I'm trying to prove that $ \mathcal{G} $ is $\sigma$ - algebra.
Let $\mathcal{U}$ be $\sigma$ - algebra. Given a collection $\mathcal{F}$ of subsets of $\Omega$ that do not form a $\sigma$-algebra, we can form $\sigma(\mathcal{F}) = \text{smallest } \sigma \text{-algebra containing } \mathcal{F}$ where
$\mathcal{F} = \{(a,b):-\infty\leq a\leq b \leq \infty\} $
Thus, $ \mathcal{F} \subset \sigma(\mathcal{F})$. We can also say if $ \mathcal{F} \subset \mathcal{U}$, then $ \sigma(\mathcal{F}) \subset \mathcal{U}$.
We recall that a function $X:(\Omega,\mathcal{U}) \rightarrow \mathbb{R}$ is a random variable if $ \{ X \in (a,b)\} \in \mathcal{U} $ for $(a,b)$ any interval, where $\{X \in A\}=X^{-1}(A)=\{\omega \in \Omega:X(\omega) \in A \} $ is the pre-image of a set.
Let $\mathcal{G} = \{A \subset \mathbb{R}:\{X \in A \} \in \mathcal{U} \} $. Use the fact that $\mathcal{U}$ is a $\sigma$-algebra to show that $\mathcal{G}$ is a $\sigma$-algebra.
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To show that $\mathcal{G}$ is a $\sigma$-algebra, we need to show that:
(a) $\emptyset \in \mathcal{G}$. How I'm viewing this is that since $\emptyset \in \mathcal{U}$ and that $\emptyset = \emptyset \in A$, then $\emptyset \in \mathcal{G}$.
(b) If $A \in \mathcal{U}$, then $A^c \in \mathcal{U}$, then I need to show that $A^c \in \mathcal{G}$, right? I'm not sure how to go about this.
(c) If $A_{n} \in \mathcal{U}$ then
$$ \bigcup_n A_n \Rightarrow X^{-1}(\bigcup_n A_n) = \bigcup_n X^{-1}(A_n) \in \mathcal{G}. $$ Right?
