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It is known that whenever $f: \mathbb{R} \rightarrow \mathbb{R}$ is Borel measurable it is not necessarily true that $f(X)$ is Borel measurable for Borel $X \subset \mathbb{R}$. However I have not seen a counterexample yet.
It is known that whenever $f$ is injective the claim holds.
Question: Does there exists a Borel function $\tilde{f}: \mathbb{R} \rightarrow \mathbb{R}$ such that $f = \tilde{f}$ a.e. and such that $\tilde{f}(X)$ is Borel measurable for each Borel $X \subset \mathbb{R}$?

abcdef
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    $f(X)$ can even be not measurable. Maybe see in the exercise of the book Real analysis by Stein and Shakarshi in chapter 1. – Walace May 15 '20 at 08:22
  • Related: https://math.stackexchange.com/questions/1302157/the-image-of-borel-set-under-measurable-mapping/1302171#1302171 –  May 15 '20 at 08:39
  • Thanks for the comments, especially the link to the other question. I realised that my conditions were to strong. However I altered the question. Does it hold now? – abcdef May 15 '20 at 08:51
  • Is $\tilde f$ allowed to depend on $X$? – harfe May 15 '20 at 09:34
  • Oh, ideally $X$ may be arbitrary (see my edit) and $f$ should work for all Borel $X$. If this does not work I'm willing to take $X = \mathbb{R}$ fixed. – abcdef May 15 '20 at 09:40
  • Gee... I thought I had it but my proof only works for... continuous $f$ :-( (and $\tilde f$ depending on $X$). Actually, I was assuming that $f$ preserved $\sigma$-compactness. Very nice question indeed. – Pedro Sánchez Terraf May 18 '20 at 18:52
  • @Walace : It would be helpful to know which exercise in chapter 1. Perhaps you mean exercise 21, though this uses the Lebesgue measure (similar to the d.k.o. link) which does not quite work here for Borel measure (such examples are easier than for the Borel measure). I found the book link here: http://www.cmat.edu.uy/~mordecki/courses/medida2013/book.pdf – Michael May 20 '22 at 15:47
  • Regarding the simpler case when $\tilde{f}$ can depend on the Borel set $X$: If $X$ is countable then $f(X)$ is countable hence Borel; if $X$ is uncountable then find an uncountable Borel-measure-0 subset $Y\subseteq X$. It is known that there is a bijective Borel function $\phi:Y\rightarrow\mathbb{R}$. Define $\tilde{f}(x) = f(x)$ if $x \notin Y$, $\tilde{f}(x)=\phi(x)$ if $x \in Y$. Then $\tilde{f}(X)=\mathbb{R}$. This link talks about finding such $Y$: https://math.stackexchange.com/questions/1742137/lebesgue-measure-0-set-which-is-not-borel – Michael Sep 02 '23 at 13:57

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