# Continuous Time Optional Sampling Theorem for Submartingales

I want to show the following result for continuous time submartingales:

Suppose $$\{X_t\}_{t \ge0}$$ is a right continuous submartingale with respect to the filtration $$\{\mathcal{F}_t\}_{t \ge 0}$$. I want to show that for stopping times $$\sigma \leq \tau$$, if EITHER

(1) $$\tau \leq C < \infty$$ or

(2) $$\{X_t\}_{t \ge0}$$ is uniformly integrable,

then $$X_\sigma \leq E(X_\tau| \mathcal{F}_\sigma)$$

My attempt/start of a proof:

It suffices to show that $$E(X_\sigma) \leq E(X_\tau)$$ for all stopping times $$\sigma, \tau$$ which satisfy either (1) or (2).

I know how to prove the discrete time theorem of the identical form, so if we define $$\tau_n \equiv \inf\{k/2^n : k, n \in \mathbb{N}\, k/2^n \ge \tau\}$$ and similarly for $$\sigma_n$$, we know that

$$X_{\sigma_n} \leq E(X_{\tau_n}| \mathcal{F}_{\sigma_n})$$

as it is easy enough to show that $$\sigma_n, \tau_n$$ are both (discrete time) stopping times.

Because $$\mathcal{F}_\sigma \subseteq \mathcal{F}_{\sigma_n}$$ for all $$n$$,

$$E(X_{\sigma_n}|\mathcal{F}_\sigma) \leq E(X_{\tau_n}| \mathcal{F}_{\sigma}) \quad \quad (1)$$ applying the tower property.

If I knew that $$X_{\sigma_n}$$ and $$X_{\tau_n}$$ were uniformly integrable and that $$X_\sigma, X_\tau$$ were both in $$L^1$$, I would be able to conclude the proof by taking expectations in (1) and subsequently taking limits because $$X_{\tau_n} \xrightarrow{a.s.} X_\tau$$ by right continuity and likewise for $$\sigma_n$$. Is the UI stuff true? I really don't know where to start from here. Any help would be massively appreciated.

Fix $$k$$. For $$n\ge k$$, one has $$\sigma_n \le \tau_n \le \tau_k$$, and so

$$X_{\sigma_n} \le E(X_{\tau_k}|\mathcal F_{\sigma_n}) =:M_n.$$

Since $$\{\mathcal F_{\sigma_n}\}_{n\ge1}$$ is a decreasing sequence of $$\sigma$$-fields with intersection $$\bigcap_{n \ge 1}\mathcal F_{\sigma_n} = \mathcal F_\sigma$$, $$\{M_n\}$$ is a backwards martingale and hence converges a.s. and in $$L^1$$ to $$E(X_{\tau_k}|\mathcal F_\sigma)$$. Thus, since $$X_{\sigma_n}\to X_{\sigma}$$ a.s. one has

$$X_\sigma \le E(X_{\tau_k}|\mathcal F_\sigma)$$

for all $$k$$. Notice now that $$X_{\tau_k} \le E(X_{\tau_0}|F_{\tau_0})$$, which is integrable (this follows by applying (1) or (2) to the discrete submartingale $$\{X_n\}_n$$). Since $$X_{\tau_k}\to X_\tau$$ a.s., it follows by Fatou's lemma that

$$\limsup_{k\to\infty}E(X_{\tau_k}|\mathcal F_\sigma) \le E(X_\tau|\mathcal F_\sigma).$$

Hence $$X_\sigma \le E(X_\tau | \mathcal F_\sigma)$$, as claimed.