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  "body": "\n## Statement\n\nLet $f \\colon (0, \\infty) \\to \\R$.\nCall $f$ completely monotone if $f$ is infinitely differentiable and $(-1)^k f^{(k)}(x) \\ge 0$ for all integers $k \\ge 0$ and all $x \u003e 0$.\nThen $f$ is completely monotone if and only if there is a positive Borel measure $\\mu$ on $[0, \\infty)$ with the following property.\nFor every $x \u003e 0$ the integral below converges and\n$$f(x) = \\int_{[0, \\infty)} e^{-tx} \\, d\\mu(t).$$\n\n## Remarks\n\nSo the completely monotone functions are exactly the [[Laplace transforms|0V49JEZW06]] of positive measures.\nThe \"if\" direction is direct: differentiating under the integral sign gives $(-1)^k f^{(k)}(x) = \\int t^k e^{-tx} \\, d\\mu(t) \\ge 0$.\nThe \"only if\" direction is the hard part; one proof approximates $f$ by sums built from its derivatives and uses a compactness argument for measures.\nExample: $f(x) = 1/x$ is completely monotone on $(0, \\infty)$, and $1/x = \\int_0^\\infty e^{-tx} \\, dt$, so $\\mu$ is Lebesgue measure on $[0, \\infty)$.\nBy $x \\mapsto -x$, the theorem also describes [[absolutely monotonic functions|QQSPWH4NG9]] on $(-\\infty, 0)$ as integrals $\\int e^{tx} \\, d\\mu(t)$.\n\n## Explanations\n\n### L1 human\n\nA function whose derivatives alternate in sign forever looks very special.\nBernstein's theorem says that these functions are exactly the mixtures, with nonnegative weights, of decaying [[exponential functions|0J0FCWV32D]] $e^{-tx}$.\n\n### L2 human\n\nThe theorem links a condition on derivatives with a positivity condition on a measure.\nIt is used in probability, where completely monotone functions are Laplace transforms of distributions on $[0, \\infty)$, and in the theory of positive definite functions.\nIf $f$ is also continuous on $[0, \\infty)$, then $\\mu$ is finite and $\\mu([0, \\infty)) = f(0)$.\n",
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