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  "body": "\n## Proof\n\nWe compare Riemann sums for the same tagged partition.\n\n1. Let $a \u003c b$, and let $f, g \\colon [a, b] \\to \\mathbb{R}$ be [[Riemann integrable|0N1JQXZXHC]] with $f \\le g$ on $[a, b]$. Let $I_f$ and $I_g$ be their [[integrals|0QM4TXT3DR]].\n2. For each tagged partition $(P, t)$ with $P = (x_0, \\dots, x_n)$, we have $f(t_i)(x_i - x_{i-1}) \\le g(t_i)(x_i - x_{i-1})$ for each $i$. Adding gives $S(f, P, t) \\le S(g, P, t)$ for the [[Riemann sums|0HBY48G8G1]].\n3. Let $\\varepsilon \u003e 0$. Take $\\delta_f \u003e 0$ such that the Riemann sums of $f$ for tagged partitions of [[mesh|04GM4NYXJB]] less than $\\delta_f$ are within $\\varepsilon$ of $I_f$.\n   Take $\\delta_g \u003e 0$ in the same way for $g$ and $I_g$. Let $\\delta = \\min(\\delta_f, \\delta_g)$.\n4. By the [[Archimedean property|0A3GWZ86AY]], take $n$ with $n \\delta \u003e b - a$. The points $x_i = a + i(b - a)/n$ with tags $t_i = x_i$ give a tagged partition $(P, t)$ of mesh less than $\\delta$.\n5. By steps 2 and 3, $I_f - \\varepsilon \u003c S(f, P, t) \\le S(g, P, t) \u003c I_g + \\varepsilon$. So $I_f \u003c I_g + 2 \\varepsilon$.\n6. This holds for each $\\varepsilon \u003e 0$. Hence $I_f \\le I_g$, that is, $\\int_a^b f(x) \\, dx \\le \\int_a^b g(x) \\, dx$.\n\n## Idea\n\nFor the same tagged partition, each Riemann sum of $f$ is at most the Riemann sum of $g$. Inequalities that hold for every approximation survive in the limit.\n",
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