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How can I show that for $A \subset B$, it must be true that $m_*(A) \leq m_*(B)$? That is, the inner measure of A is less than or equal to the inner measure of B? I understand how to show a similar proof for the outer measure, but is there an explicit way to show it for inner measure? Do I just extend the outer measure proof by showing that $\forall A$, $m_*(A) \leq m^*(A)$? How would I show that this wouldn't then violate the inner measure inequality (if, say, the inner measure of the set differs more greatly from it's outer measure than the subset does.)

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It is clear from the definition of inner measure: $m_*(A) = \sup \{m(S): S \in \Sigma, S \subseteq A\}$, since if $A \subset B$, all $S$ that are considered for $m_*(A)$ are also considered for $m_*(B)$.

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Did you mean to have the asterisks where they are in the second line? – Old John Sep 23 '12 at 23:34
Oops, I'll edit. Thanks – Robert Israel Sep 24 '12 at 2:57

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