Proof outer measure satisfies monotonicity: $A \subseteq B \implies m^*(A) \leq m^*(B)$ Theorem: $$A \subseteq B \implies m^*(A) \leq m^*(B)$$
Proof Attempt:
By definition, $m^*(B) = \inf\{\sum\limits_{k=1}^\infty |J_k||\{J_k\} \text{ is a cover of B }\}$, $m^*(A) = \inf\{\sum\limits_{k=1}^\infty |I_k||\{I_k\} \text{ is a cover of A }\}$
Then since $A \subset B \implies \bigcup_{k =1}^\infty I_k \subseteq \bigcup_{k=1}^\infty J_k$, then $\sum\limits_{k=1}^\infty |I_k| \leq \sum\limits_{k=1}^\infty |J_k|$, so we have $m^*(A) \leq m^*(B)$
Can someone check if everything checks out?
 A: I am just a little bit unsighted by your proof, hence I want to give you a proof which is roughly the same, but a little more sharpened so that you can see how get rid of small discrepancies in proofs.
So $m^*(B) = \inf \{ \sum_1^n|J_k|, \{ J_k\} \text{ is an open interval cover of B}\}$.
Similarly, $m^*(A) = \inf \{ \sum_1^n|J_k|, \{ J_k\} \text{ is an open interval cover of A}\}$.
Note that $A \subset B$,whence if $\{J_k\}$ covers $B$, it also covers $A$. Hence, the following is true:
$$
\{ \sum_1^n|J_k|, \{ J_k\} \text{ an open interval cover of B}\} \subset \{ \sum_1^n|J_k|, \{ J_k\} \text{ an open interval cover of A}\}
$$
Now, the infimum of the subset of a family is larger than the infimum of the family, because the family could contain an element that is  not in the subset but smaller than every element in the subset.
Hence, it follows that the infimum of the left side, which is $m^*(B)$, is greater than the right side, which is $m^*(A)$.
EDIT: As other have pointed  out, there was something wrong in your proof. However, you were thinking in the right direction, and hopefully this proof will show you how to  put those efforts in a more rigorous direction.
A: I don't really think that's good sounding. The gist of it is that, since $A\subseteq  B$, any cover of $B$ is also a cover of $A$, so the infimum for $m^*(B)$ is being taken over a set 'no greater' than that of $m^*(A)$.
