# Why does the boundary of a cube in $\mathbb{R}^n$ have measure zero?

I'm trying to develop the fact that the measure of a box is equal to the volume of a box for any measure $m$, i.e., a function satisfying the standard properties of monotonicity, positivity, countable sub-additivity, translation invariance, etc.

I'm trying to show that the sets $(1,\frac{1}{q})^n$ and $[1,\frac{1}{q}]^n$ in $\mathbb{R}^n$ both have measure $\frac{1}{q^n}$, for $q>1$ an integer. First, by translating each coordinate by $k/q$ where $0\leq k\leq q-1$, one sees that $q^n$ disjoint translates of $(0,1/q)^n$ are contained in $[0,1]^n$. But normalization, monotonicity, and translation invariance, it follows that $$q^n m((0,1/q)^n)\leq 1$$ so $m((0,1/q)^n)\leq q^{-n}$. Similarly, the $q^n$ translates of $[0,1/q]^n$ cover $[0,1]^n$, so $q^nm([0,1/q]^n)\geq 1$, which implies $m([0,1/q]^n)\geq q^{-n}$.

To finish, I'd like to show that $m([0,1/q]^n\setminus(0,1/q)^n)=0$. So given $\epsilon>0$, I'm trying to cover the boundary of the cube by a family of boxes whose measure is less than $\epsilon$. But without knowing at this point what the measure of a box is, since the measure is not necessarily the Lesbegue measure, I'm stuck. How can I show the measure of the boundary of the cube is $0$? Thanks.

The axioms for $m$ are as follows: $\mathbb{R}^n$ is a Euclidean space.

• Every open and closed set is measurable.
• The complement of every measurable set is measurable.
• Finite/countable unions and intersections of measurable sets are measurable.
• $m(\emptyset)=0$.
• $0\leq m(A)\leq\infty$ for every measurable set.
• If $A\subseteq B$, then $m(A)\leq m(B)$.
• If $(A_j)_{j\in J}$ is a family of measurable sets, then $m(\bigcup A_j)\leq\sum m(A_j)$.
• If $(A_j)_{j\in J}$ is a family of disjoint measurable sets, then $m(\bigcup A_j)=\sum m(A_j)$.
• $m([0,1]^n)=1$.
• If $A$ is measurable, $m(x+A)=m(A)$ for all $x\in\mathbb{R}^n$.
• Could you please list the axioms you're requiring for $m$? We need to rule out the $n-1$-dimensional Hausdorff measure. I guess it's ruled out by normalization, which you used to get $m([0,1]^n)=1$? – Chris Culter Aug 4 '13 at 3:27
• @ChrisCulter Sure, I've added the axioms now. – Nastassja Aug 4 '13 at 4:55

Oh, okay! From the above axioms, you can show that $$m\left((-\epsilon,\epsilon)\times[0,1]^{n-1}\right)\leq2\epsilon,$$ i.e. $(n-1)$-dimensional faces can be covered by thickenings with arbitrarily small measure. This follows from stacking a bunch of translates of such a slice inside of $[0,1]^n$, which has known measure. It's the same argument you're already using for cubes, except it focuses on one dimension at a time, instead of shrinking all $n$ dimensions at once.

• I'm beginning to see it now, thanks. – Nastassja Aug 4 '13 at 7:19

I assume that the sigma-algebra you are working with is the usual (Borel) one. Then translation-invariance property of your measure implies that it is the Haar measure on the Lie group $R^n$, which is therefore unique up to scaling, see http://en.wikipedia.org/wiki/Haar_measure. Thus, your measure is a scalar multiple of the Lebesgue measure.