First suppose that $X_1,...,X_t$ are IID random variables and $P(X_1=1)=p, P(X_1=-1)=1-p=q$ for $p\in (0,1)$, then $S_t=X_1+...+X_t, S_0=0$ is the simple random walk. I have two questions:
(1) Why are $A_t=(\frac{q}{p})^{S_t}$ and $N_t=S_t-E[S_t]$ martingales w.r.t to filtration $\sigma(X_1,...,X_t)$?
I make the standard approach: $E[A_{t+1}|F_t]=E[(\frac{q}{p})^{S_t}(\frac{q}{p})^{X_{t+1}}| F_t]=(\frac{q}{p})^{S_t}E[(\frac{q}{p})^{X_{t+1}}| F_t]$ Why should tthe last term be 1?
$E[N_{t+1}|F_t]=S_t-E[E[S_{t+1}]|F_t]$. Using tower property does not lead us to $E[E[S_{t+1}]|F_t]=E[S_t]$
(2) Why is the martingale $S^\tau$ not converging in $L^1$ where $\tau=\inf_\{t\in\mathbb Z^+: S_t=0\}$ is a stopping time? In this case we assume that that $S_0=1$
I have no idea here.
(1) By independence, \begin{equation*} E\Bigg[\bigg(\frac{q}{p}\biggr)^{X_{t+1}}\Bigg|\mathscr{F}_t\Biggr]=E\Bigg[\bigg(\frac{q}{p}\biggr)^{X_{t+1}}\Biggr]=\bigg(\frac{q}{p}\biggr)p+\bigg(\frac{q}{p}\biggr)^{-1}q=p+q=1. \end{equation*} Also, $E[E[S_{t+1}]|\mathscr{F}_t]=E[S_{t+1}]$ simply because $E[S_{t+1}]$ is a constant. Now \begin{align*} E[N_{t+1}|\mathscr{F}_t]&=E[S_{t+1}|\mathscr{F}_t]-E[E[S_{t+1}]|\mathscr{F}_t]\\ &= E[S_t+X_{t+1}|\mathscr{F}_t]-E[S_{t+1}]\\ &=E[S_t|\mathscr{F}_t]+E[X_{t+1}|\mathscr{F}_t]-E[S_{t+1}]\\ &=S_t+E[X_{t+1}]-E[S_t]-E[X_{t+1}]\\ &=N_t, \end{align*} the fourth equality because $S_t$ is $\mathscr{F}_t$-measurable, and so $E[S_t|\mathscr{F}_t]=S_t$, and because $X_{t+1}$ is independent of $\mathscr{F}_t$, and so $E[X_{t+1}|\mathscr{F}_t]=E[X_{t+1}]$.
As to the second question, I do not understand what you mean. If $\tau$ is a stopping time, $S_{\tau}$ is merely a random variable not a sequence, so what should converge in $L^1$?