1. Lebesgue Integral of Non-negative Simple Function

Definition 1.1 Suppose $E$ is measurable, $E_i$ are disjoint measurable sets, and \[f(x)=\sum_{i=1}^s c_i\chi_{E_i}(x), c_i \geq 0.\] The Lebesgue integral of $f$ on $E$ is \[\int_E f(x)\text{d}x=\sum_{i=1}^s c_i \cdot m(E_i).\] Suppose $A \subset E$, and $f|_A(x)=c_i$, $x \in A \cap E_i$, then \[\int_A f(x)\text{d}x=\sum_{i=1}^s c_i \cdot m(A \cap E_i).\]

Theorem 1.1 Suppose $A$ and $B$ are disjoint measurable subsets of $E$, then \[\int_{A \cup B} f(x)\text{d}x=\int_A f(x)\text{d}x+\int_B f(x)\text{d}x.\]

Proof. Since $A \cap B=\varnothing$, then \[\begin{aligned} \int_{A \cup B} f(x)\text{d}x&=\sum_{i=1}^s c_i \cdot m((A \cup B) \cap E_i) \\&=\sum_{i=1}^s c_i \cdot m((A \cap E_i) \cup (B \cap E_i)) \\&=\sum_{i=1}^s c_i \cdot (m(A \cap E_i)+m(B \cap E_i)) \\&=\sum_{i=1}^s c_i \cdot m(A \cap E_i)+\sum_{i=1}^s c_i \cdot m(B \cap E_i) \\&=\int_A f(x)\text{d}x+\int_B f(x)\text{d}x. \end{aligned}\]

$\square$

Theorem 1.2 Suppose $\{A_n\}_{n=1}^\infty$ is an increasing collection of measurable subsets of $E$, and \[E=\bigcup_{n=1}^\infty A_n=\lim_{n \to \infty} A_n.\] Then \[\int_E f(x)\text{d}x=\lim_{n \to \infty}\int_{A_n} f(x)\text{d}x.\]

Proof. We have \[\begin{aligned} \lim_{n \to \infty} \int_{A_n} f(x)\text{d}x&=\lim_{n \to \infty} \sum_{i=1}^s c_i \cdot m(A_n \cap E_i) \\&=\sum_{i=1}^s c_i \cdot \lim_{n \to \infty} m(A_n \cap E_i) \\&=\sum_{i=1}^s c_i \cdot m(E \cap E_i) \\&=\int_E f(x)\text{d}x. \end{aligned}\]

$\square$

Theorem 1.3 Suppose $\alpha$ and $\beta$ are non-negative real numbers. Then \[\int_E (\alpha f(x)+\beta g(x))\text{d}x=\alpha\int_E f(x)\text{d}x+\beta\int_E g(x)\text{d}x.\]

Proof. Suppose $c \in \mathbb{R} \geq 0$, we have $\displaystyle cf(x)=\sum_{i=1}^s cc_i \cdot \chi_{E_i}(x)$, where $cc_i \geq 0$. Then $cf$ is a non-negative simple function. Therefore, \[\begin{aligned} \int_E cf(x)\text{d}x&=\sum_{i=1}^s cc_i \cdot m(E_i) \\&=c\sum_{i=1}^s c_i \cdot m(E_i) \\&=c\int_E f(x)\text{d}x. \end{aligned}\]

Suppose $\displaystyle f(x)=\sum_{i=1}^s c_i \cdot \chi_{E_i}(x)$ and $\displaystyle g(x)=\sum_{i=1}^p d_i \cdot \chi_{\widetilde{E}_i}(x)$.

When $x \in E_i \cap \widetilde{E}_j$, $f(x)+g(x)=c_i+d_j \geq 0$. Hence, \[f(x)+g(x)=\sum_{\substack{1 \leq i \leq s \\ 1 \leq j \leq p}} (c_i+d_j) \cdot \chi_{E_i \cap \widetilde{E}_j}(x),\] and $f+g$ is a non-negative simple function. Therefore, \[\begin{aligned} \int_E (f(x)+g(x))\text{d}x&=\sum_{\substack{1 \leq i \leq s \\ 1 \leq j \leq p}} (c_i+d_j) \cdot m(E_i \cap \widetilde{E}_j) \\&=\sum_{\substack{1 \leq i \leq s \\ 1 \leq j \leq p}} c_i \cdot m(E_i \cap \widetilde{E}_j)+\sum_{\substack{1 \leq i \leq s \\ 1 \leq j \leq p}} d_j \cdot m(E_i \cap \widetilde{E}_j) \\&=\sum_{1 \leq i \leq s} c_i\sum_{1 \leq j \leq p} m(E_i \cap \widetilde{E}_j)+\sum_{1 \leq j \leq p} d_j\sum_{1 \leq i \leq s} m(E_i \cap \widetilde{E}_j) \\&=\sum_{1 \leq i \leq s} c_i \cdot m(E_i)+\sum_{1 \leq j \leq p} d_j \cdot m(\widetilde{E}_j) \\&=\int_E f(x)\text{d}x+\int_E g(x)\text{d}x. \end{aligned}\]

Hence, it is easy to show that \[\int_E (\alpha f(x)+\beta g(x))\text{d}x=\alpha\int_E f(x)\text{d}x+\beta\int_E g(x)\text{d}x.\]

$\square$

2. Lebesgue Integral of Non-negative Measurable Function

2.1. Definition

Definition 2.1 Suppose $E$ is measurable, and $f$ is a non-negative measurable function on $E$. The Lebesgue integral of $f$ on $E$ is \[\int_E f(x)\text{d}x=\sup\left\{\int_E \varphi(x)\text{d}x\right\},\] where $\varphi$ is a non-negative simple function on $E$, and $\varphi(x) \leq f(x)$.

Theorem 2.1 Suppose $\varphi$ and $\psi$ are non-negative simple functions on $E$. If $\varphi(x) \leq \psi(x)$, then \[\int_E \varphi(x)\text{d}x \leq \int_E \psi(x)\text{d}x.\]

Corollary 2.2 Suppose $f$ and $g$ are non-negative measurable functions on $E$. If $f(x) \leq g(x)$, then \[\int_E f(x)\text{d}x \leq \int_E g(x)\text{d}x.\]

Proof. For any $0 \leq \varphi(x) \leq f(x)$, where $\varphi$ is a non-negative simple function, we have $0 \leq \varphi(x) \leq g(x)$, $x \in E$. Then \[\int_E g(x)\text{d}x \geq \int_E \varphi(x)\text{d}x.\] Therefore, \[\int_E g(x)\text{d}x \geq \sup_{0 \leq \varphi(x) \leq f(x)}\left\{\int_E \varphi(x)\text{d}x\right\}=\int_E f(x)\text{d}x.\]

$\square$

Definition 2.2 If $\displaystyle \int_E f(x)\text{d}x<\infty$, then $f$ is Lebesgue integrable on $E$.

