# Is a stopped local martingale a local martingale?

Let

• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
• $(\mathcal F_t)_{t\ge0}$ be a filtration of $\mathcal A$
• $M$ be a local $\mathcal F$-martingale on $(\Omega,\mathcal A,\operatorname P)$
• $\tau$ be an $\mathcal F$-stopping time

Is $M^\tau$ a local $\mathcal F$-martingale?

With the given assumptions, this claim can be found in the proof of Lemma 15.1 in Foundations of Modern Probability by Olav Kallenberg.

I know how we're able to prove the claim using the Optional Sampling Theorem, if $\tau$ takes only countable many values or $M$ is almost surely right-continuous, but I have no idea how I can prove the claim without one of these additional assumptions.

More concretely, he states that if $(\sigma_n)_{n\in\mathbb N}$ is an $\mathcal F$-localizing sequence for $M$, then $(M^\tau)^{\sigma_n}=(M^{\sigma_n})^\tau$ is an $\mathcal F$-martingale. So, maybe the question reduces to the question if a stopped martingale is a martingale, but, again, I'm only able to proof this with the Optional Sampling Theorem and one of its additional assumptions.

To prove a stopped martingale is a martingale, please use the Th.7.29, p.135 in Kallenberg's book. Since $N=M^{\sigma_n}$ is a martingale and both $N,-N$ are submartingales also, then for $t\ge s$, we have \begin{align} \mathsf{E}[(N^\tau)_t|\mathcal{F}_s]&=\mathsf{E}[N_{\tau\wedge t }| \mathcal{F}_s] \stackrel{(16)}=N_{(\tau\wedge t)\wedge s}\\ &=N_{\tau\wedge s}=(N^\tau)_s. \end{align} This means $(M^\tau)^{\sigma_n}$ is a martingale.
• @JGWang's reference is to the second edition of Kallenberg's book. However, the submartingale discussed in that theorem is assumed to be right continuous. The issue with Kallenberg's Lemma 15.1 (Lemma 17.1 in the second edition) is that the stopped-process value at time $t$, $(M^\tau)^{\sigma_n}_t = M_{t\wedge\tau\wedge\sigma_n}$ is not guaranteed to be a random variable (let alone $\mathcal F_t$ measurable) without some supplementary condition on $M$; for example, that $M$ be progressively measurable. Jun 21, 2017 at 16:32
• @JohnDawkins 1. So you agree to me that the Lemma, as stated, is wrong. Right? 2. Sure, if $M$ is one-sided continuous and adapted, then $M$ is progressively measurable. So, I could imagine that we can show the desired result in that case. However, how do you prove it? Do you have a reference? Jun 21, 2017 at 20:38
• 1. As far as I can tell, the Lemma as stated lacks a necessary regularity hypothesis on the martingale. 2. If $M$ is right continuous then the argument found in Kallenberg is sound. (And when the filtration is right continuous, any martingale admits a right-continuous (and left-limited) martingale modification.) Jun 21, 2017 at 22:36