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[parent] properties of the Lebesgue integral of nonnegative measurable functions (Theorem)
Theorem   Let % latex2html id marker 439 $ (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 % latex2html id marker 445 $ A,B \in \mathfrak{B}$. Then the following properties hold:
  1. $ \displaystyle \int_A f \, d\mu \ge 0$
  2. If $ f \le g$, then $ \displaystyle \int_A f \, d\mu \le \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 \le \int_B f \, d\mu$.
  5. If $ c \ge 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 \le s \le f$. Let $ \displaystyle s=\sum_{k=1}^n c_k \chi_{A_k}$ for $ c_k \in [0,\infty]$ and % latex2html id marker 494 $ 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$.
  2. 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$.
  3. 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 % latex2html id marker 528 $ 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$.

  4. 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$.
  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~$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 ... ...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 \le s \le f$. Let $ \displaystyle s=\sum_{k=1}^n c_k \chi_{A_k}$ for $ c_k \in [0,\infty]$ and % latex2html id marker 584 $ 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{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}
  9. Let $ E=\{x \in A:f(x)=g(x)\}$. Since $ f$ and $ g$ are measurable functions and % latex2html id marker 616 $ A \in \mathfrak{B}$, it must be the case that % latex2html id marker 618 $ E \in \mathfrak{B}$. Thus, % latex2html id marker 620 $ 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\... ...g \, d\mu +0=\int_E g \, d\mu +\int_{A \setminus E} g \, d\mu =\int_A g \, d\mu$.
$ \qedsymbol$



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See Also: properties of the Lebesgue integral of Lebesgue integrable functions


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Cross-references: hypothesis, Lebesgue's monotone convergence theorem, pointwise, sequence, simple function, almost everywhere, characteristic function, properties, measurable functions, measure space
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This is version 19 of properties of the Lebesgue integral of nonnegative measurable functions, born on 2006-09-09, modified 2007-06-27.
Object id is 8331, canonical name is PropertiesOfTheLebesgueIntegralOfNonnegativeMeasurableFunctions.
Accessed 1829 times total.

Classification:
AMS MSC28A25 (Measure and integration :: Classical measure theory :: Integration with respect to measures and other set functions)
 26A42 (Real functions :: Functions of one variable :: Integrals of Riemann, Stieltjes and Lebesgue type)

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