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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.

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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.

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