How can two finite measure on the same measurable space have different completions? I am given the following question:
Let $\mu$ and $\nu$ be finite measures on a measurable space $(\Omega, \mathcal{A})$. Let $\hat{\mathcal{A}}_{\mu}$ and $\hat{\mathcal{A}}_{\nu}$ be the completion of the two measures, i.e.,
$$
\begin{aligned}
&\hat{\mathcal{A}}_{\mu} \equiv\left\{A: A_{1} \subset A \subset A_{2} \text { with } A_{1}, A_{2} \in \mathcal{A} \text { and } \mu\left(A_{2} \backslash A_{1}\right)=0\right\} \\
&\hat{\mathcal{A}}_{\nu} \equiv\left\{A: A_{1} \subset A \subset A_{2} \text { with } A_{1}, A_{2} \in \mathcal{A} \text { and } \nu\left(A_{2} \backslash A_{1}\right)=0\right\}
\end{aligned}
$$
(a). Show by example that $\hat{\mathcal{A}}_{\mu}$ and $\hat{\mathcal{A}}_{\nu}$ need not be equal.
(b). Prove or disprove: $\hat{\mathcal{A}}_{\mu}=\hat{\mathcal{A}}_{\nu}$ implies $\mu$ and $\nu$ have exactly the same sets of measure zero. That is, if the conclusion is correct, prove it. If not, provide an example.
(c). Prove or disprove: that $\mu$ and $\nu$ have exactly the same sets of measure zero implies $\hat{\mathcal{A}}_{\mu}=\hat{\mathcal{A}}_{\nu} .$ That is, if the conclusion is correct, prove it. If not, provide an example.
For a) I am having a hard time imagining any concrete example where $ A_1 \subset A \subset A_2$ are all defined to be 0 under the completion of two measures and are different from each other.
for b) Also, if $\hat{\mathcal{A}}_{\mu}=\hat{\mathcal{A}}_{\nu}$, would $\mu $ and $\nu$ have exactly the same sets of measure zero? Intuitively I think the answer should be yes but again I don't know how to prove it. The idea is that if the completion of two measures is identical, then all measure 0 sets are defined, all measure 0 sets are defined and the competition is the same, the measure zero sets must also be identical.
c) I suspect the answer is wrong because a) says the two there are examples where two completions need not be equal. Honestly, if I know how to approach a) I may know how to approach this question too...
I think this question reveals I don't quite understand what exactly is a completion. I only intuitively understand that if completion means all subsets of a measure 0 set are also defined to have measure 0. I don't see much use of these statements and any significant implication. I am struggling with the definition and ay implications and important intuition I have overlooked.
I would be grateful if someone could point out my blindspot of my understanding and maybe guide me how to solve this problem and any other related concepts I should pay closer attention, too. (I suspect if I have no idea and clue for this question I must have misunderstood and overlooked any other important concepts as well).
Thank you so much for the help!
 A: (a) Consider the measures in $\nu$ and $\mu$ and $(\mathbb{R},\mathscr{B}(\mathbb{R})$ given by $\mu(dx)=\mathbb{1}_{(0,1]}(x)\,dx$ and $\nu(dx)=\mathbb{1}_{(1,2]}(x)\,dx$ It is easy to check that these two measures have different completions.
For example, any subset $A\subset (1,2]$ belongs to the completion of $\mu$ but not every subset of $(1,2]$ belongs to the completion of $\nu$.
Other examples of these type can be obtained by considering measures $\mu$ and $\nu$ on $(\Omega,\mathscr{F})$ which are mutually singular ($\mu\perp\nu)$), that is, there is $E\in\mathcal{A}$ such that $\mu(E)=\nu(\Omega\setminus E)=0$

(c) Let $\mathcal{N}_\mu=\{N\subset \Omega: \text{there is}\, B\in\mathcal{A}\,\text{with}\, N\subset B\, \text{and}\,\mu(B)=0\}$, that is $\mathcal{N}_\mu$ is the collection of all $\mu$-negligible sets. It is not difficult to check that
$$\sigma(\mathcal{A}\cup\mathcal{N}_\mu)=\hat{\mathcal{A}}_\mu$$
Indeed, clearly $\mathcal{N}_\mu\subset\hat{\mathcal{A}}_\mu$ and so,  $\sigma(\mathcal{A}\cup\mathcal{N}_\mu)\subset\hat{\mathcal{A}}_\mu$. Conversely, if $A\in\hat{\mathcal{A}}_\mu$ then there are some $E,F\in\mathcal{A}$ such that $E\subset A\subset F$ such that $\mu(F\setminus E)=0$. Then $A=E\cup (A\setminus E)\in \sigma(\mathcal{A}\cup\mathcal{N}_\mu)$.
From this, it follows that  if $\mu$ and $\nu$ are measures on $(\Omega,\mathcal{A})$ and $\mu$ and $\nu$ have the same negligible sets ($\mathcal{N}_\mu=\mathcal{N}_\nu$), then $\mu$ and $\nu$ have the same completion.
Comment: Another interesting representation of $\hat{\mathcal{A}}_\mu$ is
$$\hat{\mathcal{A}}_\mu=\{A\subset \Omega: \text{there is} \quad B\in\mathcal{A}\quad\text{with}\quad A\triangle B\in\mathcal{N}_\mu\}$$
where $A\triangle B:=(A\setminus B)\cup (B\setminus A)$.
Indeed, if $A\in\hat{\mathcal{A}}_\mu$, there are $E,F\in\mathcal{A}$ with $E\subset A\subset F$ such that $\mu(F\setminus E)=0$; hence $A\triangle F=F\setminus A\in\mathcal{N}_\mu$. Conversely,  if $A\triangle B\in\mathcal{N}_\mu$ and $B\in\mathcal{A}$, then there is $M\in\mathcal{A}$ such that $A\triangle B\subset M$ and $\mu(M)=0$. Then
$$B\setminus M\subset B\setminus (B\setminus A) = A\cap B\subset B$$
and so,
$$ B\setminus M\subset (B\cap A) \cup (A\setminus B)\subset B\cup M$$
Since $B\setminus M,B\cup M\in\mathcal{A}$ and $\mu((B\cup M)\setminus (B\setminus N)))\leq\mu(M)=0$, it follows that $A\in\hat{\mathcal{A}}_\mu$.
From this representation one can also answer (c) in the positive.

(b) If $\hat{\mathcal{A}}_\mu=\hat{\mathcal{A}}_\nu$, then $\mathcal{N}_\mu\subset\hat{\mathcal{A}}_\nu$ and $\mathcal{N}_\nu\subset\hat{\mathcal{A}}_\mu$; however $\mathcal{N}_\mu=\mathcal{N}_\nu$ may not hold.

*

*Consider the Lebesgue measure $\lambda$ on the real line, and let $\delta_0$ be the measure on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$ that gives mass $1$ to every Borel set that contains $0$. Then $\lambda$ and $\lambda+\delta_0$ have the same completions; however $\{0\}$  has $\lambda$-measure $0$ but has $(\lambda+\delta_0)$-measure $1$.


*For a simpler example, consider the measure $\delta_1$ on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$ (the measure that assigns mass $1$ to any Borel set containing $1$). The completions of $\delta_1$ and $\delta_0$ are the same, namely the power set of $\mathbb{R}$; however $0=\delta_0(\{1\})\neq\delta_1(\{1\})=1$.
