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  "body": "\n## Definition\n\nLet $n \\ge 1$, let $\\lambda$ be [[Lebesgue measure|2Z2QVWZ262]] on $\\R^n$, and let $f \\colon \\R^n \\to \\R$ be locally [[integrable|X7GTF3D98F]].\nFor $x \\in \\R^n$ and $r \u003e 0$, let $B(x, r)$ be the [[open ball|0YE5868HZB]] of centre $x$ and radius $r$.\nThe *Hardy–Littlewood maximal function* of $f$ is the function $Mf \\colon \\R^n \\to [0, \\infty]$ given by\n$$Mf(x) = \\sup_{r \u003e 0} \\frac{1}{\\lambda(B(x, r))} \\int_{B(x, r)} \\abs{f(y)} \\, d\\lambda(y).$$\n\n## Remarks\n\nExample: let $n = 1$ and $f = \\mathbf{1}_{[0, 1]}$, and let $x \u003e 1$.\nFor $r \\ge x$ the average is $1/(2r) \\le 1/(2x)$.\nFor $x - 1 \u003c r \u003c x$ the average is $(r - x + 1)/(2r)$, which increases with $r$ and tends to $1/(2x)$.\nFor $r \\le x - 1$ the average is $0$.\nSo $Mf(x) = 1/(2x)$.\n\nNon-example of integrability: this shows that $Mf$ need not be integrable even when $f$ is, since $\\int_1^\\infty dx/(2x) = \\infty$.\nIn fact $Mf$ is integrable on $\\R^n$ only when $f = 0$ almost everywhere.\n\nVariants use balls containing $x$ instead of balls centred at $x$ (the uncentred maximal function), or cubes instead of balls.\nThey are comparable to $Mf$ up to constants depending on $n$.\n",
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