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  "body": "\n## Counterexample\n\nLet $S \\subseteq [0, 1]$ be the [[Smith–Volterra–Cantor set|HZTRH10Z8Y]].\nIt is closed, contains no interval, and has positive [[Lebesgue measure|2Z2QVWZ262]].\nLet $g(t) = t^2 \\sin(1/t)$ for $t \\ne 0$ and $g(0) = 0$.\nThen $g$ is differentiable on $\\R$, with $g'(t) = 2t \\sin(1/t) - \\cos(1/t)$ for $t \\ne 0$ and $g'(0) = 0$.\n\nThe complement $[0, 1] \\setminus S$ is a union of disjoint open intervals.\nOn each such interval $(a, b)$, let $t_0$ be the largest $t \\in (0, (b - a)/2]$ with $g'(t) = 0$.\nDefine $V(x) = g(x - a)$ for $a \u003c x \\le a + t_0$, $V(x) = g(t_0)$ for $a + t_0 \\le x \\le b - t_0$, and $V(x) = g(b - x)$ for $b - t_0 \\le x \u003c b$.\nSet $V(x) = 0$ for $x \\in S$.\n\nThen $V$ is differentiable at every point of $[0, 1]$, and its derivative $V'$ is bounded.\nAt each point of $S$, one has $V'(x) = 0$, but $V'$ is not continuous there.\nSo $V'$ is discontinuous on a set of positive measure, and by [[Lebesgue's criterion|0565EGK6XP]] it is not [[Riemann integrable|0N1JQXZXHC]].\n\n## Remarks\n\nThe number $t_0$ exists because $g'$ has zeros at points of $(0, \\varepsilon)$ for every $\\varepsilon \u003e 0$.\nNear an endpoint $a$, the derivative $V'(x) = g'(x - a)$ keeps taking values close to $1$ and close to $-1$.\nEvery point of $S$ is a limit of such endpoints, which gives the discontinuity of $V'$ on $S$.\nThe example shows that a function can have a bounded [[derivative|0F1WR9C83K]] that has no Riemann integral.\nSo the [[fundamental theorem of calculus|0NHGY0N0JB]] in the Riemann form $\\int_a^b f' = f(b) - f(a)$ needs the hypothesis that $f'$ is Riemann integrable.\nThe Lebesgue integral repairs this: a bounded derivative is Lebesgue integrable, and the formula holds.\n\n## Explanations\n\n### L1 human\n\nVolterra's function is differentiable everywhere, and its slope stays bounded.\nYet the slope jumps around wildly on a fat Cantor set, a set with no intervals but with positive length.\nBecause the bad set is so large, the Riemann integral cannot handle the derivative.\nThe function shows that the Riemann integral is too weak to undo every differentiation.\n",
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