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. $\\square$

support of integrable function is σ-finite

support of integrable function is $\\sigma$ -finite