Volume function is additive set-function I have included some definitions right before the theorem, for convenience.
$\newcommand{\cA}{ {\cal A} }$
$\newcommand{\bR}{ {\mathbb{R}} }$
$\newcommand{\cP}{ {\cal P} }$
$\newcommand{\cM}{ {\cal M} }$

Definition. Let $\cA$ be a set of subsets of $\bR^n$.
  Then $\cA$ is a field of subsets of $\bR^n$ when 
  it has the following properties:
(FS-1) The empty set $\emptyset$ belongs to $\cA$.
(FS-2) If $A,B \in \cA$, then the set-difference 
         $A \setminus B$ still belongs to $\cA$.
(FS-3) If $A_1, \ldots , A_k \in \cA$, then the union 
  $A_1 \cup \cdots \cup A_k$ and the intersection 
  $A_1 \cap \cdots \cap A_k$ 
Definition. Volume function 
  $\mbox{vol}_o : \cP_n \to [0, \infty )$, where for 
  $P = ( a_1, b_1 ] \times \cdots \times ( a_n, b_n ] \in \cP_n$,
  $\mbox{vol}_o (P) := (b_1 -a_1) \cdots (b_n -a_n)$.
Definition. Half-open rectangle in $\bR^n$ is a set of the form
  $P = ( a_1, b_1 ] \times \cdots \times ( a_n, b_n ]
\subseteq \bR^n$,
  with $a_1 < b_1, \ldots , a_n < b_n$ in $\bR$.
  The collection of all half-open rectangles in $\bR^n$ is denoted by $\cP_n$.
Definition. Let $A \subseteq \bR^n$ 
  be a Jordan measurable set.
  Consider the real non-negative numbers $V^{-} (A)$ and $V^{+} (A)$ 
  defined as follows:
  $
V^{-}(A) := \sup \Bigl\{ \sum_{i=1}^r \mbox{vol}_o (P_i) \ 
\begin{array}{cc}
\lvert  &  P_1, \ldots , P_r \in \cP_n, \mbox{ with }
                 \cup_{i=1}^r P_i \subseteq A           \\
\lvert  &  \mbox{and with $P_i \cap P_j = \emptyset$ for 
                                          $i \neq j$}
\end{array}  \Bigr\} .$
$V^{+}(A) := \inf \Bigl\{ \sum_{j=1}^s \mbox{vol}_o (Q_j) \mid
Q_1, \ldots , Q_s \in \cP_n, \mbox{ with }
 \cup_{j=1}^s Q_j \supseteq A  \Bigr\} .
$
Then $V^{-}(A) = V^{+}(A)$.  The volume of $A$ is defined
  as
  $
\mbox{vol} (A) := V^{-}(A) = V^{+}(A). 
$
Definition. Let $\cA$ be a field of subsets of $\bR^n$. A function $\mu : \cA \to [0, \infty )$ is said to be an
  additive set-function when it has the property that:
  $
\left(   \begin{array}{c}
A_1, \ldots , A_k \in \cA  \\
\mbox{with $A_i \cap A_j = \emptyset$ for $i \neq j$}
\end{array}  \right) \ \Longrightarrow \
\Bigl( \, \mu (A_1 \cup \cdots \cup A_k) 
    = \sum_{i=1}^k \mu (A_i) \, \Bigr) .
$
Theorem. Consider the function 
  $\mbox{vol} : \cM_n \to [0, \infty )$, as defined above, where $\cM_n$ is the collection of all Jordan measurable sets on $\bR^n$. Prove that $\mbox{vol} : \cM_n \to [0, \infty )$ is an 
  additive set-function.

I want to prove the theorem stated above. Please let me know if my approach is OK. I'm concerned that it is not rigorous enough.
Let $A_1, \dots, A_k\in \cM_n$, with $A_i \cap A_j = \emptyset$ if $i\ne j$. Then $V^-(A_1\cup \dots \cup A_k)=\sup\left\{ \sum\limits_{i=1}^r\mbox{vol}_0(P_i): P_1, \dots, P_r\in \cP_n, \text{ with } \bigcup\limits_{i=1}^r P_i \subset A_1\cup \dots\cup A_k \text{ and } P_i\cap P_j = \emptyset \text{ for } i\ne j \right\}$.
