Suppose that $A \subseteq \mathbb{R}$, and consider the Lebesgue outer measure $m: \mathcal{P}(\mathbb{R}) \to [0,\infty]$ on $\mathbb{R}$. Denote by $A^o$, $\overline{A}$ and $\partial A$ the interior, closure and boundary of $A$, respectively.

Suppose that $m(A^o) = m(\overline{A})$. Is it then true that $m(\partial A) = 0$?

This is certainly true if $m(A) < \infty$. Since open sets and closed sets are Lebesgue measurable, we have that $A^o$ and $\overline{A}$ are Lebesgue measurable, and hence so is $\partial A = \overline{A} \setminus A^o$. Therefore, we can write $m(\overline{A}) = m(A^o) + m(\partial A)$. By the monotonicity of the Lebesgue outer measure, we also have that $m(A^o) < \infty$. Hence, we can cancel to conclude. But what about the case $m(A) = \infty$?

If we take for example $A = \bigcup_{n \geq 1} (n - 1, n)$, then $m(A^0) = m(\overline{A}) = \infty$, but $m(\partial A) = m(\mathbb{N}) = 0$. Are there any counter examples?

  • $\begingroup$ Choose $A$ a measurable subset of $[0, +\infty[$ such that $m(A^o) = m(\overline A) = +\infty$, and let B be any measurable subset of $]-\infty, 0[$. If your theorem were true, then by application to $A \cup B$ it would follow that $m(\partial B) = 0$ even though we had no assumptions on $B$. $\endgroup$
    – Stef
    Jun 7 at 10:13

2 Answers 2


Let $A$ be the set you have constructed and let $C$ be a fat Cantor set, i.e. a Cantor like set of positive measure. The $C \cup A$ would satisfy your requirements.

  • $\begingroup$ Nice example. Just to clarify: since $C$ has empty interior, the boundary of $A \cup C$ is $ \partial A \cup C$. Therefore $m(\partial (A \cup C)) = m(C) > 0$. Is this correct? $\endgroup$
    – user848964
    Nov 30, 2020 at 10:20
  • $\begingroup$ @Pyronaut Yes, that is correct. $\endgroup$ Nov 30, 2020 at 11:28

Consider $q_\bullet:\Bbb N\setminus\{0\}\to\Bbb Q\cap (0,1)$ a bijective map, and call $$B=\bigcup_{n=1}^\infty \left(q_n-\min\left\{\frac123^{-k}, \lvert q_n\rvert\right\};q_n+\min\left\{\frac123^{-k}, \lvert 1-q_n\rvert\right\}\right)$$

Then, $B=B^\circ$, $\overline B=[0,1]$ and $\partial B=[0,1]\setminus B$. It is clear that $$0<m(B^\circ)\le\sum_{k=1}^\infty 3^{-k}=\frac12\\ m(\partial B)\ge\frac12$$ Now, call $A=\bigcup\limits_{n\in\Bbb Z}n+B$ and you'll see that $A^\circ =A$, $\overline A=\Bbb R$ and $\partial A=\Bbb R\setminus A=\bigcup\limits_{n\in\Bbb Z}n+\partial B$. Notice that $A$ is a disjoint union, and that $\mu((n+\partial B)\cap(m+\partial B))=0$ for all $m\ne n$. Therefore $m(A)=m(A^\circ)=\infty\times m(B)=\infty$, $m(\overline A)=m(\Bbb R)=\infty$ and $m(\partial A)=\infty\times m(\partial B)=\infty$.


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