Show that $\operatorname{vol_n}(I)\le\sum_{k=1}^m\operatorname{vol_n}(J_k)$

Question

In the book of Analysis III of Amann and Escher the first exercise on page 29 says

Prove

(a) $I,J\in\Bbb J(n)\implies I\cap J\in\Bbb J(n)$

(b) If $I,J_k\in\Bbb J(n)$ and $I\subset\bigcup_{k=1}^m J_k$ then $$\operatorname{vol_n}(I)\le\sum_{k=1}^m\operatorname{vol_n}(J_k)$$ (Dont use the definition of the Lebesgue outer-measure.)

Im stuck in the part (b). The notation: $\Bbb J(n)$ is the set of intervals in $\Bbb R^n$. An interval in $\Bbb R^n$ is defined as the cartesian product of intervals in the real line (the intervals can be empty), and the function $\operatorname{vol_n}$ is the standard volume of these boxes, but this function is, in this context, only defined on $\Bbb J(n)$.

I can show that for $I,J\in\Bbb J(n)$ when $I\subset J$ then $\operatorname{vol_n}(I)\le\operatorname{vol_n}(J)$.

Now Im trying to show that if $I\cup J\in\Bbb J(n)$ and $I\cap J=\emptyset$ then

$$\operatorname{vol_n}(I\cup J)=\operatorname{vol_n}(I)+\operatorname{vol_n}(J)$$

what will almost finish the proof. For this task Im trying to get some identity of the set theoretic union

$$(A\times B)\cup (C\times D)=E\times F\implies (A\times B)\cup (C\times D)=(A\cup C)\times (B\cup D)$$

and working various cases, but it get long and messy. There is other more simple approach that doesn't use the theory related to the Lebesgue outer-measure?

Answer 1

$(A \times B) \cup (C \times D) =(E \times F)$ implies $E=A \cup C$ and $F=B \cup D$ as a simple verification shows. Just look at first coordinates and second coordinates on both sides. You only have to assume that the four sets are nonempty to draw this conclusion. Here is an example of how you argue: let $x \in E$. take any point $y \in F$. Then $(x,y) \in (E \times F)$ so either $(x,y) \in (A \times B)$ or $(x,y) \in (C \times D)$. In the first case $x \in A$ and in the second case $x \in C$. We have proved that $E \subset A \cup C$. Can you now complete the argument?

Answer 2

There is another path that uses a discretization. The idea would be that the volume of a parallelotope would be approximately the number of points of a dense lattice $L_N\colon = (\frac{1}{N}\mathbb{Z})^n$ multiplied by the volume of the fundamental region of that lattice $\frac{1}{N}^n$. That is $$\text{vol}(J) = \frac{1}{N^n} \cdot \lim_{N\to \infty} \#(J \cap L_N)$$

Since $J\mapsto \#(J \cap L_N)$ is subadditive by set theoretical reasons, the same will be true for the limits. This appears to be due to Von Neumann, although the idea seems classical.

Note that a proof of the equality $$\text{vol}(J) = \sum_k \text{vol}(J_k)$$ if $J_k$'s are "partitioning" $J$ can also use the same argument and some care. But a clearer reason for the equality comes from understanding how different parts cover a cartesian product. One is keeping track carefully of all the cutting and piecing together intervals and gets the proof- mostly all is set theoretical.