Optional Sampling Theorem in discrete setting I have a question about proving the optional sampling theorem in discrete setting. I dont know if what I am doing is mathematical justified. Can someone help me with this?
Defenition: Let $\tau$ be a stopping time, then $\mathcal{F}_{\tau}=\{F\subset \Omega: \forall n \in N \cup \{\infty \}, F\cap(\tau\leq n)\in \mathcal{F}_{n} \}$ is a sigma-algebra.
Defenition: Let $[M_{k}, k \in \mathcal{N}]$ be a martingale w.r.t. a filtration. Let $\tau$ be a stopping time that is never $\infty.$ Then $M_{\tau}$ is a function defined by $\forall \omega \in \Omega: M_{\tau}(\omega)=M_{\tau(\omega)}(\omega)$
Defenition Indicator function: Let $F\in \mathcal{F}$, where $\mathcal{F}$ is a sigma algebra. Then the indicator function is a function from $\Omega$ to $[0,1]$. $1_{F}(\omega)=1$ if $\omega \in F$, else $1_{F}(\omega)=0$
Question 1) Prove that $M_{\tau}(\omega)\in m\mathcal{F}_{\tau}$
Question 2) When in addition is given that $\tau \leq K$, where $K$ is a given positive integer, prove then that $E(M_{\tau})=M_{0}$. Here it might help to write $1_{\Omega}=\sum_{k=0}^{K}1_{\tau=k}$.

My attempts to answer these questions. 
(Q1)
Since $\tau(\omega) \neq \infty, \forall \omega \in \Omega$, we can write for some $t \in N$
$M_{\tau}=\sum_{i=0}^{t}1_{\tau=i}M_{i} \in \mathcal{F}_{t}$, Since the indicator and the martingale are both random variables, and products of random variables are also random variables.
Then $\forall n \in N: [\sum_{i=0}^{n}1_{\tau=i}M_{i}]^{-1}(B(\Re))\cap(\tau \leq n) \in \mathcal{F}_{n}$. Since both parts of the intersections are in $\mathcal{F}_{n}$ and hence $M_{\tau}(\omega)\in m\mathcal{F}_{\tau}$
(Q2)
$E(M_{\tau})=E(M_{\tau}1_{\Omega})=E(M_{\tau}\sum_{k=0}^{K}1_{\tau=k})=\sum_{k=0}^{K}E(M_{k}1_{\tau=k})=\sum_{k=0}^{K}E(M_{k}1_{\tau=k}|\mathcal{F}_{0})=\sum_{k=0}^{K}M_{0}E(1_{\tau=k})=\sum_{k=0}^{K}M_{0}P(1_{\tau=k})=M_{0}$
Can someone assure me whether my attempts are legit? Or give legit proofs?

I have a new attempt to answer question 1. I think this one is more general.
Given is that $M:\Omega \times \mathcal{T} \rightarrow \Re$ such that $M(n,\cdot) \in m \mathcal{F}_{n}, \forall n \in \mathcal{N}$
$\tau: \Omega \rightarrow \mathcal{N}$
$\mathcal{F}_{\tau}=\{ F \subset \Omega: \forall n \in \mathcal{N} \cup\{ \infty \}, F \cap (\tau \leq n) \in \mathcal{F}_{n} \}$
Claim: $M_{\tau} \in m \mathcal{F}_{\tau}$
Proof:
$M_{\tau}(\omega)=M(\tau(\omega),\omega) \forall \omega \in \Omega$.
Since $M(n,\cdot) \in m \mathcal{F}_{n}, M_{n}^{-1}(\mathcal{B}(\Re))\subseteq \mathcal{F}_{n}$ and $\forall \omega \in \Omega: \tau(\omega) \in \mathcal{N}$
it follows that
$M_{\tau(\omega)}^{-1}(\mathcal{B}(\Re)) \subseteq \mathcal{F}_{\tau(\omega)}$ 
$\rightarrow M_{\tau(\omega)}^{-1}(B) \in \mathcal{F}_{\tau(\omega)}, \forall B \in \mathcal{B}(\Re)$ 
$\rightarrow M_{\tau(\omega)}^{-1}(B) \cap (\tau \leq n) \in \mathcal{F}_{n}, \forall B \in \mathcal{B}(\Re)$
$\rightarrow M_{\tau}^{-1}(B) \in \mathcal{F}_{\tau}, \forall B \in \mathcal{B}(\Re)$
$\rightarrow M_{\tau} \in m \mathcal{F}_{\tau}$.
Hence $M_{\tau}$ is a random variable on $\mathcal{F}_{\tau}$.
 A: Note: I just saw your updated answer for question (1), and it is essentially the same as the one I have written here. Maybe the alternate presentation will be useful anyways.
Question 1:
So, an equivalent definition for $\mathcal{F}_{\tau}$ is:
$$ \mathcal{F}_{\tau} = \{ F \subset \Omega: F \cap \{\tau = n\} \in \mathcal{F}_n, \; \forall n \in \mathbb{N} \cup \{\infty\}\} $$
To show that $M_{\tau}$ is adapted to $\mathcal{F}_{\tau}$, we need to show that for an arbitrary Borel set $B$, the event $\{M_{\tau} \in B\}$ is $\mathcal{F}_{\tau}$-measurable. By the definition of $\mathcal{F}_{\tau}$, this is equivalent to showing that:
$$ \{M_n \in B\} \cap \{\tau = n\} \in \mathcal{F}_n $$
$\{M_n\}$ is a martingale, so $\{M_n \in B\} \in \mathcal{F}_n$. Also, $\tau$ is a stopping time so $\{\tau = n\} \in \mathcal{F}_n$ also. Thus their intersection is also $\in \mathcal{F}_n$.
Question 2:
OK, I'm not sure about the hints you've given, but here is one easy method. I'm also assuming that $\mathcal{F}_0 = \{\emptyset, \Omega\}$ but you can correct me if I'm wrong. If it's wrong then... I'm pretty sure there's some information still missing about the problem.
Denote the process stopped at time $n$ by $M_{\tau \wedge n}$. As Did mentioned in the comments, you have previously shown that $\{M_{\tau \wedge n}\}$ is a martingale. You can verify that $E M_{\tau \wedge n} = M_0$ for all $n$. Then since $\tau \leq K$, setting $n = K$ gives the result.
