I solved this problem, but since my understanding of Borel sets and $X^{-1} \in \mathcal{B}$ is still not polished, I decided to ask it.
On a probability triple with Lebesgue measure on $[0,1]$, a random variable is defined such that $$ X(\omega) = \left\{ \begin{array}{l} 1 & \omega \in [0,\frac{1}{4})\\ 2\omega^2 & \omega \in [\frac{1}{4}, \frac{3}{4})\\ w^2 & \omega \in [\frac{3}{4},1] \end{array} \right. $$ If $A=[0,1]$, what's the $P(X \in A$). So I split $A$ into $4$ disjoint subsets, $A_1 \ldots A_4$: $$ A=[0,\frac{1}{4}),[\frac{1}{4},\frac{3}{4}), [\frac{3}{4},1), \{1\} $$ when $A=1, X$ is a simple random variable, so $$ A_4 = \{\omega:X(\omega)=1\} \rightarrow P(X \in A_4) = \frac{1}{4}-0 $$ For $A_1$, I think $P(X \in A_1)=0$, because $\{ \omega: 0 < X(\omega)<\frac{1}{4}\}$ is not defined. \
For $A_3, P(X \in [\frac{1}{4}, \frac{3}{4})) = \frac{\sqrt{3}-1}{\sqrt{8}}$
Lastly, $P(X \in [\frac{3}{4}, 1)) = 1-\frac{\sqrt{3}}{2}$
Putting it all together, $P(X \in A) = \frac{1}{4} + \frac{\sqrt{3}-1}{\sqrt{8}} + 1-\frac{\sqrt{3}}{2} \approx 0.64$
If that's correct, what would be the probability $P(X \in A^c) = P(X>1)$? From the definition of $X(\omega)$, it should be $$ A^c = \{\omega:1 < X(\omega)\leq \frac{9}{8}\} \Leftrightarrow X^{-1}((1,\frac{9}{8}]) = \bigg[\frac{1}{\sqrt{2}}, \frac{3}{4}\bigg] \Rightarrow P(X^{-1}) = \frac{3}{4}-\frac{1}{\sqrt{2}} \approx 0.04 $$ so they don't sum to $1$! Then either my calculation for $P(X \in A)$ is incorrect, or definition of $A^c$.
Verify that $X \notin A$ iff $\frac 1 {\sqrt 2} <\omega <\frac 3 4 $ and $X \in A$ iff $\omega >\frac 3 4 $ or $\omega <\frac 1 {\sqrt 2}$. ( I am ignoring possible equalities because Lebesgue measure of a single point is $0$).
Hence $P(X \in A)=\frac 14 +\frac 1 {\sqrt 2 }$ and $P(X \notin A)=\frac 3 4 -\frac 1 {\sqrt 2}$ and these add up to $1$.