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  "body": "\n## Definition\n\nLet $(X, \\mathcal{A})$ be a measurable space and let $T \\colon X \\to X$ be a [[measurable map|YYNCJSR0PJ]].\nA [[measure|PR11BZQQPG]] $\\mu$ on $(X, \\mathcal{A})$ is *$T$-invariant* if\n$$\\mu(T^{-1}(A)) = \\mu(A) \\quad \\text{for every } A \\in \\mathcal{A}.$$\nLet a family of measurable maps be given, such as a flow $(\\varphi_t)$ or a group action.\nThen $\\mu$ is *invariant* if it is invariant under each map of the family.\n\n## Remarks\n\nEquivalently, the [[pushforward measure|3N66TM3TSQ]] $T_* \\mu$ equals $\\mu$, or $T$ is a [[measure-preserving transformation|556WBDAHNP]] of $(X, \\mathcal{A}, \\mu)$ to itself.\nFor an integrable function $g$, invariance gives $\\int g \\circ T \\, d\\mu = \\int g \\, d\\mu$.\nThe invariant probability measures of $T$ form a convex set, whose extreme points are the ergodic ones.\n\nExample: the doubling map $T(x) = 2x \\bmod 1$ on $[0, 1)$ and Lebesgue measure $\\lambda$.\nFor an interval $[a, b) \\subseteq [0, 1)$, one has $T^{-1}([a, b)) = [a/2, b/2) \\cup [(a+1)/2, (b+1)/2)$, of length $b - a$.\nSince intervals generate the Borel sets and are closed under intersection, $\\lambda$ is $T$-invariant.\n\nNon-example: the map $T(x) = x^2$ on $[0, 1]$ with Lebesgue measure $\\lambda$.\nOne has $T^{-1}([0, 1/4]) = [0, 1/2]$, so $\\lambda(T^{-1}([0, 1/4])) = 1/2 \\ne 1/4$, and $\\lambda$ is not $T$-invariant.\nThe point masses at $0$ and at $1$ are $T$-invariant, since $0$ and $1$ are fixed points.\n",
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