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  "body": "\n## Definition\n\nLet $(X, \\mathcal{A}, \\mu)$ and $(Y, \\mathcal{B}, \\nu)$ be [[measure spaces|P15VQS5DJ9]].\nA map $T \\colon X \\to Y$ is *quasi-measure-preserving* if two conditions hold.\n\n1. $T$ is [[measurable|YYNCJSR0PJ]]: $T^{-1}(B) \\in \\mathcal{A}$ for every $B \\in \\mathcal{B}$.\n2. If $B \\in \\mathcal{B}$ and $\\nu(B) = 0$, then $\\mu(T^{-1}(B)) = 0$.\n\n## Remarks\n\nCondition 2 says that the pushforward $T_* \\mu$ is [[absolutely continuous|P0P21KDV6E]] with respect to $\\nu$, that is, preimages of [[null sets|VQQM4Y550T]] are null sets.\nEvery [[measure-preserving|556WBDAHNP]] map satisfies it.\nExample: $T(x) = 2x$ on $\\R$ with Lebesgue measure $\\lambda$.\nFor a Borel set $B$ one has $\\lambda(T^{-1}(B)) = \\lambda(B)/2$, so null sets pull back to null sets, but $T$ is not measure-preserving.\nNon-example: the constant map $T(x) = 0$ from $[0,1]$ to $\\R$, both with Lebesgue measure, is measurable.\nBut $\\lambda(\\set{0}) = 0$ while $\\lambda(T^{-1}(\\set{0})) = \\lambda([0,1]) = 1$, so condition 2 fails.\n",
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