Prove the additivity of the volume of intervals, namely, if \(A\) is subdivided,
in any manner, into \(m\) mutually disjoint subintervals \(A_{1}, A_{2}, \ldots,
A_{m}\) in \(E^{n}\), then
$$
v A=\sum_{i=1}^{m} v A_{i}
$$
(This is true also if some \(A_{i}\) contain common faces). [Proof outline: For
\(m=2\), use Problem 8 Then by induction, suppose additivity holds for any
number of intervals smaller than a certain \(m\) \((m>1) .\) Now let
$$
A=\bigcup_{i=1}^{m} A_{i} \quad\left(A_{i} \text { disjoint }\right)
$$
One of the \(A_{i}\) (say, \(\left.A_{1}=[\bar{a}, \bar{p}]\right)\) must have
some edge-length smaller than the corresponding edge-length of
\(A\left(\right.\) say \(\left., \ell_{1}\right) .\) Now cut all of \(A\) into
$$
P=[\bar{a}, \bar{d}] \text { and } Q=A-P \text { (Figure 4) }
$$
by the plane \(x_{1}=c\left(c=p_{1}\right)\) so that \(A_{1} \subseteq P\) while
\(A_{2} \subseteq Q\). For simplicity, assume that the plane cuts each \(A_{i}\)
into two subintervals \(A_{i}^{\prime}\) and \(A_{i}^{\prime \prime}\). (One of
them may be empty.) Then
$$
P=\bigcup_{i=1}^{m} A_{i}^{\prime} \text { and } Q=\bigcup_{i=1}^{m}
A_{i}^{\prime \prime}
$$
Actually, however, \(P\) and \(Q\) are split into fewer than \(m\) (nonempty)
intervals since \(A_{1}^{\prime \prime}=\emptyset=A_{2}^{\prime}\) by
construction. Thus, by our inductive assumption,
$$
v P=\sum_{i=1}^{m} v A_{i}^{\prime} \text { and } v Q=\sum_{i=1}^{m} v
A_{i}^{\prime \prime}
$$
where \(v A_{1}^{\prime \prime}=0=v A_{2}^{\prime},\) and \(v A_{i}=v
A_{i}^{\prime}+v A_{i}^{\prime \prime}\) by Problem \(8 .\) Complete the
inductive proof by showing that
$$
\left.v A=v P+v Q=\sum_{i=1}^{m} v A_{i} .\right]
$$
