support of integrable function is $\sigma$-finiteSupportOfIntegrableFunctionIsSigmaFinite2008-12-26 16:40:382008-12-26 16:40:38Theoremsummable function$L^p$ functions have $\sigma$-finite support% this is the default PlanetMath preamble. as your knowledge
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{\bf 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 \PMlinkname{$\sigma$-finite}{SigmaFinite}.
It follows easily from this result that any function in an \PMlinkname{$L^p$-space}{LpSpace}, with $1 \leq p < \infty$, must have $\sigma$-finite support.
{\bf \emph{\PMlinkescapetext{Proof}:}} 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
\begin{align*}
\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.
\end{align*}
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. $\square$