---
tag: 3B4P26FTP2
kind: proof
name: Proof of the formula for the volume of a box
msc: ["28A75"]
proves: 0VEKNG02W0
uses: [0565DC8RMK, 0NKBMFC18R, 0GTBGMZXV8]
sources:
  - url: https://terrytao.files.wordpress.com/2011/01/measure-book1.pdf
    locator: Section 1.1
---

## Proof

We compare $Q$ with half-open boxes inside and around it, and we compute the Riemann sums of the constant $1$.

1. Let $Q = [a_1, b_1] \times \dots \times [a_n, b_n]$ with $a_i < b_i$, and let $V = (b_1 - a_1) \cdots (b_n - a_n)$. We use the boxes, simple sets and measure $m$ of the definition of [[Jordan measure|0565DC8RMK]]. Boxes there are half-open.
2. Lemma: if $S \subseteq T$ are simple sets, then $m(S) \le m(T)$. Write $S$ and $T$ as finite unions of pairwise disjoint boxes, and drop the empty ones.
3. For each coordinate $i$, list all $i$-th endpoints of all these boxes, in increasing order. They cut each axis into consecutive half-open intervals. The products of such intervals are pairwise disjoint half-open boxes, the cells.
4. Each interval of a box of $S$ or $T$ is a union of consecutive intervals of step 3. So each box of $S$ or $T$ is the disjoint union of the cells it contains. Two different cells are disjoint, so a cell meets a box only if it lies in it.
5. The definition says that $m$ of a simple set does not depend on how we split it into disjoint boxes. So $m(S)$ is the sum of $\abs{C}$ over the cells $C \subseteq S$, and $m(T)$ is the sum of $\abs{C}$ over the cells $C \subseteq T$.
6. Each cell in $S$ is also in $T$, since $S \subseteq T$. So $m(S) \le m(T)$, which proves the lemma.
7. The half-open box $Q_0 = [a_1, b_1) \times \dots \times [a_n, b_n)$ is a simple set with $Q_0 \subseteq Q$ and $m(Q_0) = V$. So the inner measure satisfies $m_*(Q) \ge V$.
8. For $\varepsilon > 0$, the box $T_\varepsilon = [a_1, b_1 + \varepsilon) \times \dots \times [a_n, b_n + \varepsilon)$ contains $Q$. So the outer measure satisfies $m^*(Q) \le (b_1 - a_1 + \varepsilon) \cdots (b_n - a_n + \varepsilon)$.
9. Letting $\varepsilon \to 0$ in step 8, we get $m^*(Q) \le V$, since the product is continuous in $\varepsilon$.
10. If $S \subseteq Q \subseteq T$ with $S, T$ simple, then $m(S) \le m(T)$ by the lemma. Taking the supremum over $S$ and the infimum over $T$ gives $m_*(Q) \le m^*(Q)$.
11. By steps 7, 9 and 10, $V \le m_*(Q) \le m^*(Q) \le V$. So $Q$ is Jordan measurable with $\operatorname{vol}(Q) = m(Q) = V$.
12. The Riemann integral over a box in $\R^n$ is defined as for [[double integrals|0NKBMFC18R]] and [[triple integrals|0GTBGMZXV8]]. One partitions each $[a_i, b_i]$ and forms sums over the resulting closed sub-boxes.
13. Let $a_i = t_{i,0} < \dots < t_{i,k_i} = b_i$ be partitions. A sub-box has volume $\prod_{i=1}^n (t_{i, l_i} - t_{i, l_i - 1})$ for some indices $l_1, \dots, l_n$.
14. Every Riemann sum of the constant function $1$ is the sum of the volumes of all sub-boxes. Expanding the product of sums gives $\sum_{l_1, \dots, l_n} \prod_{i} (t_{i, l_i} - t_{i, l_i - 1}) = \prod_{i} \sum_{l} (t_{i, l} - t_{i, l-1}) = \prod_i (b_i - a_i) = V$.
15. So every Riemann sum of $1$ equals $V$, for every choice of partitions and points. Hence $1$ is integrable on $Q$ with $\int_Q 1 \, dV = V$.

## Idea

The half-open box with the same edges sits inside $Q$, and slightly larger half-open boxes cover $Q$. Their measures squeeze the inner and outer measures of $Q$ to the product of the edge lengths. For the integral, every Riemann sum of $1$ adds up the volumes of the pieces of a grid, and this sum always gives back the product.