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...

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    $\begingroup$ It's worth noting (though not important to the problem) that not every continuous injective function is a homeomorphism from its domain onto its image. This is a special property that holds in this case because it is a function between spaces whose topologies are induced by total orders. $\endgroup$ – Cameron Buie Dec 23 '13 at 23:06
  • $\begingroup$ Sure, but for the moment I'm only interested in functions from $\mathbb{R}$ to $\mathbb{R}$, for which it is true... $\endgroup$ – zcn Dec 23 '13 at 23:08
  • $\begingroup$ I assumed as much. $\endgroup$ – Cameron Buie Dec 23 '13 at 23:25
  • $\begingroup$ No answers? Should I ask this on MathOverflow? Not sure what the etiquette is here... $\endgroup$ – zcn Dec 26 '13 at 21:32
  • $\begingroup$ It's probably a better fit here. Sometimes, for whatever reason, a question just doesn't grab much attention, and those who notice it are either disinclined or unable to answer. Your best bet is probably to offer a bounty, but I'd wait a few more days before resorting to that. (You have to wait at least 3, anyway.) $\endgroup$ – Cameron Buie Dec 26 '13 at 22:06

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