properties of the Lebesgue integral of nonnegative measurable functions

Theorem.

Let $\mathrm{(}X\mathrm{,}\mathrm{B}\mathrm{,}\mu \mathrm{)}$ be a measure space, $f\mathrm{:}X\mathrm{\to }\mathrm{[}\mathrm{0}\mathrm{,}\mathrm{\infty }\mathrm{]}$ and $g\mathrm{:}X\mathrm{\to }\mathrm{[}\mathrm{0}\mathrm{,}\mathrm{\infty }\mathrm{]}$ be measurable functions, and $A\mathrm{,}B\mathrm{\in }\mathrm{B}$. Then the following properties hold:

1. ${\displaystyle {\int }_A}f𝑑\mu \ge 0$

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 $A\subseteq B$, then ${\displaystyle {\int }_A}f𝑑\mu \le {\displaystyle {\int }_B}f𝑑\mu$.

5. If $c\ge 0$, then ${\displaystyle {\int }_A}cf𝑑\mu =c{\displaystyle {\int }_A}f𝑑\mu$.

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

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

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

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

Proof.

1. Let $s$ be a simple function with $0\le s\le f$. Let $s={\displaystyle \sum _{k=1}^n}{c}_k{\chi }_{A_k}$ for $c_k\in [0,\mathrm{\infty }]$ and $A_k\in 𝔅$. Then ${\displaystyle {\int }_A}s𝑑\mu ={\displaystyle \sum _{k=1}^n}{c}_k\mu (A\cap A_k)\ge 0$. By definition, ${\displaystyle {\int }_A}f𝑑\mu \ge {\displaystyle {\int }_A}s𝑑\mu$. It follows that ${\displaystyle {\int }_A}f𝑑\mu \ge 0$.

2. Let $s$ be a simple function with $0\le s\le f$. Since $f\le g$, $0\le s\le g$. By definition, ${\displaystyle {\int }_A}s𝑑\mu \le {\displaystyle {\int }_A}g𝑑\mu$. Since this holds for every simple function $s$ with $0\le s\le f$, it follows that ${\displaystyle {\int }_A}f𝑑\mu \le {\displaystyle {\int }_A}g𝑑\mu$.

3. Let $s$ be a simple function with $0\le s\le f$. Then $0\le {\chi }_{A}s\le {\chi }_{A}f$. Let $s={\displaystyle \sum _{k=1}^n}{c}_k{\chi }_{A_k}$ for $c_k\in [0,\mathrm{\infty }]$ and $A_k\in 𝔅$. Then

   Thus, ${\displaystyle {\int }_A}f𝑑\mu \le {\displaystyle {\int }_X}{\chi }_{A}f𝑑\mu$.

   Let $t$ be a simple function with $0\le t\le {\chi }_{A}f$. Then ${\chi }_{A}t=t$. Thus, ${\displaystyle {\int }_X}t𝑑\mu ={\displaystyle {\int }_X}{\chi }_{A}t𝑑\mu ={\displaystyle {\int }_A}t𝑑\mu$. Therefore, ${\displaystyle {\int }_X}{\chi }_{A}f𝑑\mu ={\displaystyle {\int }_A}{\chi }_{A}f𝑑\mu$. Since ${\chi }_{A}f\le f$, ${\displaystyle {\int }_A}{\chi }_{A}f𝑑\mu \le {\displaystyle {\int }_A}f𝑑\mu$ by property 2. Hence, ${\displaystyle {\int }_A}f𝑑\mu \le {\displaystyle {\int }_X}{\chi }_{A}f𝑑\mu ={\displaystyle {\int }_A}{\chi }_{A}f𝑑\mu \le {\displaystyle {\int }_A}f𝑑\mu$. It follows that ${\displaystyle {\int }_A}f𝑑\mu ={\displaystyle {\int }_X}{\chi }_{A}f𝑑\mu$.

4. Since $A\subseteq B$, ${\chi }_A\le {\chi }_B$. Thus, ${\chi }_{A}f\le {\chi }_{B}f$. By property 2, ${\displaystyle {\int }_X}{\chi }_{A}f𝑑\mu \le {\displaystyle {\int }_X}{\chi }_{B}f𝑑\mu$. By property 3, ${\displaystyle {\int }_A}f𝑑\mu ={\displaystyle {\int }_X}{\chi }_{A}f𝑑\mu \le {\displaystyle {\int }_X}{\chi }_{B}f𝑑\mu ={\displaystyle {\int }_B}f𝑑\mu$.

5. If $c=0$, then ${\displaystyle {\int }_A}cf𝑑\mu ={\displaystyle {\int }_A}0𝑑\mu =0=0{\displaystyle {\int }_A}f𝑑\mu =c{\displaystyle {\int }_A}f𝑑\mu$.

   If $c>0$, let $S=\{s:X\to [0,\mathrm{\infty }]\mid s\text{is simple and }s\le cf\}$ and

   $T=\{t:X\to [0,\mathrm{\infty }]\mid t\text{is simple and }t\le f\}$. Then ${\displaystyle {\int }_A}cf𝑑\mu =\underset{s\in S}{sup}{\displaystyle {\int }_A}s𝑑\mu =\underset{s\in S}{sup}{\displaystyle {\int }_A}c\cdot {\displaystyle \frac{s}{c}}𝑑\mu =c\underset{s\in S}{sup}{\displaystyle {\int }_A}{\displaystyle \frac{s}{c}}𝑑\mu =c\underset{t\in T}{sup}{\displaystyle {\int }_A}t𝑑\mu =c{\displaystyle {\int }_A}f𝑑\mu$.

6. Let $s$ be a simple function with $0\le s\le f$. Let $s={\displaystyle \sum _{k=1}^n}{c}_k{\chi }_{A_k}$ for $c_k\in [0,\mathrm{\infty }]$ and $A_k\in 𝔅$. Then ${\displaystyle {\int }_A}s𝑑\mu ={\displaystyle \sum _{k=1}^n}{c}_k\mu (A\cap A_k)={\displaystyle \sum _{k=1}^n}{c}_k\cdot 0=0$. Thus, ${\displaystyle {\int }_A}f𝑑\mu =0$.

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 $n$, ${\displaystyle {\int }_A}(s_n+t_n)𝑑\mu ={\displaystyle {\int }_A}{s}_n𝑑\mu +{\displaystyle {\int }_A}{t}_n𝑑\mu$. By Lebesgue’s monotone convergence theorem, ${\displaystyle {\int }_A}(f+g)𝑑\mu ={\displaystyle {\int }_A}f𝑑\mu +{\displaystyle {\int }_A}g𝑑\mu$.

9. 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, $A\setminus E\in 𝔅$. 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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