I was working through an exercise in Stein's 3rd Princeton Lectures in Analysis book and was looking for some clarification regarding my approach and its validity.
The problem is precisely showing that for some arbitrary subset E, the normal exterior measure $m_\ast(E)$ is equivalent to $m_\ast^R(E)$, where the latter is defined as $inf \sum_{j = 1}^{\infty} |R_{j}|$, where the infinum is taken over all countable closed rectangle covers.
It is of course readily seen that $m_\ast^R(E) \leq m_\ast(E)$, given that all cube coverings are rectangle coverings. As for the reverse inequality, Stein seems fairly suggestive to utilize a lemma in the book that if a rectangle is the finite union of almost disjoint rectangles, its volume is the sum of the constituent rectangles. I anticipate the idea here is to render each closed rectangle as a countable union of closed cubes and thus get a cube covering of the same total volume and so forth, but I had trouble retooling this lemma from the finite to the countable case and furthermore finding a way to represent a closed rectangle as a countable union of closed cubes.
With this is mind, I considered a sort of estimation argument which certainly seems to be not the prettiest, but I am of course most concerned with its validity and only secondarily any notes on how one may attain the result in the way the author clearly intended.
First, we assume for the sake of contradiction that $m_\ast(E) > m_\ast^R(E)$ (and that the rectangle exterior measure is finite, as contradiction is otherwise straightforward). Then we can find some rectangle covering with volume less than $m_\ast^R(E) + \epsilon$ for arbitrary positive epsilon, say here that epsilon is perhaps $\frac{m_\ast(E) - m_\ast^R(E)}{2}$.
We remark then the corresponding countable covering by closed rectangles, say $\{ \bar R_{j} \}$. For each closed rectangle $\bar R$, we have naturally some open rectangle $R$ both containing $\bar R$ and with $|R| < |\bar R| + \epsilon$ for arbitrary positive epsilon (this is intuitively clear and I won't right down by proof for this result, but it can be established by induction on dimension fairly quickly). Thus, for the family $\{ \bar R_{j} \}$, there is some corresponding collection of open rectangles $\{ R_{j}\}$, such for each $i$ we have $|R_{i}| < |\bar R_{i}| + \frac{\epsilon}{2^{n}}$, and this collection is still a cover (this is my main concern, whether or not this reasoning is justified). It follows: $$ \Big ( \sum_{i = 0}^{\infty} |\bar R_{i}| \Big ) + \epsilon = \sum_{i = 0}^{\infty} |\bar R_{i}| + \frac{\epsilon}{2^{i}} \geq \sum_{i = 0}^{\infty} |R_{i}|. $$ But then, as we can represent any open set as a countable union of closed cubes, we have the following where the set $\{ Q_{j}^{i} \}$ represents the cubes that compose the corresponding open rectangle: $$ \sum_{i = 0}^{\infty} |R_{i}| = \sum_{i = 0}^{\infty} \sum_{j = 0}^{\infty} |Q_{j}^{i}| = \sum_{k = 0}^{\infty} |Q_{k}| $$ Of course then, we now can generate cube coverings with volume sums arbitrarily close to the rectangle exterior measure, generating an easy contradiction. My main qualm lies however in constructing that family of open rectangles and whether or not that is valid. Thanks for the assistance.