$U:[0,1]\rightarrow[0,1]$ is a continuous function, $U(1)>U(0)$ and $U'$ exists almost everywhere. Let $\max U([0,1])=\overline{v}$. Pick $\underline{v} \in (U(0), \overline{v})$. Define $\psi: [\underline{v},\overline{v}] \rightarrow [0,1]$ as $\psi(v)=\min\limits_{\theta \in [0,1]}\{U(\theta)=v\}$.
Claim: $\psi$ is strictly increasing.
Proof: By continuity of $U$, either $U(\theta)<v$ or $U(\theta)>v$ for all $\theta<\psi(v)$. $U(0)<\underline{v} \leq v$. Therefore the former is the only possibility. For any $\theta<\psi(v) \implies U(\theta)<v'\;\forall\;v'>v$. $\therefore \psi(v') \geq \psi(v)$. But $\psi(v') \neq \psi(v)$ whenever $v' \neq v$, because $U(\theta)$ is obviously unique for any $\theta$. Hence the claim follows.
Now comes the problem.
$U(\psi(v))=v\;\forall\;v \in [\underline{v},\overline{v}]$. $U'$ exists almost everywhere. $\psi$ is strictly increasing therefore $\psi'$ exists almost everywhere as well. Therefore differentiating both sides of the aforementioned equation wrt $v$, $U'(\psi(v))\psi'(v)=1$ for almost all $v \in [\underline{v},\overline{v}]$. Hence $U'(\psi(v)) > 0$ for almost all $v \in [\underline{v},\overline{v}]$. This is a contradiction because what about the Cantor function.
I'm not able to understand which step is the mistake. I think the step where I concluded $U'(\psi(v))\psi'(v)=1$ for almost all $v \in [\underline{v},\overline{v}]$ (unless the proof of increasingness of $\psi$ is wrong), but I don't understand why.
Also, are there additional assumptions on $U$ (e.g. absolute continuity) that would make the above conclusion correct?