Let $M=(M_i)_{i \in \mathbb{Z}}$ a martingale with respect to natural filtration $\mathcal{F}_{i}=\sigma(M_j; j \leq i)$. Defines $X_{i}=M_{i}-M_{i-1}$ and, for fixed $n \in \mathbb{N}$,
\begin{equation*} Z_i=X_i{1}_{\{|X_i| > n^{1/3}\}}- \mathbb{E}\left[X_i {1}_{\{|X_i|> n^{1/3}\}}\mid \mathcal{F}_{i-1} \right]. \end{equation*} I need to show that
\begin{equation*} \mathbb{E}\left[Z_{i}^{2}\right] = \mathbb{E}\left[ X_i^{2}{1}_{\{|X_i| > n^{1/3}\}}\right]- \mathbb{E}\left[\left[\mathbb{E}\left[X_i {1}_{\{|X_i|\leq n^{1/3}\}}\mid \mathcal{F}_{i-1}\right] \right]^2\right]. \end{equation*}
$\textit{My idea:}$ Writing $A=\{|X_i|> n^{1/3}\}$ to simplify, we have
\begin{equation} \mathbb{E}\left[Z_{i}^{2}\right] = \mathbb{E}\left[ X_i^{2}{1}_{A}\right]-2\mathbb{E}\left[X_i 1_{A}\mathbb{E}\left[X_i {1}_{A}\mid \mathcal{F}_{i-1}\right]\right]+ \mathbb{E}\left[\left[\mathbb{E}\left[X_i {1}_{A}\mid \mathcal{F}_{i-1}\right] \right]^2\right]. \end{equation}
To conclude the exercise I need to show that
\begin{equation} -2\mathbb{E}\left[X_i 1_{A}\mathbb{E}\left[X_i {1}_{A}\mid \mathcal{F}_{i-1}\right]\right]+ \mathbb{E}\left[\left[\mathbb{E}\left[X_i {1}_{A}\mid \mathcal{F}_{i-1}\right] \right]^2\right] = - \mathbb{E}\left[\left[\mathbb{E}\left[X_i {1}_{A}\mid \mathcal{F}_{i-1}\right] \right]^2\right]. \end{equation}
But, for any event $A$, we have
\begin{equation} \mathbb{E}\left[X_i {1}_{A}| \mathcal{F}_{i-1}\right]=\mathbb{E}\left[X_i\mid \mathcal{F}_{i-1}\right]-\mathbb{E}\left[X_i {1}_{A^c}\mid \mathcal{F}_{i-1}\right] = - \mathbb{E}\left[X_i {1}_{A^c}\mid \mathcal{F}_{i-1}\right], \end{equation}
since $=\mathbb{E}\left[X_i\mid \mathcal{F}_{i-1}\right] =\mathbb{E}\left[M_{i}-M_{i-1}\mid \mathcal{F}_{i-1}\right]= M_{i-1} -M_{i-1}=0$, because $M$ is a martingale. Combining the last two equations, we should have
\begin{equation} \mathbb{E}\left[X_i 1_{A}\mathbb{E}\left[X_i {1}_{A}\mid \mathcal{F}_{i-1}\right]\right]= \mathbb{E}\left[\left[\mathbb{E}\left[X_i {1}_{A}\mid \mathcal{F}_{i-1}\right] \right]^2\right]. \end{equation}
Why is this true?
Thank you for your attention.
For any integrable r.v. $Y$ and any sub-sigma algebra $\mathcal G$ we have $E[YE(Y|\mathcal G)]=E[E[YE(Y|\mathcal G)]|\mathcal G]=E[E[Y|\mathcal G]E[Y|\mathcal G)]=E[(E(Y|\mathcal G)^{2}]$.
In the first step I used towering property of conditional expectation. In the second step I used the fact that $E[Y|\mathcal G]$ is already $\mathcal G$-measurable so it can be pulled out of the conditional expectation.