PlanetMath: properties of the Lebesgue integral of nonnegative measurable functions

Proof.

Let $s$ be a simple function with $0 \le s \le f$ . 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 $\displaystyle \int_A s \, d\mu =\sum_{k=1}^n c_k \mu(A \cap A_k) \ge 0$ . By definition, $\displaystyle \int_A f \, d\mu \ge \int_A s \, d\mu$ . It follows that $\displaystyle \int_A f \, d\mu \ge 0$ .
Let $s$ be a simple function with $0 \le s \le f$ . Since $f \le g$ , $0 \le s \le g$ . By definition, $\displaystyle \int_A s \, d\mu \le \int_A g \, d\mu$ . Since this holds for every simple function $s$ with $0 \le s \le f$ , it follows that $\displaystyle \int_A f \, d\mu \le \int_A g \, d\mu$ .
Let $s$ be a simple function with $0 \le s \le f$ . Then $0 \le \chi_As \le \chi_Af$ . 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
$\begin{displaymath}\begin{array}{ll} \displaystyle \int_A s \, d\mu & \displayst... ...\ \ & \displaystyle \le \int_X \chi_Af \, d\mu. \end{array}\end{displaymath}$

Thus, $\displaystyle \int_A f \, d\mu \le \int_X \chi_Af \, d\mu$ .

Let $t$ be a simple function with $0 \le t \le \chi_Af$ . Then $\chi_At=t$ . Thus, $\displaystyle \int_X t \, d\mu =\int_X \chi_At \, d\mu=\int_A t \, d\mu$ . Therefore, $\displaystyle \int_X \chi_Af \, d\mu =\int_A \chi_Af \, d\mu$ . Since $\chi_Af \le f$ , $\displaystyle \int_A \chi_Af \, d\mu \le \int_A f \, d\mu$ by property 2. Hence, $\displaystyle \int_A f \, d\mu \le \int_X \chi_Af \, d\mu =\int_A \chi_Af \, d\mu \le \int_A f \, d\mu$ . It follows that $\displaystyle \int_A f \, d\mu =\int_X \chi_Af \, d\mu$ .
Since $A \subseteq B$ , $\chi_A \le \chi_B$ . Thus, $\chi_Af \le \chi_Bf$ . By property 2, $\displaystyle \int_X \chi_Af \, d\mu \le \int_X \chi_Bf \, d\mu$ . By property 3, $\displaystyle \int_A f \, d\mu =\int_X \chi_Af \, d\mu \le \int_X \chi_Bf \, d\mu =\int_B f \, d\mu$ .
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 $S=\{s \colon X \to [0,\infty] \mid s~{is simple and }s \le cf\}$ and

$T=\{t \colon X \to [0,\infty] \mid t~{is simple and }t \le f\}$ . Then $\displaystyle \int_A cf \, d\mu =\sup_{s \in S} \int_A s \, d\mu =\sup_{s \in S} \int_A c \cdot \frac{s}{c} \, d\mu =c\sup_{s \in S} \int_A \frac{s}{c} \, d\mu =c\sup_{t \in T} \int_A t \, d\mu =c \int_A f \, d\mu$ .
Let $s$ be a simple function with $0 \le s \le f$ . 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, $\displaystyle \int_A f \, d\mu=0$ .
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, $\displaystyle \int_A (f+g) \, d\mu =\int_A f \, d\mu +\int_A g \, d\mu$ .
$\begin{displaymath}\begin{array}{ll} \displaystyle \int_{A \cup B} f \, d\mu & \... ...& \displaystyle =\int_A f \, d\mu +\int_B f \, d\mu \end{array}\end{displaymath}$
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, $\displaystyle \int_A f \, d\mu =\int_E f \, d\mu +\int_{A \setminus E} f \, d\mu =\int_E f \, d\mu +0=\int_E g \, d\mu +0=\int_E g \, d\mu +\int_{A \setminus E} g \, d\mu =\int_A g \, d\mu$ .

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