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## surjective function from non-measurable sets

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.

is there a "standard" way to construct surjective and bijective functions from non-measurable sets to measurable ones with the same cardinality?

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 What is $[x]$?? – Gerald Edgar Jan 27 2012 at 16:06 take the equivalence relation $x\sim y\Leftrightarrow x-y\in\mathbb Q$, then take the quotient set $R/\sim$. this is basically the vitali set. $[x]$ is the class of $x$ in the quotient. – alberto.bosia Jan 27 2012 at 19:15

For instance let $f:[0,1]\to[0,1]$ be the Cantor function 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 $W$.
edit. 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.
 A reformulation of the same idea: Define a function $f$ from $[0,1]$ into the Cantor set $C$ by taking the binary expansion of any $x\in[0,1]$ (in case of ambiguity, use the expansion that ends in 0's), replace all the 1's by 2's, and read the result as a ternary expansion. This bijects $[0,1]$ to a subset of $C$, so it bijects every subset of $[0,1]$ to a measurable set (of measure zero). – Andreas Blass Jan 27 2012 at 14:59