Let \(n\) be an upper bound for \(\tau \). To see that the stochastic interval is an elementary predictable set, note that it can be written as a finite disjoint union of sets

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

By Lemmas 11.11 and 11.10, \(\sigma ^k_{a, b, N}\) and \(\tau ^{k+1}_{a, b, N}\) are stopping times. Hence by Lemma 11.2, the stochastic interval \(]\! ]\sigma ^k_{a, b, N}, \tau ^{k+1}_{a, b, N}]\! ]\) is a predictable set for each \(k\). Since a finite union of predictable sets is predictable, \(A_N\) is predictable.

By Lemmas 11.11 and 11.10, \(\sigma ^k_{a, b, N}\) and \(\tau ^{k+1}_{a, b, N}\) are stopping times, and are bounded by \(N\). By Lemma 11.3, the stochastic intervals \(]\! ]\sigma ^k_{a, b, N}, \tau ^{k+1}_{a, b, N}]\! ]\) are elementary predictable sets for each \(k\). Since a finite union of elementary predictable sets is elementary predictable, \(A_N\) is elementary predictable.

Go from \(T\) to a subset of \(\mathbb {N}\) and use Lemma 11.15?

Let \(D\) be a countable dense subset of \(T\) (e.g. the rationals in \(T\)) and let \(t \in T\) and let \(S\) be a finite subset of \(D \cap [0, t]\). By Lemma 11.16, for all \(a {\lt} b\) in \(\mathbb {R}\), the number of upcrossings \(U_S[a, b]\) of the interval \([a, b]\) along \(S\) satisfies

for some elementary predictable set \(A\). Taking expectations and using the boundedness hypothesis, we obtain

where \(L\) is a bound for \(\{ \mathbb {E}[(\mathbb {1}_A \bullet X)_t] \mid A \text{ elementary predictable}\} \).

Now, let \(S_1 \subseteq S_2 \subseteq \cdots \) be an increasing sequence of finite subsets of \(D \cap [0, t]\) with \(\bigcup _n S_n = D \cap [0, t]\). The number of upcrossings \(U_{S_n}[a, b]\) is increasing in \(n\). By monotone convergence,

In particular, \(U_{D \cap [0, t]}[a, b] {\lt} \infty \) a.s. for each pair \(a {\lt} b\). Since \(\mathbb {Q}^2\) is countable, we conclude that almost surely, for all rational \(a {\lt} b\), the number of upcrossings of \([a, b]\) by \(X\) along \(D \cap [0, t]\) is finite.

By Lemma 11.9, finite upcrossings of all rational intervals implies that the left and right limits of \(s \mapsto X_s(\omega )\) along \(D\) exist at every \(t \in T\), for almost every \(\omega \). Define

whenever this limit exists, and \(\tilde{Y}_t(\omega ) = 0\) otherwise. Then \(\tilde{Y}\) has left and right limits everywhere (along \(T\)) almost surely, and is right-continuous at every point of \(D\).

We claim that the set

is countable. For each \(n \in \mathbb {N}\), consider the set

We have \(S = \bigcup _n S_n\), so it suffices to show each \(S_n\) is finite. If \(S_n\) were infinite, it would contain a monotone sequence \(t_k\) converging to some limit \(t^* \in [0, n]\). Adding \(\{ t_k\} \) to \(D\) and repeating the upcrossing argument, we would obtain that \(X_{t_k}\) converges along \(D \cup \{ t_k\} \) almost surely. But \(\tilde{Y}_{t_k}\) also converges (since \(\tilde{Y}\) has left/right limits), giving \(\tilde{Y}_{t_k} - X_{t_k} \to 0\) in probability, contradicting the definition of \(S_n\). Therefore each \(S_n\) is finite and \(S\) is countable.

Define

Then \(Y_t = X_t\) almost surely for all \(t\): for \(t \in S\) this is by definition, and for \(t \notin S\) this follows from \(\tilde{Y}_t = X_t\) a.s. Thus \(Y\) is a modification of \(X\).

Since \(\tilde{Y}\) has left and right limits everywhere almost surely, and \(Y\) differs from \(\tilde{Y}\) only on the countable set \(S\), \(Y\) also has left and right limits everywhere almost surely. Furthermore, \(Y\) agrees with \(\tilde{Y}\) on \(T \setminus S\), so \(Y\) is right-continuous on \(T \setminus S\).

Let \(Y\) be the modification of \(X\) given by Theorem 11.17. \(Y\) 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 10.42)

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 11.18, in which the boundedness condition of Theorem 11.17 holds by Lemma 11.19.

Apply Theorem 11.17, in which the boundedness condition holds by Lemma 11.19.

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 11.22, 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 11.21. \(Y\) 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 11.23. 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 11.22), 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 11.24.

See 8.2.3 of Pascucci.