Proving that Caratheodory's Criterion holds..... but How to Make it Precise? I'm really sorry for the obscenely long post. This question can be thought of as an exercise in being able to articulate carefully why something is true. If that's not what your in the mood for right now... I understand and I'll catch you on the flip side.
Problem
Let $A$ and $B$ be open boxes in $\mathbb{R}^d$. Let $m^*$ denote the Lebesgue outer measure (defined in terms of coverings by open elementary sets). Prove that 

$$m^*(B) = m^*(B \cap A) + (B \setminus A).$$

I know it might seem tedious, but this result is pivotal in the development of the theory of Lebesgue measurable sets (well, at least for the strategy in Tao's Analysis II). So, I'd really like to be able to articulate precisely why it is true
and I'm mildly concerned that I actually don't know how to do so. I would love to hear how other people would attack this problem!

Definitions
(nothing surprising I expect):
An open box in $\mathbb{R}^d$ is defined to be either the empty set,
or a set of the form $(a_1,b_1) \times \cdots \times (a_d,b_d)$ where $a_j < b_j$ for each $j\in\{1,\ldots,d\}$.
The volume of an open box $B$ in $\mathbb{R}^d$ is defined to be zero if $B = \varnothing$,
and $(b_1-a_1)\cdots(b_d-a_d)$ otherwise.
In either case it is denoted by $\text{vol}_d(B)$.
In abbreviation of the statement that $\mathcal{B}$ is a finite or countably infinite collection of open boxes in $\mathbb{R}^d$, let us write $\mathcal{B} \in \mathscr{B}_d$.
The outer measure of a subset $\Omega$ of $\mathbb{R}^d$ is defined to be the extended real number
$$
    m_d^*(\Omega)
:=
    \inf\left\{
        a \in\mathbb{R} \cup \{\pm\infty\}
    \hspace{1mm}:\hspace{1mm}
        \text{there exists a } \mathcal{B} \in \mathscr{B}_d
    \text{ such that both} \hspace{2mm}
        a = \sum_{B \in \mathcal{B}} \text{vol}_d(B)
    \hspace{2mm}\text{and}\hspace{2mm}
        \Omega \subseteq \bigcup_{B \in \mathcal{B}} B
    \right\}.
$$
For brevity, the $d$ subscript may be dropped when this does not cause ambiguity.

Facts that can be used


*

*Basic properties of the outer measure: sub-additive, monotone, $m^*_d(\varnothing) = 0$, translation invariant, etc.

*If $B$ is an non-empty open box in $\mathbb{R}^d$ and $\beta$ is any subset of the boundary of $B$, then $m^*_d(B \cup \beta) = \text{vol}_d(B)$.

*Any open set is equal to a union of countably many open boxes.

*If $\Omega$ is an arbitrary subset of $\mathbb{R}^d$, then there exists a sequence of open sets $(G_n)$ such that $\Omega \subseteq \bigcap_n G_n \subseteq \cdots \subseteq G_2 \subseteq G_1$ and $m^*(\Omega) = m^*(\bigcap_n G_n)$. Further, if $m^*(\Omega) < +\infty$, then $(G_n)$ can be chosen so that $m^*(G_n) < +\infty$ for every $n$ and $\lim_{n \to \infty} m^*(G_n) = m^*(\Omega).$

Stuff that is off limits


*

*Basically anything having to do with the theory of Lebesgue measurable sets, as the development of that theory would (in my case) depend on the result that we're currently trying to prove.

*Any open set is equal to the union of countably many closed boxes, the interiors of which are mutually disjoint (this is at least as difficult as the problem at hand).



