surjective function from non-measurable sets - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T04:50:42Z http://mathoverflow.net/feeds/question/86795 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/86795/surjective-function-from-non-measurable-sets surjective function from non-measurable sets alberto.bosia 2012-01-27T06:17:24Z 2012-02-01T11:14:37Z <p>let $V$ be the vitali set and let $g:V\to\mathbb R$ be a surjective function. then the fuction $f:\mathbb R\to\mathbb R$ such that $f(x)=g([x])$ will be a function that is surjective in any interval of the real line. i know the examples with base 9 digits, but this one would be much easier.</p> <p>is there a "standard" way to construct surjective and bijective functions from non-measurable sets to measurable ones with the same cardinality?</p> http://mathoverflow.net/questions/86795/surjective-function-from-non-measurable-sets/86802#86802 Answer by Pietro Majer for surjective function from non-measurable sets Pietro Majer 2012-01-27T07:50:21Z 2012-02-01T11:14:37Z <p>For instance let $f:[0,1]\to[0,1]$ be the <a href="http://en.wikipedia.org/wiki/Cantor_function" rel="nofollow">Cantor function</a> and define $g(x):=x+f(x)$. Then $g:[0,1]\to[0,2]$ is a homeomorphism that maps the complement of the Cantor set $C$ onto a measure one open set of $[0,2]$ (just because $g'(x)=1$ on $[0,1]\setminus C$). So $g_{|C}:C\to g(C)$ is a homeomorphism of the Cantor set onto a compact set of measure one, and if $W$ is any non-measurable subset of $g(C)$, $g$ is also a homeomorphism between the Lebesgue measurable null-set $g^{-1}(W)$ and $W$.</p> <p><strong>edit.</strong> As to the issue of finding a non-measurable subset within a Lebesgue measurable set of positive measure $S$, there is such a set of the form $S\cap (V+q)$, the trace on $S$ of a suitable translation of the Vitali set $V$. Indeed, the Vitali set $V$ does not contain any measurable subset of positive measure (reason: if $E\subset V$ then $E-E \subset V-V\subset \big(\mathbb{R}\setminus\mathbb{Q}\big)\cup \{0 \}$, while $E-E$ is always a nbd of zero for any measurable set of positive measure $E$). On the other hand, the sets $S\cap (V+q)$ for $q\in\mathbb{Q}$ are a countable cover of $S$, so one of them has positive exterior measure. </p>