This is related to this earlier question of mine.
$U:[0,1]\rightarrow[0,1]$ is an absolutely continuous function, $U(1)>U(0)$. 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 $\psi(v)=\min\limits_{\theta \in [0,1]}\{U(\theta)=v\}$.
We can show that $\psi$ is strictly increasing. (proven as part of the other question)
Claim: $U'(\psi(v)) > 0$ for almost all $v \in [\underline{v},\overline{v}]$.
First we claim: $\int\limits_{\psi([\underline{v},\overline{v}])} U'(\theta)d\theta=\overline{v}-\underline{v}$
Proof: $\int\limits^{\psi(\overline{v})}_{\psi(\underline{v})} U'(\theta)d\theta=\overline{v}-\underline{v}$. If $\lim\limits_{v' \uparrow v} \psi(v') \neq \lim\limits_{v' \downarrow v} \psi(v')$, $U(\lim\limits_{v' \uparrow v} \psi(v')) = U(\lim\limits_{v' \downarrow v} \psi(v'))=v$. $\therefore\;\int\limits^{\lim\limits_{v' \downarrow v} \psi(v')}_{\lim\limits_{v' \uparrow v} \psi(v')} U'(\theta)d\theta=0$. As $\psi$ is strictly increasing, it has an at most countable no. of discontinuities. Let them be at $\{v_n\}^{\infty}_{n=1}$. $\therefore\;\overline{v}-\underline{v}=\int\limits^{\psi(\overline{v})}_{\psi(\underline{v})} U'(\theta)d\theta=\int\limits_{\psi([\underline{v},\overline{v}])} U'(\theta)d\theta + \sum\limits^{\infty}_{n=1}\int\limits^{\lim\limits_{v' \downarrow v_n} \psi(v')}_{\lim\limits_{v' \uparrow v_n} \psi(v')} U'(\theta)d\theta=\int\limits_{\psi([\underline{v},\overline{v}])} U'(\theta)d\theta.$
Hence the set $\psi([\underline{v},\overline{v}])$ has positive Lebesgue measure.
$U(\psi(v))=v\;\forall\;v \in [\underline{v},\overline{v}]$. $U'$ exists almost everywhere. $\psi([\underline{v},\overline{v}])$ has positive Lebesgue measure, so $U'$ exists almost everywhere on $\psi([\underline{v},\overline{v}])$ as well. $\psi$ is strictly increasing therefore $\psi'$ exists almost everywhere in $[\underline{v},\overline{v}]$ 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 $v \in [\underline{v},\overline{v}]$.
I'm not sure if this reasoning is correct or somewhere I'm implicitly making additional "niceness" assumptions on $U$ unknowingly.
I think that the answer is much simpler. Assume that there is no subset of $[0,1]$ of positive measure such that $U'(x)>0$ on that set.
Therefore $U'(x)\,\leq\,0$ on $\,[0,1].$ (at all points at which $U'(x)$ exists).
Then by a standard property of absolutely continuous functions we get:
$\int_{0}^{1}U'(x)=U(1)-U(0)$.
But the integrand is $\,\,U'(x)\,\leq\,0$ and hence the integral is
non-positive i.e. $\,\,U(1)-U(0)\,\leq\,0$ which contradicts our
assumption. So we can infer that the derivative $U'(x)$ HAS to be
positive on some set of positive measure!!