Prove the additivity of the volume of intervals, namely, if is subdivided,
in any manner, into mutually disjoint subintervals in , then
(This is true also if some contain common faces). [Proof outline: For
, use Problem 8 Then by induction, suppose additivity holds for any
number of intervals smaller than a certain Now let
One of the (say, must have
some edge-length smaller than the corresponding edge-length of
say Now cut all of into
by the plane so that while
. For simplicity, assume that the plane cuts each
into two subintervals and . (One of
them may be empty.) Then
Actually, however, and are split into fewer than (nonempty)
intervals since by
construction. Thus, by our inductive assumption,
where and by Problem Complete the
inductive proof by showing that