Let $X$ a set, and $\mathcal{P}(X)$ the class of its subset's.
Let $\mathcal{A}\subset \mathcal{P}(X)$, we call a map $L: \mathcal{P}(X)\to[0, \infty]$ $\mathcal{A}$-***regular*** if for any $S\subset X$ we have $L(S)= inf\_{S\subset A\in \mathcal{A}} L(A)$.


A map $L: \mathcal{P}(X)\to[0, \infty]$ is called a *outer-measure* if


$L(\emptyset)=0$, $A\subset B\Rightarrow L(A)\leq L(B)$, $L(\bigcup\_{n\in \mathbb{N}} A\_n) \leq \sum\_{n\in \mathbb{N}} L(A\_n)$


Form a outer-measure $L$ we have the Carathéodory's construction associated measure: $(X, \mathcal{A}\_L, \mu\_L)$ where $\mathcal{A}\_L$ is the class of subset $A\subset X$ such that $\forall S\subset X: L(S)=L(S\setminus A)+ L(S\cap A)$, and $\mu\_L$ is the restriction of $L$.


Let $\mu: \mathcal{A}\to[0, \infty[$ a real finite ($\sigma$-additive) measure on a a $\sigma$-algebra $\mathcal{A}\subset \mathcal{P}(X)$, we can define the associated outer-measure $\mu^\ast: \mathcal{P}(X)\to[0, \infty]$ as $\mu^\ast(S):= inf\_{S\subset A\in \mathcal{A}} \mu(A)$. It is well know that if $(X, \mathcal{A'}, \mu')$ is its Carathéodory' constructiuon then $\mathcal{A}\subset \mathcal{A'}$ and $\mu$ is the restriction of $\mu'$, and $\mu^\ast$ is $\mathcal{A}$-regular and then $\mathcal{A'}$-regular.


***THen I ask***: If I have a outer measure $L$ as above


and its Carathéodory's construction $(X, \mathcal{A}\_L, \mu\_L)$


*is $L$ a $\mathcal{A}\_L$-regular map*?


*Motive of the question*: I try to study a (a adjoint functors pair?) correspondence between the measure and the outer measure on a set.



About the Caratheodory class.
Given a set $X$ and outer measure $L:\mathcal{P}(X)\to [0,\infty]$, is $L$ an $\mathcal{A}\_L$-regular map? The answer is No. Consider the following example:


Let $X = \mathbb{N}$ and define $L:\mathcal{P}(\mathbb{N}) \to [0,\infty]$ by $L(\emptyset) = 0$ and $L(F) = 1$ for $F\subseteq \mathbb{N}$ finite, and $L(S) = \infty$ for $S\subseteq\mathbb{N}$ infinite.


It is easy to see that $L$ is an outer measure.


**Observation:** If $F\subseteq \mathbb{N}$ is non-empty and finite, then $F\notin \mathcal{A}\_L$.


*Proof* : Let $F\subseteq \mathbb{N}$ be a non-empty finite set. Then let $F^\*:= F \cup \{\max(F) + 1\}$. So $L(F^\*) = 1$ but $L(F^\*\setminus F) + L(F^\*\cap F) = 2$.


Now consider the finite set $\{1\}$. We have $L(\{1\}) = 1$, but the observation above implies that for any $A\in \mathcal{A}\_L$ with $\{1\} \subseteq A$ we have that $A$ is infinite, therefore $L(A) = \infty$. So $\inf\{L(A): \{1\} \subseteq A\}=\infty$, therefore $1 = L(1) \neq \inf\{L(A): \{1\} \subseteq A\}$. So $L$ is not $\mathcal{A}\_L$-regular.


