Remember Caratheodory extension theorem: we get an outer measure by extending a premeasure over an algebra. We also have Caratheodory theorem, which give us a measure by restricting that outer measure to measurable set. I have find the theorems rather weird, so I hope I can understand it more by asking this.
Now this question is about the collection of sets rather than specific numerical value of the measure. We will first set the context:
Let $X$ be a set, and consider a collection of set $\mathcal{L}\subset P(X)$ such that $\emptyset\notin\mathcal{L}$ and $X\in\sigma(\mathcal{L})$. Note that $\mathcal{L}$ is not an algebra. Fix this $\mathcal{L}$. We said a measure $\mu$ to be a PF-measure if for any $S\in\mathcal{L}$ then $\mu(S)$ is defined and $0<\mu(S)<\infty$. A $\sigma$-algebra $\mathcal{A}$ is called PF-measure-definable if $\mathcal{L}\subset\mathcal{A}$ and there exist a PF-measure defined over $\mathcal{A}$. Assume that $\sigma(\mathcal{L})$ is PF-measure-definable.
Question:
Is there exist a maximum PF-measure-definable collection? That is, is there exist a PF-measure-definable $\mathcal{M}$ such that for any PF-measure-definable $\mathcal{A}$ then $\mathcal{A}\subset\mathcal{M}$.
Given a PF-measure defined over $\sigma(\mathcal{L})$, we a corresponding collection of measurable sets using Caratheodory's. Given two different PF-measure, does the collections of measurable sets corresponding to these two measure the same? If not unique, is there some sorts of reasonable sufficient/necessary assumption about $\mathcal{L}$ that if true imply uniqueness?
If the answers to the above two questions are favourable, is the maximum collection that is PF-measure-definable the same as the collection of measurable sets due to Caratheodory's? If answer to question #2 is not unique, does there exist at least a measure on $\sigma(\mathcal{L})$ such that Caratheodory's collection of measurable set would be that maximum PF-measure-definable collection?
Thank you for your input.