定义1
设\(B \subset \mathbb{R}^2\)是有界集，函数\(f: B \to \mathbb{R}\)，令
\[ f_B(\boldsymbol{p}) = \left\{ \begin{aligned} & f(\boldsymbol{p}), & \boldsymbol{p} \in B \\ & 0, & \boldsymbol{p} \in B^c \end{aligned} \right. \]
则函数\(f_B\)在全平面\(\mathbb{R}^2\)上有定义，如果限制在集合\(B\)上，\(f_B\)与\(f\)相等。
定义2
任取有界的闭矩形\(I \supset B\)，如果函数\(f_B\)在\(I\)上可积，则称函数\(f\)在\(B\)上可积，并称数值\(\displaystyle \int_I f_B \mathrm{d} \sigma\)为函数\(f\)在\(B\)上的（二重）积分，记作
\[ \iint \limits_B f(x, y) \mathrm{d} x \mathrm{d} y \quad 或 \quad \int_B f \mathrm{d} \sigma \]
定理1
设有界集\(B \subset \mathbb{R}^2\)，函数\(f: B \to \mathbb{R}\)有界，如果集合\(B\)的边界\(\partial B\)和\(f\)在\(B\)上的间断点都是零测集，那么\(f\)在\(B\)上可积。
证：取闭矩形\(I\)，满足\(I^\circ \supset \overline B\)，由于\(f_B\)在\(\overline B^c\)上处处为零，所以\(\overline B^c\)中的每个点都是\(f_B\)的连续点，在\(B^\circ\)上，\(f_B = f\)，所以在\(B^\circ\)上\(f_B\)的不连续点就是\(f\)的不连续点，从而
\[
D(f_B) \subset D(f) \cup \partial B
\]
由于\(D(f)\)与\(\partial B\)都是零测集，所以\(D(f_B)\)也是零测集，即\(f_B\)在\(I\)上可积，从而\(f\)在\(B\)上可积。
Q.E.D.
定理2
设有界集\(B \subset \mathbb{R}^2\)，\(f: B \to \mathbb{R}\)在\(B\)上可积，那么对任意常数\(c\)，函数\(cf\)在\(B\)上也可积，并且
\[ \int_B cf \mathrm{d} \sigma = c \int_B f \mathrm{d} \sigma \]
又若\(g: B \to \mathbb{R}\)在\(B\)上可积，那么\(f \pm g\)也在\(B\)上可积，且
\[ \int_B (f \pm g) \mathrm{d} \sigma = \int_B f \mathrm{d} \sigma \pm \int_B g \mathrm{d} \sigma \]
证：对于第一个等式，设闭矩形\(I \supset
B\)，有
\[
\int_B cf \mathrm{d} \sigma = \int_I cf_B \mathrm{d} \sigma = c
\int_I f_B \mathrm{d} \sigma = c \int_B f \mathrm{d} \sigma
\]
对于第二个等式，同样有
\[
\int_B (f \pm g) \mathrm{d} \sigma = \int_I (f_B \pm g_B) \mathrm{d}
\sigma = \int_I f_B \mathrm{d} \sigma \pm \int_I g_B \mathrm{d} \sigma =
\int_B f \mathrm{d} \sigma \pm \int_B g \mathrm{d} \sigma
\]
Q.E.D.
定理3
设\(B_1, B_2 \subset \mathbb{R}^2\)有界，且\(B_1 \cap B_2\)是零面积集，若函数\(f\)在\(B_1\)和\(B_2\)上都可积，那么\(f\)在\(B_1 \cup B_2\)上可积，且
\[ \int_{B_1 \cup B_2} f \mathrm{d} \sigma = \int_{B_1} f \mathrm{d} \sigma + \int_{B_2} f \mathrm{d} \sigma \]
证：易知，只要\(\boldsymbol{x} \notin B_1
\cap B_2\)，有\(f_{B_1 \cup
B_2}(\boldsymbol{x}) = f_{B_1}(\boldsymbol{x}) +
f_{B_2}(\boldsymbol{x})\)，从而使得\(f_{B_1 \cup B_2}\)与\(f_{B_1} + f_{B_2}\)不相等的集合必属于\(B_1 \cap B_2\)，即为零面积集。所以由函数积分八的定理9可知，
\[
\int_I (f_{B_1} + f_{B_2}) \mathrm{d} \sigma = \int_I f_{B_1 \cap
B_2} \mathrm{d} \sigma
\]
设矩形\(I \supset B\)，则有
\[
\int_{B_1} f \mathrm{d} \sigma + \int_{B_2} f \mathrm{d} \sigma =
\int_I f_{B_1} \mathrm{d} \sigma + \int_I f_{B_2} \mathrm{d} \sigma =
\int_I (f_{B_1} + f_{B_2}) \mathrm{d} \sigma = \int_I f_{B_1 \cap B_2}
\mathrm{d} \sigma
\]
Q.E.D.
