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        "quote": "> [!proof]- 证明\n>\n> 若 $f ( x ) = 0 { \\mathrm { ~ a . e . } }$ ，则令 $\\begin{array} { r } { h = \\sum _ { i } a _ { i } \\chi _ { A _ { i } } \\geq 0 } \\end{array}$ 为一简单函数，使得 $h \\leq f$ 每个集合 $A _ { i }$ 的测度必为零，否则 $f ( x ) > 0$ 在一个正测度集合上。因此 $\\textstyle \\int f \\mathrm { d } \\mu = \\operatorname* { s u p } \\int h \\mathrm { d } \\mu = 0 .$\n>\n> 反之，假设 $\\textstyle \\int f \\mathrm { d } \\mu = 0$ 。令 $E _ { n } = \\{ x \\mid f ( x ) \\geq 1 / n \\}$ 。这些构成一列递增的可测集， $E _ { n + 1 } \\supset E _ { n }$ ，且\n>\n> $$\n> f \\geq \\frac {1}{n} \\chi_ {A _ {n}}.\n> $$\n>\n> 因此\n>\n> $$\n> \\int \\frac {1}{n} \\chi_ {A _ {n}} \\mathrm{d} \\mu = \\frac {1}{n} \\mu (A _ {n}) \\leq \\int f \\mathrm{d} \\mu = 0,\n> $$\n>\n> 这只有在 $\\textstyle \\mu ( A _ { n } ) = 0$ 时才可能。由<a href=\"02-measure-spaces#theorem-11-3\" class=\"cross-reference\" id=\"xref-03-lebesgue-integration-8\" data-reference=\"theorem-11-3\">定理 11.3</a> 可得\n>\n> $$\n> \\mu \\big (\\{x \\mid f (x) > 0 \\} \\big) = \\mu \\left(\\bigcup_ {n = 1} ^ {\\infty} A _ {n}\\right) = \\lim _ {n \\rightarrow \\infty} \\mu \\left(A _ {n}\\right) = 0.\n> $$\n>\n> 因此 $f ( x ) = 0$ 几乎处处成立。\n>\n积分可推广到取正值或负值的实值函数。若可测函数 $f$ 的正部与负部 $f ^ { + }$ 与 $f ^ { - }$ 均关于测度 $\\mu$ 可积，则称该函数关于该测度 <u class=\"concept-term\" data-concept-identity=\"urn:uuid:7dc71b7b-0612-46b1-88a7-e838f7ee6134\" data-concept-route=\"/concept-identities#concept-7dc71b7b-0612-46b1-88a7-e838f7ee6134\" id=\"first-929d60427013a6d10773-zh\">可积(integrable)</u>。此时 $f$ 的积分定义为",
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