While my question is not high-level, I've had lots of views and no answers on Stackexchange (https://math.stackexchange.com/q/2391423 ) so I'm hoping for better luck posting it here .....
I'm trying to untangle the relationship between monotone and sub-additive functions, and would like confirmation or correction on the proposition that monotonicity and sub-additivity are independent..
If $\mu$ is a set function mapping some collection $\mathscr B$ of subsets of $X$ to $[0, \infty]$ then
$\mu$ is monotone if for $A, B \in \mathscr B $ and $A \subset B$ then $\mu(A) \le \mu(B)$
$\mu$ is sub-additve if for $A, B$ and $A \cup B \in \mathscr B $ then $\mu(A \cup B) \le \mu(A) + \mu(B)$
Proposition: monotonicity and sub-additive are independent. Proof by example. Let $\mathscr B $ be the powerset of {$a, b$} (so, $\mathscr B $ has the structure of an algebra) and $\mu$ take the values $\mu(\emptyset) = 0; \mu(\{a\}) = 1 ; \mu(\{b\}) = 1 ; \mu(\{a, b\}) = 0 $ or $3 $
Then for $\mu(\{a, b\}) = 0 $, $\mu$ is sub-additive $\mu(\{a, b\}) = 0 \le 2 = \mu(\{a\}) + \mu(\{b\}) $, but not monotone $\{a\} \subset \{a, b\}$ but $1 = \mu(\{a\}) > 0 = \mu(\{a, b\})$
While for $\mu(\{a, b\}) = 3 $, $\mu$ is monotone $\{a\} \subset \{a, b\}$ and $1 = \mu(\{a\}) < 3 = \mu(\{a, b\})$, but not sub-additive $\mu(\{a, b\}) = 3 > 2 = \mu(\{a\}) + \mu(\{b\}) $
(I found this particularity confusing since it seems that finitely monotone $\implies$ finitely sub-additive and countably monotone $\implies$ countably sub-additive. It appears that monotone (as defined above) is of a somewhat different nature to finitely monotone).
