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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?

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    But the answer I gave below does not assume the sets A,B are countably in$f$inite. I am not sure I get your point.2011-09-11

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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.

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    I really don't know how to define measures in the extended reals. In the case of standard reals, if either one of the sets is infinite, then the inequality is trivially true. Also, the answer to your questions is yes; (sorry, I don't know how to do mu-hat, so I will use $m^*$ for outer- and $m_*$ for inner). Then you are right; $m_* \leq m^*$, so the inequality does hold.2011-09-11