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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 *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. $\\mu(T^{-1}(B)) = \\nu(B)$ for every $B \\in \\mathcal{B}$.\n\n## Remarks\n\nCondition 2 says that the [[pushforward measure|3N66TM3TSQ]] $T_* \\mu$ equals $\\nu$.\nWhen $X = Y$ and $\\mu = \\nu$, one says that $T$ preserves $\\mu$, and $(X, \\mathcal{A}, \\mu, T)$ is a measure-preserving system.\nExample: the doubling map $T(x) = 2x \\bmod 1$ on $[0,1)$ with Lebesgue measure $\\lambda$.\nFor $0 \\le a \u003c b \\le 1$ one has $T^{-1}([a, b)) = [a/2, b/2) \\cup [(a+1)/2, (b+1)/2)$, of measure $b - a$.\nThe intervals form a $\\pi$-system that generates the Borel sets, so condition 2 holds for all Borel sets.\nNon-example: $T(x) = x^2$ on $[0,1]$ with Lebesgue measure is measurable but $\\lambda(T^{-1}([0, 1/4])) = \\lambda([0, 1/2]) = 1/2 \\ne 1/4$.\n",
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