5
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ Image under what kind of map? Why can't I just take my favourite non-measurable set, find a measureable set of the same cardinality, and map one to the other? $\endgroup$
    – Yemon Choi
    Dec 13, 2011 at 8:36
  • 1
    $\begingroup$ In your definition of $f$ the $r$ is $i$?. This function maps everything to a set of measure zero (a Cantor set) and therefore it has the desired property. $\endgroup$ Dec 13, 2011 at 8:57
  • $\begingroup$ @Tapio Rajala: yes, sorry. edited. And thanks for the solution. @Yemon Choi: uhm, I guess you can, but I don't see how that answers the question since you're not saying anything about the actual map? $\endgroup$
    – Ignas
    Dec 13, 2011 at 9:13
  • $\begingroup$ @Ignas: I was trying to suggest that you make the question more precise, and in particularly specify: what is the domain and range of your function $f$? Is it supposed to send every non-measurable set to a measurable one? Is it supposed to admit some kind of explicit description? As it stands, I find it hard to work out what the precise question actually is $\endgroup$
    – Yemon Choi
    Dec 13, 2011 at 9:19
  • $\begingroup$ @Yemon: Well, my question is precisely what are the minimal restrictions that we need to impose on $f$ such that it maps an arbitrary non-measurable set to a measurable set (and is not something stupid like a constant function). It doesn't have to map every non-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. $\endgroup$
    – Ignas
    Dec 13, 2011 at 9:29

2 Answers 2

10
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ It seems you need additional hypotheses on the domain of $f$. After all, we can map a measure zero set continuously to a set with positive measure, and such a function will vacuously have the property that it maps every non-measurable set to a measurable set, but fail your criterion. $\endgroup$ Dec 13, 2011 at 10:23
  • $\begingroup$ Thank you Joel, I modified the condition to take this into account. $\endgroup$ Dec 13, 2011 at 10:40
  • $\begingroup$ I have deleted my second objection, because it was incorrect. $\endgroup$ Dec 13, 2011 at 15:27
  • 2
    $\begingroup$ Following Joel's example: I have deleted my comments on the second objection, because they were correct. :) $\endgroup$ Dec 13, 2011 at 15:30
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.