Doob's Optional Stopping Theorem: $\xi_\tau$ vs $\xi_{\tau\land n}$

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I have some troubles in understanding the Optional Stopping Theorem by Doobs.

I have a bit of confusion about the following (Brzezniak, Zastawniak - Basic Stochastic Processes p. 58-59):

Let $\xi_n$ is a martingale and $\tau$ is a stopping time wrt a filtration $\mathcal{F}_{n}$ such that:

1) $\tau < \infty$ a.s. 2) $\xi_\tau$ is integrable 3) $E(\xi_n1_{\tau>n})$ goes to $0$ as n go to $\infty$.

Then $E(\xi_\tau) = E(\xi_1)$.

The proof starts saying

$\xi_\tau = \xi_{\tau\land n}+(\xi_\tau-\xi_n)1_{\tau>n}$

and then, by applying $E(.)$

$E(\xi_\tau) = E(\xi_{\tau\land n})+E(\xi_\tau1_{\tau>n})+E(\xi_n1_{\tau>n})$.

Now, the crucial point is: I do not understand the difference between $\xi_\tau$ and $\xi_{\tau\land n}$. I mean, the second one is for sure the stopped martingale at the stopping time $\tau$ while the first one is the martingale evaluated at the instant $\tau$.

But if:

$\xi_{n}=\eta_1+2\eta_2+...+2^{n-1}\eta_n$ ("game martingale") and $\tau<n$

then aren't $\xi_\tau$ and $\xi_{\tau\land n}$ the same?

The proof goes on saying $E(\xi_{\tau\land n})=E(\xi_1)$ thanks to definition of martingale (the stopped process is a martingale itself, so this is clear).

Thank you :)

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$\tau\wedge n=\min\{\tau,n\}$. Notice that $n$ is fix and $\tau$ is a random variable. So $\tau\wedge n$ is the random variable that is $\tau$ if $\tau\leq n$ and $n$ if $\tau>n$. Here, $$\xi_{\tau\wedge n}=\begin{cases}\xi_\tau&\tau\leq n\\ \xi_n&\tau>n\end{cases}.$$