2 edited body; deleted 8 characters in body

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^{-r}$3^{-i}$. Then$f(N)$is Lebesgue measurable. , since it maps any set to a Cantor-like set (Unfortunately, I am still working out on how of measure zero) (thanks to prove this, I will update this with Tapio Rajala for the solution by today hopefully.)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. 1 # 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$f(x) = 2 \sum_{i=1}^{\infty} x_i 3^{-r}$. Then$f(N)$is Lebesgue measurable. (Unfortunately, I am still working out on how to prove this, I will update this with the solution by today hopefully.)$f$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.