I'm trying to prove the following statement:
If $\mathcal{F}$ is an algebra, then it is a $\sigma$-algebra if and only if $\{A_n\} \subset \mathcal{F}$ with $A_i \cap A_j = \emptyset$ for all $i \neq j$ implies $\cup_{n=1}^\infty A_n \in \mathcal{F}$.
I did the following proof but i donno if it makes sense:
Proof of $(\Rightarrow)$ direction:
Assume $\mathcal{F}$ is a $\sigma$-algebra. Assume also $\{A_n\} \subset \mathcal{F}$ and $A_i \cap A_j = \emptyset$ for all $i \neq j$. Hence, from the definition of a $\sigma$-algebra, $\cup_{n=1}^\infty \in \mathcal{F}$.
Proof of $(\Leftarrow)$ direction:
Assume $\{A_n\} \subset \mathcal{F}$ and $A_i \cap A_j = \emptyset$ for all $i \neq j \Rightarrow \cup_{n=1}^\infty \in \mathcal{F}$. We need to show that $\mathcal{F}$ is a sigma algebra. Assume by contradiction that $\mathcal{F}$ is not a $\sigma$-algebra. Since $\mathcal{F}$ is an algebra, assuming that $\mathcal{F}$ is not a sigma algebra means assuming that $\{A_n\} \subset \mathcal{F}$ and $\cup_{n=1}^\infty \notin \mathcal{F}$. By definition of $\{A_n\} \subset \mathcal{F}$, we have that $\forall A_i \in \{A_n\} \Rightarrow A_i \in \mathcal{F}$. Moreover, since $\mathcal{F}$ is an algebra, $A_i^c \in \mathcal{F}$, $\cup_{n=1}^n A_i \in \mathcal{F}$, and $\cup_{n=1}^n A_i^c \in \mathcal{F}$. Now define $\tilde{A_1} = A_1$, $\tilde{A_2} = A_2 \cap (\cup_{i \neq 2} A_i^c)$ and so on. But then $\tilde{A_i} \cap \tilde{A_j} = \emptyset$ for all $i \neq j$. Hence, by hypothesis, $\cup_{n=1}^\infty \in \mathcal{F}$ so we reach a contradiction.
Could you please review the proof and provide me feedback?
Thanks a lot!