I saw the following characterization of martingales (without proof) in some lecture notes I found on the web and I haven't been able to produce a proof it.
Let $X$ be an adapted process. If $E[X_{\tau}] = E[X_0]$ for all bounded stopping times $\tau$, then $X$ is a martingale.
First of all, I think the integrability condition is missing here. I don't think it would follow from the given property. Anyway, there is a similar result in Protter's "Stochastic Integration". There the stopping times need not be bounded but in return for this stronger condition you also get uniform integrability of $X$. In that case the proof uses the stopping times
$$u_A = \begin{cases}u & \omega \in A \\ 0 & \text{otherwise} \end{cases}$$
for some fixed $0 \leq s$ and $A \in \mathcal{F}_s$. If $u \geq s$, then $u_A$ is indeed a stopping time (not necessarily bounded). I tried to come up with a similar stopping time construction for the bounded case but haven't succeeded so far. I would appreciate it if someone could give me some pointers or simply post the proof.
We need to show that $\Bbb E [X_s | \mathcal{F}_t] = X_t$ for $s>t$, i.e. for any $F ∈ \mathcal{F}_t$, $\Bbb EX_s\Bbb 1_F = \Bbb EX_t \Bbb 1_F$.
Define the bounded stopping time T for fixed $s,t,F$, $$ T := t\Bbb 1_F + s \Bbb 1_{F^c}$$
By assumption, $\Bbb E X_T = \Bbb E X_0$, i.e. $$\Bbb E X_0 = \Bbb E X_T(\Bbb 1_F + \Bbb 1_{F^c}) = \Bbb EX_t\Bbb 1_F + \Bbb EX_s\Bbb 1_{F^c}$$ But also, the deterministic time $s$ is a bounded stopping time, giving $$\Bbb E X_0 = \Bbb E X_s(\Bbb 1_F + \Bbb 1_{F^c}) = \Bbb EX_s\Bbb 1_F + \Bbb EX_s\Bbb 1_{F^c}$$
Comparing the two gives $\Bbb EX_t \Bbb 1_{F} = \Bbb EX_s \Bbb 1_{F}$ as required.
Regarding integrability, I think its safe to say that you are implicitly assuming $\Bbb E X_0 < ∞ $, and we only define $\Bbb E Y$ if $\Bbb E |Y| < ∞$, so for any fixed time $t$, the given condition implies $\Bbb E X_t = \Bbb E X_0< ∞$ and so $\Bbb E |X_t| < ∞$.