support of integrable function with respect to counting measure is countable

Let $(X,\\mathfrak{B},\\mu)$ be a measure space with $\\mu$ the counting measure. If $f$ is an integrable function, $\\displaystyle\\int_{X}f\\,d\\mu<\\infty$ , then it has countable (http://planetmath.org/Countable) support (http://planetmath.org/Support6).

Proof.

WLOG, we assume that $f$ is real valued and is nonnegative. Let $S_{0}$ denote the preimage of the interval $[1,\\infty)$ and, for every positive integer $n$ , let $S_{n}$ denote the preimage of the interval $\\left[\\frac{1}{n+1},\\frac{1}{n}\\right)$ . Since the integral of $f$ is bounded, each $S_{n}$ can be at most finite. Taking the union of all the $S_{n}$ , we get the support $\\displaystyle\\operatorname{supp}f=\\bigcup_{n=0}^{\\infty}S_{n}$ . Thus, $\\operatorname{supp}f$ is a union of countably many finite sets and hence is countable.∎

单词	SupportOfIntegrableFunctionWithRespectToCountingMeasureIsCountable
释义	support of integrable function with respect to counting measure is countable Let $(X,\\mathfrak{B},\\mu)$ be a measure space with $\\mu$ the counting measure. If $f$ is an integrable function, $\\displaystyle\\int_{X}f\\,d\\mu<\\infty$ , then it has countable (http://planetmath.org/Countable) support (http://planetmath.org/Support6). Proof. WLOG, we assume that $f$ is real valued and is nonnegative. Let $S_{0}$ denote the preimage of the interval $[1,\\infty)$ and, for every positive integer $n$ , let $S_{n}$ denote the preimage of the interval $\\left[\\frac{1}{n+1},\\frac{1}{n}\\right)$ . Since the integral of $f$ is bounded, each $S_{n}$ can be at most finite. Taking the union of all the $S_{n}$ , we get the support $\\displaystyle\\operatorname{supp}f=\\bigcup_{n=0}^{\\infty}S_{n}$ . Thus, $\\operatorname{supp}f$ is a union of countably many finite sets and hence is countable.∎
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