Theorem 2.3 If $f$ is Lebesgue integrable on $E$, then $f$ is finite almost everywhere on $E$, i.e., \[m(E[f=\infty])=0.\]

Proof. Since $f(x) \geq f(x) \cdot \chi_{E[f \geq n]}(x)$, then \[\infty>\int_E f(x)\text{d}x \geq \int_E f(x) \cdot \chi_{E[f \geq n]}(x)\text{d}x=\int_{E[f \geq n]} f(x)\text{d}x \geq \int_{E[f \geq n]} n\text{d}x=n \cdot m(E[f \geq n]).\] Therefore, \[m(E[f \geq n]) \leq \frac{1}{n}\int_E f(x)\text{d}x.\] Since $E[f \geq n] \supset E[f \geq n+1]$ and $\displaystyle E[f=\infty]=\bigcap_{n=1}^\infty E[f \geq n]$, then \[m(E[f=\infty])=\lim_{n \to \infty} m(E[f \geq n])=0,\] i.e., $f$ is finite almost everywhere on $E$.

$\square$

Theorem 2.4 Suppose $A$ is a measurable subset of $E$. Then \[\int_A f(x)\text{d}x=\int_E f(x) \cdot \chi_A(x)\text{d}x.\]

Proof. We have \[\begin{aligned} \int_A f(x)\text{d}x&=\sup_{\substack{\varphi(x) \leq f(x) \\ x \in A}}\left\{\int_A \varphi(x)\text{d}x\right\} \\&=\sup_{\substack{\varphi(x)\chi_A(x) \leq f(x)\chi_A(x) \\ x \in E}}\left\{\int_A \varphi(x)\text{D}x\right\} \\&=\int_E f(x)\varphi_A(x)\text{d}x. \end{aligned}\]

$\square$

Theorem 2.5 Suppose $A$ and $B$ are disjoint measurable subsets of $E$, then \[\int_{A \cup B} f(x)\text{d}x=\int_A f(x)\text{d}x+\int_B f(x)\text{d}x.\]

Proof. Take an arbitrary simple function $\varphi$ on $A \cup B$, where $0 \leq \varphi(x) \leq f(x)$ for all $x \in E$. We have \[\int_{A \cup B} \varphi(x)\text{d}x=\int_A \varphi(x)\text{d}x+\int_B \varphi(x)\text{d}x \leq \int_A f(x)\text{d}x+\int_B f(x)\text{d}x.\] Therefore, \[\int_{A \cup B} f(x)\text{d}x \leq \int_{A \cup B} \varphi(x)\text{d}x \leq \int_A f(x)\text{d}x+\int_B f(x)\text{d}x.\]

Take an arbitrary simple function $\varphi_1$ on $A$, where $0 \leq \varphi_1(x) \leq f(x)$ for all $x \in A$, and an arbitrary simple function $\varphi_2$ on $B$, where $0 \leq \varphi_2(x) \leq f(x)$ for all $x \in B$. Then \[\varphi(x)=\begin{cases} \varphi_1(x), &x \in A \\ \varphi_2(x), &x \in B \end{cases}\] is a simple function on $A \cup B$, where $0 \leq \varphi(x) \leq f(x)$. Hence, \[\int_{A \cup B} f(x)\text{d}x \geq \int_{A \cup B} \varphi(x)\text{d}x=\int_A \varphi(x)\text{d}x+\int_B \varphi(x)\text{d}x=\int_A \varphi_1(x)\text{d}x+\int_B \varphi_2(x)\text{d}x.\] Therefore, \[\int_{A \cup B} f(x)\text{d}x \geq \int_A f(x)\text{d}x+\int_B f(x)\text{d}x.\] As a consequence, \[\int_{A \cup B} f(x)\text{d}x=\int_A f(x)\text{d}x+\int_B f(x)\text{d}x.\]

$\square$

2.2. Limit and Integral

Theorem 2.6 Suppose $\alpha$ and $\beta$ are non-negative real numbers, then \[\int_E (\alpha f(x)+\beta g(x))\text{d}x=\alpha\int_E f(x)\text{d}x+\beta\int_E g(x)\text{d}x.\]

Lemma 2.6.1 (Levi's Monotonicity Theorem) Suppose $\{f_n\}_{n=1}^\infty$ is a sequence of non-negative measurable functions on $E$. For $x \in E$, and for all $n$, we have $f_n(x) \leq f_{n+1}(x)$. Then \[\lim_{n \to \infty} \int_E f_n(x)\text{d}x=\int_E \lim_{n \to \infty} f_n(x)\text{d}x.\]

Proof. Let $\displaystyle f(x)=\lim_{n \to \infty} f_n(x)$. We know $f$ is a non-negative measurable function on $E$, and $f_n(x) \leq f(x)$ for all $n \in \mathbb{N}$. Hence, \[\int_E f_n(x)\text{d}x \leq \int_E f(x)\text{d}x.\] Since \[\int_E f_n(x)\text{d}x \leq \int_E f_{n+1}(x)\text{d}x,\] we have \[\lim_{n \to \infty} \int_E f_n(x)\text{d}x \leq \int_E f(x)\text{d}x.\]

Take an arbitrary non-negative simple function $\varphi$ on $E$, where $0 \leq \varphi(x) \leq f(x)$. Take an arbitrary $0<c<1$. Let $E_n=E[f_n \geq c\varphi]$, which is a measurable subset of $E$.

Since $f_n(x) \leq f_{n+1}(x)$, then \[f_n(x) \geq c\varphi(x) \Rightarrow f_{n+1}(x) \geq c\varphi(x),\] i.e., $E_n \subset E_{n+1}$.

