Consider a probability space $(\Omega, \mathcal F, \mathbb P)$, endowed with a filtration $(\mathcal F_t)_{t\geq 0}$. A stopping time $T$ is a random variable taking values in $[0,\infty]$ such that for all $t\geq 0$ we have $\{T\leq t\}\in\mathcal F_t$.
I want to show that the sum of two stopping times $T$ and $S$ is a stopping time. I know that I can rewrite
$$\{S+T<t\} = \bigcup_{r,s \in \mathbb{Q}^+;\; r+s<t} \{S < r\} \cap \{T < s\}\,.$$
But to make it a proof I would need to show that an equivalent definition for stopping time is that $\{T<t\}\in\mathcal F_t$ for all $t\geq 0$.
Using $\{T<t\} = \bigcup_{n\in\mathbb N^+}\{T\leq t- 1/n\}$ I see that if $T$ is a stopping time than $\{T<t\}\in\mathcal F_t$. But is the reverse implication true? It seems to me that it requires right continuity of the filtration... But as far as I know the sum of two stopping times is a stopping time with no further assumption on the the filtration.
So how to prove that $S+T$ is a stopping time? Should I use a different kind of argument or am I wrong about the necessity of the right continuity?
Your observations are correct. The trick is to look at $\{S+T >u\}=\cup_r \{S>r\} \cap \{T>u-r\}$ for $u>0$ and the take complements on both sides. Also $(S+T=0) =(S=0)\cap (T=0) \in \mathcal F_0$.