Proof of Simple Properties of Volume

Let $e_{1},\ldots,e_{n}$ be vectors in $\mathbb{R}^{n}$. Define a parallelepiped $P$ to be a translate of the set

$$\left\{x\in\mathbb{R}^{n} : x=t^{1}e_{1}+\cdots+t^{n}e_{n}, 0\leq t^{i}\leq 1\right\}$$

Define the volume of $P$ by $\text{vol}(P):=\left|\det(A)\right|$, where $A$ is the matrix with columns $e_{1},\ldots,e_{n}$.

How does one proof directly from this definition of the volume of parallelepiped that if $I_{1},\ldots,I_{n}$ are nonoverlapping (i.e. disjoint interiors) parallelepipeds contained in $P$, then

$$\sum_{j=1}^{n}\text{vol}(I_{j})\leq\text{vol}(P)$$

This inequality seems geometrically evident from "cutting up" $P$, but I am having trouble making this rigorous. It seems like the easiest way to show this is to prove that this definition can be realized by Riemann integration of indicator functions using the change of variable formula and then use the additivity properties of the integral. But I want to avoid this.

As a simple case, how would one show that a parallelepiped (not necessarily of the same orientation) $I\subset P$ satisfies $\text{vol}(I)\leq\text{vol}(P)$ directly?

• What do you mean by an interval? A two dimensional line segment connecting two points? – uniquesolution Aug 31 '15 at 20:19
• I believe you mean non-overlapping parallelepipeds,and that by non-overlapping you mean that the intersection of any two of them is a pllpiped of zero volume. Also there's a typo in def'n of P : There should be $t^1,...t^n$ each independently ranging from 0 to 1. – DanielWainfleet Aug 31 '15 at 21:25
• @user254665: Yes, sorry. I was in a hurry when I wrote the question. – Nik Quine Aug 31 '15 at 22:02
• As I said elsewhere there seem to be books that make Lebesgue measure in R^n seem subtle and complicated, but it's not. Using some general results about it here would make this Q obvious. – DanielWainfleet Sep 7 '15 at 21:54
• @user254665: That would be circular though in this case. The authors assume these properties in order to prove "general results" about the Lebesgue measure. – Nik Quine Sep 12 '15 at 15:18