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For a bounded subset $B$ of $\mathbb{R}^n$, its inner Jordan measure is defined as $$ m_*(B)=\sup_{S\subset B} m (S) $$ and its outer Jordan measure as $$ m^*(B)=\inf_{S\supset B} m (S) $$ where the infimum and supremum are taken over simple sets $S$ and $m$ is the volume function of simple sets. The definitions are from Wikipedia.

I am considering this question: are Jordan inner and outer measures outer and inner measures respectively?

  • Jordan inner measure satisfies null at empty set and superadditivity, but can it fail "Limits of decreasing towers" and "Infinity must be approached"?

  • Jordan outer measure satisfies null at empty set and monotonicity, but can it fail the countable subadditivity?

If they are not outer and inner measures, what kinds of set functions are they? Thanks in advance!

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Perhaps you know $m^*(\mathbb Q \cap [0,1])$ and $m^*(\{x\})$ ... and perhaps these will answer one of your questions ??? – GEdgar Jan 20 '13 at 20:30
@GEdgar: $ m^∗(\{x\}) = 0$ because we can choose an arbitrarily small interval containing $x$? $m^∗(Q\cap [0,1]) = 0$ because for any $\epsilon >0$, we can always find finitely many intervals containing $Q\cap [0,1]$, and the volume of their union is smaller than $\epsilon$? So $m^∗(Q\cap [0,1]) = 0 = \sum_{x \in Q\cap [0,1]} m^∗(\{x\})$. It isn't a counterexample, and does it answer the question for outer measure? – Ethan Jan 20 '13 at 20:37
Your answer for $m^*(\{x\})$ is correct. Your answer for the other one is not. – GEdgar Jan 20 '13 at 20:46
@GEdgar: I see, $m^∗(Q∩[0,1]) = 1$. If I am correct, how about the question for inner measure? – Ethan Jan 20 '13 at 20:48

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