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I'm having trouble understanding one step of Stein and Shakarchi's proof that the exterior measure of a rectangle is equal to its volume. The proof I reference is part of Example 4 in section 1.2 of the Real Analysis book.

To summarize the proof, first they show $|R| \leq m_*(E)$. To show the reverse inequality, the book divides $\mathbb{R}^d$ into cubes of side length $1/k$. Then, it considers the sets $\mathcal{Q}$ and $\mathcal{Q}'$.

The claim is that if $\mathcal{Q}$ consists of the finite collection of all cubes entirely contained in rectangle $R$ and $\mathcal{Q}'$ the finite collection of all cubes that intersect the complement of $R$, then there are $O(k^{d-1})$ cubes in $\mathcal{Q}'$, where $O$ denotes standard Big-Oh notation.

From here, using the volumes of the cubes, the text uses:

$$ \sum_{Q \in (\mathcal{Q} \cup \mathcal{Q}')} |Q| \leq |R| + O(1/k) $$

to complete the proof.

Question: Why are there $O(k^{d-1})$ cubes in $\mathcal{Q}'$? I interpret $\mathcal{Q}'$ to be the set of all cubes in the complement, i.e. $\mathcal{Q} \cup \mathcal{Q}' = \mathbb{R}^d$. If this is the case, then I believe $\mathcal{Q}'$ is not a finite collection of cubes and a Big-Oh approximation of its size doesn't make sense.

3 Answers3

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I'll try to answer why the number of cubes in $\mathcal Q'$ is $O(k^{d-1})$ as $k\to\infty$. The set $\mathcal Q'$, by the way, is the set of cubes that intersect the original rectangle and its complement. In $3$-D, this would be the cubes that cover the faces of the rectangular box.

First let's consider the case where $R$ is actually a cube, so all of its edges are the same common length, which we will call $a$.

If we consider just one of these edges, then the number of cubes we need to cover the edge is almost $a/k^{-1} = ak$. This is what it would be if the cubes fit perfectly into the edge, but there may be a little space leftover, so the number of cubes we need is surely no more than $ak + 2$, adding one for each endpoint of the edge.

Thus the number of cubes in a $(d-1)$-dimensional face is no more than the product of the lengths of all the $d-1$ edges: $$ \#\{\text{cubes in a face}\} \le (ak + 2)^{d-1} \le (a+1)^{d-1}k^{d-1}, \qquad {(2\le k).} $$ Now, there are $2d$ faces, so the number of cubes in $\mathcal Q'$ is no more than $2d(a+1)^{d-1}k^{d-1}$. Thus, if $2\le k$ and if $k$ is so large that $1/k < a$, then we have the desired result.

To apply this argument to a general rectangle, let $a$ be the length of the longest edge of the rectangle. Then it is clear that the rectangle $R$ is contained in a cube of edge length $a$. The number of cubes in $\mathcal Q'$ for the rectangle can be no more than that of the larger cube, which we know to be $O(k^{d-1})$. Hence there are $O(k^{d-1})$ cubes in $\mathcal Q'$ for the rectangle, as desired.

Alex Ortiz
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You are interpreting $Q`$ wrong. What I think they mean is that forming the grid of cubes of length $1/k$, there will be finitely such cubes that are entirely contained in $R$ and then some other finite collection that have part of their volume intersecting $R$ and other part of their volume intersecting $R^c$. As such $Q\cup Q'\neq \mathbb{R}^d$

mathemagician
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One side is quite trivial, following from Example two. Another side (to prove $m_*(R) \leq |R|$): First observe that if each side of rectangle is rational then it is trivial. However, suppose not all sides are rational. Then $\forall \epsilon>0$ we know exists $r \in \mathbf{Q}$ such that $r- |R| < \epsilon$. Then try to construct a rectangle with each side rational and its volume equal to $r$. We know such rectangle exists (Since $\mathbf{Q}$ is dense in $\mathbf{R}$).

scitamehtam
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