Bounded stopping times and martingales I am trying to show that if for every bounded stopping time $\tau$ it holds that $E(X_{\tau})=E(X_{0})$ then ${X_{n}}$ is a discrete time martingale.
I have a few questions on my ideas on solving this.  
First I consider a set $A \in F_{n}.$  I define $\tau= n-1$ for $\omega \in A$ and $\tau= n$ for $\omega \in A^{c}.$  I want to show that this definition of $\tau$ is a stopping time; that is $\{ \tau \le t \}  \in F_{t}$ for all $t$.  I am not quite sure if I know how to show this correctly:  $\{ \tau \le t \}=(A\cap\{n-1 \le t\})\cup (A^c\cap\{n \le t\}) \in F_{t}.$  I am not sure if I formulated the algebra correctly here.  Any help would be appreciated.
If I can show that $\tau$ is a stopping time then I know I can show $E(X_{n-1}1_{A})=E(X_{n}1_{A}),$ where $1_{A}$ is the indicator function for the set A.  Is this enough to conclude $X_{n}$ is a martingale.  Why?
 A: First of all, note that you have to consider $A \in \mathcal{F}_{\color{red}{n-1}}$ (otherwise $\tau$ fails to be a stopping time). Although I understand very well what you want to say with the following line
$$\{\tau \leq t\} = (A \cap \{n-1 \leq t\}) \cup (A^c \cap \{n \leq t\})$$
this is not written up nicely (e.g. what exactly is $\{n-1 \leq t\}$ supposed to mean?). From my point of view it is much clearer to write
$$\{\tau \leq t\} = \begin{cases} \emptyset, & t \leq n-2, \\ A & t =n-1, \\ \Omega, & t \geq n. \end{cases}$$
Since $A \in \mathcal{F}_{n-1}$, this identity shows that $\tau$ is a stopping time.

If I can show that $\tau$ is a stopping time then I know I can show $\mathbb{E}(X_{n-1} 1_A) = \mathbb{E}(X_n 1_A)$ [...]. Is this enough to conclude $X_n$ is a martingale?

Yes. By the very definition of conditional expectation
$$\forall A \in \mathcal{F}_{n-1}: \qquad \int_A X_n \, d\mathbb{P} = \mathbb{E}(1_A X_n) = \mathbb{E}(1_A X_{n-1}) = \int_A X_{n-1} \, d\mathbb{P}$$
is equivalent to
$$\mathbb{E}(X_{n-1} \mid \mathcal{F}_{n-1}) = X_{n-1}. \tag{1}$$
Combining $(1)$ with the tower property, we find that $(X_n)_{n \in \mathbb{N}}$ is a martingale. (This is a standard exercise in martingale theory; give it a try!)
