Lebesgue integral over a subset of the measure space

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

Let $s:X\to [0,\mathrm{\infty }]$ be a simple function. Then ${\displaystyle {\int }_A}s𝑑\mu$ is defined as ${\displaystyle {\int }_A}s𝑑\mu :={\displaystyle {\int }_X}{\chi }_{A}s𝑑\mu$, where ${\chi }_A$ denotes the characteristic function of $A$.

Let $f:X\to [0,\mathrm{\infty }]$ be a measurable function and
$S=\{s:X\to [0,\mathrm{\infty }]|s\text{ is a simple function and }s\le f\}$. Then ${\displaystyle {\int }_A}f𝑑\mu$ is defined as ${\displaystyle {\int }_A}f𝑑\mu :=\underset{s\in S}{sup}{\displaystyle {\int }_A}s𝑑\mu$.

By the properties of the Lebesgue integral of nonnegative measurable functions (property 3), we have that ${\displaystyle {\int }_A}f𝑑\mu ={\displaystyle {\int }_X}{\chi }_{A}f𝑑\mu$.

Let $f:X\to [-\mathrm{\infty },\mathrm{\infty }]$ 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 ${\displaystyle {\int }_A}f𝑑\mu$ is defined as ${\displaystyle {\int }_A}f𝑑\mu :={\displaystyle {\int }_A}{f}^{+}𝑑\mu -{\displaystyle {\int }_A}{f}^{-}𝑑\mu$.

By the properties of the Lebesgue integral of Lebesgue integrable functions (property 3), we have that ${\displaystyle {\int }_A}f𝑑\mu ={\displaystyle {\int }_X}{\chi }_{A}f𝑑\mu$.