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  "body": "\n## Statement\n\nLet $(X, \\Sigma, \\mu)$ be a measure space and let $f$ be a real or complex [[integrable function|X7GTF3D98F]] on $X$.\nThen for every $\\varepsilon \u003e 0$ there is $\\delta \u003e 0$ with the following property.\nFor every $A \\in \\Sigma$ with $\\mu(A) \u003c \\delta$,\n$$\\int_A \\abs{f} \\, d\\mu \u003c \\varepsilon,$$\nwhere the integral is an [[integral of a non-negative function|DG76G5Z6XW]].\n\n## Remarks\n\nProof idea: let $f_N = \\min(\\abs{f}, N)$.\nBy the [[monotone convergence theorem|G2S6H789RG]], $\\int (\\abs{f} - f_N) \\, d\\mu \\to 0$, so one can fix $N$ with $\\int (\\abs{f} - f_N) \u003c \\varepsilon / 2$.\nThen $\\int_A \\abs{f} \\le \\varepsilon / 2 + N \\mu(A)$, and $\\delta = \\varepsilon / (2N)$ works.\n\nExample: on $[a, b]$ with Lebesgue measure, the function $F(x) = \\int_a^x f(t) \\, dt$ is [[absolutely continuous|5NB63N9W3P]].\nIndeed, a finite union of disjoint intervals of total length less than $\\delta$ has measure less than $\\delta$.\n\nIntegrability is needed.\nFor $f(x) = 1/x$ on $(0, 1]$ with Lebesgue measure, the set $(0, \\delta / 2)$ has measure less than $\\delta$ but $\\int_0^{\\delta/2} dx / x = \\infty$.\n\n## Explanations\n\n### L1 human\n\nAn integrable function cannot hide a lot of area on a tiny set.\nIf the set is small enough, the area of the function over it is as small as one wants.\n\n### L2 human\n\nIn the language of measures, $\\nu(A) = \\int_A \\abs{f} \\, d\\mu$ defines a finite measure that is absolutely continuous with respect to $\\mu$, in the $\\varepsilon$-$\\delta$ sense.\nThe Radon–Nikodym theorem gives a converse: every finite measure absolutely continuous with respect to a σ-finite measure $\\mu$ has this form.\nThe uniform version of this property, for a whole family of functions, is part of uniform integrability.\n",
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