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  "body": "\n## Definition\n\nLet $\\mathbb{D}$ be the open [[unit disc|8G5ABXRK85]] and let $f \\colon \\mathbb{D} \\to \\C$ be a bounded [[holomorphic function|0DG0Z72CQD]].\nBy [[Fatou's theorem|Q2N917JQ6A]], the radial limit $f^*(e^{it}) = \\lim_{r \\to 1^-} f(r e^{it})$ exists for [[almost every|RARTMA0ZEC]] $t \\in \\R$.\nThe function $f$ is an *inner function* if $\\abs{f^*(e^{it})} = 1$ for almost every $t$.\n\n## Remarks\n\nAn inner function satisfies $\\abs{f(z)} \\le 1$ on $\\mathbb{D}$, because $f$ is the Poisson integral of $f^*$.\n\nExample: let $S(z) = \\exp\\left(-\\frac{1 + z}{1 - z}\\right)$.\nFor $z = r e^{it}$ one has $\\operatorname{Re} \\frac{1 + z}{1 - z} = \\frac{1 - r^2}{\\abs{1 - z}^2} \u003e 0$, so $\\abs{S(z)} \u003c 1$ and $S$ is bounded.\nFor $t \\notin 2\\pi\\Z$ this real part tends to $0$ as $r \\to 1$, so $\\abs{S^*(e^{it})} = 1$ and $S$ is inner.\nIt has no zeros, so it is not a [[Blaschke product|9N6R5ADJXT]].\n\nNon-example: $f(z) = (1 + z)/2$ is bounded and holomorphic, but $\\abs{f^*(e^{it})} = \\abs{\\cos(t/2)} \u003c 1$ for $t \\notin \\pi\\Z$.\n\nEvery inner function has the form $c B S_\\mu$, where $\\abs{c} = 1$ and $B$ is a Blaschke product.\nHere $S_\\mu(z) = \\exp\\left(-\\int \\frac{e^{it} + z}{e^{it} - z} \\, d\\mu(t)\\right)$, and $\\mu$ is a finite positive Borel measure on the circle that is singular to Lebesgue measure.\nThe factor $S_\\mu$ is called a singular inner function.\n",
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