When is the image of a non Lebesgue-measurable set measurable? 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.
 A: Suppose $A \subset I = [0,1]$ is Lebegue non-measurable, $B \subseteq I$ Lebesgue measurable, and $f: I \to I$ is a measurable function with $A = f^{-1}(B)$.  By inner regularity, $B$ is the disjoint union of sets $C$ and $D$ where $C$ is an $F_\sigma$ and $D$ has measure 0.  Then $A$ is the disjoint union of $f^{-1}(C)$, which is Lebesgue measurable, and $f^{-1}(D)$.
Thus the only way an injective measurable function can map a nonmeasurable set onto a measurable one is that it maps some nonmeasurable subset to a set of measure 0.
A: 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


*

*$N$ and $M$ are well separated.

*$f^{-1}(N)$ and $f^{-1}(M)$ are well separated.

*$f^{-1}(M)$ 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?
