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In the proof of Carathéodory's extension theorem, an outer measure $\mu^*$ is constructed from a premeasure $\mu_0$ on a ring $A$ over an arbitrary set as

$$\displaystyle \mu^*(S)\equiv\inf\left\{\sum_{i=1}^\infty \mu_0(S_i): S_i\in A,\ S\subseteq\bigcup_{i=1}^\infty S_i\right\},$$

and then a complete measure space is constructed from the outer measure.

  1. I was wondering if an inner measure can be constructed from the same premeasure on the same ring, such that the the complete measure space contruscted from the outer measure is same as the class of subsets whose inner and outer measures are equal (this is inspired from how Jordan measurable subsets are defined from Jordan outer and inner measures)?

    One guess is that can we define such an inner measure in a similar way to the outer measure as

    $$\displaystyle \mu_*(S)\equiv\sup\left\{\mu_0(\cup_{i=1}^\infty S_i): S_i\in A,\ \bigcup_{i=1}^\infty S_i \subseteq S\right\},$$ or $$\displaystyle \mu_*(S)\equiv\sup\left\{\sum_{i=1}^\infty \mu_0(S_i): S_i\in A,\ \bigcup_{i=1}^\infty S_i \subseteq S, S_i\text{'s are disjoint} \right\}?$$

    For example, in the case of constructing Lebesgue measure on $\mathbb{R}^n$, Tao said that the Lebesgue inner measure cannot be constructed in the same way as Lebesgue outer measure from the premeasure on the ring:

    In analogy with the Jordan theory, we would also like to define a concept of “Lebesgue inner measure” to complement that of outer measure. Here, there is an asymmetry (which ultimately arises from the fact that elementary measure is subadditive rather than superadditive): one does not gain any increase in power in the Jordan inner measure by replacing finite unions of boxes with countable ones. But one can get a sort of Lebesgue inner measure by taking complements; see Exercise 18. This leads to one possible definition for Lebesgue measurability, namely the Carathéodory criterion for Lebesgue measurability, see Exercise 17.

    Exercise 18 (Lebesgue inner measure) Let ${E \subset {\mathbb{R}}^d}$ be a bounded set. Define the Lebesgue inner measure ${m_*(E)}$ of ${E}$ by the formula $$ \displaystyle m_*(E) := m(A) - m^*(A \backslash E) $$ for any elementary set ${A}$ containing ${E}$, where $m$ is the premeasure on the ring of elementary sets.

    I was wondering why "one does not gain any increase in power in the Jordan inner measure by replacing finite unions of boxes with countable ones"?

    Can the way in Exercise 18 for constructing the Lebesgue inner measure be generalized from $\mathbb{R}^n$ to the general case?

  2. Also, can the complete measure space contruscted from the outer measure alone be equivalently constructed from the inner measure alone?

Thanks and regards! References are also appreciated!

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If $S$ contains a set $S_0 \in A$ with $\mu_0(S_0) \gt 0$ then your proposed definition of the inner measure yields $\mu_\ast (S) = \infty$ (take $S_i = S_0$ for all $i$). –  Martin Jan 20 '13 at 8:06
    
@Martin: Thanks! How about definite it as $\displaystyle \mu_*(S)\equiv\sup\left\{\mu_0(\cup_{i=1}^\infty S_i): S_i\in A,\ \bigcup_{i=1}^\infty S_i \subseteq S\right\}?$ Or another definition which requires $S_i$'s to be disjoint: $\displaystyle \mu_*(S)\equiv\sup\left\{\sum_{i=1}^\infty \mu_0(S_i): S_i\in A,\ \bigcup_{i=1}^\infty S_i \subseteq S, S_i\text{'s are disjoint} \right\}?$ –  Tim Jan 20 '13 at 14:46
    
Consider the basic example of the algebra generated by the half-open intervals on $\mathbb{R}$ and the premeasure given by length. The set of irrational numbers contains no element of that algebra, so your new definitions will yield $0$ (or $-\infty$ depending on your conventions on empty suprema) on the irrational numbers which have infinite inner measure by the usual definition. –  Martin Jan 20 '13 at 15:27
    
A further problem is that $\mu_0(\bigcup_{i=1}^\infty S_i)$ makes no sense unless that union belongs to $A$ (this is also an issue with your latest question). –  Martin Jan 20 '13 at 15:27
    
@Martin: I see. (1) is the way of approaching the inner measure of a subset from premeasures of its subsets within not working? When the premeasure and the ring happen to be a measure and a sigma algebra, this way is working. I wonder why? (2) In Tao's exercise 18, Lebesgue inner measure on Rn is defined from the premeasure and its induced Lebesgue outer measure, but only for bounded subsets. Can this definition of inner premeasure be generalized to any premeasure on any (semi)ring? –  Tim Jan 20 '13 at 15:56
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