When is the image of a non Lebesgue-measurable set measurable? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T06:18:39Z http://mathoverflow.net/feeds/question/83324 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/83324/when-is-the-image-of-a-non-lebesgue-measurable-set-measurable When is the image of a non Lebesgue-measurable set measurable? Ignas 2011-12-13T08:28:19Z 2011-12-13T17:55:19Z <p>Hi MathOverflow,</p> <p>I'm not sure if it makes sense to ask this question in the general setting, but:</p> <p>Are there any necessary conditions for a function, such that if $N$ is a not <em>Lebesgue</em> measurable, $f(N)$ is <em>Lebesgue</em> measurable?</p> <p>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:</p> <p>Pick a non <em>Lebesgue</em> 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 $0.1 = 0.0111...$).</p> <p>Now, take $f(x) = 2 \sum_{i=1}^{\infty} x_i 3^{-i}$. Then $f(N)$ is <em>Lebesgue</em> measurable, since it maps any set to a Cantor-like set (of measure zero) (thanks to <em>Tapio Rajala</em> for the easy solution).</p> <p>$f$ 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 $f(N) \subset (0,1)$.</p> <p>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.</p> <p>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.</p> <p>Thanks.</p> http://mathoverflow.net/questions/83324/when-is-the-image-of-a-non-lebesgue-measurable-set-measurable/83325#83325 Answer by Tapio Rajala for When is the image of a non Lebesgue-measurable set measurable? Tapio Rajala 2011-12-13T08:53:58Z 2011-12-13T10:38:05Z <p>My guess is that the characterization is the following: </p> <blockquote> <p>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.</p> </blockquote> <p>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</p> <ol> <li>$N$ and $M$ are well separated.</li> <li>$f^{-1}(N)$ and $f^{-1}(M)$ are well separated.</li> <li>$f^{-1}(M)$ has positive measure.</li> </ol> <p>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 $f$.</p> <p><em>Are there more mistakes hidden somewhere?</em></p> http://mathoverflow.net/questions/83324/when-is-the-image-of-a-non-lebesgue-measurable-set-measurable/83358#83358 Answer by Robert Israel for When is the image of a non Lebesgue-measurable set measurable? Robert Israel 2011-12-13T17:55:19Z 2011-12-13T17:55:19Z <p>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.</p>