# The volume of a cover of an interval is greater than the volume of the interval

Let $$A \subset \mathbb{R}^n$$ be a $$n-$$dimensional interval, i.e. $$A = [a_1,b_1] \times [a_2,b_2] \times \cdots \times [a_n, b_n]$$, where $$[a_i, b_i]$$ is an interval in $$\mathbb{R}$$, for every $$i \in \{1,2,\cdots, n \}$$. Prove that, for every cover $$\{B_1, B_2, \cdots, B_m \}$$ of $$A$$, where $$B_i \subset \mathbb{R}^n$$ is an $$n-$$dimensional interval for every $$i \in \{1,2,\cdots,m \}$$, we have that $$|A| \leq \sum_{i=1}^m |B_i|,$$ where $$|I|$$ denotes the volume of the interval $$I$$, i.e. if $$I = [c_1, d_1] \times \cdots \times [c_n, d_n]$$, we have that $$|I| = \prod_{i=1}^n (d_i - c_i).$$

It seems very trivial to me, but I do not know how to prove it using only the definition of a cover and algebraic manipulation.

I tried using the fact that if $$\{B_1, \cdots, B_m \}$$ is a cover of $$A$$, then, by letting $$B_i = [a^{(i)}_1, b^{(i)}_1] \times \cdots \times [a^{(i)}_n, b^{(i)}_n]$$, we have that $$\sum_{j=1}^m (b^{(j)}_i - a^{(j)}_i) \geq b_i - a_i, \forall i \in \{1,2,\cdots, n \},$$ but it does not give me the desired result.

This question is typically an exercise when defining Riemann integration in $$\mathbb{R}^n$$, so I assume you're familiar with the concept of a partition of a rectangle etc. Otherwise, see Munkres' book Analysis on Manifolds or Spivak's Calculus on Manifolds. (They probably have a proof of this statement as well)
First, note that we can WLOG assume that each $$B_i$$ is contained in $$A$$ (why?). Having done this, construct a partition $$P$$ of the given rectangle $$A$$, by using the endpoints of the component intervals of the $$B_i$$'s. If you do this, then $$A, B_1, \dots, B_m$$ will all be a union of subrectangles of the partition $$P$$. This implies (try to justify yourself or refer to Munkres): \begin{align} |A| &= \sum_{S } |S| \\ |B_i| &= \sum_{S \subseteq B_i} |S| \qquad (1 \leq i \leq m), \end{align} where $$\sum \limits_{S \subseteq B_i}$$ means sum over all the subrectangles $$S$$ of the partition $$P$$ which are contained in $$B_i$$. Now the most crucial thing to notice is that every subrectangle $$S$$ of the partition $$P$$ is contained in some $$B_i$$. Or more precisely, for every $$S$$, there is an $$i \in \{ 1, \dots, m\}$$ such that $$S \subseteq B_i$$. Hence: \begin{align} |A| &= \sum_{S} |S| \\ &\leq \sum_{i=1}^m \sum_{S \subseteq B_i} |S| \\ &= \sum_{i=1}^m |B_i| \end{align} (The bold sentence tells us that every term in the first line already appears as a summand in the second line; which is why we have the $$\leq$$ sign).