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  "body": "\n## Proof\n\nWe prove the formula first for simple sets, and then squeeze $E$ between simple sets.\n\n1. Let $E$, $a \\le b$, $E_z$ and $A(z)$ be as in the statement. We use the boxes, simple sets and measure $m$ of the definition of [[Jordan measure|0565DC8RMK]], in $\\R^3$ and in $\\R^2$. So $\\operatorname{vol}(E) = m(E)$ and $A(z) = m(E_z)$.\n2. Case $a = b$. The set $E$ is bounded, so $E \\subseteq [-r, r) \\times [-r, r) \\times [a, a + \\varepsilon)$ for some $r \u003e 0$ and every $\\varepsilon \u003e 0$. Hence $m(E) \\le m^*(E) \\le 4 r^2 \\varepsilon$ for every $\\varepsilon \u003e 0$. So $\\operatorname{vol}(E) = 0 = \\int_a^a A(z) \\, dz$. From now on, let $a \u003c b$.\n3. Lemma 1: let $J$ be an interval with endpoints $c \\le d$ in $[a, b]$, and let $\\chi$ be its indicator function on $[a, b]$. Then $\\chi$ is integrable and $\\int_a^b \\chi \\, dz = d - c$.\n4. Proof of Lemma 1. Let $\\eta \u003e 0$. Take the [[partition|069P20KYQT]] of $[a, b]$ whose points are $a$, $b$ and those of $c \\pm \\eta$, $d \\pm \\eta$ that lie in $(a, b)$.\n5. Outside $[c - \\eta, d + \\eta]$ we have $\\chi = 0$, so the upper [[Darboux sum|0QRQVH0CRW]] is at most $d - c + 2\\eta$. On $[c + \\eta, d - \\eta]$ we have $\\chi = 1$, so the lower Darboux sum is at least $d - c - 2\\eta$.\n6. Since $\\eta$ is arbitrary, the upper and lower [[Darboux integrals|09FQT057T7]] of $\\chi$ are both $d - c$. By [[the agreement of Darboux and Riemann integrals|021XJAQVDC]], Lemma 1 follows.\n7. Lemma 2: let $S$ be a simple set in $\\R^3$. Write $S$ as a disjoint union of nonempty boxes $Q_l = P_l \\times I_l$, where $P_l$ is a half-open rectangle in $\\R^2$ and $I_l = [c_l, d_l)$, for $l = 1, \\dots, p$. Then for each $z \\in \\R$, $S_z = \\set{(x, y) : (x, y, z) \\in S}$ is a simple set and $m(S_z) = \\sum_l \\abs{P_l} \\chi_l(z)$, where $\\chi_l$ is the indicator function of $I_l$.\n8. Proof of Lemma 2. We have $S_z = \\bigcup_{l : z \\in I_l} P_l$. These $P_l$ are pairwise disjoint: a common point $(x, y)$ of two of them would give a common point $(x, y, z)$ of two boxes $Q_l$. So $m(S_z) = \\sum_{l : z \\in I_l} \\abs{P_l}$, which is the claim.\n9. Also $m(S) = \\sum_l \\abs{Q_l} = \\sum_l \\abs{P_l} (d_l - c_l)$, by the definition of the volume of a box.\n10. Let $S \\subseteq E$ be a simple set, written as in Lemma 2. Each $Q_l$ is nonempty and lies in $E$, so $I_l \\subseteq [a, b]$. By Lemma 1 and [[linearity of the integral|080RWVVSB6]], $z \\mapsto m(S_z)$ is integrable on $[a, b]$ with integral $\\sum_l \\abs{P_l} (d_l - c_l) = m(S)$.\n11. Let $T \\supseteq E$ be a simple set, written as a disjoint union of nonempty boxes $P'_l \\times [c'_l, d'_l)$. On $[a, b]$, the indicator of $[c'_l, d'_l)$ equals the indicator of the interval $[c'_l, d'_l) \\cap [a, b]$, whose length is at most $d'_l - c'_l$.\n12. By Lemma 1, Lemma 2 and linearity, $z \\mapsto m(T_z)$ is integrable on $[a, b]$, with $\\int_a^b m(T_z) \\, dz \\le \\sum_l \\abs{P'_l} (d'_l - c'_l) = m(T)$.\n13. Let $z \\in [a, b]$. Since $S_z \\subseteq E_z \\subseteq T_z$ and $S_z, T_z$ are simple, the definitions of inner and outer measure give $m(S_z) \\le m_*(E_z) = A(z)$ and $A(z) = m^*(E_z) \\le m(T_z)$.\n14. Simple sets $S \\subseteq E \\subseteq T$ exist, since $E$ is bounded. Fix one such $T$. By step 13, $0 \\le A(z) \\le m(T_z) \\le \\sum_l \\abs{P'_l}$, so $A$ is bounded on $[a, b]$.\n15. Let $\\underline{J}$ and $\\overline{J}$ be the lower and upper Darboux integrals of $A$ on $[a, b]$. Let $\\varphi \\le A \\le \\psi$ with $\\varphi, \\psi$ integrable. Then each lower sum of $A$ is at least that of $\\varphi$, and each upper sum of $A$ is at most that of $\\psi$. So $\\int_a^b \\varphi \\le \\underline{J}$ and $\\overline{J} \\le \\int_a^b \\psi$.\n16. By steps 10, 13 and 15, $m(S) \\le \\underline{J}$ for every simple $S \\subseteq E$. Taking the supremum, $m_*(E) \\le \\underline{J}$.\n17. By steps 12, 13 and 15, $\\overline{J} \\le m(T)$ for every simple $T \\supseteq E$. Taking the infimum, $\\overline{J} \\le m^*(E)$.\n18. Every lower sum is at most every upper sum, so $\\underline{J} \\le \\overline{J}$. Since $E$ is Jordan measurable, $m_*(E) = m^*(E) = m(E)$. Steps 16 and 17 give $m(E) \\le \\underline{J} \\le \\overline{J} \\le m(E)$.\n19. So $\\underline{J} = \\overline{J} = m(E)$. By the agreement of Darboux and Riemann integrals, $A$ is integrable on $[a, b]$ and $\\operatorname{vol}(E) = \\int_a^b A(z) \\, dz$.\n20. If two such solids have $A(z)$ equal for each $z \\in [a, b]$, the right sides agree. So the two solids have equal volume.\n\n## Idea\n\nFor a finite union of boxes, the area of the cross-section at height $z$ is a step function of $z$, and its integral adds up the volumes of the boxes. A Jordan measurable solid is squeezed between such unions, from inside and outside. Their cross-sections squeeze the cross-sections of the solid, so the integral of $A(z)$ is squeezed to the volume.\n",
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