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This problem is taken from Section 2.5 of Royden and Fitzpatrick's Real Analaysis, Fourth Edition text.

For any open interval $I = (a,b)$, let $\ell(I) = b-a$ denote the length of $I$. For any set of real numbers $A$, define the following set functions: \begin{align} m^*(A) &= \inf\left\{\sum_{i=1}^\infty \ell(I_n): \bigcup_{n=1}^\infty I_n \supseteq A, I_n \text{ is an open, bounded interval for all $n$}\right\} \\ m^{**}(A)&= \inf\{m^*(\mathcal{O}): \mathcal{O} \supseteq A, \mathcal{O} \text{ open}\} \\ m^{***}(A)&= \sup\{m^*(F): F \subseteq A, F \text{ closed}\} \end{align}

The text asks the reader to determine how these set functions are related. I think I can show that $m^* = m^{**}$ for all subsets of $\mathbf{R}$. However, I am only able to show $m^*(A) = m^{***}(A)$ if $m^{***}(A) = \infty$ or $A$ is measurable. Is this the correct answer? In particular, is it possible to find a set such that $m^{***}(A) < m^*(A)$?

David
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1 Answers1

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Your results are correct . $m^*,$ also written $m^o,$ is called Lebesgue outer measure. $m^{***},$ also written $m^i,$ is called Lebesgue inner measure. Some authors take the def'n of a measurable set as any $A$ such that $m^o(A)=m^i(A).$ ...I dk which def'n of measurable set you are using.

Some standard examples of a non-measurable set include the Vitali set, and a Bernstein set. A Berstein set is any $B\subset \mathbb R$ such that $B\cap C \ne \emptyset \ne(\mathbb R$ \ $B)\cap C$ for every closed uncountable $C\subset \mathbb R.$

So any closed subset of a Berstein set $B,$ or closed subset of $\mathbb R$ \ $B$, is countable, and $m^i(S)=m^o(S)=0$ whenever $S$ is countable. So $m^i(B)=m^i(\mathbb R$ \ $B)=0.$

On the other hand, let $U\supset B$ where $U$ is open. Then $V=\mathbb R$ \ $U$ is a closed subset of $\mathbb R$ \ $B,$ so $V$ is countable. Now for any $S \subset \mathbb R$ and any countable $V\subset \mathbb R$ we have $m^o(S)=m^o(S\cup V).$ So $m^o(U)=m^o(U\cup V)=m^o(\mathbb R)=\infty.$ Hence $m^o(B)=\infty.$

DanielWainfleet
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