I am just curious about examples of measurable functions $f:[0,1]\to[0,1]$ such that $f[0,1]$ is not measurable.

This is motivated by the question Is measure preserving function almost surjective?, that asks whether such maps have image of inner measure one. The solution is trivial if $f[0,1]$ is measurable, however, I have not succeeded understanding ``how bad'' it could be assuming that $f[0,1]$ is measurable.

I think it is not possible to construct any such $f$ avoiding the use of the axiom of choice (because if not, it would be possible to construct a non measurable set $A=f[0,1]$ without it). On the other hand, if we start from a non-measurable set $V,$ and we try to find $f$ such that $f[0,1]=V,$ then I do not know how to make such $f$ to be measurable.

What about examples examples of measurable functions $f:[0,1]\to[0,1]$ such that $f[0,1]=A\cup B$ where $A$ is not Lebesgue, $B$ has Lebesgue measure zero and $f^{-1}A$ has strictly positive Lebesgue measure?

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    $\begingroup$ If your function doesn't have to be measure preserving, you can just take a function that maps a measure zero set with the cardinality of the continuum (the Cantor set, say) onto some nonmeasurable set and maps everything else to $0$. That function will be Lebesgue measurable but have nonmeasurable range. $\endgroup$ – Michael Greinecker Apr 18 '18 at 19:54
  • $\begingroup$ This is a good idea. Is the non Lebesgue measurable set a Borel set? If it is not, then I understand your example, if it is, then I do not understand it... $\endgroup$ – Italo Cipriano Apr 18 '18 at 20:30
  • $\begingroup$ @user39115 If $V$ is not Lebesgue measurable, then it cannot be Borel since all Borel sets are Lebesgue measurable. $\endgroup$ – Piotr Hajlasz Apr 18 '18 at 20:32
  • $\begingroup$ If I understand correctly, basically the examples are: send a Lebesgue measurable set of measure zero of cardinality continuum onto a non-measurable set and the complement onto a Lebesgue measurable set of measure zero. $\endgroup$ – Italo Cipriano Apr 18 '18 at 21:17
  • $\begingroup$ A note: your function $f$ cannot be Borel. The image of $[0,1]$ (or any Borel set) under any Borel function is an analytic set, and analytic sets are always Lebesgue measurable (indeed, universally measurable), though they need not be Borel. In particular, modifying your function on a null set can make the image measurable. So such a function has to map a null set to a non-measurable set. $\endgroup$ – Nate Eldredge Apr 18 '18 at 21:51

A measurable function $f:[0,1]\to\mathbb{R}$ maps Lebesgue measurable sets to measurable sets if and only if it has a Lusin property: the image of a set of measure zero has measure zero.

Here is an example when the Lusin property is violated. Take a Cantor set of positive Lebesgue measure in $[0,1]$. This set contains a non-measurable subset $E$. Let $F$ be a subset of the ternary Cantor set that is homeomorphic to $E$. Since the ternary Cantor set has measure zero, $F$ is measurable (and of measure zero). Let $f:[0,1]\to [0,1]$ be defined as a homeomorphism of $F$ onto $E$ and a constant map (with the value in $E$) on the complement of $F$. The mapping $f$ is measurable and $f([0,1])=E$ is not measurable.

EDIT: I have just realized that my answer is very similar to the comment of Michael Greinecker.

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