Proof of Doob's inequality by Durrett

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This is Doob's inequality given in Durrett's Probability: Theory and Examples. However, in the proof, I don't understand why $X_N \ge \lambda$ on $A$, since if $N=n$, then all we can say is $X_N=\max_{0\le m\le n}X_m^+\ge \lambda$, and so without any assumption on nonnegativity of $X_n$, I don't see how the proof makes sense. I would greatly appreciate any help.

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Theorem 5.4.1. If $X_n$ is a submartingale and $N$ is a stopping time with $P(N\le k)=1$ then $$EX_0\le EX_N \le EX_k.$$

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On the set $A$ we have $ 0<\lambda\leq \bar X_n=\max_{0\leq m\leq n} X_m^+.$ that is, at least one of the $X_m^+$'s exceeds $\lambda$ for $0\leq m\leq n$. For such an $m$ value we have $X_m=X_m^+$.