Suppose $f$ is a real-valued function on $\mathbb R$ such that $f^{−1}(c)$ is measurable for each number $c$. Is $f$ necessarily measurable?
Must $f$ be measurable if each $f^{-1}(c)$ is?
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real-analysis
measure-theory
lebesgue-integral
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0What does the notation 1(c) mean? – 2012-12-24
3 Answers
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Hint: Every injective function satisfies the hypothesis. You can take any nonmeasurable set, map it injectively into $(0,\infty)$, and map its complement injectively into $(-\infty,0)$ (for example using $e^x$ for a simple formula).
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0@user54144: Not every non-measurable set is a Vitali set. There is a plethora of non-measurable sets. – 2012-12-24
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Hint: Note that the Vitali set has size $2^{\aleph_0}$ and therefore there is a bijection between $\mathbb R$ and the Vitali set.
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0Because a measurable set (of positive measure) can always be written as the union of two non measurable sets. That's not a big deal. But $f^{-1}([0,1])=[0,1]$, and $[0,1]$ is clearly a measurable set. – 2015-10-03
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Consider for example a set $E \subset [0,1]$; then the function $ f(x)=\begin{cases} x & x \in E\\ -x & x \in [0,1]\setminus E \end{cases} $ is measurable if and only if $E$ is measurable. On the other hand, $f$ is injective, hence $f^{-1}(c)$ is either empty, either a singleton (in both cases, anyway, $f^{-1}(c)$ is measurable).
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0No, I'm sorry, but I can't explain more. There's nothing I can add. You simply have to study definitions. – 2012-12-25