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Let $A$ be a set and define

$m^{**}(A) \in [0,\infty] $ by $m^{**} (A)= \inf\{m^{*}(\mathcal{O}) \ | \ A \subset \mathcal{O}, \mathcal{O} \ \text{open} \} $.

And

$m^{***}(A) = \sup \{m^*(F) \mid F \subset A, F$ is closed $\}$.

Show how are $m^{**}(E)$ and $m^{***}(E)$ related to $m^{*}(E)$?

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    Is your $m^*$ the Lebesgue measure? – 2017-02-01

2 Answers 2

-1

Solution: Let $E \in B( \Bbb R)$ let $C$ be an arbitary closed set and $O$ be an arbitrary open set in $\Bbb R$ such that $C \subset E \subset O$. So $m^*(C) \leq m^*(E) \leq m^*(O).$

Since $C$ is arbitary, we have $m^*(E) \leq \inf\{m^{*}(O \ | \ E \subset O, O \ \text{open} \} = m^{**} (E)$.

And since $O$ is arbitrary we have, $m^{***}(E) = \sup \{m^*(C) \mid C \subset E, C \ \text{is closed } \} \leq m^*(E)$

2

By the Regularity theorem for Lebesgue measure, we have $m^{*}=m^{**}=m^{***}$.