When is it true that $m^*(\overline A) = m^*(A)$, $A \subset \mathbb{R}$? Let $m^*$ be an outer measure on a set $\mathbb{R}$
Let $A \subset \mathbb{R}$, when is it true that $m^*(\overline A) = m^*(A)$?
My thought, 
$\overline A = \text{int}(A) \sqcup \partial A$
$m^*(\overline A) \leq m^*(\text{int} A) + m^*(\partial A)$
If $\partial A$ is a zero (null) set, then we are good. (Note: actually we are not, how can we guarantee that $\leq$ turns into =? ) I can't think of any situation where $\partial A$ would not be a zero set...
If we had $A = (a,b)$, the closure of which is $[a,b]$, then $\partial A$ are just two points, which is a zero (null) set.
Can someone give a more general condition? Does it still hold when we replace $m^*$ with $m$
 A: Your equality trivially holds when $m^*(A)=\infty$.
If $m^*(A)<\infty$, consider that $$\operatorname{int}(A)\subseteq A\subseteq \overline A=\operatorname{int}(A)\cup \partial A$$ 
So it always holds $$m^*(\operatorname{int}(A))\leq m^*(A)\leq m^*(\overline A)\leq m^*(\operatorname{int}(A))+ m^*(\partial A)$$
By your reasoning, a sufficient condition for your equality to hold is "if $m^*(\partial A)=0$". If $A$ is also $m^*$-measurable, this condition is necessary.
Of course, $m^*(\partial A)=0$ may not hold. For instance, for $A=\Bbb Q$ and $m^*$ the external Lebesgue measure. Or, if you prefer everything to be of finite measure, $A=\Bbb Q\cap[0,1]$.
This tells you what happens for $m$ as well.
Related interesting stuff: This question studies sufficient conditions guaranteeing that $\partial A\subseteq \Bbb R^d$ is Lebesgue-negligible (and some nastier counterexamples). Reading more carefully the question, though, it is a bit of a waste for the case $d=1$, because in $\Bbb R$ convexity and connectedness are equivalent (so all the intermediate hypothesis cited there are trivial). :)
