Is the collection $\mathcal{M}$ of $\mu$-measurable sets maximal so that $\mu|_{\mathcal{M}}$ is a measure? Let $\mu:2^{X} \to [0, \infty]$ be an outer measure. The collection $\mathcal{M}$ of $\mu$-measurable sets are then defined as those sets $A$ satisfying $\mu(S)=\mu(S \cap A) + \mu(S \setminus A)$ for each $S \subset X$. It's proven in any measure theory course that $\mathcal{M}$ is a $\sigma$-algebra, with $\mu|_{\mathcal{M}}$ a complete measure.
I've always found this to be an elusive definition. This has been discussed elsewhere.
A more natural approach to defining the collection $\mathcal{M}$ of $\mu$-measurable sets, in my mind, is that we want it to satisfy the following property: Let $\mathcal{U}_{\mu} \subset 2^{2^{X}}$ denote the collection of $\sigma$-algebras $\mathcal{F}$ of $X$ with the property that $\mu|_{\mathcal{F}}$ is a complete measure on $\mathcal{F}$. Then $\mathcal{M}$ is inclusion-wise maximal in $\mathcal{U}_{\mu}$. In simple terms, we want $\mathcal{M}$ to be the largest set possible on which $\mu$ is a measure.
This isn't a good definition for $\mathcal{M}$, unfortunately, since we have no guarantee a priori that $\mathcal{U}_{\mu}$ has a unique maximal element.
So I have some questions:

Is $\mathcal{M}$, as defined in the first paragraph, in fact inclusion-wise maximal in $\mathcal{U}_{\mu}$?


Does $\mathcal{U}_{\mu}$ has a unique inclusion-wise maximal element?


If the question to the latter question is "no", then what can we say about inclusion-wise maximal members of $\mathcal{U}_{\mu}$ distinct from $\mathcal{M}$ as defined in paragraph one?

 A: 
Is $\mathcal{M}$, as defined in the first paragraph, in fact inclusion-wise maximal in $\mathcal{U}_{\mu}$?

Yes, provided subsets of $X$ can be approximated by members of $\mathcal M$, that is for each $S\subset X$ and each $\varepsilon>0$ there exists $S’\in\mathcal M$, $S’\supset S$ such that $\mu(S’)<\mu(S)+\varepsilon$. Indeed, if $A\subset X$ but $A\not\in\mathcal M$ then there exists $S\subset X$ such that $\mu(S)<\mu(S \cap A) + \mu(S \setminus A)$. Pick $S’\in\mathcal M$, $S’\supset S$ such that $\mu(S’)< \mu(S \cap A) + \mu(S \setminus A)$. Then $\mu(S’)< \mu(S’\cap A) + \mu(S’\setminus A)$, so the restriction of $\mu$ on the algebra generated by $\mathcal M$ and $S$ is not additive.
On the other hand, if $\mu$ has no above approximation property then $\mathcal{U}_{\mu}$ can fail to have a unique inclusion-wise maximal element. Indeed, let $X=\{1,2,3\}$. For each $A\subset X$ put
$$\mu(A)=\cases{
0, \mbox{ if }A=\varnothing,\\
1,\mbox{ if }1\le |A|\le 2,\\ 
2, \mbox{ if }A=X.}
$$
Then $\mathcal M=\{\varnothing, X\}$ but $\mathcal{U}_{\mu}$ has three maximal elements:
$\{\varnothing,\{1\},\{2,3\},X\}$,   $\{\varnothing,\{2\},\{1,3\},X\}$, and $\{\varnothing,\{3\},\{1,2\},X\}$.
