家里蹲大学数学杂志第7卷第481期一道实分析题目参考解答
2016-07-23 09:01
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(1) Define what it means for a set $A\subset \bbR^2$ to have zero content.
(2) Prove the following result: Let $g:[a,b]\to\bbR$ be bounded and integrable. Show that its graph $$\bex graph(g)=\sed{(x,g(x));x\in[a,b]} \eex$$ has zero content.
Proof:
(1) If $$\bex \inf\sed{\sum_{i=1}^\infty |I_i|;\ A\subset \cup_{i=1}^\infty I_i}=0, \eex$$ then $A$ is said to have zero content. Here, $\sed{I_i}_{i=1}^\infty$ are rectangles with $|I_i|$ being their areas.
(2) Since $g$ is (Riemann) integrable, we have $$\bex \lim_{\sen{T}\to 0}\sum_{i=1}^n (M_i-m_i)\lap x_i=0, \eex$$ where $$\bex T:\ a=x_0<x_1<\cdots<x_n=b, \eex$$ $$\bex \sen{T}=\max_i \lap x_i,\ \lap x_i=x_i-x_{i-1}, \eex$$ $$\bex M_i=\sup_{x\in [x_{i-1},x_i]}f(x),\quad m_i=\inf_{x\in [x_{i-1},x_i]}f(x). \eex$$ Thus (by the $\ve-\del$ definition of limit), $$\bex \forall\ \ve>0,\ \exists\ T,\st graph(f)\subset \cup_{i=1}^n [x_{i-1},x_i]\times [m_i,M_i], \eex$$ $$\bex |[x_{i-1},x_i]\times [m_i,M_i]| =\sum_{i=1}^n (M_i-m_i)\lap x_i<\ve. \eex$$ Consequently, $$\bex \inf\sed{\sum_{i=1}^\infty |I_i|;\ graph(f)\subset \cup_{i=1}^\infty I_i}=0 \eex$$ This yields the desired result.
(2) Prove the following result: Let $g:[a,b]\to\bbR$ be bounded and integrable. Show that its graph $$\bex graph(g)=\sed{(x,g(x));x\in[a,b]} \eex$$ has zero content.
Proof:
(1) If $$\bex \inf\sed{\sum_{i=1}^\infty |I_i|;\ A\subset \cup_{i=1}^\infty I_i}=0, \eex$$ then $A$ is said to have zero content. Here, $\sed{I_i}_{i=1}^\infty$ are rectangles with $|I_i|$ being their areas.
(2) Since $g$ is (Riemann) integrable, we have $$\bex \lim_{\sen{T}\to 0}\sum_{i=1}^n (M_i-m_i)\lap x_i=0, \eex$$ where $$\bex T:\ a=x_0<x_1<\cdots<x_n=b, \eex$$ $$\bex \sen{T}=\max_i \lap x_i,\ \lap x_i=x_i-x_{i-1}, \eex$$ $$\bex M_i=\sup_{x\in [x_{i-1},x_i]}f(x),\quad m_i=\inf_{x\in [x_{i-1},x_i]}f(x). \eex$$ Thus (by the $\ve-\del$ definition of limit), $$\bex \forall\ \ve>0,\ \exists\ T,\st graph(f)\subset \cup_{i=1}^n [x_{i-1},x_i]\times [m_i,M_i], \eex$$ $$\bex |[x_{i-1},x_i]\times [m_i,M_i]| =\sum_{i=1}^n (M_i-m_i)\lap x_i<\ve. \eex$$ Consequently, $$\bex \inf\sed{\sum_{i=1}^\infty |I_i|;\ graph(f)\subset \cup_{i=1}^\infty I_i}=0 \eex$$ This yields the desired result.
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