I've come across the following proposition and its proof.
Given $f(x)=x+V(x)$, where $V(x)$ is the Cantor-Vitali function on $[0,1]$:
- $f(x)$ is an homeomorphism from $[0,1]$ to $[0,2]$.
- $\lambda(f(\mathcal{C}))=1$, where $\lambda$ is Lebesgue measure and $\mathcal{C}$ is Cantor set.
- Setting $g=f^{-1}$, there is a measurable set $A\subset[0,1]$ whose preimage under $g$ is non-measurable.
Proof for the first two points is straightforward. About the third one, it says
$f([0,1]\setminus \mathcal{C})$ is a countable union of disjoint open intervals, hence $f(\mathcal{C})$ is a countable union of disjoint closed intervals $T_j$. There exists a $T_k$ such that $\lambda(T_k)>0$ and, since $T_k$ is a closed interval, there's a non-measurable subset $E\subset T_k$.
Then $A=g(E)\subset\mathcal{C}$ is a measurable set because $\lambda$ is complete and $\lambda(\mathcal{C})=0$.
Although this seems reasonable to me, wouldn't it imply that $g(T_k)\subset \mathcal{C}$? But, being $g$ continuous, it would mean that $\mathcal{C}$ contains an interval, and that's false.
Where am I wrong? Or is the excerpt from the proof incorrect? If so, how would I prove the third point?
Best Answer
You are absolutely correct, for the reason you gave, and the "proof" sketched is false.
What we know is that $f(C)$ is a compact set of positive measure. And any set of positive measure contains a nonmeasurable subset. Thus $f(C)$ contains a nonmeasurable set $E.$ Therefore $f^{-1}(E)$ is a subset of $C$ that gets mapped to $E$ by $f.$ Because $\lambda (C)=0$ and $\lambda$ is complete, $f^{-1}(E)$ is measurable. Thus we have found a measurable set that is mapped to a nonmeasurable set via the homeomorphism $f.$