Very often I find difficult to prove what it seems obvious. This is the problem:
Let $R\subseteq \mathbb{R}^d$ a compact rectangle and $R_1,\ldots,R_n$ open rectangles such that $R\subseteq R_1\cup\ldots\cup R_n$. Then show that $v(R) \le v(R_1)+\ldots + v(R_n)$
Now my guess is, since $\cup R_i$ is open then we can write it like a union of nonoverlapping cubes and then we take volumes. Thanks in advance for your help.
Take a rectangle $Q$ that contains the rectangles $R_1, \ldots, R_n$. Using the endpoints of the component intervals for each rectangle $R_k$, that is $a_{k,1}, \ldots a_{k,d}$ and $b_{k,1}, \ldots b_{k,d}$ where $R_k = (a_{k,1},b_{k,1}) \times \ldots \times (a_{k,d},b_{k,d})$ form partition $P$ of $Q$ with subrectangles $Q_1, \ldots Q_m$.
By construction, each of the (closed) rectangles $\overline{R_k}$ is a union of some of the rectangles $Q_1, \ldots, Q_m$ and we also have that the original compact rectangle $R$ is a union of some of the partition subrectangles
$$R = \bigcup_{Q \subset R}Q$$
I suggest drawing a picture to understand this.
Since each partition subrectangle $Q$ is contained in one of the closed covering rectangles $\overline{R_1}, \ldots, \overline{R_n}$, it follows that
$$\sum_{Q \subset R}v(Q) \leqslant \sum_{k=1}^n\sum_{Q' \subset \overline{R_k}}v(Q'),$$
since the sum on the RHS may double count.
Since partition subrectangles are non-overlapping it follows that
$$v(R) = \sum_{Q \subset R} v(Q) \leqslant \sum_{k=1}^n\sum_{Q' \subset \overline{R_k}}v(Q') = \sum_{k=1}^n v(\overline{R_k}) = \sum_{k=1}^n v(R_k)$$