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  "body": "\n## Definition\n\nLet $K$ be a closed subset of $\\R$.\nLet $\\mu$ be a positive Borel [[measure|PR11BZQQPG]] on $\\R$ with support in $K$, such that every [[moment|01XAHPADBA]] $m_n = \\int x^n \\, d\\mu(x)$, $n \\ge 0$, is finite.\nThe [[moment problem|CS7TWHY7JM]] for $\\mu$ on $K$ is *indeterminate* if there is a positive Borel measure $\\nu \\ne \\mu$ with the following properties.\nThe support of $\\nu$ lies in $K$, and $\\int x^n \\, d\\nu(x) = m_n$ for all $n \\ge 0$.\nEquivalently, it is not [[determinate|H0GY2J14CA]].\n\n## Remarks\n\nExample: the log-normal density $f(x) = (x \\sqrt{2\\pi})^{-1} e^{-(\\ln x)^2/2}$ on $(0, \\infty)$.\nThe substitution $x = e^{u}$ shows $\\int_0^\\infty x^n f(x) \\sin(2\\pi \\ln x) \\, dx = e^{n^2/2} \\int_{\\R} \\frac{e^{-v^2/2}}{\\sqrt{2\\pi}} \\sin(2\\pi v) \\, dv = 0$, with $v = u - n$, since the last integrand is odd.\nSo $\\nu$ with density $f(x)(1 + \\sin(2\\pi \\ln x))$ is a different positive measure on $[0, \\infty)$ with the same moments.\nNon-example: the uniform distribution on $[0, 1]$ is not indeterminate, since a measure on a compact set is determined by its moments.\nEvery measure with finitely many atoms is determinate, so an indeterminate measure has infinite support; its support must also be unbounded.\n",
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