Since $\displaystyle \lim_{n \to \infty} f_n(x)=f(x)$, and $c\varphi(x)<f(x)$, then when $n \gg 0$, $c\varphi(x) \leq f_n(x)$. Hence, \[\bigcup_{n=1}^\infty E_n=E.\]

Therefore, \[\int_E \varphi(x)\text{d}x=\lim_{n \to \infty} \int_{E_n} \varphi(x)\text{d}x.\] Since \[\lim_{n \to \infty} \int_E f_n(x)\text{d}x \geq \lim_{n \to \infty} \int_{E_n} f_n(x)\text{d}x \geq \lim_{n \to \infty} \int_{E_n} c\varphi(x)\text{d}x=c\lim_{n \to \infty} \int_{E_n} \varphi(x)\text{d}x,\] then \[\lim_{n \to \infty} \int_E f_n(x)\text{d}x \geq c\int_E \varphi(x)\text{d}x.\] Take $c \to 1$, we have \[\lim_{n \to \infty} \int_E f_n(x)\text{d}x \geq \int_E \varphi(x)\text{d}x.\] Hence, \[\lim_{n \to \infty} \int_E f_n(x)\text{d}x \geq \int_E f(x)\text{d}x.\]

Therefore, \[\lim_{n \to \infty} \int_E f_n(x)\text{d}x=\int_E \lim_{n \to \infty} f_n(x)\text{d}x.\]

$\square$

Proof. Suppose $c \in \mathbb{R} \geq 0$. When $c>0$, we have \[\int_E cf(x)\text{d}x=\sup\left\{\int_E \varphi(x)\text{d}x\right\},\] where $0 \leq \varphi(x) \leq cf(x)$, and $\varphi$ is a simple function on $E$. Let $\displaystyle \psi(x)=\frac{1}{c}\varphi(x)$, which is a simple function on $E$, then we have \[\int_E cf(x)\text{d}x=\sup\left\{\int_E c\psi(x)\text{d}x\right\}=\sup\left\{c\int_E \psi(x)\text{d}x\right\}=c\sup\left\{\int_E \psi(x)\text{d}x\right\},\] where $0 \leq \psi(x) \leq f(x)$. Hence, \[\int_E cf(x)\text{d}x=c\int_E f(x)\text{d}x.\]

When $c=0$, it is obvious that \[\int_E cf(x)\text{d}x=c\int_E f(x)\text{d}x.\]

There exist two sequences of non-negative simple function $\{\varphi_n\}$ and $\{\psi_n\}$ on $E$ such that \[\lim_{n \to \infty} \varphi_n(x)=f(x), \lim_{n \to \infty} \psi_n(x)=g(x),\] where for all $x \in E$, $\varphi_n(x) \leq \varphi_{n+1}(x) \leq f(x)$, and $\psi_n(x) \leq \psi_{n+1}(x) \leq g(x)$. We know $\varphi_n+\psi_n$ is a non-negative simple function on $E$, $\varphi_n(x)+\psi_n(x) \leq \varphi_{n+1}(x)+\psi_{n+1}(x) \leq f(x)+g(x)$, and \[\lim_{n \to \infty} (\varphi_n(x)+\psi_n(x))=f(x)+g(x).\] By Levi's monotonicity theorem, we have \[\begin{aligned} \int_E \lim_{n \to \infty} (\varphi_n(x)+\psi_n(x))\text{d}x&=\int_E (f(x)+g(x))\text{d}x \\&=\lim_{n \to \infty} \int_E (\varphi_n(x)+\psi_n(x))\text{d}x \\&=\lim_{n \to \infty} \left(\int_E \varphi_n(x)\text{d}x+\int_E \psi_n(x)\text{d}x\right) \\&=\lim_{n \to \infty} \int_E \varphi_n(x)\text{d}x+\lim_{n \to \infty} \int_E \psi_n(x)\text{d}x \\&=\int_E \lim_{n \to \infty} \varphi_n(x)\text{d}x+\int_E \lim_{n \to \infty} \psi_n(x)\text{d}x \\&=\int_E f(x)\text{d}x+\int_E g(x)\text{d}x. \end{aligned}\]

Hence, it is easy to show that \[\int_E (\alpha f(x)+\beta g(x))\text{d}x=\alpha\int_E f(x)\text{d}x+\beta\int_E g(x)\text{d}x.\]

$\square$

Theorem 2.7 Suppose $\{f_n\}_{n=1}^\infty$ is a sequence of non-negative measurable functions on $E$. Then \[\int_E \left(\sum_{n=1}^\infty f_n(x)\right)\text{d}x=\sum_{n=1}^\infty \int_E f_n(x)\text{d}x.\]

Proof. Let $\displaystyle g_n(x)=\sum_{k=1}^n f_k(x)$, which is a non-negative measurable function. Besides, \[g_{n+1}(x)=g_n(x)+f_{n+1}(x) \geq g_n(x).\] By Levi's monotonicity theorem, we have \[\lim_{n \to \infty} \int_E g_n(x)\text{d}x=\int_E \lim_{n \to \infty} g_n(x)\text{d}x=\int_E \left(\sum_{n=1}^\infty f_n(x)\right)\text{d}x.\] Since \[\lim_{n \to \infty} \int_E g_n(x)\text{d}x=\lim_{n \to \infty} \sum_{k=1}^n \int_E f_k(x)\text{d}x=\sum_{n=1}^\infty \int_E f_n(x)\text{d}x,\] then \[\int_E \left(\sum_{n=1}^\infty f_n(x)\right)\text{d}x=\sum_{n=1}^\infty \int_E f_n(x)\text{d}x.\]

$\square$

Theorem 2.8 (Fatou's Lemma) Suppose $\{f_n\}_{n=1}^\infty$ is a sequence of non-negative measurable functions on $E$. Then \[\int_E \varliminf_{n \to \infty} f_n(x)\text{d}x \leq \varliminf_{n \to \infty} \int_E f_n(x)\text{d}x.\]

Proof. Since $\displaystyle \left\{\inf_{k \geq n} f_k\right\}$ is an increasing sequence of non-negative measurable functions on $E$. By Levi's monotonicity theorem, we have \[\int_E \varliminf_{n \to \infty} f_n(x)\text{d}x=\int_E \left(\lim_{n \to \infty} \inf_{k \geq n} f_k(x)\right)\text{d}x=\lim_{n \to \infty} \int_E \inf_{k \geq n} f_k(x)\text{d}x.\] Since \[\int_E \inf_{k \geq n} f_k(x)\text{d}x \leq \inf_{k \geq n} \int_E f_k(x)\text{d}x,\] then \[\lim_{n \to \infty} \int_E \inf_{k \geq n} f_k(x)\text{d}x \leq \lim_{n \to \infty} \inf_{k \geq n} \int_E f_k(x)\text{d}x=\varliminf_{n \to \infty} \int_E f_n(x)\text{d}x.\] Hence, \[\int_E \varliminf_{n \to \infty} f_n(x)\text{d}x \leq \varliminf_{n \to \infty} \int_E f_n(x)\text{d}x.\]

