The "measure" I am referring to is similar to the inner Jordan measure, and defined as:
$m^{**}(A) = \sup\{\sum_{n=1}^{N}\ell(I_n) : (I_n)$ is a finite sequence of pairwise disjoint open intervals with $\cup_{n=1}^{N}I_n \subseteq A \}$
Prove that $m^{**}(A) = m(A^{\circ})$, where $m$ is the Lebesgue measure and $A^{\circ}$ is the interior of A.
Intuitively, it makes sense why this is true. However, I cannot figure out how to show it rigorously. Here is what I have so far:
If A is open:
$m^{**}(A)$
$= \sup\{\sum_{n=1}^{N}\ell(I_n) : (I_n)$ finite seq. of pairwise disjoint open intervals, $\cup_{n=1}^{N}I_n \subseteq A \}$
$= \inf\{\sum_{n=1}^{N}\ell(I_n) : (I_n)$ finite seq. of pairwise disjoint open intervals, $A \subseteq \cup_{n=1}^{N}I_n \} $
$\geq m(A)$
$ = m(A^{\circ})$
But obviously this isn't the full proof for that particular case, let alone considering all cases.
Any insight is appreciated.