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.

everynon-measurable set to a measurable one? Is it supposed to admit some kind ofexplicitdescription? As it stands, I find it hard to work out what the precise question actuallyis– Yemon Choi Dec 13 '11 at 9:19everynon-measurable, but if one can find necessary conditions for that, it would be interesting too. So yes, my question is quite open, but I inteded it to be so. See Tapio's answer for example. – Ignas Dec 13 '11 at 9:29