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  "body": "\n## Definition\n\nLet $(X, \\Sigma)$ be a [[measurable space|N4KDT83V3S]] and let $\\mu$ and $\\nu$ be [[measures|PR11BZQQPG]] on it.\nThe measure $\\nu$ is *absolutely continuous* with respect to $\\mu$, written $\\nu \\ll \\mu$, if for every $A \\in \\Sigma$,\n$$\\mu(A) = 0 \\implies \\nu(A) = 0.$$\nFor a [[signed|VG2P7R91Z7]] or complex measure $\\nu$, one says $\\nu \\ll \\mu$ if its [[total variation|EPBND51CR5]] satisfies $\\abs{\\nu} \\ll \\mu$.\n\n## Remarks\n\nExample: let $f \\colon X \\to [0, \\infty]$ be measurable and $\\nu(A) = \\int_A f \\, d\\mu$.\nIf $\\mu(A) = 0$, then $f \\mathbf{1}_A = 0$ $\\mu$-almost everywhere, so $\\nu(A) = 0$; hence $\\nu \\ll \\mu$.\n\nNon-example: on $\\R$ with the Borel sets, the Dirac measure $\\delta_0$ is not absolutely continuous with respect to Lebesgue measure $\\lambda$.\nThe set $A = \\set{0}$ has $\\lambda(A) = 0$ but $\\delta_0(A) = 1$.\n\nIf $\\nu$ is finite, then $\\nu \\ll \\mu$ holds if and only if for every $\\varepsilon \u003e 0$ there is $\\delta \u003e 0$ with $\\nu(A) \u003c \\varepsilon$ whenever $\\mu(A) \u003c \\delta$.\nIf $\\mu$ and $\\nu$ are σ-finite, the [[Radon–Nikodym theorem|1ZSZS5G1Q4]] shows that $\\nu \\ll \\mu$ if and only if $\\nu$ has a density $f$ with respect to $\\mu$.\n",
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