Assume that $ \Omega =[0,2] $ and P is geometrical probability on $ \Omega $. Find the distribution function of the random variable defined as: $$ X(\omega) = \omega, 0 \leq \omega \lt 1$$ $$ X(\omega) = \omega-1, 1 \leq \omega \leq 2$$
So
$t = \omega$
$t = \omega-1$ => $\omega = t+1$
I know that the answer is
$$ F_x(t) = 0, t<0$$ $$ F_x(t) = 1, t \geq1$$, but I'm not sure how it should be for $$ 0\leq t \lt 1$$
The phrase "$P$ is a geometrical distribution on $\Omega = [0,2]$" seems to be non-standard terminology. Under the interpretation that $P$ is the uniform distribution on $\Omega$, it's pdf can be written $p(\omega) = 1/2$ for $\omega \in \Omega$, i.e. $P(\omega \lt t) = \frac{1}{2}t$ for all $t \in [0,2]$.
Fix $t \in [0, 1]$
$$P(X(\omega) < t) = P(X(\omega) < t \ |\ \omega \in [0,1))P(\omega \in [0,1)) + P(X(\omega) < t \ |\ \omega \in [1,2))P(\omega \in [1,2))$$ $$ = P(X(\omega) < t \ |\ \omega \in [0,1))\frac{1}{2} + P(X(\omega) < t \ |\ \omega \in [1,2))\frac{1}{2}$$
And using the definition of $X$,
$$ = P(\omega < t \ |\ \omega \in [0,1))\frac{1}{2} + P(\omega - 1 < t \ |\ \omega \in [1,2))\frac{1}{2}$$
$$ = t\frac{1}{2} + t\frac{1}{2} = t$$