Showing that stopped filtration $\mathcal{F}_{S \vee T} = \sigma(\mathcal{F}_S,\mathcal{F}_T)$ Just to be sure we're on the same page, we define the stopped filtration of the stopping time $T$, $\mathcal{F}_T$, to be the $\sigma$-algebra which satisfies the following:
$$
A \in \mathcal{F}_T \Leftrightarrow A \cap \{T \leq n\} \in \mathcal{F}_n \;\forall n \geq 1
$$
Let $S$ and $T$ be two stopping times (i.e. $\{S = n\} \in \mathcal{F}_n$ $\forall n \geq 1$, same for $T$). I want to show that $\mathcal{F}_{S \vee T} = \sigma(\mathcal{F}_S, \mathcal{F}_T)$. 

When attempting this problem, my first observation is the following:
$$
\begin{align*}
&A \cap \{S \vee T \leq n\} \\
&=  A \cap \{S \leq n \text{ and } T \leq n\}\\
&= A \cap \{S \leq n\} \cap \{T \leq n\} \in \mathcal{F}_n
\end{align*}
$$
I know that all of $\{S = n\}, \{S \leq n\}, \{S < n\}, \{S \geq n\}, \{S > n\}$ lie in $\mathcal{F}_n$ (same for $T$), so I believe that the main idea of showing this is to play around with unions/intersections/complements etc. of the sets. However, I find myself facing the following problems.


*

*I don't fully understand what is inside $\sigma(\mathcal{F}_S, \mathcal{F}_T)$. I know it's meant to be the "union" of two $\sigma$-algebras, and the $\sigma(\mathcal{F}_S, \mathcal{F}_T)$ is to make sure that it is indeed a $\sigma$-algebra. However, when taking $A \in \sigma(\mathcal{F}_S, \mathcal{F}_T)$, I cannot assume that $A \in \mathcal{F}_S$ or $A \in \mathcal{F}_T$, but a set formed by the countable union/intersection/complement etc. of sets in $\mathcal{F}_S$ and $\mathcal{F}_T$. Therefore, I have no idea where to start when trying to show $\sigma(\mathcal{F}_S, \mathcal{F}_T) \subseteq \mathcal{F}_{S \vee T}$.

*For showing that $\mathcal{F}_{S \vee T} \subseteq \sigma(\mathcal{F}_S, \mathcal{F}_T)$, while I can start since I know what is in $\mathcal{F}_{S \vee T}$, now I am not sure how to show that $A \in \mathcal{F}_{S \vee T} \Rightarrow A \in \sigma(\mathcal{F}_S, \mathcal{F}_T)$. I thought of showing that $A \in \mathcal{F}_S$ or $A \in \mathcal{F}_T$, but this seems like a strange way to approach since I foresee it going something along the line of $A \cap \{S \leq n\} \notin \mathcal{F_n} \Rightarrow A \cap \{T \leq n\} \in \mathcal{F_n}$. Once again, as I don't know that $\sigma(\mathcal{F}_S, \mathcal{F}_T)$ contains, I can't seem to show this direction directly as well.


Thanks for any help in advance.
 A: 
$\sigma(\mathcal{F}_S,\mathcal{F}_T) \subseteq \mathcal{F}_{S \vee T}$

$\sigma(\mathcal{F}_S,\mathcal{F}_T)$ is the smallest $\sigma$-algebra which contains $\mathcal{F}_S$ and $\mathcal{F}_T$. In order to prove the inclusion, it therefore suffices to show that $\mathcal{F}_S \subseteq \mathcal{F}_{S \vee T}$ and $\mathcal{F}_T \subseteq \mathcal{F}_{S \vee T}$. Why? Well, then $\mathcal{F}_{S \wedge T}$ is one $\sigma$-algebra containing $\mathcal{F}_S$ and $\mathcal{F}_T$. Since $\sigma(\mathcal{F}_S,\mathcal{F}_T)$ is the smallest $\sigma$-algebra containing $\mathcal{F}_S$ and $\mathcal{F}_T$, it follows that $\sigma(\mathcal{F}_S,\mathcal{F}_T) \subseteq \mathcal{F}_{S \vee T}$.
Consequently, it remains to show $\mathcal{F}_S \subseteq \mathcal{F}_{S \vee T}$ and $\mathcal{F}_T \subseteq \mathcal{F}_{S \vee T}$. Fix, say, $A \in \mathcal{F}_S$. Then, by your "first observation", $$A \cap \{S \vee T \leq n\} = (A \cap \{S \leq n\}) \cap \{T \leq n\}.$$ Since $A \in \mathcal{F}_S$, we have $A \cap \{S \leq n\} \in \mathcal{F}_n$. Moreover, $\{T \leq n\} \in \mathcal{F}_n$ since $T$ is a stopping time, and so $A \cap \{S \vee T \leq n\} \in \mathcal{F}_n$. By the very definition of $\mathcal{F}_{S \vee T}$, this proves $A \in \mathcal{F}_{S \vee T}$. Hence, we have shown $\mathcal{F}_S \subseteq \mathcal{F}_{S \wedge T}$. The other inclusion is proved analogously (just swap the roles of $S$ and $T$ in the above reasoning).

$\mathcal{F}_{S \vee T} \subseteq \sigma(\mathcal{F}_S,\mathcal{F}_T)$

Let $A \in \mathcal{F}_{S \vee T}$, and write $$A = \underbrace{(A \cap \{S \leq T\})}_{=:A_T} \cup \underbrace{(A \cap \{T \leq S\})}_{=:A_S}.$$
Then
$$A_S \cap \{S \leq n\} = \underbrace{(A \cap \{S \vee T \leq n\})}_{\in \mathcal{F}_n} \cap (\{T \leq S\} \cap \{S \leq n\}).$$
where we used $A \in \mathcal{F}_{S \vee T}$. Since $S$ and $T$ are stopping times, it is not difficult to see that $\{T \leq S\} \cap \{S \leq n\} \in \mathcal{F}_n$. Hence, $A_S \cap \{S \leq n\} \in \mathcal{F}_n$, i.e. $A_S \in \mathcal{F}_S$. In a similar fashion, we get $A_T \in \mathcal{F}_T$ (again, swap the roles of $S$ and $T$ in the above reasoning). Hence,
$$A = A_T \cup A_S \in \sigma(\mathcal{F}_S,\mathcal{F}_T).$$
