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本帖最后由 elim 于 2020-7-27 02:38 编辑
这个函数很有趣:
\(\,\forall x>0\,\exists\,n\in\mathbb{N}^+: (\frac{1}{5^n}\le x,\;f(x)\ge f(\frac{1}{5^n})=\frac{1}{2^n}>0)\)
对\(\,\frac{1}{5^{n+1}}< x<\frac{1}{5^n},\;f(\frac{1}{5}-x)\le f(\frac{1}{5}-\frac{1}{5^{n+1}})=\frac{1}{2}f(1-\frac{1}{5^n})\)
\(\quad=\frac{1}{2}(1-\frac{1}{2^n})< \frac{1}{2},\;\;\therefore\;f(\frac{4}{5}+x)=1-f(\frac{1}{5}-x)>\frac{1}{2}\)
\(\therefore\;f^{-1}(\{\frac{1}{2}\})=[\frac{1}{5},\frac{4}{5}],\;\boxed{f^{-1}(\{{\small\frac{1}{2^n}}\})=[{\small\frac{1}{5^n},\frac{4}{5^n}}],\small\;n=1,2,\ldots}\)
于是\(\,f^{-1}(\{1-\frac{1}{2^n}\})=[1-\frac{4}{5^n},1-\frac{1}{5^n}],\)
\(\qquad f^{-1}(\{\frac{1}{2}(1-\frac{1}{2^n})\})=[\frac{1}{5}-\frac{4}{5^{n+1}},\frac{1}{5}-\frac{1}{5^{n+1}}]\)
\(\qquad f^{-1}(\{\frac{1}{2}(1+\frac{1}{2^n})\})=[\frac{4}{5}(1+\frac{1}{5^n}),\frac{1}{5}(4-\frac{1}{5^n})]\)
\(\qquad f^{-1}(\{\frac{1}{2^2}(1+\frac{1}{2^n})\})=[\frac{1}{5^2}(4-\frac{1}{5^n}),\frac{4}{5^2}(1+\frac{1}{5^n})]\)
\(\cdots\cdots\)
如此可知函数在[0,1]的一个可数两两不交的闭区间序列上取有理数值,这些区间的并的测度是多少?
更基本的问题是: 满足这组性质的函数是否存在? 或者说, 这组性质是否相容? |
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