My attempt so far
Fix $\varepsilon>0$. Fix $d \in \mathbb{Z}^+$.
Let $A$ and $B$ be open boxes.
For what it's worth, the fact that $m^*(B) = m^*(\text{cl}(B))$ gives a little more room, at least in principle.
By definition of the outer measure, we have a finite or infinite sequence $\{B_n\}$ of open boxes which cover $\text{cl}(B)$ and satisfy
$
    \sum_{n} \text{vol}(B_n) \leq m^*(\text{cl}(B)) + \varepsilon.
$
In fact, since $\text{cl}(B)$ is compact, the same can be said of a finite sub-collection $\{B_1, \ldots, B_N\}$ of $\{B_n\}$. That is, for some finite number of open boxes $\{B_1, \ldots, B_N\}$
$$
    \sum_{n=1}^N \text{vol}(B_n)
\hspace{1mm} \leq \hspace{1mm}
     m^*(\text{cl}(B)) + \varepsilon.
$$
On the other hand,
$$
    Q_1 := \{B_n : n \leq N, B_n \cap A \neq \varnothing\}
$$
covers $\text{cl}(B) \cap A$ (does it cover $\text{cl}(B) \cap \text{cl}(A)$? what about $\text{cl}(B \cap A)$?) and
$$
    Q_2 := \{B_n : n \leq N, B_n \setminus A \neq \varnothing\}
$$
covers $\text{cl}(B) \setminus A$. So, using monotonicity and then sub-additivity of the outer measure,
$$
    m^*(\text{cl}(B) \cap A) + m^*(\text{cl}(B) \setminus A)
\hspace{1mm} \leq \hspace{1mm}
    \sum_{B_n \in Q_1} m^*(B_n) + \sum_{B_n \in Q_2} m^*(B_n).
$$
But
$$
    \sum_{B_n \in Q_1} m^*(B_n) + \sum_{B_n \in Q_2} m^*(B_n) - \sum_{B_n \in Q_1 \cap Q_2} m^*(B_n)
\hspace{1mm} =\hspace{1mm}
    \sum_{n = 1}^N m^*(B_n).
$$
Therefore,
$$
    m^*(\text{cl}(B) \cap A) + m^*(\text{cl}(B) \setminus A)
\hspace{1mm} \leq \hspace{1mm}
    m^*(B) + \varepsilon + \sum_{B_n \in Q_1 \cap Q_2} m^*(B_n)
$$
where $Q_1 \cap Q_2$ are precisely those boxes among our cover for $\text{cl}(B)$ that intersect both $A$ and $A^c$.
Since this reasoning applies to any collection of boxes $\{B_n\}$ that cover cl$(B)$, it seems like we should be able to refine the given $\{B_n\}$ so that each box in $B_n \in Q_1 \cap Q_2$ is made smaller and smaller, eventually either being shrunk out of the class $Q_1 \cap Q_2$, or being made so small that $\text{vol}(B_n) < \varepsilon / 2^n$.
But, I really can't grasp how to make this precise and it's bothering me... Will you join me in this indulgence of detail?

Miscellaneous other notes
Breaking things into cases, it is tractable to prove by brute force that the claim is true if $d = 1$ (I've already written this proof, actually). Is it then possible to induct on $d$?
(unrelated) We can assume WLOG that $B \cap A \neq \varnothing$, but I've yet to find this helpful.
(unrelated) The following is why I have used the "alternative proof" tag. The approach taken by Tao is to prove two lemmas which lead to an even stronger result in the end: 

(i) Prove that an open half space $H$ (e.g., $ H = \mathbb{R} \times (0,\infty) \times \mathbb{R}$) satisfies $m^*(B) = m^*(B \cap H) + m^*(B \setminus H)$ for any open box $B$.
  Consequently (pixelating a given set into open boxes plus a tolerance of $\varepsilon$), any open half space $H$ is Lebesgue measurable (LM) by Caratheodory's criterion (which we would take as the definition).
(ii) The translate of a LM set remains LM, as does the intersection of any finite number of LM sets (these are both actually quite easy to prove).
Then, one argues that an open box $B$ can be expressed as an intersection of a finite number of translates of open half spaces. Thus, an open box is LM and the claim is true. 

