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A measurable function's derivative in the sense that $f^{-1}(]-\infty,\alpha[) \in\mathcal M \ \forall \alpha\in\Bbb R$ is measurable. But what about in the sense $f^{-1}(\text{measurable set})=\text{measurable set}$ ?

The first statement can be seen by looking at $\frac{f(x+1/n)-f(x)}{1/n}$ which is measurable for all $n$. (why does it imply it converges towards a measurable function?)

John Cataldo
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  • Are the brackets facing the other direction on purpose? If so what does it mean, if not please explain how the "weaker version" is weaker (it seems stronger). – Yanko Jun 05 '18 at 17:19
  • $]a,b[\equiv (a,b)$ means open interval Yes, it could be stronger actually thanks – John Cataldo Jun 05 '18 at 17:26
  • @JohnCataldo this might help you https://math.stackexchange.com/questions/1327081/how-to-prove-limit-of-measurable-functions-is-measurable – Yanko Jun 05 '18 at 18:23
  • @Yanko The notation $]a,b[$ is equivalent to the notation $(a,b)$; it denotes the interval with the endpoints excluded (i.e. the open interval). It is a common quirk of a Bourbakist approach, and ameliorates some of the overloading of the notation $(\cdot,\cdot)$. – Xander Henderson Jun 16 '18 at 04:09

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