Since $P_i$ are mutually disjoint and $A_m$ are mutually disjoint, if $x\in P_i$ for some $1\le i \le r$, $x\in A_m$ for some $1\le m\le k$. Reorder the indices of the sets $P_i$ in such a way as to have $\bigcup\limits_{i=1}^{r_1} P_{i} \subset A_1$, $\bigcup\limits_{i=r_1+1}^{r_2} P_{i} \subset A_2$, $\dots$, $\bigcup\limits_{i=r_s}^{r} P_{i} \subset A_k$ and consider the following:
$\sup\left\{ \sum\limits_{i=1}^r\mbox{vol}_0(P_i): P_1, \dots, P_r\in \cP_n, \text{ with } \bigcup\limits_{i=1}^r P_i \subset A_1\cup \dots\cup A_k \text{ and } P_i\cap P_j = \emptyset \text{ for } i\ne j \right\}\\
= \sup\left\{ \sum\limits_{i=1}^{r_1}\mbox{vol}_0(P_i)+\sum\limits_{i=r_1+1}^{r_2}\mbox{vol}_0(P_i) + \dots + \sum\limits_{i=r_s}^r\mbox{vol}_0(P_i): P_1, \dots, P_r\in \cP_n, \text{ with } \bigcup\limits_{i=1}^{r_1}\subset A_1,\dots, \bigcup\limits_{i=r_s}^{r}\subset A_k, \text{ and } P_i\cap P_j = \emptyset \text{ for } i\ne j \right\}\\
=\sup\left\{ \sum\limits_{i=1}^{r_1}\mbox{vol}_0(P_i): P_1, \dots, P_{r_1}\in \cP_n, \text{ with } \bigcup\limits_{i=1}^{r_1} P_i \subset A_1 \text{ and } P_i\cap P_j = \emptyset \text{ for } i\ne j \right\}\\
+\dots+\sup\left\{ \sum\limits_{i=r_s}^{r}\mbox{vol}_0(P_i): P_{r_s}, \dots, P_{r}\in \cP_n, \text{ with } \bigcup\limits_{i=r_s}^{r} P_i \subset A_k \text{ and } P_i\cap P_j = \emptyset \text{ for } i\ne j \right\}\\
=\mbox{vol}(A_1)+\dots+\mbox{vol}(A_k)$.
Q.E.D.
 A: $\newcommand{\vol}{\operatorname{vol}}\newcommand{\P}{\mathcal{P}}\newcommand{\M}{\mathcal{M}}$
Instead of using those long formulae with unproven equalities of suprema, use $\varepsilon$-arguments instead:
Take $A_1,\ldots A_k \in \M$ pairwise disjoint and let $A = \cup_i A_i$.
Suppose $\varepsilon>0$. 
Then for each fixed $i$ ($1 \le i \le k$) we have some finite pairwise disjoint family
$\mathcal{P}_i = \{P^i_1, P^i_2, \ldots, P^i_{n_i}\} \subseteq \P$, such that
$$ \sum_{j=1}^{n_i} \vol_0(P^i_j) > \vol^-(A_i) - \frac{\varepsilon}{k}$$
which can be done by the definition of $\vol^-(A_i)$ as a supremum of such sums.
Then note that $\P' = \cup_{i=1}^k \P_i$ is also a pairwise disjoint subfamily of $\P$ (as the $A_i$ are) and 
$$\sum_{P \in \P'} \vol_0(P) = \sum_{i=1}^k \sum_{j=1}^{n_i} vol_0(P^i_j) > \sum_{i=1}^k vol^-(A_i) - \varepsilon$$.
As this holds for all $\varepsilon >0$, and $\P'$ is a valid family in the definition of $\vol^-(A)$ we have 
$$\vol^-(A) \ge \sum_{i=1}^k vol^{-}(A_i)$$
and the other inequality should be clear as well:
Suppose $\P'= \{P_1, \ldots P_N\}$ is a pairwise disjoint subfamily of $\P$
such that $\cup \P' \subseteq A$. Then we can split ( gap here!) each $P \in \P'$ into a part inside $A_i$ for each $i$ and use the $\vol^-(A_i)$ upperbounds to have a bound for $\vol^-(A)$...
I think the gap can be filled with ideas from Tao's post on these matters, working with  properties of what he calls elementary sets, i.e. unions of half-open boxes, which form the smallest algebra of sets containing $\P$, and first extend $\vol_0$ to these...
Just some ideas to get you started, you see it's all quite subtle.
