Let $\{x_n,\mathcal{F}_n,n\ge 1\}$ be a martingale differences sequence. Put $x_n'=x_n \mathbf{1}_{\{ |x_n|\le a_n \}}$ with $a_n\ge 0$ and $$y_n=x_1'+\cdots+x_n'-\mathbb{E} (x_2'\mid\mathcal{F}_1) - \cdots -\mathbb{E} (x_n'\mid\mathcal{F}_{n-1}). $$ I need to proof that $\mathbb{E}(|y_n|) <\infty$ for all $n\ge 1$. I proved it as follows:
Because $x'_i$ is independent with $\mathcal{F}_{i-1}$, $i\ge 2$ so \begin{align*} \mathbb{E} (|y_n|) &= \mathbb{E} \left[|x_1'+\cdots+x_n' - \mathbb{E} (x_2'\mid\mathcal{F}_1) -\cdots- \mathbb{E}(x_n'\mid\mathcal{F}_{n-1})| \right] \\ & \le \mathbb{E} \left[ |x_1'+\cdots+x_n'| +\mathbb{E}( |x_2'|\ |\mathcal{F}_1) +\cdots+ \mathbb{E} (|x_n'| \ |\mathcal{F}_{n-1})\right] \\ & = \mathbb{E} (|x_1'+\cdots+x_n'|) + \mathbb{E} (|x_2'|) +\cdots+ \mathbb{E} (|x_n'|) \\ & \le 2\displaystyle\sum_{i=1}^n \mathbb{E} (|x_i|) <\infty \text{ almost surely ,}\\ & \ \ (\text{Since } \{ x_n \} \text{ is a martingale differences sequence so } \mathbb{E} (|x_n|)<\infty ). \end{align*} Is this wrong or right?