# 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.

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First note that $$X(\omega) = \begin{cases} 2 & \omega \in \{c,d\}\\ 3 & \omega \in \{a,b\}\end{cases}$$

Now consider $$\{\omega \in \Omega : X(\omega) \leq \alpha\}.$$

1. For $\alpha <2$, you get $\emptyset$.
2. For $\alpha \in [2,3)$, you get $\{c,d\}$.
3. For $\alpha \geq 3$, you get $\{a,b,c,d\} = \Omega$.

You want these set to be in your $\sigma$-algebra. Hence, $$\sigma(X(\omega)) = \sigma(\emptyset, \{c,d\}, \Omega) = \{\emptyset, \{a,b\}, \{c,d\}, \Omega\}$$

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am I missing something, or are the values for X(w) switched in your answer. Shouldnt the RV take the value 3 when $\omega\in\left\{a,b\right\}$? – Justin May 22 '12 at 14:30
@Justin Yes. It was a typo. Fixed it. – user17762 May 22 '12 at 15:49
Thanks. I thought that was the case, but sometimes when I'm really looking at these concepts and dont fully understand them, I hesitate to think something is a typo. Thank you for the comment and help! – Justin May 22 '12 at 17:28
If I were to extend this to determine the sigma-field for (X,Y) with Y defined similarly (but not exactly the same), would it simply be a list of all the events in $\sigma(X), \sigma(Y)$ listed as ordered pairs, such as $\sigma(X,Y)=$ {{a,b},{c,d}} etc? – Justin May 22 '12 at 20:08

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*}

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