定义3
设\(B \subset \mathbb{R}^2\)有界，若常值函数\(1\)在\(B\)上可积，那么积分\(\displaystyle \int_B 1 \mathrm{d} \sigma\)称为点集\(B\)的面积，记为\(\sigma(B)\)，这时称\(B\)是有面积的。记函数
\[ \chi_B (\boldsymbol{p}) = \left\{ \begin{aligned} 1, & \quad \boldsymbol{p} \in B \\ 0, & \quad \boldsymbol{p} \notin B \end{aligned} \right. \]
称为\(B\)的特征函数，从而\(B\)的面积也可以写为
\[ \sigma(B) = \int_B 1 \mathrm{d} \sigma = \int_I \chi_B \mathrm{d} \sigma \]
这里\(I\)是任何一个包含\(B\)的闭矩形。
定理4
设\(B \subset \mathbb{R}^2\)是一个有界集，且\(\displaystyle \sigma(B) = \int_B 1 \mathrm{d} \sigma\)存在，则
\[ D(\chi_B) = \partial B \]
证：由于
\[
\mathbb{R}^2 = B^\circ \cup \partial B \cup (B^c)^\circ
\]
而在开集\(B^\circ\)和\((B^c)^\circ\)上，\(\chi_B\)分别为\(1\)和\(0\)，所以\(\chi_B\)在\(B^\circ\)和\((B^c)^\circ\)上连续，从而
\[
D(\chi_B) \subset \partial B \tag {1}
\]
任取\(\boldsymbol{p} \in \partial
B\)，则在\(\boldsymbol{p}\)的任意小的领域内即有\(B\)中的点\(\boldsymbol{p}^\prime\)，又有\(B^c\)中的点\(\boldsymbol{p}^{\prime\prime}\)，而\(\chi_B(\boldsymbol{p}^\prime) = 1,
\chi_B(\boldsymbol{p}^{\prime\prime})=0\)，所以\(\chi_B\)在\(\boldsymbol{p}\)处不连续，所以
\[
\partial B \subset D(\chi_B) \tag {2}
\]
由\((1)(2)\)即得\(D(\chi_B) = \partial B\)。
Q.E.D.
定理5
设\(B \subset \mathbb{R}^2\)是一个有界集，则点集\(B\)为零面积集的充分必要条件是
\[ \sigma (B) = \int_B 1 \mathrm{d} \sigma = 0 \]
证：必要性。作闭矩形\(I\)使得\(I^{\circ} \supset \overline B\)，则\(B\)上的特征函数\(\chi_B\)在\(I\)上取非零值的点集正好是\(B\)，而\(B\)是零面积集，由函数积分八的定理8可知，
\(\chi_B\)在\(I\)上可积且
\[
\sigma(B) = \int_B 1 \mathrm{d} \sigma = \int_I \chi_B \mathrm{d}
\sigma = 0
\]
这表明\(B\)的面积为零。
充分性。由于\(\sigma(B) = 0\)，从而\(B^\circ = \varnothing\)，又因为\(\chi_B\)可积，从而\(D(\chi_B)\)是零测集，再由定理4可知\(\partial B\)是零测集，而\(\partial B\)又是有界闭集，从而由函数积分八的定理1可知\(\partial B\)是零面积集，由\(B \subset \partial B \cup B^\circ = \partial B\)，得\(B \subset \partial B\)，所以\(B\)也是零面积集。
Q.E.D.
