PlanetMath: properties of the Lebesgue integral of Lebesgue integrable functions

Proof.

$\displaystyle \left\vert \int_A f \, d\mu \right\vert$	$\displaystyle =\left\vert \int_A f^+ \, d\mu -\int_A f^- \, d\mu \right\vert$ by definition
	$\displaystyle \le \left\vert \int_A f^+ \, d\mu \right\vert +\left\vert \int_A f^- \, d\mu \right\vert$ by the triangle inequality
	$\displaystyle =\int_A f^+ \, d\mu +\int_A f^- \, d\mu$ by the
	properties of the Lebesgue integral of nonnegative measurable functions (property 1),
	$\displaystyle =\int_A (f^++f^-) \, d\mu$ by the
	properties of the Lebesgue integral of nonnegative measurable functions (property 7),
	$\displaystyle =\int_A \vert f\vert \, d\mu$

Since $f \le g$ , the following must hold:
- $f^+=\max\{0,f\}\le\max\{0,g\}=g^+$ ;
- $-f \ge -g$ ;
- $f^-=\max\{0,-f\}\ge\max\{0,-g\}=g^-$ .
Thus, by the properties of the Lebesgue integral of nonnegative measurable functions (property 2), $\displaystyle \int_A f^+ \, d\mu \le \int_A g^+ \, d\mu$ and $\displaystyle \int_A f^- \, d\mu \ge \int_A g^- \, d\mu$ . Therefore, $\displaystyle -\int_A f^- \, d\mu \le -\int_A g^- \, d\mu$ . Hence, $\displaystyle \int_A f^+ \, d\mu -\int_A f^- \, d\mu \le \int_A g^+ \, d\mu -\int_A f^- \, d\mu \le \int_A g^+ \, d\mu -\int_A g^- \, d\mu$ . It follows that $\displaystyle \int_A f \, d\mu \le \int_A g \, d\mu$ .
$\displaystyle \int_A f \, d\mu$ $\displaystyle =\int_A f^+ \, d\mu -\int_A f^- \, d\mu$ by definition

$\displaystyle =\int_X \chi_Af^+ \, d\mu -\int_X \chi_Af^- \, d\mu$ by the

properties of the Lebesgue integral of nonnegative measurable functions (property 3),

$\displaystyle =\int_X (\chi_Af)^+ \, d\mu -\int_X (\chi_Af)^- \, d\mu$

$\displaystyle =\int_X \chi_Af \, d\mu$ by definition
If $c \ge 0$ , then

$\displaystyle \int_A cf \, d\mu$ $\displaystyle =\int_A (cf)^+ \, d\mu -\int_A (cf)^- \, d\mu$ by definition

$\displaystyle =\int_A cf^+ \, d\mu -\int_A cf^- \, d\mu$

$\displaystyle =c\int_A f^+ \, d\mu -c\int_A f^- \, d\mu$ by the

properties of the Lebesgue integral of nonnegative measurable functions (property 5)

$\displaystyle =c\left( \int_A f^+ \, d\mu -\int_A f^- \, d\mu \right)$

$\displaystyle =c\int_A f \, d\mu$ by definition.

If , then

$\displaystyle \int_A cf \, d\mu$ $\displaystyle =\int_A (cf)^+ \, d\mu -\int_A (cf)^- \, d\mu$ by definition

$\displaystyle =\int_A (-c)f^- \, d\mu -\int_A (-c)f^+ \, d\mu$

$\displaystyle =-c\int_A f^- \, d\mu +c\int_A f^+ \, d\mu$ by the

properties of the Lebesgue integral of nonnegative measurable functions (property 5)

$\displaystyle =c\left( -\int_A f^- \, d\mu +\int_A f^+ \, d\mu \right)$

$\displaystyle =c\int_A f \, d\mu$ by definition.
Note that $\displaystyle \int_A f^+ \, d\mu=0$ and $\displaystyle \int_A f^- \, d\mu=0$ by the properties of the Lebesgue integral of nonnegative measurable functions (property 6). It follows that $\displaystyle \int_A f \, d\mu =0$ .

Let $\{s_n\}$ be a nondecreasing sequence of nonnegative simple functions converging pointwise to

and $\{t_n\}$ be a nondecreasing sequence of nonnegative simple functions converging pointwise to

. Note that, for every

, $\displaystyle \int_A s_n \, d\mu -\int_A t_n \, d\mu =\int_A (s_n-t_n) \, d\mu$ .

Since and are integrable and $\vert f+g\vert \le \vert f\vert+\vert g\vert$ , is integrable. Thus,

$\displaystyle \int_A f \, d\mu +\int_A g \, d\mu$	$\displaystyle =\int_A f^+ \, d\mu -\int_A f^- \, d\mu +\int_A g^+ \, d\mu -\int_A g^- \, d\mu$ by definition
	$\displaystyle =\int_A f^+ \, d\mu +\int_A g^+ \, d\mu -\left( \int_A f^- \, d\mu +\int_A g^- d\mu \right)$
	$\displaystyle =\int_A (f^++g^+) \, d\mu -\left( \int_A (f^-+g^-) \, d\mu \right)$ by the
	properties of the Lebesgue integral of nonnegative measurable functions (property 7)
	$\displaystyle =\lim_{n \to \infty} \int_A s_n \, d\mu -\left( \lim_{n \to \infty} \int_A t_n \, d\mu \right)$ by Lebesgue's monotone convergence theorem
	$\displaystyle =\lim_{n \to \infty} \left( \int_A s_n \, d\mu -\int_A t_n \, d\mu \right)$
	$\displaystyle =\lim_{n \to \infty} \int_A (s_n-t_n) \, d\mu$
	$\displaystyle =\int_A (f^++g^+-(f^-+g^-)) \, d\mu$ by Lebesgue's dominated convergence theorem
	$\displaystyle =\int_A (f^+-f^-+g^+-g^-) \, d\mu$
	$\displaystyle =\int_A (f+g) \, d\mu$ by definition.

$\displaystyle \int_{A \cup B} f \, d\mu$ $\displaystyle =\int_{A \cup B} f^+ \, d\mu -\int_{A \cup B} f^- \, d\mu$ by definition

$\displaystyle =\int_A f^+ \, d\mu +\int_B f^+ \, d\mu -\left( \int_A f^- \, d\mu +\int_B f^- \, d\mu \right)$ by the

properties of the Lebesgue integral of nonnegative measurable functions (property 8),

$\displaystyle =\int_A f^+ \, d\mu -\int_A f^- \, d\mu +\int_B f^+ \, d\mu -\int_B f^- \, d\mu$

$\displaystyle =\int_A f \, d\mu +\int_B f \, d\mu$ by definition
Let $E=\{x \in A:f(x)=g(x)\}$ . Since and are measurable functions and $% latex2html id marker 719 $ A \in \mathfrak{B}$$ , it must be the case that $% latex2html id marker 721 $ E \in \mathfrak{B}$$ . Thus, $% latex2html id marker 723 $ A-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, $\displaystyle \int_A f \, d\mu =\int_E f \, d\mu +\int_{A \setminus E} f \, d\... ... \, d\mu +0=\int_E g \, d\mu +\int_{A \setminus E} g \, d\mu =\int_A g \, d\mu.$

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