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  "body": "\n## Proof\n\nWe compare lower and upper sums of $f$ on a grid with the Darboux sums of the inner integral.\n\n1. Let $R$ and $f$ be as in the statement. By the definition of the [[double integral|0NKBMFC18R]], $f$ is bounded, say $\\abs{f} \\le B$. Let $I = \\iint_R f \\, dA$, and let $F(x) = \\int_c^d f(x, y) \\, dy$ for $x \\in [a, b]$.\n2. Take partitions $a = x_0 \u003c \\dots \u003c x_k = b$ and $c = y_0 \u003c \\dots \u003c y_l = d$, with rectangles $R_{ij} = [x_{i-1}, x_i] \\times [y_{j-1}, y_j]$. Let $m_{ij} = \\inf_{R_{ij}} f$ and $M_{ij} = \\sup_{R_{ij}} f$. Let $L = \\sum_{i,j} m_{ij} \\Delta A_{ij}$ and $U = \\sum_{i,j} M_{ij} \\Delta A_{ij}$.\n3. Each Riemann sum $S$ of $f$ for this grid satisfies $L \\le S \\le U$. For each $\\eta \u003e 0$, there are Riemann sums $S$ with $S \u003c L + \\eta$, and others with $S \u003e U - \\eta$. Choose the points $p_{ij}$ with $f(p_{ij})$ close to $m_{ij}$, or close to $M_{ij}$.\n4. Fix $i$ and $x \\in [x_{i-1}, x_i]$. For $y \\in [y_{j-1}, y_j]$ we have $m_{ij} \\le f(x, y) \\le M_{ij}$.\n5. The function $y \\mapsto f(x, y)$ is integrable on $[c, d]$. By [[additivity over intervals|05RCDKHZM6]], it is integrable on each $[y_{j-1}, y_j]$, and $F(x)$ is the sum of these integrals.\n6. By [[monotonicity of the integral|0MBWQKM8QR]] and step 4, $m_{ij} (y_j - y_{j-1}) \\le \\int_{y_{j-1}}^{y_j} f(x, y) \\, dy \\le M_{ij} (y_j - y_{j-1})$. Summing over $j$ gives $\\sum_j m_{ij} (y_j - y_{j-1}) \\le F(x) \\le \\sum_j M_{ij} (y_j - y_{j-1})$.\n7. In particular $\\abs{F(x)} \\le B (d - c)$, so $F$ is bounded. Let $P = (x_0, \\dots, x_k)$. Taking the infimum and the supremum of step 6 over $x \\in [x_{i-1}, x_i]$, and summing over $i$, gives $L \\le L(F, P) \\le U(F, P) \\le U$. Here $L(F, P)$ and $U(F, P)$ are the [[Darboux sums|0QRQVH0CRW]] of $F$.\n8. Let $\\underline{J}$ and $\\overline{J}$ be the lower and upper [[Darboux integrals|09FQT057T7]] of $F$. Then $L(F, P) \\le \\underline{J}$ and $\\overline{J} \\le U(F, P)$. Also $\\underline{J} \\le \\overline{J}$, since every lower sum is at most every upper sum (compare both with a common refinement).\n9. We now prove the first equality. Let $\\varepsilon \u003e 0$. Since $f$ is integrable, there is $\\delta \u003e 0$ with $\\abs{S - I} \u003c \\varepsilon$ for all Riemann sums $S$ whose rectangles have diameter less than $\\delta$.\n10. Choose the grid of step 2 with all diameters less than $\\delta$. By step 3, $L \\ge I - \\varepsilon$ and $U \\le I + \\varepsilon$. By steps 7 and 8, $I - \\varepsilon \\le \\underline{J} \\le \\overline{J} \\le I + \\varepsilon$.\n11. Since $\\varepsilon$ was arbitrary, $\\underline{J} = \\overline{J} = I$. By the theorem that [[Darboux and Riemann integrals agree|021XJAQVDC]], $F$ is integrable and $\\int_a^b F(x) \\, dx = I$.\n12. This says that the [[iterated integral|0P4WX7Y502]] $\\int_a^b \\int_c^d f(x, y) \\, dy \\, dx$ exists and equals $\\iint_R f \\, dA$.\n13. Exchange the roles of $x$ and $y$ in steps 2 to 11. This uses the hypothesis that $x \\mapsto f(x, y)$ is integrable for each $y$. It gives $\\int_c^d \\int_a^b f(x, y) \\, dx \\, dy = \\iint_R f \\, dA$.\n14. Now let $f$ be [[continuous|0S9EK5MQPE]] on $R$. The rectangle $R$ is closed and bounded, so it is compact by the [[Heine–Borel theorem|07PFYV65MB]]. So $f$ is bounded by the [[extreme value theorem|0V4DET59NW]].\n15. For each $x$, the function $y \\mapsto f(x, y)$ is continuous on $[c, d]$, so it is [[integrable|06SESV7GQY]]. In the same way, each $x \\mapsto f(x, y)$ is integrable. So steps 2 to 8 apply.\n16. By the [[Heine–Cantor theorem|06B8BNJCD6]], $f$ is uniformly continuous on $R$. Let $\\varepsilon \u003e 0$. There is $\\delta \u003e 0$ with $\\abs{f(p) - f(q)} \u003c \\varepsilon$ when $\\norm{p - q} \u003c \\delta$.\n17. For a grid whose rectangles have diameter less than $\\delta$, we get $M_{ij} - m_{ij} \\le \\varepsilon$. So $U - L \\le \\varepsilon (b - a)(d - c)$.\n18. By steps 7 and 8, $\\overline{J} - \\underline{J} \\le \\varepsilon (b - a)(d - c)$ for each $\\varepsilon \u003e 0$. So $\\underline{J} = \\overline{J}$, and $F$ is integrable. Let $I_0 = \\int_a^b F(x) \\, dx$. Then $L \\le I_0 \\le U$ for each such grid.\n19. Each Riemann sum $S$ for such a grid also satisfies $L \\le S \\le U$, by step 3. So $\\abs{S - I_0} \\le U - L \\le \\varepsilon (b - a)(d - c)$.\n20. Hence $f$ is Riemann integrable on $R$, with $\\iint_R f \\, dA = I_0$. So all hypotheses of the theorem hold, and the two equalities follow from steps 12 and 13.\n\n## Idea\n\nOn a fine grid, the lower and upper sums of $f$ are close to the double integral. On each vertical strip, the inner integral $F(x)$ is squeezed between the column sums of the infima and suprema of $f$. So the Darboux sums of $F$ are squeezed between the lower and upper sums of $f$, and $F$ integrates to the double integral.\n",
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