If U is uniformly distributed on [0, 1], is 2U - floor(2U) also?

Question

If $U \sim U(0, 1)$, does $2U - \left\lfloor 2U\right\rfloor \sim U(0, 1)$?

If $g(x) = 2x - \left\lfloor 2x\right\rfloor$, $\epsilon \in (0, 1)$, then $g(\frac{1}{2} - \epsilon) = 1 - 2\epsilon$. But $g(\frac{1}{2}) = 0$, making $g$ not continuous, so I can't use the result before Example 4 here.

Answer 1

If $U <\frac 1 2$ then $\left\lfloor 2U\right\rfloor=0$ and if $U >\frac 1 2$ then $\left\lfloor 2U\right\rfloor=1$. This gives $$P(2U-\left\lfloor 2U\right\rfloor \le x)=P(U<\frac 1 2, 2U-\left\lfloor 2U\right\rfloor \le x)+P(U>\frac 1 2, 2U-\left\lfloor 2U\right\rfloor \le x)$$ $$=\frac x 2+\frac x2 =x$$

so $2U-\left\lfloor 2U\right\rfloor$ is uniformly distributed on $(0,1)$.