How is it Related to the Outer Measure?

Question

For any set $A \subseteq \Bbb{R}$, define $\mu(A) = \sup \{m^*(F) \mid F \subseteq A \text{ is closed } \} \in [0, \infty]$, where $m^*(F) = \inf \{\sum_{k=1}^\infty \ell(I_k) \mid \{I_k\}_{k=1}^\infty \text{ is an open cover of } A \}$ is the outer measure of $F$. How is this set function related to $m^*$.

It's plain to see that $\mu (A) \le m^*(A)$, since if $F \subseteq A$, we have $m^*(F) \le m^*(A)$ by monotonicity. The difficulty lies in determining whether the reverse inequality holds. I just proved that $m^*(A)$ is equal to $\inf \{m^*(O) \mid O \supseteq A \text{ is open} \}$, so I will be working this perspective on the outer measure.

Note that $m^*(F) \le m^*(A) \le m^*(O)$ for every $F$ and $O$ of the kind we've been speaking of, and so if I can show there exists an $F$ and $O$ such that $m^*(F) = m^*(O)$, then this, from what I can tell, would prove $m^*(A) \le \mu(A)$; thus there existing such a pair is sufficient to prove $m^*(A) = \mu(A)$. I suspect that this isn't always the case, but trying to think up a counterexample has proven difficult. But I have a crazy idea: does $m^*(A) = \mu(A)$ imply there exists such a pair of $F$ and $O$; i.e., is the existence of such a pair necessary for equality? If so, I believe this would reduce the counterexample space, since I believe this would imply $m^*(A) = \min \{m^*(O) \mid O \supseteq A \text{ is open} \}$; thus, to find a counterexample to the claim, we need only find an $A$ for which $\min \{m^*(O) \mid O \supseteq A \text{ is open} \}$ doesn't exist.

Answer 1

You are correct, in that $m^{*}(A)\neq \mu(A)$ for every $A,$ and indeed, coming up with a counterexample is tricky (the one I know is $\mathcal{N},$ a set of representatives in $[0,1]$ of the equivalence classes of $\mathbb{R}$ under the equivalence relation $x\sim y$ if and only if $x-y\in\mathbb{Q}$). But there are sets $A$ for which $m^{*}(A)=\mu(A),$ but still there is no pair of $F$ and $O$: just consider $(0,1].$ For any $F\subseteq(0,1],$ $\min\{x:x\in F\}>0,$ so $m^{*}(F)<1,$ and for any $O\supset (0,1],$ $\sup\{x:x\in O\}>1,$ so $m^{*}(O)>1.$

Answer 2

No, this is not necessary.

If you had $m^*(F)=m^*(O)$, then necessarily you would have $m^*(F)=m^*(O)=m^*(A)$. This is trivially false if $m^*(A)=0$ and $A$ is nonempty, for instance (e.g. $A=\{0\}$) -- any nonempty open set has nonzero measure.

To find a counterexample, you need some form of axiom of choice (without axiom of choice, counterexamples may not exist). Indeed, you can find $A$ such that $\mu(A)$ (which is usually called the inner measure of $A$ and denoted by $m_*(A)$) is zero, while $m^*(A)$ is full (i.e. any open set containing $A$ has complement of zero measure), for example if you choose $A$ to be a Bernstein set.