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?