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Can we give an example of Lebesgue non-measurable function, for which set $\{x: f(x)=C\}~\forall C\in\mathbb{R}$ is measurable? Thanks.

Wanksta
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2 Answers2

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Let $S$ a non-measurable subset of $]0,+\infty[$. Define $$g(x)=\begin{cases} x\text{ if } x\in S\\-x\text{ if } x\notin S\end{cases}$$

$g^{-1}(y)$ is finite $\forall y\in \mathbb{R}$, but $\{ g\geq 0\}\setminus\ ]-\infty,0]=S$ is not measurable.

  • What does $]0, +\infty]$ denote? – David Zhang Apr 16 '15 at 19:37
  • @DavidZhang Editing right... NOW! –  Apr 16 '15 at 19:39
  • Huh. I'm not sure what $]0, +\infty[$ denotes either. Is it the complement of $[0, +\infty]$, perhaps? – David Zhang Apr 16 '15 at 19:43
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    I believe $]$ and $[$ are notations of open intervals. It's uncommon, but, for example, I've seen it in Brown's Topology and Groupoids. –  Apr 16 '15 at 19:45
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    @DavidZhang Oh, ok. One sometimes uses reversed square brackets to exclude the extremal point of the interval. For instance $]0,+\infty[=(0,+\infty)$.

    In Italy it's not that rare, I guess.

     –  Apr 16 '15 at 19:46
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    Its much more common in continental Europe. – Calvin Khor Apr 16 '15 at 23:03
  • $g$ is a function. What does ${g \geq 0}$ denote? – meisterluk Aug 08 '17 at 15:50
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    @meisterluk A very common shorthand notation for $${x\in (0,\infty),:, g(x)\ge 0}$$ Or, more of the same, $g^{-1}((0,\infty))$. –  Aug 08 '17 at 16:48
  • @G.Sassatelli Ah, thanks. – meisterluk Aug 09 '17 at 03:37
  • @G.Sassatelli: Can we take $g(x)=\begin{cases} x\text{ if } x\in S\-|x|\text{ if } x\notin S\end{cases}$ so that ${g\geq 0} = S$ instead of dealing with ${ g\geq 0}\setminus\ ]-\infty,0]=S,$ in which latter is, to me, more complicated. – Idonknow Mar 17 '18 at 05:54
  • @Idonknow Yes, I guess. –  Mar 17 '18 at 11:33
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    $]a,b[$ are known as the French brackets. There are many other countries in Europe (minus Britain, Ireland, Iceland) in which this notation would seem as unusual as it is in Anglosphere. I don't know if the notation due to Bourbaki, but they might've played a role. – paperskilltrees May 18 '22 at 18:19
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Take $V$ to be a non-measurable set on $[0,1]$, and consider on $[0,1]$ the function $$ f(x) := x \mathbf{1}_V(x) + (-10 -x)\mathbf{1}_{[0,1]\setminus V}(x) $$ where $\mathbf{1}_V$ is the indicator function of $V$, $$ \mathbf{1}_V(x) := \begin{cases} 1 & x∈ V \\ 0 & x \notin V \end{cases} $$Then the preimage of any $C∈ℝ$ is a singleton or empty and hence measurable.

Its also not hard to use this idea to make a non-measurable function on $ℝ$, also satisfying your criterion: $$ f(x) := e^x \mathbf{1}_V(x) + (-10 -e^x)\mathbf{1}_{ℝ \setminus V}(x)$$

Calvin Khor
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