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  "body": "\n## Proof\n\nWe compare $Q$ with half-open boxes inside and around it, and we compute the Riemann sums of the constant $1$.\n\n1. Let $Q = [a_1, b_1] \\times \\dots \\times [a_n, b_n]$ with $a_i \u003c b_i$, 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.\n2. 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.\n3. 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.\n4. 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.\n5. 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 $C \\subseteq T$.\n6. Each cell in $S$ is also in $T$, since $S \\subseteq T$. So $m(S) \\le m(T)$, which proves the lemma.\n7. 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 $m_*(Q) \\ge V$.\n8. For $\\varepsilon \u003e 0$, the box $T_\\varepsilon = [a_1, b_1 + \\varepsilon) \\times \\dots \\times [a_n, b_n + \\varepsilon)$ contains $Q$. So the outer measure satisfies $m^*(Q) \\le (b_1 - a_1 + \\varepsilon) \\cdots (b_n - a_n + \\varepsilon)$.\n9. Letting $\\varepsilon \\to 0$ in step 8, we get $m^*(Q) \\le V$, since the product is continuous in $\\varepsilon$.\n10. 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 $m_*(Q) \\le m^*(Q)$.\n11. By steps 7, 9 and 10, $V \\le m_*(Q) \\le m^*(Q) \\le V$. So $Q$ is Jordan measurable with $\\operatorname{vol}(Q) = m(Q) = V$.\n12. 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.\n13. Let $a_i = t_{i,0} \u003c \\dots \u003c t_{i,k_i} = b_i$ be partitions. A sub-box has volume $\\prod_{i=1}^n (t_{i, l_i} - t_{i, l_i - 1})$ for some indices $l_1, \\dots, l_n$.\n14. Every Riemann sum of the constant function $1$ is the sum of the volumes of all sub-boxes. Expanding the product of sums gives $\\sum_{l_1, \\dots, l_n} \\prod_{i} (t_{i, l_i} - t_{i, l_i - 1}) = \\prod_{i} \\sum_{l} (t_{i, l} - t_{i, l-1}) = \\prod_i (b_i - a_i) = V$.\n15. So every Riemann sum of $1$ equals $V$, for every choice of partitions and points. Hence $1$ is integrable on $Q$ with $\\int_Q 1 \\, dV = V$.\n\n## Idea\n\nThe 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.\n",
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