properties of the Lebesgue integral of nonnegative measurable functions

Theorem.

Let $(X,\mathfrak{B},\mu)$ be a measure space, $f\colon X\to[0,\infty]$ and $g\colon X\to[0,\infty]$ be measurable functions, and $A,B\in\mathfrak{B}$ . Then the following properties hold:

1.

$\displaystyle\int_{A}f\,d\mu\geq 0$
2.

If $f\leq g$ , then $\displaystyle\int_{A}f\,d\mu\leq\int_{A}g\,d\mu$ .
3.

$\displaystyle\int_{A}f\,d\mu=\int_{X}\chi_{A}f\,d\mu$ , where $\chi_{A}$ denotes the characteristic function of $A$
4.

If $A\subseteq B$ , then $\displaystyle\int_{A}f\,d\mu\leq\int_{B}f\,d\mu$ .
5.

If $c\geq 0$ , then $\displaystyle\int_{A}cf\,d\mu=c\int_{A}f\,d\mu$ .
6.

If $\mu(A)=0$ , then $\displaystyle\int_{A}f\,d\mu=0$ .
7.

$\displaystyle\int_{A}(f+g)\,d\mu=\int_{A}f\,d\mu+\int_{A}g\,d\mu$
8.

If $A\cap B=\emptyset$ , then $\displaystyle\int_{A\cup B}f\,d\mu=\int_{A}f\,d\mu+\int_{B}f\,d\mu$ .
9.

If $f=g$ almost everywhere with respect to $\mu$ , then $\displaystyle\int_{A}f\,d\mu=\int_{A}g\,d\mu$ .

Proof.

1.

Let $s$ be a simple function with $0\leq s\leq f$ . Let $\displaystyle s=\sum_{k=1}^{n}c_{k}\chi_{A_{k}}$ for $c_{k}\in[0,\infty]$ and $A_{k}\in\mathfrak{B}$ . Then $\displaystyle\int_{A}s\,d\mu=\sum_{k=1}^{n}c_{k}\mu(A\cap A_{k})\geq 0$ . By definition, $\displaystyle\int_{A}f\,d\mu\geq\int_{A}s\,d\mu$ . It follows that $\displaystyle\int_{A}f\,d\mu\geq 0$ .
2.

Let $s$ be a simple function with $0\leq s\leq f$ . Since $f\leq g$ , $0\leq s\leq g$ . By definition, $\displaystyle\int_{A}s\,d\mu\leq\int_{A}g\,d\mu$ . Since this holds for every simple function $s$ with $0\leq s\leq f$ , it follows that $\displaystyle\int_{A}f\,d\mu\leq\int_{A}g\,d\mu$ .
3.

Let $s$ be a simple function with $0\leq s\leq f$ . Then $0\leq\chi_{A}s\leq\chi_{A}f$ . Let $\displaystyle s=\sum_{k=1}^{n}c_{k}\chi_{A_{k}}$ for $c_{k}\in[0,\infty]$ and $A_{k}\in\mathfrak{B}$ . Then

$\begin{array}[]{ll}\displaystyle\int_{A}s\,d\mu&\displaystyle=\sum_{k=1}^{n}c_% {k}\mu(A\cap A_{k})\\ \\ &\displaystyle=\int_{X}\sum_{k=1}^{n}c_{k}\chi_{A\cap A_{k}}\,d\mu\\ \\ &\displaystyle=\int_{X}\sum_{k=1}^{n}c_{k}\chi_{A}\chi_{A_{k}}\,d\mu\\ \\ &\displaystyle=\int_{X}\chi_{A}\sum_{k=1}^{n}c_{k}\chi_{A_{k}}\,d\mu\\ \\ &\displaystyle=\int_{X}\chi_{A}s\,d\mu\\ \\ &\displaystyle\leq\int_{X}\chi_{A}f\,d\mu.\end{array}$

Thus, $\displaystyle\int_{A}f\,d\mu\leq\int_{X}\chi_{A}f\,d\mu$ .

Let $t$ be a simple function with $0\leq t\leq\chi_{A}f$ . Then $\chi_{A}t=t$ . Thus, $\displaystyle\int_{X}t\,d\mu=\int_{X}\chi_{A}t\,d\mu=\int_{A}t\,d\mu$ . Therefore, $\displaystyle\int_{X}\chi_{A}f\,d\mu=\int_{A}\chi_{A}f\,d\mu$ . Since $\chi_{A}f\leq f$ , $\displaystyle\int_{A}\chi_{A}f\,d\mu\leq\int_{A}f\,d\mu$ by property 2. Hence, $\displaystyle\int_{A}f\,d\mu\leq\int_{X}\chi_{A}f\,d\mu=\int_{A}\chi_{A}f\,d% \mu\leq\int_{A}f\,d\mu$ . It follows that $\displaystyle\int_{A}f\,d\mu=\int_{X}\chi_{A}f\,d\mu$ .
4.

