$\mathbb F$-Martingale indexed on $\mathbb Q$ and left/right limit on $\mathbb R$ along $\mathbb Q$

Question

I have a question about an article (reference below).

---

Let $(X_t)_{t\in\mathbb Q_+}$ be an $\mathbb F$-martingale (*) (Assume $\mathbb F$ is right-continuous).

Let $C_t$ be the set of all $\omega$ s.t. $X_.(\omega)$ has a left and right limit in $\mathbb R$ along $\mathbb Q$ for all $s\in[0,t]$.

I have no more assumptions (I think I have carefully read the reference below).

Questions :

1. (*) involve $\mathbb P(C_t) = 1$. My question is why ?
2. Furthermore do I get : $\mathbb E [X_t\mid \mathcal F_s]=X_s$ for $t\in\mathbb Q$ and $s\in\mathbb R$ with $t>s$ ? I think no because $X_s$ is not well defined.

Reference : Jacod - 1985 - Grossissement initial, hypothese H' et theoreme de girsanov - pages 18/19 - proof of lemma 1.8 - take a glance at (1.9) page 19.

Answer 1

Answer to question 1: To prove the existence of the limits the so-called upcrossing lemma is a very useful tool.

Definition Let $f: [0,\infty) \to \mathbb{R}$ be a function and $0 \leq t_1 < \ldots < t_n$. For $a<b$ and $D := \{t_1,\ldots,t_n\}$ we denote by $$U_D(f,[a,b]) := \sup\{k \in \mathbb{N}; \exists \tau_1 < \sigma_1 < \tau_2 < \ldots < \tau_k < \sigma_k: f(\tau_j)<a, f(\sigma_j)>b \, \, \text{for $j=1,\ldots,k$}\}$$ the number of upcrossings.

Upcrossing lemma Let $(X_t,\mathcal{F}_t)_{t \in \mathbb{Q} \cap [0,\infty)}$ be a martingale and $D = (t_j)_{j \in \mathbb{N}} \subseteq \mathbb{Q} \cap [0,\infty)$. Then we have for all $a<b$ $$(b-a) \mathbb{E}(U_{D}(X,[a,b])) \leq \sup_{t \in D} \mathbb{E}((X_t-a)^+)$$ where $$U_D(X,[a,b]) := \sup_{n \in \mathbb{N}} U_{\{t_1,\ldots,t_n\}}(X,[a,b]).$$

For a proof see e.g. Revuz & Yor, p.61. Using the upcrossing lemma, we get the following statement.

Theorem Let $(X_t,\mathcal{F}_t)_{t \in \mathbb{Q} \cap [0,\infty)}$ be a martingale. Then there exists $\Omega_0$, $\mathbb{P}(\Omega_0)=1$ such that the limits $$\lim_{\mathbb{Q} \ni r \uparrow t} X_r(\omega) \qquad \text{and} \qquad \lim_{\mathbb{Q} \ni r \downarrow t} X_r(\omega) \tag{1}$$ exist in $[-\infty,\infty]$ for all $t>0$ and $\omega \in \Omega_0$.

Proof: Fix some $t \in I := [u,v] \subseteq [0,\infty)$. Applying the upcrossing lemma, we find for any $a<b$, $a,b \in \mathbb{Q}$, $$(b-a) \mathbb{E} U_{\mathbb{Q} \cap [u,v]}(X,[a,b]) \leq \sup_{t \in \mathbb{Q} \cap [u,v]} \mathbb{E}((X_t-a)^+) \leq \mathbb{E}((X_v-a)^+) \leq \mathbb{E}(|X_v|)+a$$ where we have used that $t \mapsto (X_t-a)^+$ is a submartingale. Consequently, we get $U_{\mathbb{Q} \cap [u,v]}(X,[a,b])<\infty$ almost surely. This, however, implies that the limits $(1)$ exist almost surely for any $t \in (u,v)$. To prove this note that $$\begin{align*}&\quad \{\omega; \lim_{\mathbb{Q} \ni r \downarrow t} X_r(\omega) \, \text{does not exist in $[-\infty,\infty]$}\} \\ &= \bigcup_{a,b \in \mathbb{Q},a<b} \{\omega; \liminf_{\mathbb{Q} \ni r \downarrow t} X_r(\omega)<a<b< \limsup_{\mathbb{Q} \ni r \downarrow t} X_r(\omega)\} \\ &= \bigcup_{a,b \in \mathbb{Q},a<b} \{\omega; U_{\mathbb{Q} \cap [u,v]}(X(\omega),[a,b])=\infty\}. \end{align*}$$ A similar reasoning works for $\lim_{\mathbb{Q} \ni r \uparrow t} X_r(\omega)$. This finishes the proof.

In fact, it is possible to show that the limits in $(1)$ exist in $(-\infty,\infty)$. To this end, recall that we have the following maximal inequality for (super)martingales:

$$r \mathbb{P} \left( \sup_{t \in \mathbb{Q} \cap [0,T]} |X_t| \geq r \right) \leq \mathbb{E}(X_0) + 2 \mathbb{E}(X_T)$$

for all $r>0$. Since the right-hand side is bounded as $r \to \infty$, we get

$$\lim_{r \to \infty} \mathbb{P} \left( \sup_{t \in \mathbb{Q} \cap [0,T]} |X_t| \geq r \right)=0$$

which is equivalent to

$$\mathbb{P} \left( \sup_{t \in \mathbb{Q} \cap [0,T]} |X_t| = \infty \right)=0.$$

This implies that the limits in $(1)$ exist (almost surely) in $(-\infty,\infty)$ for all $t \geq 0$.

---

Answer to question 2: From the first part of this answer, we know that

$$\bar{X}_t := \lim_{\mathbb{Q} \ni r \downarrow t} X_r$$

is well-defined and takes values in $(-\infty,\infty)$ (up to a null set). The so-defined process is a modification of $(X_t)_{t \in \mathbb{Q} \cap [0,\infty)}$, i.e.

$$\mathbb{P}(\forall t \in \mathbb{Q} \cap [0,\infty): X_t = \bar{X}_t)=1. \tag{2}$$

Indeed: Since $\mathbb{Q}$ is countable, it suffices to prove $\mathbb{P}(X_t = \bar{X}_t)=1$ for each $t \in \mathbb{Q} \cap [0,\infty)$. Fix $t \in \mathbb{Q} \cap [0,\infty)$. By the martingale property, we have

$$\mathbb{E}(X_T \mid \mathcal{F}_s) = X_s$$

for all $s \leq T$, $s,T \in \mathbb{Q}$. If we let $s \downarrow t$, then the right-hand side converges to $\bar{X}_t$. On the other hand, the left-hand side converges by Lévy's convergence theorem and the right-continuity of the filtration to $\mathbb{E}(X_T \mid \mathcal{F}_t) = X_t$. Consequently, $X_t = \bar{X}_t$ almost surely.

Finally, we can show that

$$\mathbb{E}(X_t \mid \mathcal{F}_s) = \bar{X}_s \tag{3}$$

for all $t \in \mathbb{Q}$, $s \in \mathbb{R}$, $t > s$. To this end, note that

$$\bar{X}_s = \lim_{\mathbb{Q} \ni r \downarrow s} X_r = \lim_{\mathbb{Q} \ni r \downarrow s} \mathbb{E}(X_t \mid \mathcal{F}_r) = \mathbb{E}(X_t \mid \mathcal{F}_s)$$

where we have used again Lévy's continuity theorem and the martingale property. Using $(2)$, we actually get

$$\mathbb{E}(\bar{X}_t \mid \mathcal{F}_s) = \bar{X}_s.$$