See this chapter of [ Low ] .

In the proof, note that \(A\) can be written as a finite disjoint union of sets of the form \(\{ (t, \omega ) \mid \sigma (\omega ) {\lt} t \le \tau (\omega )\} \) for stopping times \(\sigma , \tau \). To see that each is an elementary predictable set, note that if \(T = \{ t_0, \ldots , t_n\} \) where \(t_0 {\lt} \cdots {\lt} t_n\), then it can be written as a finite disjoint union of sets

where \(\{ \sigma \le t_{k-1} {\lt} \tau \} \in \mathcal{F}_{t_{k-1}}\). Therefore \(A\) is an elementary predictable set.

TODO

Let \(Y\) be the modification of \(X\) given by Theorem 11.3. \(Y\) has left and right limits everywhere and there is a countable set \(S\) such that \(Y\) is right-continuous on \(T \setminus S\). It remains to show that \(Y\) is right-continuous at the points of \(S\). Let \(Y_{t+}\) denote the right limit of \(Y\) at \(t\). Since \(X\) is right-continuous in probability, for all \(t \in S\), \(Y_t = Y_{t+}\) almost surely. Therefore, since \(S\) is countable, we have that almost surely \(Y_t = Y_{t+}\) for all \(t \in S\) (and hence for all \(t \in T\)). \(Y\) is therefore almost surely cadlag. We can modify \(Y\) on the null set to obtain a cadlag modification of \(X\).

Since \(X\) is a submartingale, for any elementary predictable set \(A\), we have (Corollary 10.26)

As \(X_t\) and \(X_0\) are integrable, the set \(\{ \mathbb {E}[(\mathbb {1}_A \bullet X)_t] \mid A \text{ elementary predictable}\} \) is bounded.

We apply Corollary 11.4, in which the boundedness condition of Theorem 11.3 holds by Lemma 11.5.

Apply Theorem 11.3, in which the boundedness condition holds by Lemma 11.5.

By the submartingale property, for all \(n\) we have that \(X_t \le P[X_{t_n} \mid \mathcal{F}_t]\) almost surely. By Lemma 11.8, the family \(\{ X_{t_n}\} _{n \in \mathbb {N}}\) is uniformly integrable.

TODO: conclude that \(P[X_{t+} \mid \mathcal{F}_t]\) is the limit of \(P[X_{t_n} \mid \mathcal{F}_t]\).

Let \(Y\) be the modification of \(X\) given by Lemma 11.7. \(Y\) has left and right limits everywhere and there is a countable set \(S\) such that \(Y\) is right-continuous on \(T \setminus S\). It remains to show that for each \(t \in S\), \(Y\) is right-continuous at \(t\) almost surely. We will then be able to modify \(Y\) on a null set to obtain a cadlag modification of \(X\). Let \(t \in S\) and let \(Y_{t+}\) denote the right limit of \(Y\) at \(t\).

We show that \(Y_t = P[Y_{t+} \mid \mathcal{F}_t]\) almost surely, and that \(P[Y_{t+} \mid \mathcal{F}_t] = Y_{t+}\) almost surely, which will conclude the proof. For the second equality it suffices to show that \(Y_{t+}\) is \(\mathcal{F}_t\)-measurable, which follows from the right-continuity of the filtration.

For the first equality, it suffices to show that \(P[Y_{t+} \mid \mathcal{F}_t] - Y_t\) is an almost surely nonnegative random variable with zero expectation. Nonnegative follows from Lemma 11.9. The expectation is \(P[Y_{t+}] - P[Y_t]\). Let \(t_n\) be a decreasing sequence in \(T\) converging to \(t\). By right-continuity of \(t \mapsto \mathbb {E}[Y_t]\), we have that

By uniform integrability (Lemma 11.8), we have that

Therefore \(P[Y_{t+}] = P[Y_t]\), which concludes the proof.

\(X\) is in particular a submartingale, and for all \(t \in T\), we have that \(\mathbb {E}[X_t] = \mathbb {E}[X_0]\) by the martingale property. Therefore \(t \mapsto \mathbb {E}[X_t]\) is right-continuous, and we can apply Theorem 11.10.

See 8.2.3 of Pascucci.