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  "body": "\n## Definition\n\nLet $(X, \\Sigma)$ be a measurable space and let $\\mu$ and $\\nu$ be [[σ-finite measures|WEBBT18D4X]] on $\\Sigma$ with $\\nu$ [[absolutely continuous|P0P21KDV6E]] with respect to $\\mu$.\nBy the [[Radon–Nikodym theorem|1ZSZS5G1Q4]] there is a measurable $f \\colon X \\to [0, \\infty)$ with $\\nu(A) = \\int_A f \\, d\\mu$ for all $A \\in \\Sigma$, and $f$ is unique [[$\\mu$-almost everywhere|RARTMA0ZEC]].\nThe *Radon–Nikodym derivative* of $\\nu$ with respect to $\\mu$ is this function $f$, determined up to $\\mu$-null sets.\nIt is written $\\dfrac{d\\nu}{d\\mu}$.\n\n## Remarks\n\nExample: let $\\mu$ be Lebesgue measure on $\\R$ and $\\nu(A) = \\int_A 2x \\, \\mathbf{1}_{[0,1]}(x) \\, dx$.\nThen $\\nu$ is a [[measure with density|Y98C5T956Q]], and $\\frac{d\\nu}{d\\mu}(x) = 2x$ for $x \\in [0,1]$ and $0$ elsewhere, up to a null set.\n\nNon-example: the Dirac measure $\\delta_0$ has no Radon–Nikodym derivative with respect to Lebesgue measure $\\lambda$.\nIndeed $\\lambda(\\{0\\}) = 0$ but $\\delta_0(\\{0\\}) = 1$, so $\\delta_0$ is not absolutely continuous with respect to $\\lambda$.\n\nFor every measurable $g \\colon X \\to [0, \\infty]$,\n$$\\int_X g \\, d\\nu = \\int_X g \\, \\frac{d\\nu}{d\\mu} \\, d\\mu.$$\nIf $\\nu \\ll \\mu \\ll \\rho$ with all three σ-finite, then the chain rule $\\frac{d\\nu}{d\\rho} = \\frac{d\\nu}{d\\mu} \\cdot \\frac{d\\mu}{d\\rho}$ holds $\\rho$-almost everywhere.\n",
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