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  "body": "\n## Proof\n\nWe use uniform continuity and Riemann's criterion.\n\n1. Let $a \u003c b$ and let $f \\colon [a, b] \\to \\mathbb{R}$ be continuous. By the [[boundedness theorem|0ATVXC3NCC]], $f$ is bounded. So its [[Darboux sums|0QRQVH0CRW]] are defined.\n2. Let $\\varepsilon \u003e 0$. By the [[uniform continuity theorem|0W0TZDRF9C]], $f$ is [[uniformly continuous|0A3CK6DVVB]] on $[a, b]$. So there is $\\delta \u003e 0$ with $|f(x) - f(y)| \u003c \\frac{\\varepsilon}{2(b - a)}$ for all $x, y \\in [a, b]$ with $|x - y| \u003c \\delta$.\n3. By the [[Archimedean property|0A3GWZ86AY]], take $n$ with $n \\delta \u003e b - a$. Let $P$ have the points $x_i = a + i (b - a)/n$ for $i = 0, \\dots, n$. Each subinterval has length $(b - a)/n \u003c \\delta$.\n4. Fix $i$. The restriction of $f$ to $[x_{i-1}, x_i]$ is continuous. By the [[extreme value theorem|0WS1A8E2P9]], there are $c_i, d_i \\in [x_{i-1}, x_i]$ with $f(c_i) \\le f(x) \\le f(d_i)$ for all $x$ there.\n5. So $M_i = f(d_i)$ and $m_i = f(c_i)$. Since $|d_i - c_i| \\le (b - a)/n \u003c \\delta$, step 2 gives $M_i - m_i \u003c \\frac{\\varepsilon}{2(b - a)}$.\n6. Hence\n   $$U(f, P) - L(f, P) = \\sum_{i=1}^{n} (M_i - m_i)(x_i - x_{i-1}) \u003c \\frac{\\varepsilon}{2(b - a)} \\sum_{i=1}^{n} (x_i - x_{i-1}) = \\frac{\\varepsilon}{2} \u003c \\varepsilon.$$\n7. Such a partition exists for each $\\varepsilon \u003e 0$. By [[Riemann's criterion|017DH8WXTE]], $f$ is Riemann integrable on $[a, b]$.\n\n## Idea\n\nUniform continuity gives one $\\delta$ that works on the whole interval. On every subinterval shorter than $\\delta$, the largest and smallest values of $f$ differ by a tiny amount. So the upper and lower sums differ by that amount times $b - a$.\n",
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