Inner Measures of Subsets 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.)

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 $m_*(B)$.
