Let $(\omega, F, P)$ 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) $A = [0,1] = \frac{\sqrt{2}}{2} + \frac{1}{4}$
(B) $A = [\frac{1}{2},1] = \frac{\sqrt{2}}{2}$
However for this problem I need to solve it via one of these equations I'm assuming: $\mu(B) = P(X \in B) = P(X^{-1}(B)), B \in B$
$\int_\Omega f(X(\omega))P(d\omega) = \int^\infty_{-\infty} f(t)\mu(dt)$
I've been looking around on Google for anything that may be helpful and so far I've found this: http://terrytao.wordpress.com/2010/09/19/245a-notes-2-the-lebesgue-integral/