This strategy is pretty, and I cannot disrespect Tao... but getting down to brass tacks, when one goes to prove (i), I don't see how you can treat all $\mathcal{O}(d^2)$-many cases without just asserting that they're mutually similar. (fyi Tao left both lemmas as an exercise; so the lack of insight is mine alone, not his).
Again, I feel that if something is so clear, we should be able to articulate why it's true, and we should be capable of coming up with a more precise explanation than just "do a couple of cases by hand and figure that we're good", which is currently all I've got.
Edit(s): just cosmetic changes
 A: Here is an update:
Now, I'm taking a class on real analysis, and it has been pointed out to me (thanks Alan!): in order to cover $B$ with a fininte number of open boxes with an error of at most $\varepsilon$, we don't need to merely rely on the definition of the outer measure to tell us that such a collection exists. Rather, we can actually write down the explicit form of such a collection $\{B_1, \ldots, B_N\}$, and the additional information provided by this explicit form gives us the extra information that we needed.
Let $A$ and $B$ be open boxes in $\mathbb{R}^d$. Say,
$$
    B = \prod_{j=1}^d (a_j,b_j).
$$
Fix $m$ a positive integer. For each $(\ell_1,\ldots,\ell_d)\in\{1,\ldots,m\}^d$, let $\delta(\ell_1,\ldots,\ell_d)>0$ be arbitrary.
Define the open box
$$
    B((\ell_1,\ldots,\ell_d))
\hspace{1mm}:=\hspace{1mm}
        \prod_{j=1}^d \bigg(
            a_j + \frac{\ell_j-1}{m}(b_j-a_j) - \frac{\delta(\ell_1,\ldots,\ell_d)}{2}, \hspace{1mm}
            a_j + \frac{\ell_j}{m}(b_j-a_j) + \frac{\delta{(\ell_1,\ldots,\ell_d)}}{2}
        \bigg).
$$
Then, using the formula for volume, we have
$$
    \text{vol}\big(B((\ell_1,\ldots,\ell_d))\big)
\hspace{1mm}=\hspace{1mm}
    \bigg( \frac{b_1-a_1}{m} + \delta(\ell_1,\ldots,\ell_d) \bigg) \cdots \bigg( \frac{b_d-a_d}{m} +  \delta(\ell_1,\ldots,\ell_d) \bigg).
$$
Now, for each given $\ell\in\{1,\ldots,m\}^d$, we can make this bigger than vol$(B)/m^d$ by as small an error as we d*** well please. In particular, since the preceding is values of $\delta(\ell)$ were all arbitrary, we can now, for each $\ell\in\{1,\ldots,m\}^d$, choose $\delta(\ell)>0$ sufficiently small that
$$
    \text{vol}\big(B(\ell)\big)
\hspace{1mm}<\hspace{1mm}
    \frac{\text{vol}(B)}{m^d} + \frac{\varepsilon}{2^{\iota(\ell)}}.
$$
where $\iota : \{1,\ldots,m\}^d \to \{1,\ldots,m^d\}$ is any enumeration (a bijection). Finally, as alluded to previously, we now have that the collection of open boxes $\{B(\ell)\}_\ell$ covers $B$, covers $B \cap A$, and covers $B \setminus A$. Hence,
$$
    \{B(\ell) : B(\ell) \cap A \neq\varnothing\}_{\ell} \hspace{1mm}\text{ covers } \hspace{1mm} B \cap A
$$
and, likewise,
$$
    \{B(\ell) : B(\ell) \setminus A \neq\varnothing \}_{\ell} \hspace{1mm}\text{ covers } \hspace{1mm} B \setminus A.
$$
Next, using the definition of the outer measure
\begin{equation}
    m^*(B \cap A) + m^*(B \setminus A)
\hspace{1mm}\leq\hspace{1mm}
    \sum_{\substack{\ell \\  B(\ell) \cap A \neq\varnothing}} \text{vol}(B(\ell))
+
    \sum_{\substack{\ell \\  B(\ell) \setminus A \neq\varnothing}} \text{vol}(B(\ell)),
\end{equation}
Second, we have the following elementary (though, perhaps, non-trivial) fact about partitioning a finite sum based on a logical predicate:
$$
    \sum_{\ell} \text{vol}(B(\ell))
\hspace{1mm}=\hspace{1mm}
    \sum_{\substack{\ell \\  B(\ell) \cap A \neq\varnothing}} \text{vol}(B(\ell))
\hspace{1mm}+
    \sum_{\substack{\ell \\  B(\ell) \setminus A \neq\varnothing}} \text{vol}(B(\ell))
\hspace{1mm}-
    \sum_{\substack{\ell \\  B(\ell) \cap A \neq\varnothing \\  B(\ell) \setminus A \neq\varnothing}} \text{vol}(B(\ell)).
$$
So, to summarize the last two facts,
$$
    m^*(B \cap A) + m^*(B \setminus A)
\hspace{2mm}\leq\hspace{2mm}
    \sum_{\ell} \text{vol}(B(\ell))
\hspace{1mm}+
    \sum_{\substack{\ell \\  B(\ell) \cap A \neq\varnothing \\  B(\ell) \setminus A \neq\varnothing}} \text{vol}(B(\ell)).
$$
Third, using our previously established upper bound on vol$(B(\ell))$, we have
$$
    \sum_{\ell} \text{vol}(B(\ell))
\hspace{1mm}+
    \sum_{\substack{\ell \\  B(\ell) \cap A \neq\varnothing \\  B(\ell) \setminus A \neq\varnothing}} \text{vol}(B(\ell))
\hspace{2mm}<\hspace{2mm}
    \sum_{\ell} \bigg(\frac{\text{vol}(B)}{m^d} + \frac{\varepsilon}{2^{\iota(\ell)}}\bigg)
\hspace{1mm}+
    \sum_{\substack{\ell \\  B(\ell) \cap A \neq\varnothing \\  B(\ell) \setminus A \neq\varnothing}} \bigg(\frac{\text{vol}(B)}{m^d} + \frac{\varepsilon}{2^{\iota(\ell)}}\bigg).
$$
In other words,
$$
    m^*(B \cap A) + m^*(B \setminus A)
\hspace{2mm}\leq\hspace{2mm}
    2\varepsilon + \text{vol}(B) + \frac{\text{vol}(B)}{m^d}
    \sum_{\substack{\ell \\  B(\ell) \cap A \neq\varnothing \\  B(\ell) \setminus A \neq\varnothing}} 1.
$$
As noted in the original post, we know by this point in our devlopment of the theory that $\text{vol}(B)=m^*(B)$.
Finally, since $m$ is an arbitrary positive integer, the argument is reduced to estimating the number of boxes $B(\ell)$ which can intersect both $A$ and $A^c$. This is still hard to do!!!
Edit: Notation and stuff.
