Are $\mathscr{A}_\mu$ and $\mathscr{M}_{\mu^*}$ the same? I am reading Cohn and he uses the notation $\mathscr{A}_\mu$ to mean the completion of $\mathscr{A}$ under $\mu$, and he says that a set in $\mathscr{A}_\mu$ is called $\mu$-measurable.
He uses the notation $\mathscr{M}_{\mu^*}$ to represent the $\sigma$-algebra of sets that satisfy the Caratheodory criterion with respect to the outer measure $\mu^*$, and he says that a set in $\mathscr{M}_{\mu^*}$ is called $\mu^*$-measurable.
Is there a reason why he uses the same phrase to describe what appear to be two different types of sets? Are these somehow equivalent?
Edit: Note that here I mean $(X,\Sigma,\mu)$ is an arbitrary measure space, not necessarily Lebesgue measure.
 A: Yes, they are equivalent generalizations of the construction of Lebesgue Measurable Sets on $\mathbb{R}^n.$ Throughout, let $d$ be the pseudometric induced by $\mu^* .$
$\textbf{ Part 1) }$ [$\mathscr{A}_{\mu } \subset \mathscr{M}_{\mu^*}$ ] 
By countable subaddivity, for measurable $A$ and all $E \subset X$ we have $\mu^* (E ) \le \mu^* (A \cap E ) + \mu^* (A^c \cap E),$ so that we need only show the reverse inequality.
When $A \in \mathscr{A}$ , we have for any approximate outer measure $l$ of $E$ that there exists a countable $\{ E_n \}$ covering $E$ such that 
$l \ge \displaystyle\sum_{n=1}^{\infty } \mu (E_n) =$ 
$\displaystyle\sum_{n=1}^{\infty } [ \mu (E_n \cap A) + \mu (E_n \cap A^c)] = $
$\displaystyle\sum_{n=1}^{\infty }  \mu (E_n \cap A) + \displaystyle\sum_{n=1}^{\infty }  \mu (E_n \cap A^c ) =$
$\mu (E \cap A) + \mu (E \cap A^c )$
Now for $A \in \mathscr{A}_{\mu },$ consider the same notation as before.   Note that if $l$ is an approximate outer measure for $E$ then we can write $ l = \displaystyle\sum_{n=1}^{\infty } l_n + g_n , $ where each $l_n$ is an approximate outer measure for $E_n \cap A_n ,$ and $g_n$ for $E_n \cap A^c.$ Hence
$l \ge \displaystyle\sum_{n=1}^{\infty} [\displaystyle\sum_{k=1}^{\infty } \mu (A_k \cap E_n) ] + [\displaystyle\sum_{k=1}^{\infty } \mu (B_k \cap E_n) ] = \mu (E \cap A) + \mu (E \cap A^c)$
$\textbf{ Part 2) }$ [$\mathscr{M}_{\mu^* } \subset \mathscr{A}_{\mu}$ ]
Suppose $\mu^* (A) $ is finite and that $A$ satisfies the Caratheodory criterion. Then for any $\epsilon >0,$ there exists $E \in \mathscr{A_{\mu }} $ such that $\mu^* (E) - \mu^* (A) \le \epsilon .$ Hence
$d(E, A) = \mu^* ( (A-E) \cup (E-A) ) =$ 
$\mu^* (E-A ) =$ 
$ \mu^* (A^c \cap E) =$ 
$  \mu^* (E) - \mu^* (A \cap E) = $ 
$\mu^* (E) - \mu^* (A) \le \epsilon $
