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. ∎

Title	support of integrable function with respect to counting measure is countable
Canonical name	SupportOfIntegrableFunctionWithRespectToCountingMeasureIsCountable
Date of creation	2013-03-22 14:59:34
Last modified on	2013-03-22 14:59:34
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	11
Author	Wkbj79 (1863)
Entry type	Result
Classification	msc 28A12
Related topic	UncountableSumsOfPositiveNumbers
Related topic	SupportOfIntegrableFunctionIsSigmaFinite