Jordan Measures without $d(A) = \sup( \{ d(x,y) | x,y \in A \} ) < \infty$? I am trying to prove that Jordan measures satisfy with the following properties $A, B \subset \mathbb R$ and $d(A) = \sup( \{ d(x,y) | x,y \in A \} ) < \infty$, similarly for $B$:
$$\bar{\mu} (A) \leq \bar{\mu} (A \cup B) \leq \bar{\mu}(A) + \bar{\mu}(B)$$
I am uncertain of the terminology here. So I want to know why there is a restriction $d(A) = \sup( \{ d(x,y) | x,y \in A \} ) < \infty$? How does it change the problem without and with the assumption? What kind of things we have if we do not have the assumption about the limit?
 A: You can prove the inequality by using finite additivity and monotonicity; by monotonicity we mean that if $A\subset B$ , then $m^*(A)\leq m^*(B)$, and additivity means that if $A,B$ are disjoint, i.e., if $A\cap B$ is empty, them $m^*(A\cup B)=m^*(A)+ m^*(B)$. Then, given any $A,B$, you can "disjointize them" , i.e., you can find a pair of sets $A',B'$ so that $A',B'$ are disjoint, and $A\cup B=A'\cup B'$, (what we want is A'.B' disjoint., so that $m^*(A\cup B)=m^*(A'\cup B')$.)
Then the first , i.e., leftmost part of the inequality, $m^*(A)\leq m^*(A\cup B)$, follows by additivity of the measure, since $A\subset (A\cup B))$ (sorry, I don't know how to do non-strict inclusion ).
For the other part of the inequality, i.e., to show that $m^*(A\cup B) \leq m^*(A)+m^*(B) $ we can define sets A',B' as above, by just setting :$A':=A$, and $B':=B-A$ , where $'-'$ just means set complement. Then, since
$A',B'$ are disjoint, and $ A\cup B=A'\cup B'$, we have that $m^*(A\cup B)=m^*(A'\cup B')=m^*(A')+m^*(B'):=m^*(A)+m^*(B-A)$
We then have, by transitivity, that $m^*(A\cup B) \leq m^*(A)+m^*(B-A)$.  (##)
Now, use the fact that $(B-A)\subset B$, so that, by monotonicity $m^*(B-A)\leq m^*(B)$ and we substitute this in (##) above , so that $m^*(A'\cup B')=m^*(A')+m^*(B')=m^*(A)+m^*(B-A)\leq m^*(A)+m^*(B-A) \leq m^*(A)+m^*(B)$, and we are
done.  
