Can a nowhere differentiable function preserve measurability?

Question

I want to know whether a continuous nowhere differentiable function $f: \mathbf{R} \to \mathbf{R}$ can map Lebesgue measurable sets to Lebesgue measurable sets. More generally I'm interested to know if there are any necessary conditions for a continuous functions to preserve measurability.

Answer 1

The answer is "no".

As it was noted by Martin Sleziak in order to preserve measurability, your function has to satify Luzin N property. Let me show that this is not the case. That is, any continuous nowhere differentiable function $f$ maps a set of zero measure maps to a set of positive measure.

Note that for fixed $L<\infty$ and almost any $y\in f(\mathbb{I})$ there is an interval $[p,q]\subset \mathbb{I}$ such that $y\in f([p,q])$ and $$\lambda(f([p,q]))>L\cdot\lambda([p,q]),$$ so you are in the position to apply Vitali covering theorem.

Fix $\varepsilon>0$. Applying Vitali covering theorem, you can pass to a closed subset $S\subset\mathbb{I}$ formed by a finite collection of closed intervals such that $$\lambda(f(S))>(1-\varepsilon)\cdot\lambda(f(\mathbb{I}))\quad\text{and}\quad \lambda(S)<\tfrac12\cdot \lambda(\mathbb{I}),$$ where $\lambda$ denotes Lebesgue measure.

It remains to iterate this construction for a sequence $\varepsilon_n\to 0$ such that $$\prod_n(1-\varepsilon_n)>0.$$

Answer 2

Take $\Psi$ as the standard Cantor function: $\Psi(x)=0$ for $x\le 0$, $\Psi(x)=1$ for $x\ge 1$, continuous, nondecreasing, constant on each connected component of the complement of the Cantor ternary set $K$. The distribution derivative of $\Psi$ is a nonnegative Radon measure without atoms, supported on $K$.

Define now $F$ by $F(x)=x+\Psi(x)$, which is an homeomorphism. The function $F^{-1}$ is not Lebesgue-measurable: Since it can be proven that the Lebesgue measure of $F(K)$ is positive, we can find $D\subset F(K)$ which is not Lebesgue-measurable; of course $F^{-1}(D)$ is Lebesgue as a subset of $K$ which has null measure. But we have $$ (F^{-1})^{-1}\bigl(\underbrace{F^{-1}(D)}_{\text{Lebesgue-m.}}\bigr)=D\quad \text{not Lebesgue-m.}, $$ so that the continuous mapping $F^{-1}$ is not Lebesgue measurable. This means that continuity ensures Borel measurability, but not Lebesgue measurability.