Show that M is a martingale Assume that $E[(HM)_N]=0$ for predictable process H and $H_0=0$. I want to prove that $M$ is a martingale. 
$$(HM)_N= \sum_{j=1}^N H_j (M_j -M_{j-1})+ H_0M_0 = \sum_{j=1}^N H_j (M_j -M_{j-1})$$
I am stuck at showing that $M$ is a martinagle. 
I read some theorems. (If required, I can share them) but I couldn’t not show this proof. Please help me to do this. 
Many thanks. 
 A: Let $(\mathcal{F}_n)_{n \in \mathbb{N}}$ be the underlying filtration. Fix $n \in \mathbb{N}$ and pick $F \in \mathcal{F}_n$. The process
$$H_j := \begin{cases} 0, & \text{if $j \neq n+1$}, \\ 1_F, & \text{if $j=n+1$},\end{cases}$$ is predictable and $H_0=0$. We have
$$(H \bullet M)_{n+1} = \sum_{j=1}^{n+1} H_j (M_j-M_{j-1}) = H_{n+1} (M_{n+1}-M_n) = 1_F (M_{n+1}-M_n).$$
By assumption, the random variable has expectation zero, i.e.
$$0 = \mathbb{E}((H \bullet M)_{n+1}) = \mathbb{E}(1_F M_{n+1})-\mathbb{E}(1_F M_n),$$
and so
$$\mathbb{E}(M_{n+1} 1_F) = \mathbb{E}(M_n 1_F).$$
Since this holds for any $F \in \mathcal{F}_n$, $n \in \mathbb{N}$, it follows that $(M_n)_{n \in \mathbb{N}}$ is a martingale with respect to $(\mathcal{F}_n)_{n \in \mathbb{N}}$.
A: Put $Y=H\cdot M$. Since $H$ is predictable and $(M_j-M_{j-1})_j$ is adapted, clearly $Y$ is adapted. Now, for each positive integer $N$,
\begin{align}
\mathbb E[(H\cdot M)_N] &= \mathbb E\left[\sum_{j=1}^N H_j(M_j-M_{j-1}) \right]\\
&= \sum_{j=1}^N \mathbb E[H_j(M_j-M_{j-1})]\\
&=0,
\end{align}
which implies that
$$
\sum_{j=1}^N \mathbb E[H_jM_j] = \sum_{j=1}^N \mathbb E[H_jM_{j-1}].
$$
For $N=1$ this yields $\mathbb E[H_1M_1] = \mathbb E[H_1M_0]$, and by induction $\mathbb E[H_jM_j] = \mathbb E[H_jM_{j-1}]$, $j\geqslant 1$. I am not sure how to proceed from here, but hopefully this was useful.
