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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.

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  • $\begingroup$ 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
    Commented Dec 13, 2011 at 8:36
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    $\begingroup$ 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. $\endgroup$
    – Tapio Rajala
    Commented 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
    Commented 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
    Commented 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
    Commented Dec 13, 2011 at 9:29

2 Answers 2

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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?

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  • $\begingroup$ 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. $\endgroup$
    – Joel David Hamkins
    Commented Dec 13, 2011 at 10:23
  • $\begingroup$ Thank you Joel, I modified the condition to take this into account. $\endgroup$
    – Tapio Rajala
    Commented Dec 13, 2011 at 10:40
  • $\begingroup$ I have deleted my second objection, because it was incorrect. $\endgroup$
    – Joel David Hamkins
    Commented Dec 13, 2011 at 15:27
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    $\begingroup$ Following Joel's example: I have deleted my comments on the second objection, because they were correct. :) $\endgroup$
    – Tapio Rajala
    Commented Dec 13, 2011 at 15:30
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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.

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