$\square$

The equal sign in the Fatou's lemma does not always hold. For example, let \[f_n(x)=\begin{cases}n, &\displaystyle 0<x<\frac{1}{n} \\0, &\displaystyle x \geq \frac{1}{n}\end{cases}.\] For any $x \in (0, \infty)$, $\displaystyle \lim_{n \to \infty} f_n(x)=0$, and thus \[\int_{(0, \infty)} \lim_{n \to \infty} f_n(x)\text{d}x=0.\] However, \[\int_{(0, \infty)} f_n(x)\text{d}x=\int_{\left(0, \frac{1}{n}\right)} f_n(x)\text{d}x+\int_{\left[\frac{1}{n}, \infty\right)} f_n(x)\text{d}x=1,\] and thus \[\lim_{n \to \infty} \int_{(0, \infty)} f_n(x)\text{d}x=1.\]

2.3. Zero Measure Set and Integral

Theorem 2.9 Suppose $E$ is a zero measure set. Then \[\int_E f(x)\text{d}x=0.\]

Proof. By definition, \[\int_E f(x)\text{d}x=\sup\left\{\int_E \varphi(x)\text{d}x\right\},\] where $\varphi$ is a non-negative simple function on $E$, and $\varphi(x) \leq f(x)$. Since \[\int_E \varphi(x)\text{d}x=0,\] then \[\int_E f(x)\text{d}x=0.\]

$\square$

Theorem 2.10 If $f(x) \leq g(x)\ \text{a.e.}\ x \in E$, then \[\int_E f(x)\text{d}x \leq \int_E g(x)\text{d}x.\]

Proof. Let $E_1=E[f \leq g]$ and $E_2=E[f>g]$, where $m(E_2)=0$, then $E=E_1 \cup E_2$, and $E_1 \cap E_2=\varnothing$. Hence, \[\begin{aligned} \int_E f(x)\text{d}x&=\int_{E_1} f(x)\text{d}x+\int_{E_2} f(x)\text{d}x \\& \leq \int_{E_1} g(x)\text{d}x+0 \\&=\int_{E_1} g(x)\text{d}x+\int_{E_2} g(x)\text{d}x \\&=\int_E g(x)\text{d}x. \end{aligned}\]

$\square$

Corollary 2.11 If $f(x)=g(x)\ \text{a.e.}\ x \in E$, then \[\int_E f(x)\text{d}x=\int_E g(x)\text{d}x.\] Moreover, if $f(x)=0\ \text{a.e.}\ x \in E$, then \[\int_E f(x)\text{d}x=0.\]

Theorem 2.12 If \[\int_E f(x)\text{d}x=0,\] then $f(x)=0\ \text{a.e.}\ x \in E$.

Proof. Since \[0=\int_E f(x)\text{d}x \geq \int_{E\left[f \geq \frac{1}{n}\right]} f(x)\text{d}x \geq \int_{E\left[f \geq \frac{1}{n}\right]} \frac{1}{n}\text{d}x=\frac{1}{n} \cdot m\left(E\left[f \geq \frac{1}{n}\right]\right),\] then $\displaystyle m\left(E\left[f \geq \frac{1}{n}\right]\right)=0$. Hence, \[m(E[f \neq 0])=m\left(\bigcup_{n=1}^\infty E\left[f \geq \frac{1}{n}\right]\right) \leq \sum_{n=1}^\infty m\left(E\left[f \geq \frac{1}{n}\right]\right)=0.\] Therefore, $m(E[f \geq 0])=0$, i.e., $f(x)=0\ \text{a.e.}\ x \in E$.

$\square$

3. Lebesgue Integral of Measurable Function

Definition 3.1 Suppose $E$ is measurable, and $f$ is a measurable function on $E$. When at least one of $\displaystyle \int_E f^+(x)\text{d}x$ and $\displaystyle \int_E f^-(x)\text{d}x$ is finite, the Lebesgue integral of $f$ on $E$ is \[\int_E f(x)\text{d}x=\int_E f^+(x)\text{d}x-\int_E f^-(x)\text{d}x.\] If $\displaystyle \int_E f^+(x)\text{d}x$ and $\displaystyle \int_E f^-(x)\text{d}x$ are finite, then $f$ is Lebesgue integrable on $E$. The collection of all Lebesgue integrable functions on $E$ is denoted $L(E)$.

Theorem 3.1 Suppose $E \neq \varnothing$ and $m(E)=0$. Then any real function $f$ on $E$ is Lebesgue integrable, and \[\int_E f(x)\text{d}x=0.\]

Proof. Since $m(E)=0$, and $f^+$ and $f^-$ are non-negative measurable on $E$, then \[\int_E f^+(x)\text{d}x=\int_E f^-(x)\text{d}x=0.\] Hence, $f \in L(E)$, and \[\int_E f(x)\text{d}x=\int_E f^+(x)\text{d}x-\int_E f^-(x)\text{d}x=0.\]

$\square$

Theorem 3.2 Suppose $\displaystyle E=\bigcup_{n=1}^\infty E_n$, and $\{E_n\}$ is a collection of disjoint measurable subsets. If the Lebesgue integral of $f$ on $E$ exists, then \[\int_E f(x)\text{d}x=\sum_{n=1}^\infty \int_{E_n} f(x)\text{d}x.\]

Proof. Suppose $f$ is a non-negative measurable function. We have \[\int_{E_n} f(x)\text{d}x=\int_E f(x) \cdot \chi_{E_n}(x)\text{d}x\] and thus \[\sum_{n=1}^\infty \int_{E_n} f(x)\text{d}x=\sum_{n=1}^\infty \int_E f(x) \cdot \chi_{E_n}(x)\text{d}x=\int_E \sum_{n=1}^\infty f(x) \cdot \chi_{E_n}(x)\text{d}x=\int_E f(x)\text{d}x.\]

Suppose $f$ is a measurable function. We know at least one of $\displaystyle \sum_{n=1}^\infty \int_{E_n} f^+(x)\text{d}x$ and $\displaystyle \sum_{n=1}^\infty \int_{E_n} f^-(x)\text{d}x$ is convergent. Hence, \[\begin{aligned} \int_E f(x)\text{d}x&=\int_E f^+(x)\text{d}x-\int_E f^-(x)\text{d}x \\&=\left(\sum_{n=1}^\infty \int_{E_n} f^+(x)\text{d}x\right)-\left(\sum_{n=1}^\infty \int_{E_n} f^-(x)\text{d}x\right) \\&=\sum_{n=1}^\infty \left(\int_{E_n} f^+(x)\text{d}x-\int_{E_n} f^-(x)\text{d}x\right) \\&=\sum_{n=1}^\infty \int_{E_n} f(x)\text{d}x. \end{aligned}\]

$\square$

Theorem 3.3 $f \in L(E) \Leftrightarrow |f| \in L(E)$.

