Let $(\Omega, \mathcal F, P)$ be Lebesgue measeure on $[0,1]$, and set
$X(\omega) = 1$ if $0 \le \omega < \frac{1}{4}$
$X(\omega) = 2\omega^2$ if $\frac{1}{4} \le \omega < \frac{3}{4}$
$X(\omega) = \omega^2$ if $\frac{3}{4} \le \omega \le 1$
Compute $P(X \in A)$ where
(A) $A = [0,\frac{3}{4}]$
(B) $A = [\frac{1}{2},1]$
From a previous question I know that:
(A) $P(X \in [0,1]) = \frac{\sqrt{2}}{2} + \frac{1}{4}$
(B) $P(X \in [\frac{1}{2},1]) = \frac{\sqrt{2}}{2}$
This problem needs to be solved using probability density functions and integrals. How do I approach this?