proof that a stopped martingale is a martingale?

Defenition. $\mathcal{F}_{\tau}=\{F\subset \Omega: \forall n \in N \cup \{\infty\}, F\cap(\tau\leq n)\in \mathcal{F}_{n}$} is a sigma-algebra.

Defenition. $\forall \omega \in \Omega: M_{n}^{\tau}(\omega)=M_{min(\tau(\omega), n)}(\omega)$

Question Prove that this collection of function is also a martingale w.r.t. the given filtration. First try to prove that the following is true $M_{n}^{\tau}=M_{n-1}^{\tau}+1_{\tau \leq n}(M_{n}-M_{n-1})$

My Attempt First of all I am not able to prove that $M_{n}^{\tau}=M_{n-1}^{\tau}+1_{\tau \leq n}(M_{n}-M_{n-1})$. It would be great if someone could help me with that!

Secondly, if I assume that that is true, I continued as follows.

We want to show that $E(M_{n}^{\tau}|\mathcal{F}_{n-1})=M_{n-1}^{\tau}$.

By assumption (now)

$E(M_{n}^{\tau}|\mathcal{F}_{n-1})=E(M_{n-1}^{\tau}+1_{\tau \leq n}(M_{n}-M_{n-1})|\mathcal{F}_{n-1})=M_{n-1}^{\tau}+E(1_{\tau \leq n}(M_{n}-M_{n-1})|\mathcal{F}_{n-1})=^{?}M_{n-1}^{\tau}$

Can someone help me to continue and complete this proof?

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\digamma $\to$ \mathcal F. – Did Jun 26 '13 at 14:31

The hint is to prove the identity $M_{n}^{\tau}=M_{n-1}^{\tau}+1_{\tau \geqslant n}(M_{n}-M_{n-1})$ (check this when $\tau\geqslant n$ and when $\tau\leqslant n-1$), not that $M_{n}^{\tau}=M_{n-1}^{\tau}+1_{\tau \leqslant n}(M_{n}-M_{n-1})$.
Then, following your path, one is left with showing that $E(1_{\tau \geqslant n}(M_{n}-M_{n-1})|\mathcal F_{n-1})=0$. But this is clear because $1_{\tau \geqslant n}=1-1_{\tau \leqslant n-1}$ and $M_{n-1}$ are $\mathcal F_{n-1}$-measurable while $E(M_{n}|\mathcal F_{n-1})=M_{n-1}$.