Hi MathOverflow,


I'm not sure if it makes sense to ask this question in the general setting, but:


Are there any necessary conditions for a function, such that if $N$ is a not *Lebesgue* measurable, $f(N)$ is *Lebesgue* measurable?


I am working on a problem, which seems to suggest that there are no 'trivial' conditions on the function (in particular, $f$ can be injective, which is a surprise to me). The problem is a as follows:


Pick a non *Lebesgue* measurable set $N \subset (0,1) \subset \mathbb{R}$ and write $x \in (0,1)$ in an infinite binary expansion, i.e. $x = 0.x\_1x\_2...$ with $x\_i = 0$ or $1$ and infinitely many $x\_i$'s equal to $1$ (this is ok, since $0.1 = 0.0111...$).


Now, take $f(x) = 2 \sum\_{i=1}^{\infty} x\_i 3^{-i}$. Then $f(N)$ is *Lebesgue* measurable, since it maps any set to a Cantor-like set (of measure zero) (thanks to *Tapio Rajala* for the easy solution).


$f$ just takes $x$ to a base $3$ representation with no $1$'s in the expansion, thus is clearly injective. It sort of "spreads out" the elements of set $N$. Also, clearly $f(N) \subset (0,1)$.


The thing that bothers me is that this seems to suggest that this $f$ is able to transform any non-measurable set into a measurable one, without really "loosing information" about it (because it is injective), which just sounds too good to be true.


I tried to look for sources on functions applied on non-Lebesgue measurable sets, but failed to find anything, so if anyone could guide me to some I would highly appreciate it too.


Thanks.



When is the image of a non Lebesgue-measurable set measurable?
My guess is that the characterization is the following: 



> 
> A function $f$ maps every non-measurable set into a measurable set if and only if the domain or the image of $f$ has measure zero.
> 
> 
> 


One direction is trivial. For the other direction assume that the image of $f$ is positive. Take a non-measurable subset $N$ of the image and a measurable subset $M$ of the image so that


1. $N$ and $M$ are well separated.
2. $f^{-1}(N)$ and $f^{-1}(M)$ are well separated.
3. $f^{-1}(M)$ has positive measure.


Take a non-measurable subset $K$ of $f^{-1}(M)$ and consider $K \cup f^{-1}(N)$. This set is non-measurable and so is its image under $f$.


*Are there more mistakes hidden somewhere?*


