I'm working on the following problem in my probability theory class:
Let $\Omega$ be a topological space. Show that the system of all sets in $\Omega$ which are $F_\sigma$ as well as $G_\delta$ is a $\sigma$-algebra.
Here, an $F_\sigma$-set is a countable union of closed sets, and a $G_\delta$-set is a countable intersection of open sets. I'm having trouble proving that the system is closed under countable unions. Clearly a countable union of sets which are both $F_\sigma$ and $G_\delta$ is also $F_\sigma$, but how would one show this countable union is again $G_\delta$? Any suggestions?
EDIT: This problem is actually false, as the comments have pointed out.
Best Answer
Your class of sets cannot be a sigma algebra because an infinite union of $G_{\delta}$ sets may not be $G_{\delta}$
Take for instance $\mathbb{Q}=\{q_1,q_2.....\}$ in the usual topology of the real line.
Then $\mathbb{Q}=\bigcup_{n=1}^{\infty}\bigcap_{i=1}^{\infty}(q_n-\frac{1}{i},q_n+ \frac{1}{i})$ a union of $G$ delta sets.
But it is proved from Baire's category theorem that $\mathbb{Q}$ is not a $G$ delta set.
But your class of sets it is an algebra though.
Lets Denote $\Sigma$ the class of such sets
Let $A \in \Sigma$
$A$ is a $G_{\delta}$ set thus from De Morgan Laws $A^c$ is a $F_{\sigma}$ set.
$A$ is a $F_{\sigma}$ thus proceeding as before $A^c$ is a $G_{\delta}$ set.
So $A^c$ belongs in $\Sigma$.
Now let us take a collection of sets $A_1,A_2...A_N \in \Sigma$
.
In general a finite union of $G_{\delta}$ sets is again $G_{\delta}$
So the finite union of $A_n$ is in $\Sigma$
Now using the fact that in a topological space the whole space and the empty set are both open and closed
you can prove that $\Omega,\emptyset \in \Sigma$