方企勤 第六章 多元函数积分学 第5题
📝 题目
例 5 若 $f\left( x\right) \in R\left\lbrack {a,b}\right\rbrack$ . 证明: $f\left( x\right) \in R\left( {\left\lbrack {a,b}\right\rbrack \times \left\lbrack {c,d}\right\rbrack }\right)$ .
💡 答案解析
证 $\forall \varepsilon > 0$ ,由 $f\left( x\right) \in R\left\lbrack {a,b}\right\rbrack$ ,3 区间 $\left\lbrack {a,b}\right\rbrack$ 的一个分法 $\Delta$ :
$$ a = {x}_{0} < {x}_{1} < \cdots < {x}_{n} = b. $$
令
$$ {M}_{i} = \mathop{\sup }\limits_{{{x}_{i - 1} \leq x \leq {x}_{i}}}f\left( x\right) ,\;{m}_{i} = \mathop{\inf }\limits_{{{x}_{i - 1} \leq x \leq {x}_{i}}}f\left( x\right) \left( {i = 1,\cdots ,n}\right) . $$
相应的大和与小和为
$$ {S}^{ + }\left( {f,\Delta }\right) = \mathop{\sum }\limits_{{i = 1}}^{n}{M}_{i}\Delta {x}_{i},\;{S}^{ - }\left( {f,\Delta }\right) = \mathop{\sum }\limits_{{i = 1}}^{n}{m}_{i}\Delta {x}_{i} $$
$$ \left( {\Delta {x}_{i} = {x}_{i} - {x}_{i - 1},i = 1,\cdots ,n}\right) . $$
根据一元函数可积的充要条件, 我们有
$$ {S}^{ + }\left( {f,\Delta }\right) - {S}^{ - }\left( {f,\Delta }\right) < \frac{\varepsilon }{d - c}. \tag{1.1} $$
对应区间分法 $\Delta$ ,可得矩形的一个剖分 $\widetilde{\Delta } = \left\{ {{\Omega }_{1},\cdots ,{\Omega }_{n}}\right\}$ ,其中 ${\Omega }_{i} =$ $\left\{ {\left( {x,y}\right) \mid {x}_{i - 1} \leq x \leq {x}_{i},c \leq y \leq d}\right\} \left( {i = 1,\cdots ,n}\right)$ . 函数 $f$ 在 ${\Omega }_{i}$ 上的上确界与下确界仍为 ${M}_{i}$ 与 ${m}_{i}\left( {i = 1,\cdots ,n}\right)$ . 故对应剖分 $\widetilde{\Delta }$ 的大和与小和为
$$ {\widetilde{S}}^{ + }\left( {f,\widetilde{\Delta }}\right) = \mathop{\sum }\limits_{{i = 1}}^{n}{M}_{i}\Delta {x}_{i}\left( {d - c}\right) ,\;{\widetilde{S}}^{ - }\left( {f,\widetilde{\Delta }}\right) = \mathop{\sum }\limits_{{i = 1}}^{n}{m}_{i}\Delta {x}_{i}\left( {d - c}\right) . $$
由 (1.1) 式得出
$$ {\widetilde{S}}^{ + }\left( {f,\widetilde{\Delta }}\right) - {\widetilde{S}}^{ - }\left( {f,\widetilde{\Delta }}\right) < \varepsilon . $$
根据重积分可积的充要条件,知 $f$ 在矩形 $\left\lbrack {a,b}\right\rbrack \times \left\lbrack {c,d}\right\rbrack$ 上可积.