I am trying to solve an exercise from Royden's Real Analysis, 4th ed.
For any set $A \subseteq \mathbb{R}$, define $m^{***}(A)=sup\{m^*(F):F\subseteq A$ is a closed set$\}$. How is this set function $m^{***}$ related to outer measure $m^*$?
I am aware that this exercise Relation between $m^{***}$ with $m^*$? has been discussed before on SE, yet I do not find it satisfactory because I am not assuming per se that $A$ is measurable.
Relevant definitions and theorems
Keep in mind that Royden defines outer measure $m^*(A)=inf\{\sum _{i=1}^ \infty l(I_i):$ $I_1,I_2,...$ is a sequence of nonempty, bounded, open intervals whose union covers $A$}. This is defined regardless of whether $A$ is measurable or not.
It has also been proven that we have five equivalent definitions of measurability:
The following are equivalent:
- $\forall A \subseteq \mathbb{R}[m^*(A)=m^*(A\cap E)+m^*(A \cap E^c)]$
- $\forall \epsilon >0\exists$ open set $O \supseteq E$[$m^*(O-E)< \epsilon]$
- $\exists G_δ$ set $G \supseteq E[m^*(G-E)=0]$
- $\forall \epsilon >0\exists$ closed set $F \subseteq E$[$m^*(E-F)< \epsilon]$
- $\exists F_σ$ set $F \subseteq E[m^*(E-F)=0]$
Now we tackle the actual problem...
My progress so far
We always have $m^{***}(A) \leq m^*(A)$ obviously. I have formulated the following conjecture:
$m^{***}(A)=m^*(A)$ if and only if $A$ is measurable.
I have shown that $A$ measurable implies $m^{***}(A)=m^*(A)$.
For the other direction, I have managed to show that $m^{***}(A)=m^*(A)$ implies $A$ is measurable in the special case that $m^*(A)< \infty$. But if $m^*(A)= \infty$, I don't know how to show it.
I was able to prove the finite case by finding some $F_δ$ set $F\subseteq A$ having the same measure as $A$, observing that $m^*(A)=m^*(A-F)+m^*(F)=m^*(A-F)+m^*(A)$, and subtracting $m^*(A)$ from each side to obtain $m^*(A-F)=0$ (meaning that $A$ is meaurable).
If $m^*(A)= \infty$, I don't know what to since I obviously can't subtract $\infty$ from both sides.