Question about Rene Schilling's proof of the submartingale maximal inequality I am reading the proof on the submartingale maximal inequality from Rene Schilling's Measures, Integrals and Martingales. The proof below uses an equivalence theorem on submartingales. Namely, look at the second $\le$ in the last line. I don't quite get how we get this here, however. We can set $\tau = N+1$, but to get $N$ as the index, how do we bound $\sigma$ by a stopping time that becomes $N$ on the set $A$?
Let $u_n$ be a submartingale and consider the stopping time when $u_n$ exceeds the level $s$ for the first time: 
$$\sigma:= \inf \{n \le N: u_n \ge s\} \wedge (N+1)$$ and set $A:= \{\max_{1 \le n \le N} u_n \ge s\} = \cup_{n=1}^N \{u_n \ge s\} = \{\sigma \le N\} \in \mathscr{A}_\sigma.$ 
Then from the theorem on submartingale equivalence, i.e. $u_n$ is a submartingale iff $\int_A u_\sigma d\mu \le \int_A u_\tau d\mu$ for all bounded stopping times $\sigma \le \tau$ and $A \in \mathscr{A}_\sigma$, and the fact that $u_\sigma \ge s$ on $A$, we conclude 
$$\mu(\cup_{n=1}^N \{u_n \ge s\}) \le \int_A \frac{u_\sigma}{s}d\mu = \frac{1}{s} \int_A u_\sigma d\mu \le \frac{1}{s} \int_A u_N d\mu \le \frac{1}{s} \int u_N^+ d\mu.$$

One could consider the stopping time $\tau = N \mathcal{X}_A + (N+1) \mathcal{X}_{A^c}$, where $\mathcal{X}_A$ is the characteristic function of $A = \{ \sigma \leq N \}$. Clearly, we have that $\sigma \leq \tau$ a.s. and we have $$
\int_A u_\sigma d\mu \leq \int_A u_\tau d\mu
$$ but on $A$, $\tau \equiv N$, giving us the desired inequality $$
\int_A u_\sigma d\mu \leq \int_A u_N d\mu.
$$
