Question about solution to showing $(X_t,\mathcal{F}_{t+})$ is a submartingale if $(X_t,\mathcal{F}_t)$ is.

Question

Let $(X_t,\mathcal{F}_t)_{t\ge 0}$ be a submartingale with right continuous paths. Show that $(X_t,\mathcal{F}_{t+})$ is again a submartingale.

This problem is from Schilling's Brownian motion. In the below solution, why do we get $E(X_t|\mathcal{F}_{s+})\leftarrow E(X_t|\mathcal{F}_{s_n})$ as $s_n \downarrow s$ due to submartingale property?

Answer 1

Look at Section 5.6 in Durrett's "Probability: Theory and Examples". Since $s(n)\downarrow s$, $\mathcal{F}_{s(n)}\downarrow \mathcal{F}_{s+}:=\cap_{n\ge 1}\mathcal{F}_{s(n)}$ and Theorem 5.6.3 implies that $$ \mathsf{E}[X_t\mid\mathcal{F}_{s(n)}]\to\mathsf{E}[X_t\mid\mathcal{F}_{s+}] \quad\text{a.s. and in $L^1$.} $$

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In short, $Y_n=\mathsf{E}[X_t\mid\mathcal{F}_{s(n)}]$ is a backwards martingale. Thus, $Y_n$ converges to $Y_{\infty}=\mathsf{E}[X_t\mid \mathcal{F}_{s+}]$ a.s. and in $L^1$.