If $S$ and $T$ are stopping time, $S \vee T$ is $\max ({S,T})$, $F_S$ and $F_T$ are stopped sigma algebra, show that $F_{S \vee T} = \sigma(F_S,F_T)$.
My thinking : I should take a set $A$ in $F_{S \vee T}$ and show it is in $\sigma(F_S,F_T)$. I also notice that if $A$ is in $F_{S \vee T}$ then $A \cap\{S\le T\}$ is in $F_T$
Let $A \in F_{S \vee T}$ and write
$$A = \underbrace{(A \cap [S \leq T])}_{=: A_T} \cup \underbrace{(A \cap [T \leq S])}_{=:A_S}.$$
We claim that $A_S \in F_S$ and $A_T \in F_T$. Indeed,
$$A_S \cap [S \leq t] = A \cap [T \leq S] \cap [S \leq t] = \underbrace{A \cap [S \vee T \leq t]}_{\in \mathcal{F}_t} \cap \underbrace{[T \leq S] \cap [S \leq t]}_{\in \mathcal{F}_t}$$
for any $t \geq 0$. Hence, $A_S \in F_S$. Similarly, $A_T \in F_T$. Consequently, we find $$A = A_S \cup A_T \in \sigma(F_S,F_T).$$
On the other hand, if $A \in F_S$, then
$$A \cap [S \vee T \leq t] = \underbrace{A \cap [S \leq t]}_{\in \mathcal{F}_t} \cap [S \vee T \leq t] \in \mathcal{F}_t.$$
This shows that $\sigma(F_S,F_T) \subseteq F_{S \vee T}$.