What is wrong in following argumentation?
Let $M$ be a local martingale (w.r.t. the filtration $(F_t)_{t\ge0}$). And $ T_n $ the localizing sequence of stopping times, such that $ M^{Tn} $ the stopped process is a Martingale.
Let $t,s \in \mathbb{R}$ with $s \le t $ then there exists a $n' \in \mathbb{N}$ such that $t<T_{n'}$ a.s.. If not, that would mean that $T_n \le t $ a.s. $\forall n$, which would be in contradiction to $ T_n \to \infty $ a.s, since $T_n$ is a localizing sequence. So finally we have the two properties:$$ E[M_t|F_s]= E[M_t^{T_{n'}}|F_s] = M_s^{T_{n'}} = M_s$$ $$ E[|M_t|]= E[|M_t^{T_{n'}}|] \lt \infty$$ Which would mean that M is a martingale.
I found the big mistake in my Argument. Obviously the following implication is wrong:$$\neg (\exists n \in \mathbb{N} :\space t<T_{n}\space \space a.s\space) \implies t\ge T_{n}\space a.s \space \forall n \in \mathbb{N}$$ And this of course destroys the rest of the argument.
If we take the localizing sequence $T_n = inf\{t\ge 0 : |B_t|>n\}$, for some Brownian Motion B, then intuitively it is clear, that $T_n \to \infty $ a.s.. But its is also intuitively clear, that for a fixed $t \ge 0$ there doesn't exists some $n\in\mathbb{N}$ such that almost every sample path hasn't reached the level $\pm n$ until the time $t$.