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  "body": "\n## Definition\n\nLet $X$ be a [[compact|0TVJZYB6SH]] [[metric space|0VKD1TP7F3]] and let $T \\colon X \\to X$ be a continuous map.\nA [[Borel|DYR1QFNGNJ]] [[probability measure|059TJN0NZ1]] $\\mu$ on $X$ is $T$-[[invariant|B232EZ1FMB]] if $\\mu(T^{-1}A) = \\mu(A)$ for every Borel set $A \\subseteq X$.\nThe map $T$ is *uniquely ergodic* if there is exactly one $T$-invariant Borel probability measure on $X$.\n\n## Remarks\n\nBy the [[Krylov–Bogolyubov theorem|1DHZ81S0YE]] the set of invariant Borel probability measures is never empty, so unique ergodicity asks that it has a single element.\n\nExample: let $X = \\R/\\Z$, let $\\alpha$ be irrational and let $T(x) = x + \\alpha \\bmod 1$.\nIf $\\mu$ is invariant, its Fourier coefficients $\\hat\\mu(h) = \\int e^{-2\\pi i h x} \\, d\\mu(x)$ satisfy $\\hat\\mu(h) = e^{-2\\pi i h \\alpha} \\hat\\mu(h)$.\nFor $h \\ne 0$ the factor $e^{-2\\pi i h \\alpha}$ is not $1$, so $\\hat\\mu(h) = 0$.\nA measure on the circle is determined by its Fourier coefficients, so $\\mu$ is Lebesgue measure, and $T$ is uniquely ergodic.\n\nNon-example: the doubling map $T(x) = 2x \\bmod 1$ on $\\R/\\Z$ is continuous and fixes $0$.\nBoth the point mass at $0$ and Lebesgue measure are invariant, so $T$ is not uniquely ergodic.\n\nEquivalent form: $T$ is uniquely ergodic if and only if, for every continuous $f \\colon X \\to \\R$, the averages $\\frac{1}{N} \\sum_{n=0}^{N-1} f(T^n x)$ converge uniformly in $x$ to a constant.\nThe constant is then $\\int f \\, d\\mu$ for the unique invariant measure $\\mu$.\n",
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