So this question was one we were asked to proof during class. The formulation is as follows:
Now the question is about c). I was able to proof that $\mathrm{E} M_nA_n =\mathrm{E}M_{n-1}A_n$ via the fact $\mathrm{E} \sum_{k=1}^n A_k(M_k-M_{k-1})=0$. Namely, $\mathrm{E} M_nA_n - \mathrm{E}M_{n-1}A_n = \mathrm{E} \sum_{k=1}^n A_k(M_k-M_{k-1}) - \mathrm{E} \sum_{k=1}^{n-1} A_k(M_k-M_{k-1}) = 0-0=0$.
But the rest of the proof i cannot formulate. I don't know how to proceed. I found however this book, which contains the theoreom and sort of the same proof, but I don't understand the steps which are taken.
Proof is in prop. 4.3 in: http://books.google.nl/books?id=ATNy_Zg3PSsC&printsec=frontcover&source=gbs_ge_summary_r&cad=0#v=onepage&q&f=false
Can someone explain or give me the proof?
If someone would like to have the proofs of a) and b), I quess i can post them.
You have already shown that
$$\mathbb{E}(M_n A_n) = \mathbb{E}(M_{n-1} A_n).$$
Since $M_{n-1}$ is $\mathcal{F}_{n-1}$-measurable, it follows from the tower property that
$$\begin{align*} \mathbb{E}(M_n A_n) &= \mathbb{E} \big[ \mathbb{E}(M_{n-1} A_n \mid \mathcal{F}_{n-1}) \big] \\ &= \mathbb{E} \big[ M_{n-1} \mathbb{E}(A_n \mid \mathcal{F}_{n-1}) \big] \\ &= \mathbb{E}(M_{n-1} \tilde{A}_n). \end{align*}$$
Doing the same trick again (note hat $\tilde{A}_n$ is $\mathcal{F}_{n-1}$-measurable) and using that $(M_n)_{n \in \mathbb{N}}$ is a martingale, we get
$$\begin{align*} \mathbb{E}(M_n A_n)&= \mathbb{E}\big[ \mathbb{E}(\tilde{A}_n M_{n-1} \mid \mathcal{F}_{n-1}) \big] = \mathbb{E} \big[ \tilde{A}_n \underbrace{\mathbb{E}(M_{n-1}\mid \mathcal{F}_{n-1})}_{\mathbb{E}(M_n \mid \mathcal{F}_{n-1})} \big]= \mathbb{E}(M_n \tilde{A}_n). \end{align*}$$
Hence,
$$\mathbb{E}(M_n \cdot (A_n-\tilde{A}_n)) = 0. \tag{1}$$
Recall that this identity holds for any bounded martingale. For fixed $k \in \mathbb{N}$, define a martingale $(M_n)_{n \in \mathbb{N}}$ by
$$ M_n := \begin{cases} X, & k \geq n, \\ \mathbb{E}(X \mid \mathcal{F}_n), & n=0,\ldots,k-1 \end{cases}$$
where $$X := \text{sgn}\,(A_k - \tilde{A}_k).$$ For this particular martingale (and $n=k$), $(1)$ gives
$$\mathbb{E}|A_k-\tilde{A}_k| =0.$$ This finishes the proof.