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.