定理6
设有界集\(B \subset \mathbb{R}^2\)，则\(B\)有面积当且仅当\(B\)的边界\(\partial B\)是一零面积集。
证：由于\(\partial B\)是有界闭集，所以\(\partial B\)是零面积集\(\Leftrightarrow \partial B\)是零测集\(\Leftrightarrow D(\chi_B)\)是零测集\(\Leftrightarrow \sigma(B)\)存在\(\Leftrightarrow B\)有面积。
Q.E.D.
定理7
设\(B\)是\(\mathbb{R}^2\)中的有面积的点集，\(f\)在\(B\)上可积，对\(B\)的任意分割\(T\)作Riemann和，那么对任意的\(\boldsymbol{\xi}_i \in D_i(i=1,2,\cdots,m)\)有
\[ \lim_{\Vert T \Vert \to 0} \sum_{i=1}^m f(\boldsymbol{\xi}_i) \sigma(D_i) = \int_B f \mathrm{d} \sigma \]
其中\(\Vert T \Vert = \max \limits_{1 \le i \le m} \mathrm{diam}(D_i)\)。也就是说，对任意的\(\varepsilon > 0\)，存在\(\delta>0\)，只要分割\(T = \{D_1, D_2, \cdots, D_n\}\)满足\(\Vert T \Vert < \delta\)，就有
\[ \left|\sum_{i=1}^m f(\boldsymbol{\xi}_i) \sigma(D_i) - \int_B f \mathrm{d} \sigma \right| < \varepsilon \]
证：由于\(f\)在\(B\)上可积，从而对任意的\(\varepsilon > 0\)，存在矩形网的分割\(\pi_{\varepsilon} = \{J_1, J_2, \cdots, J_t
\}\)，使得
\[
\overline S(f, \pi_{\varepsilon}) - \underline S(f,
\pi_{\varepsilon}) < \frac{\varepsilon}{2}
\]
即
\[
\sum_{i=1}^t (\sup f(J_i) - \inf f(J_i)) \sigma(J_i) <
\frac{\varepsilon}{2}
\]
这里\(J_i \subset B
(i=1,2,\cdots,t)\)，现将子矩形\(J_i\)的每一边平行地向内部收缩同一距离\(\delta > 0\)，作成一个开矩形\(\tilde {J_i} \subset J_i
(i=1,2,\cdots,t)\)。记
\[
K = I \bigcap (\bigcup \limits_{i=1}^t \tilde J_i)^c
\]
这里\(I\)是所有包含在\(B\)中的闭子矩形\(J_i\)的并，显然\(K\)是闭集，现取\(\delta > 0\)充分小，使得\(\sigma(K) < \varepsilon /
2\omega\)，这里\(\omega\)是\(f\)在\(B\)上的振幅。对这个\(\delta > 0\)，任取分割\(T = \{D_1, D_2, \cdots, D_m\}\)，使得\(\Vert T \Vert < \delta\)，记\(\displaystyle A = \int_B f \mathrm{d}
\sigma\)，则有
\[
A = \sum_{i=1}^m \int_{D_i} f \mathrm{d} \sigma
\]
设\(f\)在\(D_i\)上的上、下确界分别为\(M_i, m_i\)，则
\[
m_i \sigma(D_i) \le \int_{D_i} f \mathrm{d} \sigma \le M_i
\sigma(D_i)
\]
记\(\displaystyle \mu_i =
\frac{1}{\sigma(D_i)} \int_{D_i} f \mathrm{d} \sigma\)，从而\(m_i \le \mu_i \le M_i\)，又有
\[
\begin{aligned}
\left| A - \sum_{i=1}^m f(\boldsymbol{\xi}_i)\sigma(D_i) \right|
&= \left| \sum_{i=1}^m (\mu_i - f(\boldsymbol{\xi}_i)) \sigma(D_i)
\right| \\
& \le \sum_{i=1}^m \left| \mu_i - f(\boldsymbol{\xi}_i) \right|
\sigma(D_i) \\
& \le \sum_{i=1}^m (M_i - m_i) \sigma(D_i) \\
