Consider the probability space $(\mathbb{R},\mathcal{B}(\mathbb{R}),\mathbb{P})$, where $\mathbb{P}=0.1\delta_{-2}+0.7\delta_1+0.2\delta_{10}.$
Given $X(\omega)=2\omega^3 I_{(-\infty,3]}(\omega)$, calculate $\mathbb{P}_X([0,2])$.
Can somebody please explain how to calculate $\mathbb{P}_X([0,2])$?
I'm going to try and guide you to the answer. You're asked to find $\mathbb{P}_X ([0,2])$, which is the same as $\mathbb{P}(X^{-1}([0,2]))$. Notice that the set $X^{-1}([0,2])$ is the set of all the points $\omega \in \mathbb{R}$ such that $X(\omega)$ lies in the interval $[0,2]$ here. You should try to find it explicitly as an exercise. Notice however that your measure $\mathbb{P}$ is supported only on three points, so you can be lazy and just ask the question: which of $X(-2),X(1),X(10)$ lies in $[0,2]$? Knowing that, it should be straightforward to compute the answer.