I am reading "Calculus II" by Shizuo Miyajima.
Let $I := [a_1, b_1] \times [a_2, b_2] \times \cdots \times [a_n, b_n]$.
Let $|I| := \prod_{i=1}^{n} (b_i - a_i)$.
Let $I_i = [c_{i1}, d_{i1}] \times [c_{i2}, d_{i2}] \times \cdots \times [c_{in}, d_{in}]$ for $i \in \{1, \cdots, m\}$.
Let $|I_i| := \prod_{k=1}^{n} |d_{ik}-c_{ik}|$ for $i \in \{1, \cdots, m\}$.
Let $I_k^\circ \cap I_l^\circ = \emptyset$ for $k \ne l$. ($I^\circ$ is an interior of $I$.)
The author wrote that the following fact was obvious, but I cannot prove that:
If $I = \cup_{i=1}^{m} I_i$, then $|I| = \sum_{i=1}^{m} |I_i|$.
Please tell me the proof.