TITLE: Does homeomorphism on $\mathbb R^n$ preserves Lebesgue measurability?
QUESTION [6 upvotes]: Let $f: \mathbb R^n \to \mathbb R^n$ be a homeomorphism; then is it true that $f(A)$ is a Lebesgue measurable set for every Lebesgue measurable set $A$ ? If this is not true for all $n$, then is this true for at least one $n$ ?
I know that for Lebesgue measurable set $A$, we can write $A=B\cup C$ , where $B$ is a $F_{\sigma}$ set and $C$ has Lebesgue measure $0$. I can see that $f(B) $ is again an $F_{\sigma}$ set, hence Lebesgue measurable set; but I don't know about $f(C)$ . 
Please help .

REPLY [7 votes]: Consider the Cantor set $C\subset\Bbb R$. There are also "fat" Cantor
sets, $C'\subset\Bbb R$ which are homeomorphic (even up an order-preserving
homeomorphism) with positive Lebesgue measure. One can extend
an order-preserving homeomorphism $\phi:C'\to C$ to a homeomorphism
from $\Bbb R$ to itself. Then $C'$ has a non-measurable subset $A$, but
$\phi(A)\subset C$ is measurable, as every subset of a measure zero
set is Lebesgue measurable.