📝 题目
例 6 在 ${\mathbf{R}}^{2}$ 上给定 $\mathbf{\Omega } = \{ \left( {x,y}\right) \mid 0 \leq x \leq 1,0 \leq y \leq 1\} \cup \{ \left( {x,y}\right) \mid$ $x = 0,1 \leq y \leq 2\}$ ,及函数
$$ f\left( {x,y}\right) = \left\{ \begin{array}{ll} 1, & 0 \leq x \leq 1,0 \leq y \leq 1, \\ \frac{1}{2 - y}, & x = 0,1 < y < 2, \\ 1, & x = 0,y = 2. \end{array}\right. $$
💡 答案与解析
证明: 无界函数 $f\left( {x,y}\right)$ 在 $\Omega$ 上可积.
证 作 $\Omega$ 的剖分 $\Delta = \left\{ {{\Omega }_{1},\cdots ,{\Omega }_{n}}\right\}$ ,令 $\parallel \Delta \parallel = \delta < 1$ ,则
$$ 1 = \mathop{\sum }\limits_{{i = 1}}^{n}V\left( {\Omega }_{i}\right) \leq \mathop{\sum }\limits_{{i = 1}}^{n}f\left( {{x}_{i},{y}_{i}}\right) V\left( {\Omega }_{i}\right) \;\left( {\left( {{x}_{i},{y}_{i}}\right) \in {\Omega }_{i}}\right) . $$
为了估计上界,把 ${\Omega }_{i}$ 分成三类: $以\left( {0,1}\right)$ 为心,以 $\delta$ 为半径的圆记作 $U$ ,整个落在 $\bar{U}$ 内的 ${\Omega }_{i}$ 归作第二类. 不完全落在 $\bar{U}$ 内的 ${\Omega }_{i}$ ,要么整个落在正方形 $0 \leq x \leq 1,0 \leq y \leq 1$ 内,归作第一类; 要么整个落在 $x = 0,1 \leq y \leq 2$ 上,归作第三类. 容易看出
$$ 1 \leq \mathop{\sum }\limits_{{i = 1}}^{n}f\left( {{x}_{i},{y}_{i}}\right) V\left( {\Omega }_{i}\right) $$
$$ = \mathop{\sum }\limits_{1}f\left( {{x}_{i},{y}_{i}}\right) V\left( {\Omega }_{i}\right) + \mathop{\sum }\limits_{2}f\left( {{x}_{i},{y}_{i}}\right) V\left( {\Omega }_{i}\right) $$
$$ + \mathop{\sum }\limits_{i}f\left( {{x}_{i},{y}_{i}}\right) V\left( {\Omega }_{i}\right) $$
$$ \leq 1 + \frac{1}{2 - 1 - \delta } \cdot \pi {\delta }^{2} + 0. $$
令 $\parallel \Delta \parallel = \delta \rightarrow 0$ ,得
$$ \mathop{\lim }\limits_{{\parallel \Delta \parallel \rightarrow 0}}\mathop{\sum }\limits_{{i = 1}}^{n}f\left( {{x}_{i},{y}_{i}}\right) V\left( {\Omega }_{i}\right) = 1, $$
故 $f\left( {x,y}\right) \in R\left( \Omega \right)$ .