Proof. ($\Leftarrow$) Suppose $f \in L(E)$. Then $\displaystyle \int_E f^+(x)\text{d}x<\infty$ and $\displaystyle \int_E f^-(x)\text{d}x$, and thus \[\int_E |f(x)|\text{d}x=\int_E (f^+(x)+f^-(x))\text{d}x=\int_E f^+(x)\text{d}x+\int_E f^-(x)\text{d}x<\infty,\] i.e., $|f| \in L(E)$.

($\Rightarrow$) Suppose $|f| \in L(E)$. Then $f^+(x) \leq |f(x)|$ and $f^-(x) \leq |f(x)|$. Hence, \[\int_E f^+(x)\text{d}x \leq \int_E |f(x)|\text{d}x<\infty\] and \[\int_E f^-(x)\text{d}x \leq \int_E |f(x)|\text{d}x<\infty.\] Therefore, $f \in L(E)$.

$\square$

Theorem 3.4 Suppose $|f(x)| \leq g(x)\ \text{a.e.}\ x \in E$, and $g$ is a non-negative Lebesgue integrable function on $E$. Then $f \in L(E)$, and \[\left|\int_E f(x)\text{d}x\right| \leq \int_E |f(x)|\text{d}x \leq \int_E g(x)\text{d}x.\]

Proof. Since $|f(x)| \leq g(x)\ \text{a.e.}\ x \in E$, $|f|$ and $g$ are non-negative measurable functions on $E$, and $g$ is Lebesgue integrable, then \[\int_E |f(x)|\text{d}x \leq \int_E g(x)\text{d}x<\infty,\] i.e., $|f| \in L(E)$. Hence, $f \in L(E)$.

In addition, \[\begin{aligned} \left|\int_E f(x)\text{d}x\right|&=\left|\int_E f^+(x)\text{d}x-\int_E f^-(x)\text{d}x\right| \\&\leq \left|\int_E f^+(x)\text{d}x\right|+\left|\int_E f^-(x)\text{d}x\right| \\&=\int_E f^+(x)\text{d}x+\int_E f^-(x)\text{d}x \\&=\int_E (f^+(x)+f^-(x))\text{d}x \\&=\int_E |f(x)|\text{d}x. \end{aligned}\]

$\square$

Theorem 3.5 Suppose $f \in L(E)$, and $A$ is a measurable subset of $E$, then \[\forall \varepsilon>0, \exists \delta>0\ \text{s.t.}\ m(A)<\delta \Rightarrow \left|\int_A f(x)\text{d}x\right| \leq \int_A |f(x)|\text{d}x<\varepsilon.\]

Proof. Take an arbitrary $\varepsilon>0$. Suppose $f$ is a non-negative simple function. We know $f(x) \leq M$, for all $x \in A$. Take $\displaystyle \delta=\frac{\varepsilon}{M+1}$, when $m(A)<\delta$, we have \[\int_A f(x)\text{d}x \leq M \cdot m(A)<\varepsilon.\]

Suppose $f$ is a measurable function, then $|f|$ is a non-negative measurable function. There exists a non-negative simple function such that $\varphi(x) \leq |f(x)|$ and \[\int_E \varphi(x)\text{d}x \leq \int_E |f(x)|\text{d}x \leq \int_E \varphi(x)+\frac{\varepsilon}{2}.\]

We take a $\delta$ such that when $m(A)<\delta$, $\displaystyle \int_A \varphi(x)\text{d}x<\frac{\varepsilon}{2}$. Hence, \[\begin{aligned} \int_A |f(x)|\text{d}x&=\int_A \varphi(x)\text{d}x+\int_A (|f(x)|-\varphi(x))\text{d}x \\&\leq \int_A \varphi(x)\text{d}x+\int_E(|f(x)|-\varphi(x))\text{d}x \\&=\int_A \varphi(x)\text{d}x+\int_E |f(x)|\text{d}x-\int_E \varphi(x)\text{d}x \\&<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. \end{aligned}\]

$\square$

Theorem 3.6 Suppose $f$ and $g$ are Lebesgue integrable functions, and $\alpha$ and $\beta$ are real numbers. Then $\alpha f+\beta g \in L(E)$, and \[\int_E (\alpha f(x)+\beta g(x))\text{d}x=\alpha\int_E f(x)\text{d}x+\beta\int_E g(x)\text{d}x.\]

Proof. Suppose $c \in \mathbb{R}$. When $c=0$, it is obvious that $cf \in L(E)$, and \[\int_E cf(x)\text{d}x=c\int_E f(x)\text{d}x.\]

When $c>0$, $(cf)^+=cf^+$ and $(cf)^-=cf^-$.

Since $f \in L(E)$, then $\displaystyle \int_E f^+(x)\text{d}x<\infty$ and $\displaystyle \int_E f^-(x)\text{d}x<\infty$, and thus \[\int_E (cf)^+(x)\text{d}x=\int_E cf^+(x)\text{d}x=c\int_E f^+(x)\text{d}x<\infty\] and \[\int_E (cf)^-(x)\text{d}x=\int_E cf^-(x)\text{d}x=c\int_E f^-(x)\text{d}x<\infty.\] Therefore, $cf(x) \in L(E)$, and \[\begin{aligned} \int_E cf(x)\text{d}x&=\int_E (cf)^+(x)\text{d}x-\int_E (cf)^-(x)\text{d}x \\&=c\int_E f^+(x)\text{d}x-c\int_E f^-(x)\text{d}x \\&=c\left(\int_E f^+(x)\text{d}x-\int_E f^-(x)\text{d}x\right) \\&=c\int_E f(x)\text{d}x. \end{aligned}\]

When $c<0$, $(cf)^+=|c|f^-$ and $(cf)^-=|c|f^+$. Similarly, we can show that $cf \in L(E)$, and \[\int_E cf(x)\text{d}x=c\int_E f(x)\text{d}x.\]

In addition, we know $|f|$ and $|g|$ are Lebesgue integrable, and \[\int_E(|f|+|g|)(x)\text{d}x=\int_E |f|(x)\text{d}x+\int_E |g|(x)\text{d}x<\infty,\] i.e., $|f|+|g| \in L(E)$. Since $|f+g| \leq |f|+|g|$, then $|f+g| \in L(E)$.

