$X_n,n\geq1$ is a martingale w.r.t to $\mathcal{G}=\mathcal{\{G}_n\}_{n=1}^{\infty}$.
Let $\mathcal{F}=\mathcal{\{F}_n\}_{n=1}^{\infty}$. Where $F_n=\sigma(X_1,X_2, \ldots,X_n)$.
Show that $\mathcal{F}_n \subseteq \mathcal{G}_n$ and that $X_n$ is a martingale w.r.t $\mathcal{F}$.
Is the same true for any filtration $\mathcal{H}=\mathcal{\{H}_n\}_{n=1}^{\infty}$ such that $\mathcal{H}_n \subseteq \mathcal{G}_n$ that is, is $X_n$ a martingale w.r.t to $\mathcal{H}$?
So in the first question I assume that $X_n$ is a martingale w.r.t. $\mathcal{F}$ because of the way $\mathcal{F}_n$ is constructed which leads to $X_n$ being $\mathcal{F}_n$-measurable and as $\mathcal{F}_n$ is part of the collection of sigma algebras in $\mathcal{F}$ thus $X_n$ is $\mathcal{F}$-measurable and through that a martingale w.r.t $\mathcal{F}$?
Also if $X_n$ is a martingale w.r.t to $\mathcal{G}$ must mean that $X_n$ is $\mathcal{G}$-measurable which means that there exists a sigma algebra that is part of the collection $\mathcal{G}$ that is constructed the same way $\mathcal{F}_n$? Thus leading to?
$$\mathcal{F}_n \subseteq \mathcal{G}_n$$
In the second question the set $\mathcal{H}_n$ is equal to or a subset of $\mathcal{G}_n$ which kinda puts me in a situation where I can't really say much.
ps.
I'm kinda new to all of this so I'm trying.
For the second point, by choosing $H_n = \sigma\{ \emptyset \}$, you have that $H_n \subseteq G_n$ but $X_n$ is constant for all $n$. Hopefully, there are some nonconstant martingales.