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  "body": "\n## Definition\n\nLet $(X, \\mathcal{F})$ be a [[measurable space|N4KDT83V3S]], and let $\\nu \\colon \\mathcal{F} \\to [0, \\infty)$ be a [[fuzzy measure|CRFQEGJDA9]].\nLet $f \\colon X \\to [0, \\infty)$ be [[measurable|YYNCJSR0PJ]].\nThe *Choquet integral* of $f$ with respect to $\\nu$ is\n$$(C)\\!\\int f \\, d\\nu = \\int_0^\\infty \\nu(\\set{x \\in X : f(x) \\ge t}) \\, dt.$$\nThe function $t \\mapsto \\nu(\\set{f \\ge t})$ is non-increasing, so the right side is a well-defined Lebesgue integral with value in $[0, \\infty]$.\n\n## Remarks\n\nIf $f$ takes the values $0 = a_0 \u003c a_1 \u003c \\cdots \u003c a_k$, then $(C)\\!\\int f \\, d\\nu = \\sum_{i=1}^{k} (a_i - a_{i-1}) \\, \\nu(\\set{f \\ge a_i})$.\nIf $\\nu$ is a measure, the Choquet integral equals the Lebesgue integral, by the [[layer cake representation|H78RXJC1K1]].\nExample: let $X = \\set{1, 2}$ with all subsets, $\\nu(\\set{1}) = \\nu(\\set{2}) = 0$ and $\\nu(X) = 1$.\nThen $\\nu$ is monotone with $\\nu(\\emptyset) = 0$, and the indicators $1_{\\set{1}}$ and $1_{\\set{2}}$ have Choquet integral $0$.\nTheir sum $1_X$ has Choquet integral $\\int_0^1 \\nu(X) \\, dt = 1$, so the integral is not additive.\nNon-example: if $f$ is not measurable, some set $\\set{f \\ge t}$ lies outside $\\mathcal{F}$, so $\\nu$ is not defined on it and the formula makes no sense.\nSome authors use $\\set{f \u003e t}$; this changes the integrand at only countably many $t$, so the integral is the same.\n",
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