Let $T>0$.
Let $x(t):[0,T] \to [0,1]$ be a Lipschitz continuous function with $x(0)=1$.
Let $E$ be an open subset of $[0,T]$ (with $\mu(E)< T$ for a meaningful problem).
We are given that $x(t)$ is differentiable on $E$ and it satisfies $\dot x(t)= 1-2 x(t)$ for all $t \in E$.
Furthermore, $\dot x(t)=0$ for almost all $t \in [0,T]/E$.
does the following hold: $x(t)>1/2$ for all $t \in [0,T]$?
I can see that it intuitively holds, but I found a difficulty in articulating a rigorous proof when $E$ is complicated (e.g, $E$ is an open set with a boundary of positive measure)
Since $x$ is Lipschitz, it is absolutely continuous, and $x(t) = x(0) + \int_0^t \dot{x}(s)\; ds$. Noting that $\dot{x}(s) = (1 - 2 x(s)) I_E(s)$ for almost every $s$ (where $I_E$ is the indicator function of $E$), we have $$ x(t) = x(0) + \int_0^t (1 - 2 x(s)) I_E(s)\; ds $$
EDITED: Let $$m(t) = \int_0^t I_E(s)\; ds$$ i.e. the Lebesgue measure of $[0,t] \cap E$, and $$y(t) = \frac{1 + \exp(-2 m(t))}{2} $$ Then $y(t)$ also has $y(0)=0$ and $\dot{y}(s) = (1 - 2 y(s)) I_E(s)$ for almost every $s$ (i.e. every member of $E$ and every Lebesgue point of $E^c$), and the same argument shows $$ y(t) = y(0) + \int_0^t (1 - 2 y(s)) I_E(s)\; ds$$ Thus $z(t)=x(t)-y(t)$ satisfies $$ z(t) = -2 \int_0^t z(s) I_E(s)\; ds, \ z(0)=0 $$ So $$\max_{0 \le s \le t} |z(s)| \le 2 t \max_{0 \le s \le t} |z(s)|$$ which for $0 < t < 1/2$ says $\max_{0 \le s \le t} |z(s)|=0$, i.e. $z(s)=0$ for $0 \le t < 1/2$. Repeat the argument on each interval $[n/2, (n+1)/2]$ to get $z(s)=0$ on $[0,T]$, i.e. $x(t) = y(t)$, and in particular $x(t) > 1/2$.