Why the Lebesgue outer measure of the boundary of rectangle in $\mathbb{R^n}$ is zero?

Question

Let $A$ be a closed rectangle in $\mathbb{R^n}$ and let $m^*$ be Lebesgue outer measure. And let $\partial A$ be the boundary of $A$.

Then, prove that $m^* (\partial A)=0.$

Since $A$ is a closed rectangle, I can write $A$ as $A=[a_1, b_1] \times [a_2 , b_2] \times \cdots \times [a_n, b_n]$.

But I don't know what $\partial A$ is written as and why the Lebesgue measure of $\partial A$ is zero.

I'd like you to give me some ideas.

Answer 1

If you want a formal argument while making things easier, you can note that measure is rotation invariant so you can rotate the hypercube so the faces are each of the form $[a_1,b_1]\times ... \times[a_{n-1},b_{n-1}]\times \{c_n\} $ (with which position the constant point is in permuting. Then just take small enough intervals around the $c_n$ point to get the sum of the faces smaller than any $\epsilon$ you want

Answer 2

Hint: If you can use the fact that an $n-1$ dimensional hyperplane in $\mathbb{R}^n$ has outer measure zero, and a subset of an outer measure zero set has measure zero, then just need to note that the boundary is made out of $2n$ pieces each obtained by fixing one of the $2n$ endpoints of the product of closed intervals you mention, and letting the other $n-1$ components vary in their intervals. This will then be a subset of an $n-1$ dimensional hyperplane.

To make this rigorous one would need to introduce a bit of notation for the steps.