I wish to prove that for stopping times $S \leq T$ and $X_n$ a uniformly integrable martingale, $E(X_T | \mathcal{F}_S) = X_S$. I am well aware this is a standard result, but part of the proof I have been trying to learn confuses me. It goes as follows:
$X_n$ being UI implies $X_n \rightarrow X \quad a.s.$ and in $L^1$ for some $X \in L^1$(martingale convergence)
We define $Y_n \equiv X_{S \wedge n}$. Clearly $Y_n$ is a UI martingale with respect to the filtration $\{\mathcal{G_n}\} = \{\mathcal{F}_{S\wedge n}\}$ as:
(1) $E(X_{S \wedge n} | \mathcal{G}_m) = E(X_{S \wedge n} | \mathcal{F}_{S \wedge m}) = X_{S \wedge m}$ for $m \leq n$ by bounded stopping times, and
(2) $X_{S \wedge n} = E(X_{n}|\mathcal{F}_{S \wedge n}) = E(E(X|\mathcal{F}_n)|\mathcal{F}_{S \wedge n}) = E(X | \mathcal{F}_{S \wedge n})$ again by martingale convergence, the properties of bounded stopping times and the tower property.
The same argument can be made for $Y_n \equiv X_{T \wedge n}$, so $X_{T \wedge n} \rightarrow X_T \quad a.s.$ and in $L^1$.
We now apply the above arguments to $Y_{S \wedge n}$:
(**) $E(X_T | \mathcal{F}_{S \wedge n}) = Y_{S \wedge n} = X_{T \wedge S \wedge n} = X_{S \wedge n}$ (as $S \leq T$)
Taking limits and invoking Levy's theorem, we obtain that
$X_S = lim_{n \rightarrow \infty} E(X_T|\mathcal{F}_{S \wedge n}) = E(X_T|\sigma(\cup_{n=0}^\infty \mathcal{F}_{S \wedge n}))$
It is somehow supposed to be "obvious" $\sigma(\cup_{n=0}^\infty \mathcal{F}_{S \wedge n})) = \mathcal{F}_S$. I don't see this AT ALL.
It is indeed obvious that since $\mathcal{F}_{S \wedge n} \subseteq \mathcal{F}_S$, $\sigma(\cup_{n=0}^\infty \mathcal{F}_{S \wedge n})) \subseteq \mathcal{F}_S$. The other direction is not clear to me, and I tried addressing in this post: Stopping Times and Filtrations to no avail. Please help me clarify the end of this proof. It is driving me insane.
