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 .
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.