Suppose $\mathcal{F}_n$ is a filtration of $\sigma$-algebras (that means $F_1\subset F_2\subset...\subset F_n$), prove that $X_n=E(X|\mathcal{F}_n)$ is a martingale.
My motivation was to use the law of total expectation with this fact that I proved earlier:
For $\sigma$- fields $\mathcal{F}_{1}\subset \mathcal{F}_2 $ and r.v X, $ E(E(X|\mathcal{F}_2)|\mathcal{F}_1)=E(X|\mathcal{F}_1)$
sub 1,2 for n-1,n and we get:
$$E(X_n|X_1,....X_{n-1})\overset{*}{=}E(X_n|\mathcal{F}_{n-1})=E(E(X|\mathcal{F}_{n})|\mathcal{F}_{n-1})=E(X|\mathcal{F}_{n-1})=X_{n-1}$$
Where the only thing I need to prove that I don't know how to justify is why $$E(X_n|X_1,....X_{n-1})=E(X_n|\mathcal{F}_{n-1})$$ (if that is even true). Can anyone help me with that?
Your calculation (using the tower property of conditional expectation) shows that $E[X_k\mid \mathcal F_{k-1}]=X_{k-1}$ for $k=2,3,\ldots,n$. This means that $(X_k)_{k=1}^n$ is an $(\mathcal F_k)_{k=1}^n$-martingale.
Now define the filtration $\mathcal F_k^X:=\sigma(X_1,\ldots, X_k)$, $k=1,2,\ldots,n$. Clearly $\mathcal F^X_k\subset\mathcal F_k$ for $k=1,2,\ldots,n$. A second application of the tower property shows that $E[X_k\mid\mathcal F^X_{k-1}]=X_{k-1}$ for $k=1,2,\ldots,n$. That is, $(X_k)_{k=1}^n$ is also an $(\mathcal F^X_k)_{k=1}^n$-martingale. If no particular filtration is being used, this latter notion is what is meant if one says that "$(X_k)_{k=1}^n$ is a martingale".