support of integrable function is $\sigma$ -finite

Theroem - Let $(X,\mathcal{B},\mu)$ be a measure space and $f:X\to\mathbb{C}$ a measurable function. If $f$ is integrable, then the support of $f$ is $\sigma$ -finite (http://planetmath.org/SigmaFinite).

It follows easily from this result that any function in an $L^{p}$ -space (http://planetmath.org/LpSpace), with $1\leq p<\infty$ , must have $\sigma$ -finite support.

: Let $A_{0}:=[1,\infty[$ , and for each $n\in\mathbb{N}$ let $A_{n}:=[\frac{1}{n+1},\frac{1}{n}[$ . Since $f$ is integrable, we must necessarily have $\mu\big{(}|f|^{-1}(A_{n})\big{)}<\infty$ for each $n\in\mathbb{N}\cup\{0\}$ , because

\displaystyle\mu\big{(}|f|^{-1}(A_{n})\big{)}\cdot\frac{1}{n+1}\leq\int_{|f|^{% -1}(A_{n})}|f|\;d\mu\leq\int_{X}|f|\;d\mu<\infty.

Since $f$ and $|f|$ have the same support, and the the support of the latter is $\displaystyle\mathrm{supp}\,|f|=\bigcup_{n=0}^{\infty}|f|^{-1}(A_{n})$ , it follows that the support of $f$ is $\sigma$ -finite. $\square$

Title	support of integrable function is $\sigma$ -finite
Canonical name	SupportOfIntegrableFunctionIssigmafinite
Date of creation	2013-03-22 18:38:47
Last modified on	2013-03-22 18:38:47
Owner	asteroid (17536)
Last modified by	asteroid (17536)
Numerical id	4
Author	asteroid (17536)
Entry type	Theorem
Classification	msc 26A42
Classification	msc 28A25
Related topic	SupportOfIntegrableFunctionWithRespectToCountingMeasureIsCountable
Defines	$L^{p}$ functions have $\sigma$ -finite support

support of integrable function is σ-finite

support of integrable function is $\sigma$ -finite