& = \sum_{i=1}^m \omega_i \sigma(D_i) \\
& = \sum \nolimits_1 + \sum \nolimits_2
\end{aligned}
\]
其中
\[
\sum \nolimits_1 = \sum_{D_i \subset K} \omega_i \sigma(D_i) \quad
\sum \nolimits_2 = \sum_{D_i \nsubseteq K}\omega_i \sigma(D_i)
\]
而
\[
\sum \nolimits_1 \le \omega \sum_{D_1 \subset K} \sigma(D_i) \le
\omega_i \sigma(K) < \frac{\varepsilon}{2}
\]
对于\(\sum_2\)，由于\(D_i \nsubseteq K\)，从而\(D_i\)必与某个\(\tilde {J_j}\)相交，所以必有\(D_i \subset J_j\)，从而有
\[
\begin{aligned}
\sum \nolimits_2 = \sum_{j=1}^t \sum_{D_i \subset J_j} \omega_i
\sigma(D_i) & \le \sum_{j=1}^t (\sup f(J_j) - \inf f(J_j)) \sum_{D_i
\subset J_j} \sigma(D_i) \\
& \le \sum_{j=1}^t (\sup f(J_j) - \inf f(J_j)) \sigma(J_i) <
\frac{\varepsilon}{2}
\end{aligned}
\]
所以有
\[
\left| A - \sum_{i=1}^m f(\boldsymbol{\xi}_i)\sigma(D_i) \right|
\le \sum \nolimits_1 + \sum \nolimits_2 < \varepsilon
\]
Q.E.D.
定理8：积分平均值定理
设\(K\)是\(\mathbb{R}^2\)中的有线条光滑曲线围成的有界闭区域，函数\(f, g: K \to \mathbb{R}\)连续且\(g\)在\(K\)上不变号，于是存在一点\(\boldsymbol{\xi} \in K\)，满足
\[ \int_K fg \mathrm{d} \sigma = f(\boldsymbol{\xi}) \int_K g \mathrm{d} \sigma \]
证：连续函数\(g\)与\(fg\)在\(K\)上都可积，因为\(K\)是紧致集，所以连续函数\(f\)在\(K\)上取得最小值\(f(\boldsymbol{a})\)与最大值\(f(\boldsymbol{b})\)，不妨设在\(K\)上\(g \le
0\)，于是
\[
f(\boldsymbol{a})g(\boldsymbol{p}) \le
g(\boldsymbol{p})g(\boldsymbol{p}) \le
f(\boldsymbol{b})g(\boldsymbol{p})
\]
对一切\(\boldsymbol{p} \in
K\)都成立，从而有
\[
f(\boldsymbol{a})\int_K g \mathrm{d} \sigma \le \int_K fg \mathrm{d}
\sigma \le f(\boldsymbol{b}) \int_K g \mathrm{d} \sigma
\]
若\(\displaystyle \int_K g \mathrm{d} \sigma =
0\)，这时\(g =
0\)，定理自然成立。现设\(\displaystyle
\int_K g \mathrm{d} \sigma > 0\)，于是
\[
f(\boldsymbol{a}) \le \left( \int_k g \mathrm{d} \sigma \right)^{-1}
\int_K fg \mathrm{d} \sigma \le f(\boldsymbol{b})
\]
由于\(K\)是连通集，而\(f\)在\(K\)上连续，从而由介值定理可知，存在一点\(\boldsymbol{\xi} \in K\)，使得
\[
f(\boldsymbol{\xi}) = \left( \int_K g \mathrm{d} \sigma \right)^{-1}
\int_K fg \mathrm{d} \sigma
\]
Q.E.D.
定理9
设\(K\)是\(\mathbb{R}^2\)中的有界闭区域，函数\(f\)在\(K\)上连续，那么存在一点\(\boldsymbol{\xi} \in K\)，使得
\[ \int_K f \mathrm{d} \sigma = f(\boldsymbol{\xi}) \sigma(K) \]
证：由定理8，令\(g = 1\)即得。