The question is:
Suppose that $\Delta_1, \Delta_2,...$ are independent random variables with mean $0$. Let $X_1= \Delta_1$ and $X_{n+1}= X_n + \Delta_{n+1} f_n(X_1,X_2,...,X_n)$, and suppose that $X_n$ are integrable. Show that $\{X_n\}$ is martingale.
My attempt: Firstly, by induction I proved that $X_n$ is a function of $\Delta_1, \Delta_2,...$ .
Then I denote $\mathcal{F_n}$ the natural filtration of $X_n$.
Suppose that $\mathcal{G_n} = \sigma(\Delta_1, \Delta_2,...)$ and in this case $f_{n}(X_1,X_2,..X_n)$ are $\mathcal{G_n}$- measurable.
$E(\Delta_{n+1} f_{n}(X_1,X_2,..X_n)|\mathcal{G_n}) = f_{n}(X_1,X_2,..X_n)E(\Delta_{n+1}|\mathcal{G_n})=0$.
This implies that
$E(X_{n+1} -X_{n} | \mathcal{G_n})=0$ $\implies$ $E(X_{n+1}| \mathcal{G_n})=X_{n}$.
Hence, $X_n$ is martingale w..r.t. $\mathcal{G_n}$.
We know that $X_n$ is martingale w..r.t. $\mathcal{G_n}$ then $X_n$ is martingale w..r.t. natural filtration $\mathcal{F_n}$.
Is this attempt correct?