Should inscirbed intervals of $\mathbb{R}^n$ be subsets of the interior of the set whose volume they are used to define? This question pertains to the proof of Theorem IV-2.1 of C.H. Edwards, Jr.'s Advanced Calculus of several variables.  Page 216.  The following is my adaptation of the discussion:

Defintion: A closed interval $\mathscr{I}$ of $\mathbb{R}^{n}$
is the Cartesian product of $n$ closed intervals $[a_{i},b_{i}]\subset\mathbb{R}$.

Defintion: The volume $\mathscr{v}[\mathscr{I}]$ of the closed
interval $\mathscr{I}$ of $\mathbb{R}^{n}$ is $\mathscr{v}[\mathscr{I}]=\prod_{i=1}^{n}(b_{i}-a_{i})$.
Definition: A collection of intervals whose interiors are mutually disjoint is nonoverlapping..
Definition: The closed subset $\mathscr{A}\subset\mathbb{R}^{n}$
is a contended set with volume $\mathscr{v}[\mathscr{A}]$
if and only if, given $\varepsilon>0$, there exist
(a) nonoverlapping closed intervals $\mathscr{I}_{1},\dots,\mathscr{I}_{p}\subset\mathscr{A}$
such that $\sum_{i=1}^{p}\mathscr{v}[\mathscr{I}_{i}]>\mathscr{v}[\mathscr{A}]-\varepsilon$,
and
(b) closed intervals $\mathscr{J}_{1},\dots,\mathscr{J}_{q}$ such
that $\mathscr{A}\subset\cup_{j=1}^{q}\mathscr{J}_{j}$ and $\sum_{j=1}^{q}\mathscr{v}[\mathscr{J}_{j}]<\mathscr{v}[\mathscr{A}]+\varepsilon$.
Definition: The boundary $\partial\mathscr{A}$ of $\mathscr{A}\subset\mathbb{R}^{n}$
is the set of all points in $\mathbb{R}^{n}$ which are limit points
of both $\mathscr{A}$ and $\mathbb{R}^{n}-\mathscr{A}$.
Definition: A contended set $\mathscr{A}$ is negligible if
and only if $\mathscr{v}[\mathscr{A}]=0$.
Definition: A partition of the closed interval $\mathscr{R}$
is a finite collection of nonoverlapping closed intervals whose union
is $\mathscr{R}$.
Theorem: The bounded set $\mathscr{A}$ is contended if and only if
its boundary is negligible.
Proof: Let $\mathscr{R}$ be a closed interval containing $\mathscr{A}$. 
Assume $\mathscr{A}$ is contended. Given $\varepsilon>0$, choose
closed intervals $\mathscr{I}_{1},\dots,\mathscr{I}_{p}$ and $\mathscr{J}_{1},\dots,\mathscr{J}_{q}$
as described in (a) and (b) of the definition of $\mathscr{v}[\mathscr{A}]$.
Let $P$ be a partition of $\mathscr{R}$ such that each $\mathscr{I}_{i}$
and $\mathscr{J}_{j}$ is a union of closed intervals of $P$. Let
$\mathscr{R}_{1},\dots,\mathscr{R}_{k}$ be those intervals of $P$
which are contained in $\cup_{i=1}^{p}\mathscr{I}_{i}$, and $\mathscr{R}_{k+1},\dots,\mathscr{R}_{k+l}$
be the remaining intervals of $P$ which are contained in $\cup_{j=1}^{q}\mathscr{J}_{j}$.
Then
$\partial\mathscr{A}\subset\cup_{i=k+1}^{k+l}\mathscr{R}_{i}$, etc.

It seems to me that the last assertion is not justified on the basis of the definitions and arguments given.  In particular, it is possible that some of the boundary points of $\mathscr{A}$ could be elements of the $\mathscr{I}_{1},\dots,\mathscr{I}_{p}$ and, therefore, not contained in any of the $\mathscr{R}_{k+1},\dots,\mathscr{R}_{k+l}$.  
I contend that a further stipulation that $\cup_{i=1}^{p}\mathscr{I}_{i}\subset(\mathscr{A}-\partial\mathscr{A})$ is needed.  That is, the closed intervals described in part (a) of the definition of $\mathscr{v}[\mathscr{A}]$ shall not contain any boundary points of $\mathscr{A}$.
Is my critique valid?
 A: That definition of volume is additive. In other words, if $A,B$ are disjoint closed sets then we have $\mathcal{V}(A \cup B) = \mathcal{V}(A) + \mathcal{V}(B)$. Observe that the volume of an open set $U=\prod\limits_i (a_i,b_i)$ is the same at its closure $\overline{U} = \prod\limits_i [a_i,b_i]$. Now just observe that $A = \textbf{int}(A) \cup \partial(A)$ where this union is a disjoint one and $\partial(A)$ is closed with dimension $n-1$ and so $\mathcal{V}(A) = \mathcal{V}(\textbf{int}(A)) + \mathcal{V}(\partial(A))$. The last thing you have to show is that any $n-1$ dimensional set has zero volume. 
Sketch: $\partial(A)$ is relatively closed i.e $\partial(A) = V \cap A$ where $V$ is closed ball in $\mathbb{R}^n$. Hence you can just arrange a partition $\{V_j \cap A\}$ of the boundary such that the central cross-section of each ball has sufficiently small radius, say $\epsilon/2^j$ then,
$$\mathcal{V}(\partial(A))< \sum_j \frac{\epsilon}{2^j} = \epsilon$$

Making the diameter sufficiently small amounts to making the interval $[a_i,b_i]$ sufficiently small.
A: Edwards is correct.  The $\mathscr{I}_{i}$ and $\mathscr{J}_{j}$ can share boundary points.
