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  "body": "\n## Statement\n\nLet $a \u003c b$ be real numbers, and let integrals on $[a, b]$ be taken with respect to [[Lebesgue measure|2Z2QVWZ262]].\n\n1. Let $g \\colon [a, b] \\to \\R$ be [[integrable|X7GTF3D98F]] and let $F(x) = \\int_a^x g(t) \\, dt$ for $x \\in [a, b]$. Then $F$ is [[absolutely continuous|ZT6F4C2YHK]], and $F$ is [[differentiable|0V1EJCT8G6]] with $F'(x) = g(x)$ for [[almost every|RARTMA0ZEC]] $x \\in [a, b]$.\n2. Let $f \\colon [a, b] \\to \\R$ be absolutely continuous. Then $f$ is differentiable at almost every point of $[a, b]$, the derivative $f'$ is integrable on $[a, b]$, and $f(x) = f(a) + \\int_a^x f'(t) \\, dt$ for every $x \\in [a, b]$.\n\n## Remarks\n\nPart 1 follows from the Lebesgue differentiation theorem in dimension one; absolute continuity of $F$ comes from the absolute continuity of the integral.\nFor part 2, an absolutely continuous function has bounded variation, so it is a difference of two increasing functions and is differentiable almost everywhere with $f' \\in L^1$.\nThe function $h = f - f(a) - \\int_a^x f'$ is then absolutely continuous with $h' = 0$ almost everywhere, and an absolutely continuous function with zero derivative almost everywhere is constant.\n\nExample: $f(x) = \\abs{x}$ on $[-1, 1]$ is Lipschitz, hence absolutely continuous.\nIts derivative is $-1$ on $(-1, 0)$ and $1$ on $(0, 1)$, undefined at $0$, and $\\abs{x} = 1 + \\int_{-1}^{x} f'(t) \\, dt$ for every $x$.\n\nAbsolute continuity cannot be weakened to continuity in part 2.\nThe [[Cantor function|NPBYYK63FE]] $c$ is continuous and increasing on $[0, 1]$ with $c' = 0$ almost everywhere, but $c(1) - c(0) = 1 \\ne 0 = \\int_0^1 c'(t) \\, dt$.\n\n## Explanations\n\n### L1 human\n\nIntegration and differentiation undo each other also for the Lebesgue integral, provided the functions are of the right kind.\nIntegrating an integrable function and then differentiating gives back the function, except on a set of length zero.\nDifferentiating and then integrating gives back the function exactly when the function is absolutely continuous.\n\n### L2 human\n\nThe theorem shows that the absolutely continuous functions on $[a, b]$ are exactly the indefinite Lebesgue integrals.\nIt is the one-dimensional prototype of Sobolev spaces: absolutely continuous functions are the functions in $W^{1,1}(a, b)$, up to the choice of representative.\nIn the language of measures, it says that the Lebesgue–Stieltjes measure $df$ of an absolutely continuous $f$ has density $f'$ with respect to Lebesgue measure.\n",
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