# 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.

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math.stackexchange.com/q/68031 – user42744 Sep 26 '12 at 14:44

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

$d(E, A) = \mu^* ( (A-E) \cup (E-A) ) =$ $\mu^* (E-A ) =$ $\mu^* (A^c \cap E) =$

$\mu^* (E) - \mu^* (A \cap E) =$ $\mu^* (E) - \mu^* (A) \le \epsilon$

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Sorry, I meant $(X,\Sigma,\mu)$ to be an arbitrary measure space, not Lebesgue measure. Is it true in a more general setting? – nullUser Sep 26 '12 at 13:11
This works for an arbitrary measure space. – user17794 Sep 26 '12 at 17:12
"When $A$ is a finite union of intervals" doesn't make sense in an arbitrary measure space. What do you mean? – nullUser Sep 26 '12 at 19:05
Ah, duh. The natural analogue - see the edits. – user17794 Sep 26 '12 at 19:44