Since $f+g=(f+g)^+-(f+g)^-=f^+-f^-+g^+-g^-$, then \[(f+g)^++f^-+g^-=(f+g)^-+f^++g^+.\] Hence, \[\int_E ((f+g)^+(x)+f^-(x)+g^-(x))\text{d}x=\int_E ((f+g)^-(x)+f^+(x)+g^+(x))\text{d}x,\] and thus \[\int_E (f+g)^+(x)\text{d}x+\int_E f^-(x)\text{d}x+\int_E g^-(x)\text{d}x=\int_E (f+g)^-(x)\text{d}x+\int_E f^+(x)\text{d}x+\int_E g^+(x)\text{d}x.\] Therefore, \[\int_E (f+g)^+(x)\text{d}x-\int_E (f+g)^-(x)\text{d}x=\int_E f^+(x)\text{d}x-\int_E f^-(x)\text{d}x+\int_E g^+(x)\text{d}x-\int_E g^-(x)\text{d}x,\] i.e., \[\int_E (f(x)+g(x))\text{d}x=\int_E f(x)\text{d}x+\int_E g(x)\text{d}x.\]

Hence, it is easy to show that $\alpha f+\beta g \in L(E)$, and \[\int_E (\alpha f(x)+\beta g(x))\text{d}x=\alpha\int_E f(x)\text{d}x+\beta\int_E g(x)\text{d}x.\]

$\square$

3.1. Lebesgue's Dominated Convergence Theorem

Theorem 3.7 (Lebesgue's Dominated Convergence Theorem, DCT) Suppose $\{f_n\}$ is a sequence of measurable functions on $E$, and $\displaystyle \lim_{n \to \infty} f_n(x)=f(x)\ \text{a.e.}\ x \in E$. If there exists a non-negative Lebesgue integrable function $F$ on $E$ such that for all $n \in \mathbb{N}$, $|f_n(x)| \leq F(x)\ \text{a.e.}\ x \in E$, then \[\lim_{n \to \infty} \int_E f_n(x)\text{d}x=\int_E f(x)\text{d}x.\]

Proof. Since $|f_n(x)| \leq F(x)\ \text{a.e.}\ x \in E$, then $|f(x)| \leq F(x)\ \text{a.e.}\ x \in E$. Since $F$ is Lebesgue integrable, then $f$ and $f_n$ are Lebesgue integrable, and thus $F$, $f$ and $f_n$ are finite almost everywhere.

Assume without loss of generality that $F$, $f$ and $f_n$ are finite. Let $g_n(x)=|f_n(x)-f(x)| \geq 0$. We know $g_n(x) \leq |f_n(x)|+|f(x)| \leq 2F(x)\ \text{a.e.}\ x \in E$, then $2F-g_n$ is a non-negative measurable function.

By Fatou's lemma, \[\int_E \varliminf_{n \to \infty} (2F(x)-g_n(x))\text{d}x \leq \varliminf_{n \to \infty} \int_E (2F(x)-g_n(x))\text{d}x.\] Hence, \[\int_E 2F(x)\text{d}x-\int_E \varlimsup_{n \to \infty} g_n(x)\text{d}x \leq \int_E 2F(x)\text{d}x-\varlimsup_{n \to \infty} \int_E g_n(x)\text{d}x,\] and thus \[\varlimsup_{n \to \infty} \int_E g_n(x)\text{d}x \leq \int_E \varlimsup_{n \to \infty} g_n(x)\text{d}x.\] Since $\displaystyle \varlimsup_{n \to \infty} g_n(x)\text{d}x=0\ \text{a.e.}\ x \in E$, then \[\varlimsup_{n \to \infty} \int_E g_n(x)\text{d}x=0,\] and thus \[\lim_{n \to \infty} \int_E g_n(x)\text{d}x=\lim_{n \to \infty} \int_E |f_n(x)-f(x)|\text{d}x=0.\] Since \[\left|\int_E (f_n(x)-f(x))\text{d}x\right| \leq \int_E |f_n(x)-f(x)|\text{d}x,\] then \[\lim_{n \to \infty} \int_E (f_n(x)-f(x))\text{d}x=0,\] i.e., \[\lim_{n \to \infty} \int_E f_n(x)\text{d}x=\int_E f(x)\text{d}x.\]

$\square$

Levi's monotonicity theorem, Fatou's lemma and Lebesgue's dominated convergence theorem are equivalent to each other.

Theorem 3.8 Suppose $\{f_n\}$ is a sequence of measurable functions on $E$, and $f_n \Rightarrow f$. If there exists a non-negative Lebesgue integrable function $F$ on $E$ such that for all $n \in \mathbb{N}$, $|f_n(x)| \leq F(x)\ \text{a.e.}\ x \in E$, then \[\lim_{n \to \infty} \int_E f_n(x)\text{d}x=\int_E f(x)\text{d}x.\]

Assume without loss of generality that $F$, $f$ and $f_n$ are finite. Suppose \[\int_E |f_n(x)-f(x)|\text{d}x \not\to0,\] then we can find a subsequence $\{f_{n_k}\}_{k=1}^\infty$ such that $\displaystyle \lim_{k \to \infty} \int_E |f_{n_k}(x)-f(x)|\text{d}x=\alpha>0$.

By Riesz's theorem, there exists a subsequence $\{f_{n_{k_i}}\}_{i=1}^\infty$ such that $f_{n_{k_i}}(x) \to f(x)\ \text{a.e.}\ x \in E$, and thus \[\lim_{i \to \infty} \int_E |f_{n_{k_i}}(x)-f(x)|\text{d}x=0,\] which is a contradiction.

