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?
Best Answer
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.