# 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. 1.

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

2. 2.

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

3. 3.

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

4. 4.

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

5. 5.

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

6. 6.

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

7. 7.

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

8. 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. 9.

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

###### Proof.
1. 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. 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. 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. 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. 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. 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. 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. 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. 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 PropertiesOfTheLebesgueIntegralOfNonnegativeMeasurableFunctions 2013-03-22 16:13:50 2013-03-22 16:13:50 Wkbj79 (1863) Wkbj79 (1863) 22 Wkbj79 (1863) Theorem msc 26A42 msc 28A25 PropertiesOfTheLebesgueIntegralOfLebesgueIntegrableFunctions