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  "body": "\n## Definition\n\nLet $(X, \\Sigma, \\mu)$ be a [[measure space|P15VQS5DJ9]] and let $0 \u003c p \u003c \\infty$.\nFor a [[measurable function|YYNCJSR0PJ]] $f$ on $X$ with real or complex values, let\n$$\\norm{f}_{p,\\infty} = \\sup_{t \u003e 0} \\, t \\, \\mu(\\set{x \\in X : \\abs{f(x)} \u003e t})^{1/p}.$$\nThe *weak $L^p$ space* $L^{p,\\infty}(\\mu)$ is the set of measurable functions $f$ with $\\norm{f}_{p,\\infty} \u003c \\infty$, modulo equality [[almost everywhere|RARTMA0ZEC]].\n\n## Remarks\n\nExample: on $\\R^n$ with Lebesgue measure, $f(x) = \\abs{x}^{-n/p}$ lies in $L^{p,\\infty}$.\nThe set $\\set{\\abs{f} \u003e t}$ is the open ball of radius $t^{-p/n}$, with measure $\\omega_n t^{-p}$, where $\\omega_n$ is the volume of the unit ball.\nSo $\\norm{f}_{p,\\infty} = \\omega_n^{1/p}$, while $\\int \\abs{f}^p \\, dx = \\int \\abs{x}^{-n} \\, dx = \\infty$.\n\nNon-example: on $(0, 1)$ with Lebesgue measure, $g(x) = x^{-2}$ is not in $L^{1,\\infty}$.\nFor $t \u003e 1$, $\\set{g \u003e t} = (0, t^{-1/2})$, so $t \\, \\mu(\\set{g \u003e t}) = t^{1/2}$, which is unbounded.\n\nChebyshev's inequality $t^p \\mu(\\set{\\abs{f} \u003e t}) \\le \\int \\abs{f}^p \\, d\\mu$ gives $\\norm{f}_{p,\\infty} \\le \\norm{f}_p$, so $L^p \\subseteq L^{p,\\infty}$.\nThe map $\\norm{\\cdot}_{p,\\infty}$ is a quasinorm, not a norm.\nThe triangle inequality holds only with a constant: $\\norm{f + g}_{p,\\infty} \\le \\max(2, 2^{1/p}) (\\norm{f}_{p,\\infty} + \\norm{g}_{p,\\infty})$.\nIt follows from $\\set{\\abs{f + g} \u003e t} \\subseteq \\set{\\abs{f} \u003e t/2} \\cup \\set{\\abs{g} \u003e t/2}$.\n",
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