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Homesupport of integrable function with respect to counting measure is countable

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# 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 support.

###### 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. ∎

## Mathematics Subject Classification

28A12*no label found*

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