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  "body": "\n## Proof\n\nLet $a \u003c b$ and let $f \\colon [a, b] \\to \\mathbb{R}$ be bounded. Fix $K \u003e 0$ with $|f(x)| \\le K$ for all $x \\in [a, b]$.\nWrite $U(P)$ and $L(P)$ for the upper and lower [[Darboux sums|0QRQVH0CRW]] of $f$ for a [[partition|069P20KYQT]] $P$.\nWrite $S(P, t)$ for the [[Riemann sum|0HBY48G8G1]] with tags $t$.\nLet $U^*$ and $L^*$ be the upper and lower [[Darboux integrals|09FQT057T7]].\nWe first prove four facts, then the two directions.\n\n1. Let $P = (x_0, \\dots, x_n)$ with tags $t_i \\in [x_{i-1}, x_i]$. Then $m_i \\le f(t_i) \\le M_i$ for each $i$. Multiply by $x_i - x_{i-1} \u003e 0$ and add: $L(P) \\le S(P, t) \\le U(P)$.\n2. Let $R$ be a [[refinement|0EEKF7YFBW]] of a partition $Q = (y_0, \\dots, y_m)$. Each subinterval of $R$ lies in exactly one subinterval $[y_{j-1}, y_j]$ of $Q$, because no point of $Q$ lies strictly between two consecutive points of $R$.\n   The subintervals of $R$ inside $[y_{j-1}, y_j]$ have total length $y_j - y_{j-1}$.\n3. A [[supremum|0BTMQ04556]] over a subset is at most the supremum over the whole set, and an [[infimum|04KFTA34C5]] over a subset is at least the infimum. Group the terms of $U(R)$ by step 2. This gives $U(R) \\le U(Q)$, and in the same way $L(R) \\ge L(Q)$.\n4. For any partitions $P$ and $Q$, let $R$ consist of the points of both. By steps 3 and 1, $L(P) \\le L(R) \\le U(R) \\le U(Q)$. Taking the supremum over $P$ and then the infimum over $Q$ gives $L^* \\le U^*$.\n5. For each $\\delta \u003e 0$ there is a partition with [[mesh|04GM4NYXJB]] less than $\\delta$. Indeed, take $n$ with $n \\delta \u003e b - a$, by the [[Archimedean property|0A3GWZ86AY]], and $x_i = a + i (b - a)/n$.\n\nFirst assume that $f$ is [[Riemann integrable|0N1JQXZXHC]] with [[integral|0QM4TXT3DR]] $I$.\n\n6. Let $\\varepsilon \u003e 0$. Take $\\delta \u003e 0$ such that $|S(P, t) - I| \u003c \\varepsilon / 2$ for each tagged partition of mesh less than $\\delta$. By step 5, fix such a partition $P = (x_0, \\dots, x_n)$.\n7. For each $i$, the number $M_i - \\frac{\\varepsilon}{2(b - a)}$ is not an upper bound of $f$ on $[x_{i-1}, x_i]$. So there is $t_i$ there with $f(t_i) \u003e M_i - \\frac{\\varepsilon}{2(b - a)}$.\n8. Multiply by $x_i - x_{i-1}$ and add. This gives $S(P, t) \u003e U(P) - \\varepsilon/2$. So $U^* \\le U(P) \u003c S(P, t) + \\varepsilon / 2 \u003c I + \\varepsilon$.\n9. In the same way, choose tags $t'_i$ with $f(t'_i) \u003c m_i + \\frac{\\varepsilon}{2(b - a)}$. This gives $L^* \\ge L(P) \u003e S(P, t') - \\varepsilon/2 \u003e I - \\varepsilon$.\n10. So $I - \\varepsilon \u003c L^* \\le U^* \u003c I + \\varepsilon$ for each $\\varepsilon \u003e 0$, using step 4. Hence $L^* = U^* = I$.\n\nNow assume that $L^* = U^*$, and call this number $D$.\n\n11. Let $\\varepsilon \u003e 0$. Since $D + \\varepsilon / 8$ is not a lower bound of the upper sums, there is $Q_1$ with $U(Q_1) \u003c D + \\varepsilon/8$. In the same way, there is $Q_2$ with $L(Q_2) \u003e D - \\varepsilon / 8$.\n12. Let $Q$ consist of the points of $Q_1$ and $Q_2$. By step 3, $U(Q) - L(Q) \\le U(Q_1) - L(Q_2) \u003c \\varepsilon / 4$. Let $m$ be the number of subintervals of $Q$, and put $\\delta = \\frac{\\varepsilon}{8 m K}$.\n13. Let $P = (x_0, \\dots, x_n)$ have mesh less than $\\delta$, and let $P'$ consist of the points of $P$ and $Q$. Then $P'$ refines both $P$ and $Q$.\n14. Call a subinterval $[x_{k-1}, x_k]$ of $P$ split if a point of $Q$ lies strictly inside it. Only the $m - 1$ points of $Q$ other than $a$ and $b$ can do this. Each lies inside at most one subinterval of $P$. So at most $m - 1$ subintervals are split.\n15. Group the terms of $U(P')$ by the subintervals of $P$, as in step 2. A subinterval of $P$ that is not split is also a subinterval of $P'$, and its terms cancel in $U(P) - U(P')$.\n16. For a split subinterval $J$ of length $|J| \u003c \\delta$, its term in $U(P)$ is at most $K |J|$. Its terms in $U(P')$ add up to at least $-K |J|$, since every supremum is at least $-K$. So it adds at most $2 K |J| \u003c 2 K \\delta$ to $U(P) - U(P')$.\n17. By steps 14 to 16, $U(P) - U(P') \\le 2 (m - 1) K \\delta \u003c \\varepsilon / 4$. With step 3 for $P'$ and $Q$, $U(P) \u003c U(P') + \\varepsilon/4 \\le U(Q) + \\varepsilon / 4$.\n18. The same argument with infima gives $L(P) \u003e L(P') - \\varepsilon/4 \\ge L(Q) - \\varepsilon / 4$.\n19. By steps 17, 18 and 12, $U(P) - L(P) \u003c U(Q) - L(Q) + \\varepsilon / 2 \u003c \\varepsilon$.\n20. By the definitions of $L^*$ and $U^*$, we have $L(P) \\le D \\le U(P)$. By step 1, also $L(P) \\le S(P, t) \\le U(P)$ for all tags $t$. So $|S(P, t) - D| \\le U(P) - L(P) \u003c \\varepsilon$.\n21. This holds for every tagged partition of mesh less than $\\delta$. So $f$ is Riemann integrable, and its integral is $D$.\n\nSteps 10 and 21 prove both directions. In both cases the Riemann integral equals the common value of the Darboux integrals.\n\n## Idea\n\nA Riemann sum always lies between the lower and the upper Darboux sum. If Riemann sums converge, tags near the suprema and infima force the Darboux sums together. Conversely, a fine partition differs from a good fixed partition $Q$ only on the few short subintervals that contain points of $Q$. So its Darboux sums are also close together.\n",
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