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So consider a measurable function $f: (X, \mathcal{F}) \rightarrow (Y, \mathcal{G})$. The measurablility tells me that the pre-image of every measurable set $A\in \mathcal{G}$ is in $\mathcal{F}$, i.e., $f^{-1}(A)\in\mathcal{F}$. My question is now: If I start with some set $B\in\mathcal{F}$, is that set alwlays mapped to a set in $\mathcal{G}$, i.e., if $f(B)\in\mathcal{G}$ for every $B\in\mathcal{F}$?

Personally, I would say this is not the case, because I dont see that there would be a contradiction with the Def of measurability if $f(B)$ was mapped to a subset of $Y$ that is not in $\mathcal{F}$.

guest1
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    There are very easy counter-examples. Take $\mathcal G={\emptyset,Y}$ and $f$ to be a constant function. – Kavi Rama Murthy Aug 02 '23 at 09:51
  • For the interesting case $(X,\mathcal F)=(Y,\mathcal G)=(\Bbb R,\mathcal B(\Bbb R))$, we have $C^\infty$ functions that do not send all Borel sets to Borel sets. https://math.stackexchange.com/questions/415759/is-the-image-of-a-borel-subset-of-0-1-under-a-differentiable-map-still-a-bor?rq=1 – Sassatelli Giulio Aug 02 '23 at 10:06
  • Here is an easy example of a measurable function $f:(\Bbb R,\text{Lebesgue})\to (\Bbb R,\mathcal B(\Bbb R))$ with non-measurable (hence non-Borel) image. https://mathoverflow.net/questions/298220/measurable-functions-with-non-measurable-image

    If you look closely, $f$ is actually measurable $(\Bbb R,\text{Lebesgue})\to(\Bbb R,\text{Lebesgue})$.

     – Sassatelli Giulio Aug 02 '23 at 10:10
  • thanks for your answers! – guest1 Aug 02 '23 at 11:02
  • @geetha290krm is your counter example still a counter example if also $\mathcal{F}$ consists only of the empty set and $X$? – guest1 Aug 02 '23 at 11:08
  • No, it is not. But you just need one other set in $\mathbb F$. Also, $Y$ should be a singleton. – Kavi Rama Murthy Aug 02 '23 at 11:20

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