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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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I think that the restriction just say that your sets $A$ and $B$ are bounded. This is because the Jordan measure is only defined for bounded sets. See – leo Sep 9 '11 at 23:51
Well, if either A or B are $\infty$, then the inequality will always become trivial, with $\infty$ in each term. If either of the sets is infinite, it can only be covered by the whole of $\mathbb R$, whose measure is $\infty$ – gary Sep 10 '11 at 5:55
hhh: BTW, I answered the question in the other post; let me know if you have more questions. – gary Sep 10 '11 at 6:45
@gary: sorry but this question plays with uncountable sets, the latter question played with countable sets here. – hhh Sep 10 '11 at 15:33
But the answer I gave below does not assume the sets A,B are countably infinite. I am not sure I get your point. – gary Sep 11 '11 at 3:24

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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Think about the case $\underline{\mu}(A) \leq \underline{\mu}(A \cup B) \leq \underline{\mu}(A) + \bar{\mu}(B)$. It works similarly and the last point with upper measure due to $\bar{\mu} \geq \underline{\mu}$? – hhh Sep 10 '11 at 15:18
Yes if $d(A) = \sup( \{ d(x,y) | x,y \in A \} ) < \infty$ holds but, without the constraint, I am unsure about the latter because the sets can be infite. With extended number systems such as $\mathbb R \cup \infty$, we can consider $\infty$ as a point and things will not get broken, or? – hhh Sep 10 '11 at 15:27
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. – gary Sep 11 '11 at 3:06

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