Definition of completion of a measure space On a measure space $(\Omega,\mathcal{A},\mu)$ a completion of a measure is defined as:
$\{A: A_1\subset A\subset A_2$ with $A_1,A_2\in\mathcal{A}$ and $\mu(A_2\backslash A_1)=0\}$
I'm trying to show that this set is equivalent to:
$\{A\cup N:A\in\mathcal{A}$, and $N\subset B\in\mathcal{A}$ for some $B$ with $\mu(B)=0\}$
and
$\{A\triangle N:A\in\mathcal{A}$, and $N\subset B\in\mathcal{A}$ for some $B$ with $\mu(B)=0\}$
where $\triangle$ is the symmetric difference.
To show that the first definition is equivalent to the second, I showed that some $A_a$ in the first set is in the second set (which was pretty easy since $A_a\in\mathcal{A}$ implies that it is in $\mathcal{A}\cup N$. However, I am having trouble going the other direction.
For the third set, I'm having a hard time seeing the picture.
Thanks for your help.
 A: I call these families $\mathcal{A}_1,\mathcal{A}_2,$ and $\mathcal{A}_3$.
$\mathcal{A}_1\subset\mathcal{A}_2$: 
Let $A_1\subset A\subset A_2$ with $\mu(A_2\backslash A_1)=0$. Then $A=A_1\cup N$ where $N=A\cap (A_2\backslash A_1)$. So $A\in\mathcal{A}_2$.
$\mathcal{A}_2\subset\mathcal{A}_3$: 
Take $A\cup N$, where $N$ is a subset of a measurable set $B$ of measure $0$. Let $N'=N\backslash A\subset B\backslash A$. Clearly, $\mu(B\backslash A)=0$. Now $$A\triangle N'=(A\backslash N')\cup (N'\backslash A)=A\cup N'=A\cup N\backslash A=A\cup N.$$
$\mathcal{A}_3\subset\mathcal{A}_1$: 
Take $A\triangle N$, where $N$ is a subset of a measurable set $B$ of measure $0$. Let $A_1=A\backslash B$ and $A_2=A\cup B$. Now $A_1=A\backslash B\subset A\backslash N\subset A\backslash N\cup N\backslash A=A\triangle N\subset A\cup N\subset A\cup B=A_2$. Now $A_2\backslash A_1=(A\cup B)\backslash (A\backslash B)=B$, so $\mu(A_2\backslash A_1)=0$.
A: Another way of approaching this is to prove that each of these families ($\mathcal A_1$, $\mathcal A_2$ and $\mathcal A_3$) is equal to $\sigma(\mathcal A \cup \mathcal N)$ which is the sigma algebra generated by $\mathcal A \cup \mathcal N$ where $\mathcal N$ is the collection of subsets of null sets in $\mathcal A$. This approach, which can be more work than the direct approach of simply showing $\mathcal A_1 \subset \mathcal A_2 \subset \mathcal A_3 \subset \mathcal A_1$,, has a benefit of making one see motivations for each of the three definitions.
Let's start with $\mathcal A_1$. To show that this is equal to $\sigma(\mathcal A \cup \mathcal N)$, one only needs to show that $\mathcal A_1$ is a sigma algebra containing $\mathcal A \cup \mathcal N$ and that each element of $\mathcal A_1$ can be obtained by applying countable union, countable intersection, and complement several times to some elements in $\mathcal A \cup \mathcal N$.
It is easy to show that each element of $\mathcal A_1$, $\mathcal A_2$ and $\mathcal A_3$ can be obtained by applying countable union, countable intersection, and complement several times to some elements in $\mathcal A \cup \mathcal N$. Therefore, each of $\mathcal A_1$, $\mathcal A_2$ and $\mathcal A_3$ is a subset of $\sigma(\mathcal A \cup \mathcal N)$.
It is also easy to show that each of $\mathcal A_1$, $\mathcal A_2$ and $\mathcal A_3$ contains $\mathcal A \cup \mathcal N$.
Now it only remains to show that they are sigma algebras. For that, read the following thread first:
Completion of a measure space
The question in that thread is concerned with a collection that you can easily see is equal to $\mathcal A_3$ (Equality follows from $A \Delta B = C \iff B \Delta C = A \iff C \Delta A = B$). 
Michael Greinecker's answer in that thread contains a proof of the fact that $\mathcal A_2$ is a sigma algebra.
Jisang Yoo's answer in that thread contains a proof of the fact that $\mathcal A_3$ is closed under finite union, but you can easily generalize the proof to show that it is closed under countable union as well. 
As for how to prove that $\mathcal A_1$ is closed under countable union, let $(A_i)$ be a sequence of elements in $\mathcal A_1$. For each $i$, there are $A_i'$ and $A_i''$ in $\mathcal A$ that approximate $A_i$ from below and above in the sense that $A_i' \subset A_i \subset A_i''$ and $\mu(A_i'' \setminus A_i') = 0$.
You'd expect that the union $\bigcup_i A_i$ would be approximated by $\bigcup_i A_i'$ and $\bigcup_i A_i''$ in the same sense. Is it true that $\mu(\bigcup_i A_i'' \setminus \bigcup_i A_i') = 0$ holds? Yes because the final total exception set $\bigcup_i A_i'' \setminus \bigcup_i A_i'$ must be covered by the union of the exception sets we started with, namely $A_i'' \setminus A_i'$, but these are null sets.
