$\hat{\mu}(A \cup N)= \mu (A)\geq 0$

*Let $\{A'_n\}^\infty_1 \in \hat{ \mathcal{A}_\mu}$ disjoint. We have shown there is an $\{A_n\}^\infty_1\in \mathcal{A}$ and $\{N_n\}^\infty_1 \subset B \in \mathcal{N}$ such that $A'_n = A_n \cup N_n$ for all n. Hence $\hat{\mu}(\bigcup^\infty_1 A'_n )=\hat{\mu}(\bigcup^\infty_1(A_n \cup N_n))=\hat{\mu}(\bigcup^\infty_1(A_n) \cup \bigcup^\infty_1(N_n))=\mu(\bigcup^\infty_1A_n)=\sum^\infty_1\mu(A_n)=\sum^\infty_1\hat{\mu}(A_n \cup N_n)=\sum^\infty_1\hat{\mu}(A'_n)$
Claim: $\hat{\mu}$ is complete.

*

*Let $E \subset \Omega$, $ F \in \hat{ \mathcal{A}_\mu}$ such that $E \subset F$. If $\hat{\mu}(F)=0$, $\hat{\mu}(F)=\hat{\mu}(A \cup N)=\mu(A)=0$. Notice that $A\in \mathcal{A}, N \subset B \in \mathcal{A} \Rightarrow \mu(A \cup B)=0$.  $E=E \cup \emptyset, \emptyset \in \mathcal{A}$, $E\subset F= A \cup N \subset A \cup B \in \mathcal{A} $. $E \subset A \cup B $ with $\mu(A\cup B)=0$ hence $E \in \hat{ \mathcal{A}_\mu}$.

*I think showing that from the definition $ \hat{ \mathcal{A}_\mu}$ includes all null set may also be a valid proof but I am not quite sure:


Notice also that $ \hat{ \mathcal{A}_\mu}$ also contained all null sets, since for any $\mu$ set $$ there’s is $B \in \mathcal{A}$ with $\mu(B)=0$ and $\emptyset \subset N \subset B$.

I am concerned with the existence of $B$ (Not sure if my concern makes sense of not).
Claim: $\hat{\mu}| \mathcal{A}=\mu$

*

*Let $A\in \mathcal{A}, A=A \cup \emptyset = A \cup N $ where $N \subset B \in \mathcal{A}$. Moreover, $\mu(B)=0$ therefore $A \in \hat{ \mathcal{A}_\mu}$ and $\hat{\mu}(A)=\hat{\mu}(A\cup \emptyset)=\mu (A)$

Claim: The extension is unique.

*

*Denote $\nu$ as an extension of $ \hat{ \mathcal{A}_\mu}$. Then $\nu (A \cup N) =\mu(A)$ for all $A \in \mathcal{A}, N\subset B\in \mathcal{A}$ such that $\mu(B)=0$. If $\nu$ is not unique, we have $\nu(A \cup N)> \mu(A)$.
\begin{equation*} 
\begin{split}
\mu(A) & < \nu(A \cup N)  \\ 
 & = \nu(A \cup A^cN) \\
 & =\nu(A)+\nu(A^cN) \\
 & \leq \nu(N) +\nu(B) \\
 & = \mu(A) + \mu(B),
\end{split}
\end{equation*}
So we have $\mu(B)>0$, contradicting to what we have supposed.


Other questions which help me understand the theorem:
Completion of measure spaces - uniqueness
Complete measure space

Proceeding Question (Unsolved):
Exercise 2.3 (Prove Approximation lemma by Halmos ) Probability for Statistician by Galen R. Shorack
 A: There are several characterizations of the completion of $\mathcal{A}$ with respect to the measure $\mu$ on $(X,\mathcal{A})$.
You can start with say
$$
\begin{align}
\mathcal{A}_\mu=\{A\subset X: A_1\subset A\subset A_2,\,\text{for some}\quad A_1,A_2\in\mathcal{A} \quad\text{with} \mu(A_2\setminus A_1)=0\}\tag{1}\label{one}
\end{align}
$$
Clearly $\mathcal{A}\subset\mathcal{A}_\mu$.
To check that $\mathcal{A}_\mu$ is a $\sigma$-algebra suppose $A\in\mathcal{A}_\mu$ and $\{A_m:m\in\mathbb{N}\}\subset\mathcal{A}_\mu$.

*

*Let $A_1,A_2\in\mathcal{A}$ such that $A_1\subset A\subset \mathcal{A}_2$ with $\mu(A_2\setminus A_2)=0$. Then
$$A^c_2=X\setminus A_2\subset X\setminus A\subset X\setminus A_1=A^c_1$$
and $\mu\big(A^c_1\setminus A^c_2)=\mu(A_2\setminus A_1)=0$. Hence $X\setminus A\in\mathcal{A}_\mu$.


*For each $m$ suppose $A_{1m},A_{2m}\in\mathcal{A}$ such that $A_{m1}\subset A_m\subset A_{2m}$ and $\mu(A_{2m}\setminus A_{1m})=0$. Then
$$ \bigcup_mA_{1m}\subset \bigcup_mA_m\subset\bigcup_mA_{2m}$$
and $\mu\Big(\big(\bigcup_mA_{2m}\big)\setminus\big(\bigcup_{\ell}A_{2\ell}\big)\Big)\leq \mu\Big(\bigcup_n(A_{m2}\setminus A_{1m})\big)\leq\sum_m\mu(A_{2m}\setminus A_{1m})=0$. Hence $\bigcup_mA_m\in\mathcal{A}_\mu$.
This concludes the proof that $\mathcal{A}_\mu$ is a $\sigma$-algebra. Notice also that $A_\mu$ also contained all null sets, since for any $\mu$ set $N$ there’s is $B\in \mathcal{A}$ with $\mu(B)=0$ and $\emptyset\subset N\subset B$.
Now it remains to show that the other descriptions of $\mathcal{A}_\mu$ are indeed equivalent. I will leave many of the details.  Notice that if $A\in\mathcal{A}$ and $N$ is a $\mu$ null set, then there is $B\in\mathcal{A}$ with $N\subset B$ and $\mu(B)=0$ and so
$$ A\subset A\cup N\subset A\cup B$$
Since $(A\cup B)\setminus A=B\setminus A\subset B$,
we can conclude that $A\cup B\in \mathcal{A}_\mu$. Thus
$$\{A\cup N:A\in\mathcal{A},\,N\,\text{ is a $\mu$}-null set\}\subset\mathcal{A}_\mu$$
Conversely,  if $A\in\mathcal{A}_\mu$  and  $A_1\subset A\subset A_2$, $A_1, A_2\in\mathcal{A}$, and $\mu(A_2\setminus A_1)=0$, then
$$ A= A_1\cup(A\setminus A_1)$$
Being tha  $A\setminus A_1\subset A_2\setminus A_1\in \mathcal{A}$ and $\mu(A_2\setminus A_2)=0$, we obtained that
$$\mathcal{A}_\mu\subset\{A\cup N: A\in \mathcal{A}, N \text{ is a $\mu$ null set}\}$$
WE have shown that if $\mathcal{N}_\mu$ denotes the collection of all $\mu$-null sets, then
$$\{A\cup N: A\in\mathcal{A},\,N\in\mathcal{N}_\mu\}=\mathcal{A}_\mu=\sigma(\mathcal{A}\cup\mathcal{N}_\mu)$$
The last identity, namely $\mathcal{A}_\mu=\{A\subset X: A\triangle A'\in\mathcal{N}_\mu\,\,\text{for some}\,\, A'\in\mathcal{A}\}$ can be proven along the same lines.
