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  "body": "\n## Proof\n\nWe compute the Darboux sums for a partition into equal pieces and use Riemann's criterion.\n\n1. Let $a \u003c b$ and let $f \\colon [a, b] \\to \\mathbb{R}$ be [[monotone|0WFF6AZPBP]]. First assume that $f$ is increasing, that is, $x \\le y$ implies $f(x) \\le f(y)$.\n2. Then $f(a) \\le f(x) \\le f(b)$ for all $x \\in [a, b]$. So $f$ is bounded, and its [[Darboux sums|0QRQVH0CRW]] are defined.\n3. Let $\\varepsilon \u003e 0$. By the [[Archimedean property|0A3GWZ86AY]], take $n \\ge 1$ with $n \\varepsilon \u003e (b - a)(f(b) - f(a))$. Let $P$ have the points $x_i = a + i (b - a)/n$ for $i = 0, \\dots, n$.\n4. On $[x_{i-1}, x_i]$ we have $f(x_{i-1}) \\le f(x) \\le f(x_i)$, and both bounds are values of $f$ there. So $M_i = f(x_i)$ and $m_i = f(x_{i-1})$.\n5. Each subinterval has length $(b - a)/n$. So the sum telescopes:\n   $$U(f, P) - L(f, P) = \\frac{b - a}{n} \\sum_{i=1}^{n} \\big(f(x_i) - f(x_{i-1})\\big) = \\frac{b - a}{n} \\big(f(b) - f(a)\\big) \u003c \\varepsilon.$$\n6. Such a partition exists for each $\\varepsilon \u003e 0$. By [[Riemann's criterion|017DH8WXTE]], $f$ is Riemann integrable.\n7. Now assume that $f$ is decreasing. Then $f(b) \\le f(x) \\le f(a)$, so $f$ is bounded. Let $\\varepsilon \u003e 0$.\n   Take $n \\ge 1$ with $n \\varepsilon \u003e (b - a)(f(a) - f(b))$, and let $P$ be the partition into $n$ equal pieces as in step 3.\n8. Now $M_i = f(x_{i-1})$ and $m_i = f(x_i)$. As in step 5, $U(f, P) - L(f, P) = \\frac{b - a}{n} (f(a) - f(b)) \u003c \\varepsilon$. As in step 6, $f$ is Riemann integrable.\n\n## Idea\n\nFor an increasing function, the supremum on a piece is the value at the right end and the infimum is the value at the left end. So the gaps between upper and lower sums telescope to $(f(b) - f(a))(b - a)/n$, which is small for large $n$.\n",
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