I came across the following theorem in some online lecture notes. Let $(\Omega, \mathcal{G}, P)$ be a probability space
Thereom: If $(X_t)_{t\geq0}$ is a supermartingale with right-continuous paths (a.s.) w.r.t. a filtration $(\mathcal{G}_t)$, then it is also a supermartingale w.r.t. the filtration $(\mathcal{G}_{t+}) = \cap_{s > t}\mathcal{G_s}$.
The proof given goes like this:
Proof: For any $u$ between $s$ and $t$ \begin{equation} \mathbb{E}[X_t \rvert \mathcal{G}_u] \leq X_u \; \; \text{a.s.}, \end{equation} since $(X_t)$ is a supermartingale. So for any $A \in \mathcal{G_{s+} \subset \mathcal{G_u}}$, \begin{equation} \mathbb{E}[1_A X_t] \leq \mathbb{E}[1_A X_u]. \end{equation} Letting $u \searrow s$, $X_u \rightarrow X_s$ (by r-c) and \begin{equation} \mathbb{E}[1_A X_t] \leq \mathbb{E}[1_A X_s], \tag{a} \end{equation} for any $A \in \mathcal{G}_{s+}$.
My question: How are the limit and expectation interchanged in (a)? Clearly Monotone convergence does not apply, so it must be Dominated convergence? But I can't see why. Thanks in advance.
Let $(u_n)$ be a decreasing sequence in $(s,t)$ with limit $s$. Then $$ \Bbb E[X_t\mid\mathcal G_{u_n}]\le X_{u_n}, $$ for each $n$. As $n\to\infty$ the right side converges to $X_s$ by right continuity. The left converges (a.s. and in $L^1$) to $\Bbb E[X_t\mid\mathcal G_{s+}]$ by the backward-martingale convergence theorem.