properties of the Lebesgue integral of Lebesgue integrable functions

Theorem.

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

1. $\left|{\displaystyle {\int }_A}f𝑑\mu \right|\le {\displaystyle {\int }_A}|f|𝑑\mu$

2. If $f\le g$, then ${\displaystyle {\int }_A}f𝑑\mu \le {\displaystyle {\int }_A}g𝑑\mu$.

3. ${\displaystyle {\int }_A}f𝑑\mu ={\displaystyle {\int }_X}{\chi }_{A}f𝑑\mu$, where ${\chi }_A$ denotes the characteristic function of $A$

4. If $c\in ℝ$, then ${\displaystyle {\int }_A}cf𝑑\mu =c{\displaystyle {\int }_A}f𝑑\mu$.

5. If $\mu (A)=0$, then ${\displaystyle {\int }_A}f𝑑\mu =0$.

6. ${\displaystyle {\int }_A}(f+g)𝑑\mu ={\displaystyle {\int }_A}f𝑑\mu +{\displaystyle {\int }_A}g𝑑\mu$.

7. If $A\cap B=\mathrm{\varnothing }$, then ${\displaystyle {\int }_{A\cup B}}f𝑑\mu ={\displaystyle {\int }_A}f𝑑\mu +{\displaystyle {\int }_B}f𝑑\mu$.

8. If $f=g$ almost everywhere with respect to $\mu$, then ${\displaystyle {\int }_A}f𝑑\mu ={\displaystyle {\int }_A}g𝑑\mu$.

Proof.

2. 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 (http://planetmath.org/PropertiesOfTheLebesgueIntegralOfNonnegativeMeasurableFunctions) (property 2), ${\displaystyle {\int }_A}{f}^{+}𝑑\mu \le {\displaystyle {\int }_A}{g}^{+}𝑑\mu$ and ${\displaystyle {\int }_A}{f}^{-}𝑑\mu \ge {\displaystyle {\int }_A}{g}^{-}𝑑\mu$. Therefore, $-{\displaystyle {\int }_A}{f}^{-}𝑑\mu \le -{\displaystyle {\int }_A}{g}^{-}𝑑\mu$. Hence, ${\displaystyle {\int }_A}{f}^{+}𝑑\mu -{\displaystyle {\int }_A}{f}^{-}𝑑\mu \le {\displaystyle {\int }_A}{g}^{+}𝑑\mu -{\displaystyle {\int }_A}{f}^{-}𝑑\mu \le {\displaystyle {\int }_A}{g}^{+}𝑑\mu -{\displaystyle {\int }_A}{g}^{-}𝑑\mu$. It follows that ${\displaystyle {\int }_A}f𝑑\mu \le {\displaystyle {\int }_A}g𝑑\mu$.

4. If $c\ge 0$, then

   If , then

5. 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 (http://planetmath.org/PropertiesOfTheLebesgueIntegralOfNonnegativeMeasurableFunctions) (property 6). It follows that ${\displaystyle {\int }_A}f𝑑\mu =0$.

6. Let $\{s_n\}$ be a nondecreasing sequence of nonnegative simple functions converging pointwise to $f^{+}+g^{+}$ and $\{t_n\}$ be a nondecreasing sequence of nonnegative simple functions converging pointwise to $f^{-}+g^{-}$. Note that, for every $n$, ${\displaystyle {\int }_A}{s}_n𝑑\mu -{\displaystyle {\int }_A}{t}_n𝑑\mu ={\displaystyle {\int }_A}(s_n-t_n)𝑑\mu$.

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

8. Let $E=\{x\in A:f(x)=g(x)\}$. Since $f$ and $g$ are measurable functions and $A\in 𝔅$, it must be the case that $E\in 𝔅$. Thus, $A-E\in 𝔅$. By hypothesis, $\mu (A\setminus E)=0$. Note that $E\cap (A\setminus E)=\mathrm{\varnothing }$ and $E\cup (A\setminus E)=A$. Thus, ${\displaystyle {\int }_A}f𝑑\mu ={\displaystyle {\int }_E}f𝑑\mu +{\displaystyle {\int }_{A\setminus E}}f𝑑\mu ={\displaystyle {\int }_E}f𝑑\mu +0={\displaystyle {\int }_E}g𝑑\mu +0={\displaystyle {\int }_E}g𝑑\mu +{\displaystyle {\int }_{A\setminus E}}g𝑑\mu ={\displaystyle {\int }_A}g𝑑\mu .$

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