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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 $m$

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  • outer measure ? does it mean that $m^({x_0}) = 0$ for every isolated point $x_0$ ? with this definition wiki/Outer_measure#Method_I or with the translation invariance, it is obvious. hence with the countably additive property, it is enough that $A$ having a countably finite number of boundary points to get $m^(\overline{A}) = m^*(A)$. – reuns Mar 26 '16 at 22:52

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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). :)

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