# support of integrable function with respect to counting measure is countable

Let $(X,\U0001d505,\mu )$ be a measure space^{} with $\mu $ the counting measure. If $f$ is an integrable function, $$, 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 $[1,\mathrm{\infty})$ and, for every positive integer $n$, let ${S}_{n}$ denote the preimage of the interval $[\frac{1}{n+1},\frac{1}{n})$. 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 $\mathrm{supp}f={\displaystyle \bigcup _{n=0}^{\mathrm{\infty}}}{S}_{n}$. Thus, $\mathrm{supp}f$ is a union of countably many finite sets^{} and hence is countable.
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Title | support of integrable function with respect to counting measure is countable |
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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 |