Let $\sigma$ and $\tau$ be two stopping times in $\mathscr{F}_t$ and let this filtration satisfy all the usual conditions.
Question: Is $\sigma + \tau$ a stopping time?
Attempt at a solution:
I need to demonstrate that $\{ \sigma + \tau \leq t\}\in \mathscr{F}_t$, or that $\{\sigma \leq t - \tau \} \in \mathscr{F}_t$.
Since $\sigma$ is a stopping time we have that $\{\sigma \leq t - \tau\} \in \mathscr{F}_{t - \tau}$, where $t - \tau \in [0,t]$.
Since $t > t - \tau$, we have that $\mathscr{F}_{t-\tau} \subseteq \mathscr{F}_t$ by the definition of $\mathscr{F}$.
This implies that $\{\sigma \leq t - \tau\} \in \mathscr{F}_t$, and that $\sigma + \tau$ is a stopping time.
Is my attempt correct?
Revised answer:
I found it easier to look at the complements instead. Then we might as well show that $\{\tau+\sigma>t\}\in\mathscr{F}_t$ for all $t$. For a stopping time $\tau$ we know that $\{\tau<t\}\in\mathscr{F}_t$ and also $\{\tau=t\}\in\mathscr{F}_t$. Now we write our set as
$$ \{\tau+\sigma>t\}=\{\tau=0,\,\tau+\sigma>t\}\cup\{0<\tau<t,\,\tau+\sigma>t\}\cup\{\tau\geq t,\, \tau+\sigma>t\}\\ =\{\tau=0,\,\sigma>t\}\cup\{0<\tau<t,\,\tau+\sigma>t\}\cup\{\tau>t,\,\sigma=0\}\cup\{\tau\geq t,\,\sigma>0\}. $$
Then $\{\tau=0,\,\sigma>t\}\in\mathscr{F}_t$ and $\{\tau>t,\,\sigma=0\}\in\mathscr{F}_t$, since $\tau$ and $\sigma$ are stopping times. Furthermore $\{\tau\geq t,\,\sigma>0\}\in\mathscr{F}_t$ because $\{\sigma>0\}=\{\sigma=0\}^c\in\mathscr{F}_0$ and $\{\tau\geq t\}=\{\tau<t\}^c\in\mathscr{F}_t$. At last we have that $$ \{0<\tau<t,\tau+\sigma>t\}=\bigcup_{r\,\in\, (0,t)\cap\,\mathbb{Q}}\{r<\tau<t,\,\sigma>t-r\}\in\mathscr{F}_t. $$
I hope that this last equality with the union now holds.