Since $A\subseteq B$ , $\chi_{A}\leq\chi_{B}$ . Thus, $\chi_{A}f\leq\chi_{B}f$ . By property 2, $\displaystyle\int_{X}\chi_{A}f\,d\mu\leq\int_{X}\chi_{B}f\,d\mu$ . By property 3, $\displaystyle\int_{A}f\,d\mu=\int_{X}\chi_{A}f\,d\mu\leq\int_{X}\chi_{B}f\,d% \mu=\int_{B}f\,d\mu$ .
5.

If $c=0$ , then $\displaystyle\int_{A}cf\,d\mu=\int_{A}0\,d\mu=0=0\int_{A}f\,d\mu=c\int_{A}f\,d\mu$ .

If $c>0$ , let $S=\{s\colon X\to[0,\infty]\mid s~{}\text{is simple and }s\leq cf\}$ and

$T=\{t\colon X\to[0,\infty]\mid t~{}\text{is simple and }t\leq f\}$ . Then $\displaystyle\int_{A}cf\,d\mu=\sup_{s\in S}\int_{A}s\,d\mu=\sup_{s\in S}\int_{% A}c\cdot\frac{s}{c}\,d\mu=c\sup_{s\in S}\int_{A}\frac{s}{c}\,d\mu=c\sup_{t\in T% }\int_{A}t\,d\mu=c\int_{A}f\,d\mu$ .
6.

Let $s$ be a simple function with $0\leq s\leq f$ . Let $\displaystyle s=\sum_{k=1}^{n}c_{k}\chi_{A_{k}}$ for $c_{k}\in[0,\infty]$ and $A_{k}\in\mathfrak{B}$ . Then $\displaystyle\int_{A}s\,d\mu=\sum_{k=1}^{n}c_{k}\mu(A\cap A_{k})=\sum_{k=1}^{n% }c_{k}\cdot 0=0$ . Thus, $\displaystyle\int_{A}f\,d\mu=0$ .
7.

Let $\{s_{n}\}$ be a nondecreasing sequence of nonnegative simple functions converging pointwise to $f$ and $\{t_{n}\}$ be a nondecreasing sequence of nonnegative simple functions converging pointwise to $g$ . Then $\{s_{n}+t_{n}\}$ is a nondecreasing sequence of nonnegative simple functions converging pointwise to $f+g$ . Note that, for every $n$ , $\displaystyle\int_{A}(s_{n}+t_{n})\,d\mu=\int_{A}s_{n}\,d\mu+\int_{A}t_{n}\,d\mu$ . By Lebesgue’s monotone convergence theorem, $\displaystyle\int_{A}(f+g)\,d\mu=\int_{A}f\,d\mu+\int_{A}g\,d\mu$ .
8.

$\begin{array}[]{ll}\displaystyle\int_{A\cup B}f\,d\mu&\displaystyle=\int_{X}% \chi_{A\cup B}f\,d\mu\\ \\ &\displaystyle=\int_{X}\left(\chi_{A}+\chi_{B}-\chi_{A\cap B}\right)f\,d\mu\\ \\ &\displaystyle=\int_{X}\left(\chi_{A}+\chi_{B}-\chi_{\emptyset}\right)f\,d\mu% \\ \\ &\displaystyle=\int_{X}\left(\chi_{A}+\chi_{B}-0\right)f\,d\mu\\ \\ &\displaystyle=\int_{X}\left(\chi_{A}f+\chi_{B}f\right)\,d\mu\\ \\ &\displaystyle=\int_{X}\chi_{A}f\,d\mu+\int_{X}\chi_{B}f\,d\mu\\ \\ &\displaystyle=\int_{A}f\,d\mu+\int_{B}f\,d\mu\end{array}$
9.

Let $E=\{x\in A:f(x)=g(x)\}$ . Since $f$ and $g$ are measurable functions and $A\in\mathfrak{B}$ , it must be the case that $E\in\mathfrak{B}$ . Thus, $A\setminus E\in\mathfrak{B}$ . By hypothesis, $\mu(A\setminus E)=0$ . Note that $E\cap(A\setminus E)=\emptyset$ and $E\cup(A\setminus E)=A$ . Thus, $\displaystyle\int_{A}f\,d\mu=\int_{E}f\,d\mu+\int_{A\setminus E}f\,d\mu=\int_{% E}f\,d\mu+0=\int_{E}g\,d\mu+0=\int_{E}g\,d\mu+\int_{A\setminus E}g\,d\mu=\int_% {A}g\,d\mu$ .

∎

Title	properties of the Lebesgue integral of nonnegative measurable functions
Canonical name	PropertiesOfTheLebesgueIntegralOfNonnegativeMeasurableFunctions
Date of creation	2013-03-22 16:13:50
Last modified on	2013-03-22 16:13:50
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	22
Author	Wkbj79 (1863)
Entry type	Theorem
Classification	msc 26A42
Classification	msc 28A25
Related topic	PropertiesOfTheLebesgueIntegralOfLebesgueIntegrableFunctions