# Lebesgue integral over a subset of the measure space

Let $(X,\mathfrak{B},\mu)$ be a measure space and $A\in\mathfrak{B}$.

Let $s\colon X\to[0,\infty]$ be a simple function. Then $\displaystyle\int_{A}s\,d\mu$ is defined as $\displaystyle\int_{A}s\,d\mu:=\int_{X}\chi_{A}s\,d\mu$, where $\chi_{A}$ denotes the characteristic function of $A$.

Let $f\colon X\to[0,\infty]$ be a measurable function and
$S=\{s\colon X\to[0,\infty]~{}~{}|~{}~{}s\text{ is a simple function and }s\leq f\}$. Then $\displaystyle\int_{A}f\,d\mu$ is defined as $\displaystyle\int_{A}f\,d\mu:=\sup_{s\in S}\int_{A}s\,d\mu$.

By the properties of the Lebesgue integral of nonnegative measurable functions (property 3), we have that $\displaystyle\int_{A}f\,d\mu=\int_{X}\chi_{A}f\,d\mu$.

Let $f\colon X\to[-\infty,\infty]$ be a measurable function such that not both of $\displaystyle\int_{A}f^{+}\,d\mu$ and $\displaystyle\int_{A}f^{-}\,d\mu$ are infinite. (Note that $f^{+}$ and $f^{-}$ are defined in the entry Lebesgue integral.) Then $\displaystyle\int_{A}f\,d\mu$ is defined as $\displaystyle\int_{A}f\,d\mu:=\int_{A}f^{+}\,d\mu-\int_{A}f^{-}\,d\mu$.

By the properties of the Lebesgue integral of Lebesgue integrable functions (property 3), we have that $\displaystyle\int_{A}f\,d\mu=\int_{X}\chi_{A}f\,d\mu$.

Title Lebesgue integral over a subset of the measure space LebesgueIntegralOverASubsetOfTheMeasureSpace 2013-03-22 16:13:54 2013-03-22 16:13:54 Wkbj79 (1863) Wkbj79 (1863) 7 Wkbj79 (1863) Definition msc 26A42 msc 28A25