Lebesgue outer Measure of a face of rectangle in $\Bbb R^{n}$

Question

Show that the outer measure of a face $I_1 \times \dots \times I_{i-1} \times \{a\} \times I_{i+1} \times \dots \times I_n$ of a rectangle $I_1 \times \dots \times I_n \subset \Bbb R^{n}$ is zero. Here $I_1, \dots, I_n$ are open intervals in $\Bbb R$ and $a$ is one of the endpoints of $I_i$.

My idea: I know intuitively that the Lebesgue outer measure of a face of rectangle in $\Bbb R^{3}$ is zero. This is because in $\Bbb R^{3}$ a rectangle has only 2 dimensions, which means its volume will be zero. This intuition can be extended to n-dimensions as well and hence it is obvious that that the outer measure of a face $I_1 \times \dots \times I_{i-1} \times \{a\} \times I_{i+1} \times \dots \times I_n$ of a rectangle $I_1 \times \dots \times I_n \subset \Bbb R^{n}$ is zero.

But I want a formal construction of the proof. Please suggest.

Answer 1

Let $A := I_1 \times \cdots \times I_{i-1}\times \{a\} \times I_{i+1}\times \cdots \times I_n$, and let $C$ be the product of the measures of $I_j$ for $j\neq i$. Given $\epsilon > 0$, the sequence

$$R_k := I_1 \times \cdots \times I_{i-1} \times \left(a - \frac{\epsilon}{C2^{k+1}}, a + \frac{\epsilon}{C2^{k+1}}\right)\times I_{i+1}\times \cdots \times I_n\quad (k = 1,2,\ldots)$$

is a collection of open rectangles which covers $A$ and

$$\sum_{k = 1}^\infty \mu^*(R_k) = C\sum_{k = 1}^\infty \frac{\epsilon}{C2^k} = C\cdot\frac{\epsilon}{C} = \epsilon.$$ By countable subadditivity of the outer measure,

$$\mu^*(A) \le \sum_{k = 1}^\infty \mu^*(R_k) < \epsilon.$$

Since $\epsilon$ was arbitrary, $\mu^*(A) = 0$.

Answer 2

The way to think about this is to check that the face is contained in rectangles of arbitrarily small measure. This is intuitively true since you can shrink the $i$-th coordinate to be as thin as you want.

To make your proof formal, you should check that for any $\epsilon >0$, you have some set of the form $(a_1, b_1) \times \dots \times (a_n, b_n)$ containing your face such that $(b_1-a_1)(b_2-a_2)\dots(b_n-a_n) < \epsilon$.