Let me consider $(\Omega, F,(F_n)_n, \Bbb{P})$. Let $(X_n)_n$ be an adapted process at $0$. I need to show that $(X_n)_n$ is a martingale iff $\Bbb{E}(X_T)=0$ for all bounded stopping times $T$.
My idea was the following:
$\Rightarrow$ Let me assume $X_n$ is a martingale and pick a bounded stopping time $T$. Then from the stopping theorem we can immediately deduce that $$\Bbb{E}(X_T)=\Bbb{E}(X_0)=\Bbb{E}(0)=0$$
$\Leftarrow$Now in the other direction I have some problems. I know that I want to show that $\Bbb{E}(X_{n+1}|F_n)=X_n$ but this is by definition equivalent to show that for all $B_n\in F_n$ we have $$\Bbb{E}(X_{n+1}\Bbb{1}_{B_n})=\Bbb{E}(X_{n}\Bbb{1}_{B_n})~~~~~~~~~~(1)$$
So it would be enough to show $(1)$. But now I don't see how to do this. I somehow need to construct a bounded stopping time to use that $\Bbb{E}(X_T)=0$. Could maybe someone help me further?
What I found out in the mean time is that if $(B_n)$ is a disjoint collection of elements in $(F_n)_n$ then $T=\sum_{l=0}^\infty k\Bbb{1}_{B_k}$ is a stopping time.
Best Answer
Your first proof is true. For the second part let $n \in \mathbb{N}$ and $A \in F_n$. Consider:
$$T = n \mathrm{1}_{A} + (n+1) \mathrm{1}_{A^c}$$
$T$ is a bounded stopping time so that $\mathbb{E}(X_T) = \mathbb{E}(X_0)=0$. This implies that
$$\mathbb{E}(X_n \mathrm{1}_{A}) + \mathbb{E}(X_{n+1} \mathrm{1}_{A^c}) = 0$$
Likewise $(n+1)$ is a bounded stopping time so that $\mathbb{E}(X_{n+1}) = 0$. Therefore
\begin{align*} \mathbb{E}(X_{n+1}\mathrm{1}_{A}) &= \mathbb{E}(X_{n+1}) - \mathbb{E}(X_{n+1} \mathrm{1}_{A^c}) \\ &= - \mathbb{E}(X_{n+1} \mathrm{1}_{A^c})\\ &= \mathbb{E}(X_{n} \mathrm{1}_{A}) \end{align*}