Let $f : (0,1) \to \mathbb{R}$ be a continuous injective function (i.e., a homeomorphism onto its image). As such, $f$ has an inverse, and $f$ is differentiable almost everywhere. Let $E_f = \{x \mid f'(x) = 0\}$, and let $\lambda$ denote Lebesgue measure. If the preimages of measure zero sets under $f$ are again measure zero, then by Sard's Theorem, $\lambda(f(E_f)) = 0$, so $\lambda(E_f) \le \lambda(f^{-1}(f(E_f)) = 0$. Is the converse true, i.e.
If $\lambda(E_f) = 0$, is the preimage of a measure zero set under $f$ a measure zero set?
To recast the question in different terms, notice that $f$ pulling back measure zero sets to measure zero sets is the same as saying that images of measure zero sets under $f^{-1}$ have measure zero, i.e. $f^{-1}$ satisfies Lusin's condition $N$. Since $f^{-1}$ is already continuous and of bounded variation, this is the same as saying that $f^{-1}$ is absolutely continuous. Thus, proving the statement above is equivalent to proving the following statement:
$f'$ vanishes only on a set of measure zero iff $f^{-1}$ is absolutely continuous.
Addendum: in proving this, one can assume that there exists $\epsilon > 0$ with $f' > \epsilon$ almost everywhere, as this implies the general case (by writing $\{f' > 0\} = \cup_{n=1}^\infty \{f' > \frac{1}{n}\}$). Even with this stronger hypothesis though, I don't see how to continue...