$\square$

Theorem 3.9 Suppose $\{f_n\}$ is a sequence of Lebesgue integrable functions on $E$. If \[\sum_{n=1}^\infty \int_E |f_n(x)|\text{d}x<\infty,\] then $\displaystyle \sum_{n=1}^\infty f_n(x)<\infty\ \text{a.e.}\ x \in E$, $\displaystyle \sum_{n=1}^\infty f_n$ is Lebesgue integrable on $E$, and \[\int_E \left(\sum_{n=1}^\infty f_n(x)\right)\text{d}x=\sum_{n=1}^\infty \int_E f_n(x)\text{d}x.\]

Proof. Let \[F(x)=\sum_{n=1}^\infty |f_n(x)|.\] We have \[\int_E F(x)\text{d}x=\sum_{n=1}^\infty \int_E |f_n(x)|\text{d}x<\infty.\] Hence, $F$ is a non-negative Lebesgue integrable function on $E$, and thus $F$ is finite almost everywhere on $E$, i.e., $\displaystyle \sum_{n=1}^\infty |f_n(x)|<\infty\ \text{a.e.}\ x \in E$. Therefore, $\displaystyle \sum_{n=1}^\infty f_n(x)<\infty\ \text{a.e.}\ x \in E$.

Let $\displaystyle g_n(x)=\sum_{k=1}^n f_k(x)$ and $\displaystyle g(x)=\sum_{n=1}^\infty f_n(x)=\lim_{n \to \infty} g_n(x)$. We have \[|g_n(x)|=\left|\sum_{k=1}^n f_k(x)\right| \leq \sum_{k=1}^n |f_k(x)| \leq F(x).\] By Lebesgue's dominated convergence theorem, $g \in L(E)$, and \[\lim_{n \to \infty} \int_E g_n(x)\text{d}x=\int_E g(x)\text{d}x,\] i.e., $\displaystyle \sum_{n=1}^\infty f_n$ is Lebesgue integrable on $E$, and \[\sum_{n=1}^\infty \int_E f_n(x)\text{d}x=\lim_{n \to \infty} \sum_{k=1}^n \int_E f_k(x)\text{d}x=\lim_{n \to \infty} \int_E \sum_{k=1}^n f_k(x)\text{d}x=\int_E \left(\sum_{n=1}^\infty f_n(x)\right)\text{d}x.\]

$\square$

Theorem 3.10 Suppose $f$ is a real function on $E \times (a, b)$. If for all $t \in (a, b)$, $f(x, t)$ (as a function of $x$) is Lebesgue integrable on $E$, for $\text{a.e.}\ x \in E$, $f(x, t)$ (as a function of $t$) is differentiable on $(a, b)$, and there exists a non-negative Lebesgue integrable function $F$ on $E$ such that \[\left|\frac{\partial}{\partial t}f(x, t)\right| \leq F(x),\] then $\displaystyle \frac{\partial}{\partial t}f(x, t)$ is Lebesgue integrable on $E$, $\displaystyle \int_E f(x, t)\text{d}x$ is differentiable on $(a, b)$, and \[\frac{\text{d}}{\text{d}t}\int_E f(x, t)\text{d}x=\int_E \frac{\partial}{\partial t}f(x, t)\text{d}x.\]

Proof. Fix a $t \in (a, b)$ and take an arbitrary sequence $\{h_n\}$, where $h_n \to 0$ and $h_n \neq 0$. Let \[g_n(x)=\frac{f(x, t+h_n)-f(x, t)}{h_n},\] then \[\lim_{n \to \infty} g_n(x)=\frac{\partial}{\partial t}f(x, t),\] and \[|g_n(x)|=\left|\frac{\partial}{\partial t}f(x, t+\theta_nh_n)\right| \leq F(x),\] where $0 \leq \theta_n \leq 1$.

Since $\displaystyle \frac{\partial}{\partial t}f(x, t)$ is Lebesgue integrable on $E$, and by Lebesgue's dominated convergence theorem, \[\lim_{n \to \infty} \frac{\displaystyle \int_E f(x, t+h_n)\text{d}x-\int_E f(x, t)\text{d}x}{h_n}=\lim_{n \to \infty} \int_E g_n(x)\text{d}x=\int_E \lim_{n \to \infty} g_n(x)\text{d}x=\int_E \frac{\partial}{\partial t}f(x, t)\text{d}x<\infty.\] Therefore, $\displaystyle \int_E f(x, t)\text{d}t$ is differentiable on $(a, b)$, and \[\frac{\text{d}}{\text{d}t}\int_E f(x, t)\text{d}x=\int_E \frac{\partial}{\partial t}f(x, t)\text{d}x.\]

$\square$

4. Riemann Integral and Lebesgue Integral

Recall 4.1 Suppose $f$ is bounded on $[a, b]$. A partition sequence of $[a, b]$, $\{P^{(n)}\}$ is a finite sequence of values $x_i$ such that \[a=x_0^{(n)}<x_1^{(n)}<\cdots<x_{k_n}^{(n)}=b.\] Let $M_i^{(n)}=\sup\{f(x): x_{i-1}^{(n)} \leq x \leq x_i^{(n)}\}$ and $m_i^{(n)}=\inf\{f(x): x_{i-1}^{(n)} \leq x \leq x_i^{(n)}\}$. The upper Darboux sum of $f$ with respect to $P^{(n)}$ is \[S(f, P^{(n)})=\sum_{i=1}^{k_n} M_i^{(n)}(x_i^{(n)}-x_{i-1}^{(n)}).\] The lower Darboux sum of $f$ with respect to $P^{(n)}$ is \[s(f, P^{(n)})=\sum_{i=1}^{k_n} m_i^{(n)}(x_i^{(n)}-x_{i-1}^{(n)}).\] The upper Darboux integral of $f$ is \[\overline{\int_a^b} f(x)\text{d}x=\inf_{P^{(n)}} S(f, P^{(n)})=\lim_{n \to \infty} S(f, P^{(n)}).\] The lower Darboux integral of $f$ is \[\underline{\int_a^b} f(x)\text{d}x=\sup_{P^{(n)}} s(f, P^{(n)})=\lim_{n \to \infty} s(f, P^{(n)}).\] $f(x)$ is Riemann integrable on $[a, b]$ if and only if \[\overline{\int_a^b} f(x)\text{d}x=\underline{\int_a^b} f(x)\text{d}x.\]

Recall 4.2 The amplitude at $x$ is \[\omega(x)=\lim_{\delta \to 0^+}\sup\{|f(y)-f(z)|: y, z \in (x-\delta, x+\delta) \cap [a, b]\}.\]

Theorem 4.1 A bounded function $f$ is Riemann integrable on $[a, b]$ if and only if $f$ is continuous almost everywhere on $[a, b]$.

