Outer measure equals length of cell - proof explanation This is a proof from Bartle's Text - I summed up the definitions in the bottom. What I am confused are parts (1) (meaning of "face" and "extending") and (2) (the inequality) labelled below in the proof. May someone elaborate? Thank you so much. 


Theorem: If $I \subseteq \mathbb{R}^p$ is any cell then $m^{*}(I)= l(I)$. 
Proof: It is clear that $m^{*}(I) \le l(I)$. Now let $(I_k)$ be a covering of $I$ such that 
  $$\sum_{k=1}^{\infty}l(I_k) \le m^*(I) + \varepsilon.$$ 
  Let $J$ be a closed cell such that $J \subseteq I$ and $l(I) - \varepsilon < l(J)$. By the Heine-Borel Theorem there is an $m \in \mathbb{N}$ such that $J \subseteq \bigcup_{k=1}^m I_k$. 
$\color{blue}{(1)} $We now divide the space $\mathbb{R}^p$ into a finite number closed intervals by extending the $(p-1)$ dimensional hyperplanes that contain a face of one of the cells $I_1, \ldots, I_m$ and of $J$. $\color{blue}{*}$ Let $K_1, \ldots, K_n$ be the distinct closed cells into which the cells (closures of the cells) $\bar{I_1}, \ldots, \bar{I_m}$ are divided by these hyperplanes; further let $J_1, \ldots, J_r$ be the closed cells which $J$ is divided. 
$\color{blue}{(2)}$  Therefore, we have, 
  $$ l(J) = \sum_{j=1}^{r} l(J_j) \le \sum_{k=1}^n l(K_k) \le \sum_{k=1}^m l(I_k) \le m^{*}(I) + \varepsilon. \color{blue}{*}$$ 
  Hence, we are done. 


My thoughts:
I attempted to visualize the problem in 3D. Suppose our cells are $I_1 = (0,1] \times (0,1/2] \times [0,1]$ and $I_2 = [1/2,3/2) \times [0,1] \times [0,1]$. And we suppose $I = (1/2,1) \times (0,1) \times (0,1)$. Let $J = [1/2 + 1/16, 1 - 1/16] \times [1/16 , 1 - 1/16] \times [1/16, 1 - 1/16]$ (used $1/16$ instead of $\epsilon$ for illustration). 
(1): 
$ \color{blue}{ \text {What exactly is happening by "extending the hyperplanes...", i.e. what are $K_1, \ldots, K_n$?} } $
My understanding is as follows. For $I_1$ in my example we generate six hyperplanes. 
$$H_1 = [0,0] \times \mathbb{R} \times \mathbb{R}, \quad H_2 = [1,1] \times \mathbb{R} \times \mathbb{R}, \ldots, \quad H_6 = \mathbb{R} \times \mathbb{R} \times [1,1] $$
and similarly for the other cells. 
(2): $\color{blue}{ \text{ My problem here is how we got the line of inequalities. } } $ 
In regards to my interpretation of (1), $J$ is divided into disjoint intervals. I see why $l(J)= \sum_{j=1}^r l(J_j) \le \sum_{k=1}^n l(K_n)$. But I don't see how we obtain
$$ \sum_{k=1}^{n}l(K_k) \le \sum_{k=1}^m l(I_k) $$ 
Shouldn't this be an equality (or in fact $\ge$?) As the family $\{ K_k \}$ is just a distinct representation of $\{ \bar{I_k} \} $? 


If $E \subseteq \mathbb{R}^p$, then we define the outer measure of $E$ to be $$m^{*} (E) = \inf \sum_{k=1}^\infty l (I_k)  $$ where the infimum is taken over all family of cells $(I_k)$ of $\mathbb{R}^p$ such that $E \subseteq \bigcup_{k=1}^{\infty} I_k$.  
A cell in $\mathbb{R}^p$ is of the form $I = I_1 \times \cdots \times I_p$, where $I_k$ are bounded intervals in $\mathbb{R}$. We define $l(I) = \prod_{k=1}^p l(I_k)$, where $l(I_k)$ is length of interval in $\mathbb{R}$ (so $l((a,b))=l((a,b]) = b-a$ for example.) 
A closed cell (open cells) in $\mathbb{R}^p$ is one where $I_k$ are all closed (open) bounded intervals in $\mathbb{R}$. So my three hyperplanes which satisfies the condition are 

 A: Here is a different (& more tedious) approach that might help:
To clarify what we are doing, we want to show that if $I_k$ is a cover of
$I$, then $l(I) \le \sum_k l(I_k)$ from which we will conclude $l(I) \le m^*(I)$.
Let $\epsilon>0$.
As above, we can choose a closed cell $J\subset I$  such that
$l(J) > l(I) -\epsilon$.
We can choose open cells $O_k \supset I_k$ such that $l(O_k) < l(I_k) + {1 \over 2^k} \epsilon$. Note that $\sum_k l(O_k) < \sum_k l(I_k) + \epsilon$.
Since $J$ is compact, a finite number of the $O_k$ cover $J$, so, by renumbering appropriately, we have
$J \subset \cup_{k=1}^n O_k$ for some $n$.
Let $H_k$ be the 'half open' cell corresponding to $O_k$. That is, if $O_k = \prod_i (a_i,b_i)$ then $H_k = \prod_i [a_i,b_i)$. Note that $l(O_k) = l(H_k)$.
Similarly, let $K$ be the 'half open' cell corresponding to $J$. Note that
$l(J)=l(K)$ and
$K \subset \cup_{k=1}^n H_k$.
If we can show that $l(K) \le \sum_{k=1}^n l(H_k)$, then this shows that
$l(I) -\epsilon < \sum_{k} l(I_k) + \epsilon$, and since $\epsilon>0$ was
arbitrary, we are finished.
Note that if we have a cell $H=\prod_i [a_i, b_i)$ and we pick some $\xi_{1} \in (a_{1}, b_{1})$, then $l(H) = l([a_1,x_1) \times \prod_{i>1} [a_i, b_i)) + l([x_1, b_1) \times \prod_{i>1} [a_i, b_i)) $. It should be clear that
this partitioning holds for any other index as well, and we can repeat this
a finite number of times.
Take the (finite) set of $i$th coordinates of the cells
$K, H_1,...,H_n$ 
order them into an increasing sequence $k \mapsto x_i(k)$.
Let ${\cal H} $ be the set of 'half open' cells of the form
$\prod_i [x_i(k),x_i(k+1))$.
This is the key point in the proof: Note that each of the sets $K, H_1,...,H_n$
is the finite union of elements of ${\cal H}$ and furthermore,
if we let ${\cal H}_A = \{H \in {\cal H} | H \subset A \}$ (the elements
of ${\cal H}$ contained in $A$), then the previous note shows that
$l(A) = \sum_{H \in {\cal H}_A} l(H)$ where $A$ is any of the sets $K, H_1,...,H_n$.
Since $K \subset \cup_{k=1}^n H_k$, we see that if $A \in {\cal H}_K$,
then $A \in {\cal H}_{H_j}$ for some $j$.
Hence 
$l(K) = \sum_{H \in {\cal H}_K} l(H) \le \sum_{H \in {\cal H}_{H_1} 
\cup \cdots \cup {\cal H}_{H_n}} l(H) \le \sum_{k=1}^n \sum_{H \in {\cal H}_{H_k} } l(H) = \sum_{ki=1}^n l(H_k)$.
