# Lebesgue measure of $\partial A$ given $m(A^o) = m(\overline{A})$

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?

• 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 $B$.
– Stef
Jun 7 at 10:13

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.
• 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?
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 $$0 Now, call $$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$$.