Proof. Let \[h_n(x)=\begin{cases} M_i^{(n)}-m_i^{(n)}, &x_{i-1}^{(n)}<x<x_i^{(n)} \\ 0, &x=x_{i-1}^{(n)} \vee x=x_i^{(n)} \end{cases}.\] We know $h_n$ is a non-negative measurable function. Suppose $|f(x)| \leq M$, $x \in [a, b]$, then \[|h_n(x)| \leq M_i^{(n)}+m_i^{(n)} \leq 2M.\] Since $g(x)=2M$ is Lebesgue integrable on $[a, b]$, then \[\begin{aligned} \overline{\int_a^b} f(x)\text{d}x-\underline{\int_a^b} f(x)\text{d}x&=\lim_{n \to \infty} (S(f, P^{(n)})-s(f, P^{(n)})) \\&=\lim_{n \to \infty} \sum_{i=1}^{k_n} (M_i^{(n)}-m_i^{(n)})(x_i^{(n)}-x_{i-1}^{(n)}) \\&=\lim_{n \to \infty} \int_{[a, b]} h_n(x)\text{d}x \\&=\int_{[a, b]} \lim_{n \to \infty} h_n(x)\text{d}x & \text{(DCT)} \\&=\int_{[a, b]} \omega(x)\text{d}x=0. \end{aligned}\]

Therefore, $f$ is Riemann integrable on $[a, b]$ if and only if \[\begin{aligned} &\quad \quad \int_{[a, b]} \omega(x)\text{d}x=0 \\&\Leftrightarrow \omega(x)=0\ \text{a.e.}\ x \in [a, b] \\&\Leftrightarrow \text{$f$ is continuous almost everywhere on $[a, b]$.} \end{aligned}\]

$\square$

Theorem 4.2 If a bounded function $f$ is Riemann integrable on $[a, b]$, then $f$ is Lebesgue integrable on $[a, b]$, and \[\int_{[a, b]} f(x)\text{d}x=\int_a^b f(x)\text{d}x.\]

Proof. Since $f$ is Riemann integrable on $[a, b]$, then $f$ is continuous almost everywhere on $[a, b]$. Hence, $f$ is a bounded measurable function on $[a, b]$, and thus $f \in L([a, b])$.

Let \[g_n(x)=\begin{cases} M_i^{(n)}, &x_{i-1}^{(n)}<x<x_i^{(n)} \\ 0, &x=x_{i-1}^{(n)} \vee x=x_i^{(n)} \end{cases},\] where $\displaystyle \lim_{n \to \infty} g_n(x)=f(x)\ \text{a.e.}\ x \in [a, b]$, and $g_n$ is bounded. We have \[\begin{aligned} \int_a^b f(x)\text{d}x&=\lim_{n \to \infty} \sum_{i=1}^{k_n} M_i^{(n)}(x_i^{(n)}-x_{i-1}^{(n)}) \\&=\lim_{n \to \infty} \int_{[a, b]} g_n(x)\text{d}x \\&=\int_{[a, b]} \lim_{n \to \infty} g_n(x)\text{d}x \\&=\int_{[a, b]} f(x)\text{d}x. \end{aligned}\]

$\square$

Theorem 4.4 Suppose $f$ is a non-negative real function on $[a, b]$. If for all $A>a$, $f$ is Riemann integrable on $[a, A]$, and $\displaystyle \int_a^{\infty} f(x)\text{d}x<\infty$, then $f$ is Lebesgue integrable on $[a, \infty)$, and \[\int_{[a, \infty)} f(x)\text{d}x=\int_a^\infty f(x)\text{d}x.\]

Proof. Take an arbitrary sequence $\{A_n\}$ such that $A_n>a$ and $\displaystyle \lim_{n \to \infty} A_n=\infty$. Hence, \[\int_{[a, \infty)} f(x)\text{d}x=\lim_{n \to \infty} \int_{[a, A_n]} f(x)\text{d}x=\lim_{n \to \infty}\int_a^{A_n} f(x)\text{d}x=\int_a^\infty f(x)\text{d}x.\]

In addition, if $\displaystyle \int_a^\infty f(x)\text{d}x$ diverges, then $\displaystyle \int_{[a, \infty)} f(x)\text{d}x=\infty$.

$\square$

If $f$ is not non-negative function, then Lebesgue integral is not a generalization of Riemann improper integral.

5. Geometric Interpretation of Lebesgue Integral

Definition 5.1 Suppose $f$ is a non-negative function on $E \subset \mathbb{R}^n$. We define \[G(E, f)=\{(x, z): x \in E, 0 \leq z<f(x)\} \subset \mathbb{R}^{n+1}.\]

Theorem 5.1 Suppose $f$ is a non-negative function on $E \subset \mathbb{R}^n$, then

(1) $f$ is measurable on $E$ if and only if $G(E, f)$ is measurable;

(2) If $f$ is measurable on $E$, then \[\int_E f(x)\text{d}x=m(G(E, f)).\]

6. Fubini's Theorem

Theorem 6.1 (Fubini's Theorem) Suppose $A \subset \mathbb{R}^p$ and $B \subset \mathbb{R}^q$ are measurable. If $f$ is a non-negative measurable function on $A \times B$, then for $\text{a.e.} x \in A$, $f(x, y)$ (as a function of $y$) is measurable on $B$, and \[\int_{A \times B} f(x, y)\text{d}x\text{d}y=\int_A \left(\int_B f(x, y)\text{d}y\right)\text{d}x.\] If $f$ is Lebesgue integrable on $A \times B$, then for $\text{a.e.} x \in A$, $f(x, y)$ (as a function of $y$) is Lebesgue integrable on $B$, $\displaystyle \int_B f(x, y)\text{d}y$ (as a function of $x$) is Lebesgue integrable on $A$, and \[\int_{A \times B} f(x, y)\text{d}x\text{d}y=\int_A \left(\int_B f(x, y)\text{d}y\right)\text{d}x.\]

Example 6.1 Let $\displaystyle f(x, y)=\frac{x^2-y^2}{(x^2+y^2)^2}$ and $E=(0, 1) \times (0, 1)$. We have \[\int_{(0, 1)} \left(\int_{(0, 1)} f(x, y)\text{d}y\right)\text{d}x=\frac{\pi}{4}\] and \[\int_{(0, 1)} \left(\int_{(0, 1)} f(x, y)\text{d}x\right)\text{d}y=-\frac{\pi}{4}.\] By Fubini's theorem, $f$ is not Lebesgue integrable.