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  "body": "\n## Definition\n\nLet $(X, \\Sigma, \\mu)$ be a [[measure space|P15VQS5DJ9]].\nLet $s = \\sum_{i=1}^{n} a_i \\mathbf{1}_{A_i}$ be a non-negative [[simple function|7WMBX7M4P6]], with $a_i \\in [0, \\infty)$ and pairwise disjoint $A_i \\in \\Sigma$.\nIts integral is\n$$\\int_X s \\, d\\mu = \\sum_{i=1}^{n} a_i \\, \\mu(A_i),$$\nwith the convention $0 \\cdot \\infty = 0$.\nLet $f \\colon X \\to [0, \\infty]$ be a [[measurable function|YYNCJSR0PJ]].\nIts *integral* is the [[supremum|0BTMQ04556]], in the [[extended real line|0GSZ7B0CGR]],\n$$\\int_X f \\, d\\mu = \\sup \\set{\\int_X s \\, d\\mu : s \\text{ simple}, \\ 0 \\le s \\le f}.$$\n\n## Remarks\n\nExample: let $X = \\N$ with the counting measure and $f(n) = 2^{-n}$ for $n \\ge 1$, $f(0) = 1$.\nThe simple functions $s_N = f \\cdot \\mathbf{1}_{\\set{0, \\dots, N}}$ lie below $f$, and their integrals $\\sum_{n=0}^{N} 2^{-n}$ increase to $2$.\nEvery simple $s \\le f$ has integral at most $\\sum_n f(n) = 2$, so $\\int f \\, d\\mu = 2$.\n\nNon-example: the formula does not apply to $g(x) = \\sin x$ on $\\R$, because $g$ takes negative values.\nSuch functions need the splitting $g = g^+ - g^-$ of the Lebesgue integral.\n\nThe value of $\\int_X s \\, d\\mu$ for a simple function does not depend on the way $s$ is written.\nFor $A \\in \\Sigma$ one writes $\\int_A f \\, d\\mu = \\int_X f \\mathbf{1}_A \\, d\\mu$.\nIf $f = g$ almost everywhere, then $\\int_X f \\, d\\mu = \\int_X g \\, d\\mu$.\n",
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