# Do paths of a continuous time martingale always have a left limit?

I'm looking at this theorem:

Let $(X_t, \mathcal{F}_t)_{t \in \mathbb{R}_+}$ be a submartingale, where $\mathcal{F}$ is right-continuous and complete, and the function $\mu(t) := E[X_t]$ is right-continuous. Then there exists a submartingale $(Y_t, \mathcal{F}_t)_{t \in \mathbb{R}_+}$ satisfying $X_t = Y_t$ a.s. for all $t$ whose paths $t \mapsto Y_t(\omega)$ are rcll (right continuous with left limit) a.s.

The proof constructs $Y_t$ such that $Y_t = \underset{s\in \mathbb{Q}_+, s \searrow t}{\lim X_s}$ a.s., which shows that the paths of $Y_t$ are right-continuous. But it never proves the existence of the left limit. Is the family $(X_r, \mathcal{F}_r)_{r<t}$ a uniformly integrable submartingale? If it is, I can see that we can apply the Submartingale Convergence Theorem to show that the left limit exists. I apologize if this is an obvious question. Here's my attempt so far:

1) $E[X_r] \leq E[X_t] < \infty$ for all $r < t \implies \sup_{r<t} E[X_r] < \infty$

2) Let $\epsilon > 0$. We want to show that there exists $\delta > 0$ such that for all A \in \mathcal{F}_t$$\mathbb{P}(A) < \delta \implies \sup_{r<t} E[X_r\mathbb{1}_A] < \epsilon. Since (X_t) is a submartingale, we have E[X_t|X_r] \geq X_r a.s., and so E[X_r\mathbb{I}_A] \leq E[X_t\mathbb{I}_A] for all A \in \mathcal{F}_r. And I'm stuck... ## 1 Answer Basically your idea is correct: Since$$\mathbb{E}(X_t \mid \mathcal{F}_r) \geq X_r, \qquad r \leq t, $$the sequence (X_r)_{r \leq t} is "dominated" by the conditional expectations of X_t and therefore uniformly integrable. But, in order to show uniform integrability, we have to estimate |X_r| from above and this makes things a bit more complicated: Fix t \geq 0 and a sequence (t_n)_{n \in \mathbb{N}} with t_n \downarrow t. It follows from the right-continuity that$$M := \sup_{n \in \mathbb{N}} \mathbb{E}|X_{t_n}|<\infty.$$For given \epsilon>0 we can pick N \in \mathbb{N} such that$$\mathbb{E}X_{t_n} \geq \mathbb{E}X_{t_N}- \epsilon \qquad \text{for all} \, \, n \geq N \tag{1}$$(as \mathbb{E}X_{t_n} is decreasing as n \to \infty). The submartingale-property implies$$\begin{align*} \int_{|X_{t_n}| > c} |X_{t_n}| \, d\mathbb{P} &= \int_{X_{t_n}<c} (-X_{t_n}) \, d\mathbb{P} + \int_{X_{t_n}>c} X_{t_n} \, d\mathbb{P} \\ &= \int_{X_{t_n} \geq -c} - \mathbb{E}X_{t_n} + \int_{X_{t_n}>c} X_{t_n} \, d\mathbb{P} \\ &\leq \int_{X_{t_n} \geq -c} X_{t_N} \, d\mathbb{P} - \mathbb{E}X_{t_n} + \int_{X_{t_n} >c} X_{t_N} \,d \mathbb{P}. \end{align*}$$By (1),$$\begin{align*} \int_{|X_{t_n}| > c} |X_{t_n}| \, d\mathbb{P} &\leq \int_{X_{t_n} \geq -c} X_{t_N} \, d\mathbb{P}-\mathbb{E}X_{t_N} + \epsilon + \int_{X_{t_n} >c} X_{t_N} \, d\mathbb{P} \\ &\leq \int_{|X_{t_n}| \geq c} X_{t_N} \, d\mathbb{P}+ \epsilon. \end{align*}$$Hence, by Markov's inequality$$\begin{align*} \int_{|X_{t_n}| > c} |X_{t_n}| \, d\mathbb{P} &\leq \int_{\{|X_{t_n}| \geq c\} \cap \{|X_{t_N} > R\}} X_{t_N} \, d\mathbb{P} + \int_{\{|X_{t_n}| \geq c\} \cap \{|X_{t_N} \leq R\}} X_{t_N} \, d\mathbb{P} +\epsilon \\ &\leq \int_{|X_{t_N}| \geq R} X_{t_N} \, d\mathbb{P} + R \mathbb{P}(|X_{t_n}| \geq c) + \epsilon \\ &\leq \int_{|X_{t_N}| \geq R} X_{t_N} \, d\mathbb{P} + R \frac{M}{c} + \epsilon \\ &\stackrel{c \to \infty}{\to} \int_{|X_{t_N}| \geq R} X_{t_N} \, d\mathbb{P}+\epsilon. \end{align*}$$Letting R \to \infty and \epsilon \to 0 finishes the proof. Remark: Note that the existence of the limit$$Y_t = \lim_{\mathbb{Q} \ni s \downarrow t} X_s\$ also follows directly from Doob's upcrossing estimate. Applying Doob's upcorssing estimate is much more easier than proving uniform integrability.

• Wow, thank you for taking the time answer this lengthy question. It was very helpful! Feb 13, 2015 at 0:12