Question:
Let $\mathscr{X}=\mathbb{R}, \mathscr{C}=\{C:C=(a,b), a<b, (a,b) \in \mathbb{R}^2\}=\text{open intervals}, \mathscr{G}=\text{open sets}, \mathscr{D}=\text{closed intervals},$ and $\mathscr{F}=\text{closed sets}$. Show that $\sigma(\mathscr{C})=\sigma(\mathscr{G})=\sigma(\mathscr{D})=\sigma(\mathscr{F})$.
My work:
So, it seems that the main idea in showing that sigma algebras are equivalent is to show they are subsets of each other. To do this, one would take an element of one $\sigma$-algebra, show that element belongs to the other $\sigma$-algebra, which then proves the first $\sigma$-algebra is a subset of the other.
So, let $G \in \mathscr{G}=\text{open sets}$. We can define $G=\cup C_n$, which for some countable collection, implies that $G=\cup C_n \in \mathscr{C}$. Then, we have $G \in \sigma(\mathscr{C})$ and $\sigma(\mathscr{G}) \subset \sigma(\mathscr{C})$. To show the converse, let $C \in \mathscr{C}$. In 2-dimensions (and higher), an interval is a subset of a set. (I don’t exactly know how to show this.) So, $C \in \mathscr{C} \subset \mathscr{G} \implies C \in \mathscr{G} \implies C \in \sigma(\mathscr{G}) \implies \sigma(\mathscr{C}) \subset \sigma(\mathscr{G})$. Therefore, $\sigma(\mathscr{C}) = \sigma(\mathscr{G})$.
We would use a similar argument for showing $\sigma(\mathscr{C}) = \sigma(\mathscr{D})$. Let $C \in \mathscr{C}$. Then, $C=(a,b)=\cup_n[a+\frac{c}{n},b-\frac{c}{n}] \in \mathscr{D}$. So, $\sigma(\mathscr{C}) \subset \sigma(\mathscr{D})$. Additionally, let $D\in\mathscr{D}$. Then, $D=[a,b]=\cap_n(a-\frac{1}{n},b+\frac{1}{n})\in\mathscr{C}$. So, we have $\sigma(\mathscr{D}) \subset \sigma(\mathscr{C})$ and, consequently, $\sigma(\mathscr{C}) = \sigma(\mathscr{D})$.
Lastly, we just need to show $\sigma(\mathscr{D}) = \sigma(\mathscr{F})$. Let $D\in \mathscr{D}$. Clearly, a closed interval is a subset of a closed set, although I don’t know how to mathematically show this. So, $D\in \mathscr{F} \implies \sigma(\mathscr{D}) \subset \sigma(\mathscr{F})$. Lastly, let $F \in \mathscr{F}$. Then, $F=\cup D_n \implies \sigma(\mathscr{F}) \subset \sigma(\mathscr{D})$. So, $\sigma(\mathscr{D}) = \sigma(\mathscr{F})$.
Is my thought process correct? I am learning measure theory through self-study in preparation for probability theory. Thanks for your time.
Here is a sketch. Consider the open intervals with rational endpoints $\mathscr{C}_{rat}$. For every $U \in \mathscr{G}$ we have the union $$U=\bigcup_{I\in \mathscr{C}_{rat},I \subset U}I$$ Indeed, by definition we have $\supset $. To show that $\subset $ take arbitrary $x \in U$. As $U$ is open there is a neighborhood around it contained in $U$; and inside it an open interval with rational endpoints which contains $x$. There are at most countably many $I$'s in the union above so we can assert that $$\mathscr{G}\subset \sigma(\mathscr{C}_{rat}) \implies \sigma(\mathscr{G})\subset \sigma(\mathscr{C}_{rat})$$ But it is straightforward to show that $\sigma(\mathscr{G})\supset \sigma(\mathscr{C}) \supset \sigma(\mathscr{C}_{rat})$, so $\sigma(\mathscr{G})= \sigma(\mathscr{C}) = \sigma(\mathscr{C}_{rat})$.
We can show that $\sigma(\mathscr{C})=\sigma(\mathscr{D})$ using countable intersections and unions of respective elements.
So now we have $\sigma(\mathscr{C})=\sigma(\mathscr{D})=\sigma(\mathscr{G})$.
We can show that $\sigma(\mathscr{G})=\sigma(\mathscr{F})$ by arguing that closed sets are complements of open sets, and viceversa.
Therefore $\sigma(\mathscr{C})=\sigma(\mathscr{D})=\sigma(\mathscr{G})=\sigma(\mathscr{F})$.