Measurability question for martingale (Is $S(X_i, Z_i)- E(S(X_i,Z_i) \mid \mathcal{F}_i)$ $\mathcal{F}_i=\{X_1, \dots, X_i \}$-measurable?)

Question

One condition for a martingale $M_k$ with a general filtration $\mathcal{F}_k$ is that the involved random variables $M_k$ are $\mathcal{F}_k$-measurable.

Now I have $M_n=Y_1+\dots +Y_n$ and $Y_i=S(X_i, Z_i)- E(S(X_i,Z_i) \mid \mathcal{F}_i)$, $\mathcal{F}_i=\{X_1, \dotsc, X_i\}$, which satisfies the martingale property $E(M_{n+1} | \mathcal{F}_n) = M_n$.

$S_i:\Omega \to R, \omega \mapsto S(X_i(\omega), Z_i(\omega))$, $S:R \times R \rightarrow R$.

I know that the conditional expectation is $\mathcal{F}_i$-measurable by definition, so the question is when is $S_k$ $\mathcal{F}_k$-measurable? I would have for example that $S_k$ is $f(X_k+Z_k)/Z_k$ where f is differentiable and $Y_k \neq 0$ a.s. and iid. (This is a follow-up question to this solution) So maybe it just boils down to: Is $Z_i$ $X_i$-measurable? And I am afraid that the answer should be no, generally not. But if it is this way maybe still $M_n$ is $\mathcal{F}_n$-measurable?

Answer 1

$S_k$ is $\mathcal{F}_k=\sigma(X_1,...,X_k)$-measurable if and only if there exists a Borel-measurable function $f$ such that $S_k=f(X_1,...,X_k)$.

This is known as Doob's lemma (or sometime Doob-Dynkin's Lemma).

Regards