Generated $\sigma(X)$ where $\Omega \neq \mathbb{R}$ Simple question here. I am trying to enumerate the sigma field generated by the random variable: $$X(\omega)=2+1_{\left\{a,b\right\}}(\omega)$$ where $\Omega=\left\{a,b,c,d\right\}$.
I think what is confusing me is that I am used to looking for the pre-images under a function in $\mathbb{R}$ or similar. I know the simple discrete space should make this exercise easier, but it just isn't clicking in my head.
 A: First note that $$X(\omega) = \begin{cases} 2 & \omega \in \{c,d\}\\ 3 & \omega \in \{a,b\}\end{cases}$$
Now consider $$\{\omega \in \Omega : X(\omega) \leq \alpha\}.$$


*

*For $\alpha <2$, you get $\emptyset$.

*For $\alpha \in [2,3)$, you get $\{c,d\}$.

*For $\alpha \geq 3$, you get $\{a,b,c,d\} = \Omega$.


You want these set to be in your $\sigma$-algebra. Hence, $$\sigma(X(\omega)) = \sigma(\emptyset, \{c,d\}, \Omega) = \{\emptyset, \{a,b\}, \{c,d\}, \Omega\}$$
A: I will include another pretty easy approach since I think the reply of Marvis had a typo in the function (not that it would make any major difference in this particular case, but I think this detail made you question the final result).
Note that $X$ takes only two values in $\mathbb{R}$: $2$ and $3$. For this reason, the preimage of any Borel set $B\subset \mathbb{R}$ is equals the preimage of either the singleton $\{2\}$ or $\{3\}$. The function $X$ in fact is $X(w)=3$ if $w\in\{a,b\}$ and $X(w)=2$ if $w\in\{c,d\}$. So 
\begin{align*}
\sigma(X)=\sigma(X^{-1}\{2\},X^{-1}\{3\})=\sigma(\{a,b\},\{c,d\})=\{\emptyset,\{a,b\},\{c,d\},\Omega\}.
\end{align*}
