Understanding Kolmogorov's inequality for submartingales I'm reading the Ross & Peköz book called "Secon Course in Probability Theory" and they mention a version of the Kolmogorov's inequality for submartingales:

Suppose $Z_n,$ $n ≥ 1$, is a nonnegative submartingale, then for $a > 0$
$P(max\{Z_1,...,Z_n\} ≥ a) ≤ E[Z_n]/a$

And then proceed to name a new variable and use the Marköv inequality

Proof: Let $N$ be the smallest $i$, $i ≤ n$ such that $Z_i ≥ a$, and let it
equal $n$ if $Z_i < a$ for all $i = 1, . . . , n$. Then
$P(max\{Z_1,...,Z_n\} ≥ a) = P(Z_N ≥ a)$
$≤ E[ZN ]/a\hspace{1cm}$ by Markov’s inequality
$≤ E[Zn]/a$ $\hspace{1cm}$since $N ≤ n$

But I can't see why the equality $P(max\{Z_1,...,Z_n\} ≥ a) = P(Z_N ≥ a)$ is true, since I know that the monotony of the $Z_n$ can be anything.
Any help is appreciated, thanks.

Let $E=\left\{\max \{Z_1,\ldots,Z_n\}\ge a\right\}=\{\exists i\in[n] : Z_i\ge a\}$ and $F=\{Z_N\ge a\}$.

*

*If there exists $i$ such that $Z_i\ge a$, we have by construction $Z_N\ge a$. So $E\subseteq F$.

*If for all $i$, $Z_i<a$, then in particular $Z_N<a$. So $\overline{E}\subseteq \overline{F}$.

It follows that $E=F$, so $P(E)=P(F)$.
