## You are here

HomeLebesgue integral over a subset of the measure space

## Primary tabs

# 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 $\displaystyle\int_{A}f\,d\mu=\int_{X}\chi_{A}f\,d\mu$.

## Mathematics Subject Classification

26A42*no label found*28A25

*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff
- Corrections