Suppose $u_n$ is a submartingale and define $\tau_0:=0$ and recursively for $k=1,2,3,\dots, $ $$\sigma_k:= \inf\{n > \tau_{k-1}: u_n \le a\} \wedge N\; \text{and}\; \tau_k:= \inf\{n>\sigma_k:u_n \ge b\} \wedge N;$$ as usual we set $\inf \emptyset = +\infty$. In this case, we want to prove that both are stopping times. Below is the solution.
However, I am not sure this solution is correct. Indeed, why do we have $\{\sigma_k>l\} = \{\tau_{k-1} \le l\}$ and $\{\sigma_k=N\} = \emptyset$? I think this is incorrect and we should instead be looking at sets $\{\sigma_k \le l\}$.
First, if $1 \le l < N$, then $\{\sigma_k \le l\}=\bigcup_{j\le l} \{\tau_{k-1}<j\} \cap \{u_j \le a\}. $ For $l\ge N$, $\{\sigma_k \le l\} = \Omega$.
For $\tau_k$, we should reverse the roles of $\sigma$ and $\tau$ and look at $\{u_j \ge b\}$ instead.
Is this the correct justification?
