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  "body": "\n## Proof\n\nLet $a \u003c b$ and let $f \\colon [a, b] \\to \\mathbb{R}$ be bounded. Write $U(P)$ and $L(P)$ for the upper and lower [[Darboux sums|0QRQVH0CRW]] of $f$, and $U^*$ and $L^*$ for the upper and lower [[Darboux integrals|09FQT057T7]].\n\n1. Let $R$ be a [[refinement|0EEKF7YFBW]] of a partition $Q$. Each subinterval of $R$ lies in exactly one subinterval $J$ of $Q$, and the subintervals of $R$ inside $J$ have total length $|J|$.\n2. On a smaller set the [[supremum|0BTMQ04556]] is no larger and the [[infimum|04KFTA34C5]] is no smaller. Grouping the terms by step 1 gives $U(R) \\le U(Q)$ and $L(R) \\ge L(Q)$.\n3. For any partitions $P$ and $Q$, let $R$ consist of the points of both. Then $L(P) \\le L(R) \\le U(R) \\le U(Q)$ by step 2, since $m_i \\le M_i$ on each subinterval.\n4. Taking the supremum over $P$ and then the infimum over $Q$ in step 3 gives $L^* \\le U^*$.\n\nFirst assume that $f$ is [[Riemann integrable|0N1JQXZXHC]], with integral $I$.\n\n5. By the theorem that [[Darboux and Riemann integrals agree|021XJAQVDC]], $L^* = U^* = I$.\n6. Let $\\varepsilon \u003e 0$. The number $I + \\varepsilon/2$ is not a lower bound of the upper sums, so there is $P_1$ with $U(P_1) \u003c I + \\varepsilon / 2$. In the same way, there is $P_2$ with $L(P_2) \u003e I - \\varepsilon/2$.\n7. Let $P$ consist of the points of $P_1$ and $P_2$. By step 2, $U(P) - L(P) \\le U(P_1) - L(P_2) \u003c \\varepsilon$.\n\nNow assume that for each $\\varepsilon \u003e 0$ there is a partition $P$ with $U(P) - L(P) \u003c \\varepsilon$.\n\n8. Let $\\varepsilon \u003e 0$ and take such a $P$. By the definitions of $L^*$ and $U^*$ and by step 4, $L(P) \\le L^* \\le U^* \\le U(P)$.\n9. So $0 \\le U^* - L^* \\le U(P) - L(P) \u003c \\varepsilon$. This holds for every $\\varepsilon \u003e 0$, so $U^* = L^*$.\n10. By the theorem that [[Darboux and Riemann integrals agree|021XJAQVDC]], $f$ is Riemann integrable on $[a, b]$.\n\n## Idea\n\nThe upper and lower Darboux integrals are squeezed between $L(P)$ and $U(P)$ for every partition $P$. So they are equal exactly when the gap $U(P) - L(P)$ can be made as small as we like. Equality of the Darboux integrals is the same as Riemann integrability.\n",
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