So I am learning about Martingales and looking at the definition of a Martingale it is stochastic process that satisfies three conditions.
Definition: Let $\mathbf{X} = (X_n)^{\infty}_{n=1}$ be a sequence of random >variables on $(\Omega,\mathcal{F},\mathbf{P})$. Then we call $\mathbf{X}$ a martingale with respect to the filtration $(\mathcal{F}_n)_{n\geq 1}$, if
- $E [| X_n |] < \infty$ for all $n$.
- $X_n$ is measurable with respect to $\mathcal{F}_n$ for each $n$.
For $n \geq 1$ the martingale property holds:
$E [X_{n+1} | \mathcal{F}_n] = X_n$
Now, how do I know when $X_n$ is measurable to $\mathcal{F}_n$? Take this problem for example.
Let $(\mathcal{F}_n)_{n≥1}$ be a filtration and $E [|X|] < \infty$. Set $X_n = E [X | \mathcal{F}_n]$. Show that $(X_n)_{n≥1}$ is a martingale with respect to the filtration $(\mathcal{F}_n)_{n≥1}$.
How do I show that the second condition is satisfied? Can I use some expected value conditioned on $\mathcal{F}_n$ on both sides of the equation to show that, if $X_n$ is $\mathcal{F}_n$-measurable, then the condition falls off? (Now, if this is wrong in general, I apologize, I have not entirely grasped the concept of sigma-algebras).
Also, for the first and third conditions, I believe it is understood in the context that $X_n \rightarrow X$ when $n \rightarrow \infty$ (in what convergence mode I don't know, maybe almost surely?). Does this imply that first condition is satisfied? Does it follow that the third condition is satisfied since $X$ is the limit of the sequence, thus $X=X_{n+1}$? (This last equation/thought I am unsure how to express).
//Best regards