Martingale Transforms and quadratic variation Let $M$ be a martingale with $\mathbb{E}M_{n}^2<\infty$ for all $n$. Let $C$ be a bounded predictable process and set $X=M\cdot C$. Show that $\mathbb{E}X_{n}^2<\infty$ for all $n$ and that $\langle X\rangle =C^2\cdot \langle M\rangle$.
$\langle M\rangle$ is the quadratic variation of $M$.
Now for the first part, my reasoning is the following:
Since $C$ is bounded, we have that $\exists B_i$, such that: $B_i<\infty$ and $C_i(\omega)\leq B_i$, $\forall i$.
So now I have that $X_n(\omega)^2=(C\cdot M)^{2}_n(\omega)=(\sum_{i=1}^{n}C_i(\omega)\Delta M_i(\omega))^2=(\sum_{i=1}^{n}C_i(\omega)(M_i-M_{i-1}(\omega))^2\leq \sum_{i=1}^n(B_i(M_i-M_{i-1}(\omega))^2$.
Therefore $\mathbb{E}X_{n}^2\leq\mathbb{E}(\sum_{i=1}^nB_i(M_i-M_{i-1}))^2=\sum_{i=1}^nB_i^2\mathbb{E}(M_i-M_{i-1})^2=\sum_{i}B_i^2\mathbb{E}(M_i^2-M_iM_{i-1}+M_{i-1}^2)=\sum_iB_i(\mathbb{E}M_i^2-\mathbb{E}M_iM_{i-1}+\mathbb{E}M_{i-1}^2)<\infty$.
However, I have some difficulties proving the second part of the exercise, namely proving that: $\langle X\rangle =C^2\cdot \langle M\rangle$.
We have that: $\langle X\rangle_n=X_0+\sum_{i=1}^{n}\Delta\langle X\rangle_i$.
Now $\Delta \langle X\rangle _n=\mathbb{E}[X_n-\mathbb{E}(X_n|\mathcal{F}_{n-1})^2|\mathcal{F}_{n-1})=\mathbb{E}[(X_n^2-X_{n-1})^2|\mathcal{F}_{n-1}]=\mathbb{E}[((C\cdot M)_n-(C\cdot M)_{n-1})^2|\mathcal{F}_{n-1}]=\mathbb{E}[(\sum_{i=1}^nC_k\Delta M_k-\sum_{i=1}^{n-1} C_k\Delta M_k)^2|\mathcal{F}_{n-1}]=\mathbb{E}((C_n\Delta M_n)^2|\mathcal{F}_{n-1})$.
On the other hand, we have that: $C^2\cdot\langle M\rangle=\sum_{k=1}^nC_k\Delta\langle M\rangle_k=\sum_{k=1}^nC_k\mathbb{E}[(M_k-M_{k-1})^2|\mathcal{F}_{k-1}]=\sum_{k=1}^nC_k\mathbb(M_k^2-M_{k-1}^2|\mathcal{F}_{n-1})=\sum_{k=1}^nC_k\mathbb{E}((\Delta M_k)^2|\mathcal{F}_{k-1})$. 
Now however, I am unable to proceed any further, since I can not find any way to connect these two things.
 A: Except for several typos your calculation is "roughly" correct.
So let's start from the beginning: For a (square-integrable) martingale, the compensator $\langle M \rangle$ equals
$$\langle M \rangle_n = \sum_{k=1}^n \mathbb{E}(M_k^2 - M_{k-1}^2 \mid \mathcal{F}_{k-1}) = \sum_{k=1}^n \mathbb{E}((M_k-M_{k-1})^2 \mid \mathcal{F}_{k-1}) \tag{1}$$
where the second equality follows from the martingale property. Applying this to $C \bullet M$, we get
$$\begin{align*} \langle C \bullet M \rangle_n &= \sum_{k=1}^n \mathbb{E}((C \bullet M_k - C \bullet M_{k-1})^2 \mid \mathcal{F}_{k-1}) \\ &\stackrel{\text{def}}{=} \sum_{k=1}^n \mathbb{E}(C_k^2 (M_k-M_{k-1})^2 \mid \mathcal{F}_{k-1})  \\ &= \sum_{k=1}^n C_k^2 \mathbb{E}((M_k-M_{k-1})^2 \mid \mathcal{F}_{k-1}) \end{align*}$$
where we have used in the last step that $C_k$ is $\mathcal{F}_{k-1}$-measurable. Having a close look at $(1)$ yields
$$\langle C \bullet M \rangle_n = \sum_{k=1}^n C_k^2 (\langle M \rangle_k - \langle M \rangle_{k-1})  \stackrel{\text{def}}{=} C^2 \bullet \langle M \rangle_n.$$
