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  "body": "\n## Statement\n\nLet $(X, \\Sigma, \\mu)$ be a [[measure space|P15VQS5DJ9]] and let $1 \\le p \u003c \\infty$.\nLet $S$ be the set of [[simple functions|7WMBX7M4P6]] $s = \\sum_{i=1}^{n} a_i \\mathbf{1}_{A_i}$ with $A_i \\in \\Sigma$ and $\\mu(A_i) \u003c \\infty$ for each $i$.\nThen $S$ is a [[dense|0BSJSX7ME2]] subset of [[$L^p(\\mu)$|SASP14MA1Z]].\n\n## Remarks\n\nFor $f \\ge 0$ in $L^p$, take simple functions $0 \\le s_n \\le f$ that increase pointwise to $f$.\nEach $s_n$ vanishes outside a set of finite measure, because $s_n^p \\le f^p$ is integrable.\nSince $\\abs{f - s_n}^p \\le f^p$, the [[dominated convergence theorem|BYNV84PF86]] gives $\\norm{f - s_n}_p \\to 0$.\nA general $f$ is split into the positive and negative parts of its real and imaginary parts.\n\nThe hypothesis $p \u003c \\infty$ is necessary.\nOn $\\R$ with [[Lebesgue measure|2Z2QVWZ262]], the constant function $1$ lies in $L^\\infty$, but every $s \\in S$ vanishes outside a set of finite measure, so $\\norm{1 - s}_\\infty \\ge 1$.\nIn $L^\\infty(\\mu)$ the set of all simple functions, without the condition $\\mu(A_i) \u003c \\infty$, is dense.\n\nOn $\\R^n$ the result is the first step towards the density of step functions and of continuous functions with compact support in $L^p$, $1 \\le p \u003c \\infty$.\n\n## Explanations\n\n### L1 human\n\nEvery function in $L^p$ can be approximated as well as one wants by a staircase function that takes only finitely many values.\nThe steps of the staircase sit on sets of finite measure.\nThis lets one prove a fact about all of $L^p$ by checking it on indicator functions of sets and then passing to the limit.\n\n### L2 human\n\nThe standard way to prove an inequality or an identity on $L^p$ is to check it on a dense class and extend it by continuity.\nSimple functions are the first such class; they are used in the proof of the [[dual of Lp|Q1V7ZNS332]] and in interpolation theorems such as the [[Riesz–Thorin theorem|AGRKRG1TMJ]].\nFor $p = \\infty$ the density of all simple functions is a uniform approximation, which is why the finite measure condition must then be dropped.\n",
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