Define \begin{align} \epsilon_j = \mathbb{1}_{A_j} - \mathbb{P}(A_j) = \begin{cases} 1- \mathbb{P}(A_j) \qquad &\text{if } \omega \in A_j\\ - \mathbb{P}(A_j) \qquad &\text{otherwise }, \end{cases} \end{align} for independent events $(A_i)_{i \geq 1}$ and $\mathcal{F}_n = \sigma(\epsilon_1, \ldots , \epsilon_n)$ for $n \geq 0$.
Clearly, $(\epsilon_i)_{i \geq 1}$ are independent random variables with \begin{align} \mathbb{E}[\epsilon_j] &= (1 - \mathbb{P}(A_j)) \cdot \mathbb{P}(A_j) - \mathbb{P}(A_j) \cdot \mathbb{P}(A_j^c) \\ &= (1 - \mathbb{P}(A_j)) \cdot \mathbb{P}(A_j) - \mathbb{P}(A_j) \cdot (1 - \mathbb{P}(A_j)) \\ &= 0. \end{align} Now, let $M_n := \sum_{j=1}^n \epsilon_j$.
I am wondering why $(M_n)_{n\geq0}$ is a martingale. Let $m \leq n$, \begin{align} \mathbb{E}[M_n \mid \mathcal{F}_m] &= \mathbb{E}[ \sum_{j=1}^n \epsilon_j \mid \mathcal{F}_m] \\ &=^{?} \sum_{j=1}^n \mathbb{E}[\epsilon_j \mid \mathcal{F}_m] \end{align} How to find that $\mathbb{E}[M_n \mid \mathcal{F}_m] = \sum_{j=1}^m \epsilon_j$?
If $j>m$ then $\mathbb{E}[\varepsilon_j|\mathcal{F_m}]=\mathbb{E}[\varepsilon_j]=0$ since $\varepsilon_j$ is independent of $\mathcal{F_m}$. And if $j\leq m$ then $\mathbb{E}[\varepsilon_j|\mathcal{F_m}]=\varepsilon_j$ since $\varepsilon_j$ is $\mathcal{F_m}$-measurable.