This is a really basic problem for advanced probability, but there are some bits I can't get my head around. If $(\Omega, \mathcal{F}, P)$ is a probability triple (Lebesgue measure on $[0,1]$) with $\mathcal{F}$ Borel $\sigma-$algebra and rv $X(\omega)$ defined as $$ X(\omega) = \Bigg\{ \begin{align} \omega \ \text{if } 0 \leq \omega < \frac{1}{2}\\ \omega^2 \ \ \ \text{if } \frac{1}{2} \leq \omega\leq 1 \end{align} $$ If $A \in \mathcal{F}$, s.t. $A=[\frac{1}{4}, \frac{3}{4}]$, I want to find $P(X \in A)$. From the definition of $A, \ P(X \in A) = P(\{\omega: X(\omega) \leq x\})$.
Splitting the set into two intervals, finding the preimages $X^{-1}$, and using Lebesgue measure on $\Omega$, \begin{align} P(X \in A) &= P(\frac{1}{4}\leq X(\omega) \leq \frac{1}{2}) + P( \frac{1}{2}\leq X(\omega) \leq \frac{3}{4}) \\ &= \frac{1}{2}-\frac{1}{4} + \frac{1}{\sqrt{2}} - \frac{1}{2} +\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\\ &= \frac{2\sqrt{3}-1}{4} \end{align} I'm quite confident this is correct, but when I consider $A=[0,1]$, I think it must be that $P(X \in A)=1$ because $A$ partitions $\Omega$, and when I use the same calculation as above, I get $$ P(X \in A) = \frac{1}{2}-0 + \frac{1}{\sqrt{2}} - \frac{1}{2} + 1-\frac{1}{\sqrt{2}} = 1 $$ Nevertheless, I'd like to know if there are any issues with my logic here.
Edit: $$ \begin{align} X^{-1}([1/4,3/4])&=\Big(X^{-1}([1/4,3/4])]\cap[0,1/2)\Big)\cup\Big(X^{-1}[1/4,3/4])\cap[1/2,1]\Big)\\ &=[1/4,1/2)\cup[1/2,\sqrt{3}/2] \end{align} $$
Hence
\begin{aligned} \lambda\big([X^{-1}([1/4,3/4])\big) =\lambda([1/4,1/2)) +\lambda([1/2,\sqrt{3}/2])=\frac14 +\frac{\sqrt{3}-1}{2} \end{aligned}