Question about Protter's proof that a Cadlag, locally square integrable local martingale is a semimartingale

This is Corollary $$1$$ in Chapter $$2$$ of Protter's Stochastic Integration and Differential Equations.

Theorem 8 states that each $$L^2$$ martingale (martingales $$X$$ such that $$X_0 = 0$$ and $$E[X_\infty^2]<\infty$$) with cadlag paths is a total semimartingale.

The Corollary to Theorem 6 states that : If $$X$$ is a process and there exists a sequence $$T_n$$ of stopping times increasing to $$\infty$$ a.s. such that $$X^{T_n}$$ ( or $$X^{T_n} 1_{\{T_n>0\}}$$) is a semimartingale for each $$n$$ then $$X$$ is a semimartingale.

But I don't see how the proof is so straightforward here.

First, let $$X$$ be a cadlag, locally square integrable local martingale.

Then does this mean that $$X$$ is locally a square integrable martingale, i.e. we have a fundamental sequence $$T_n$$ such that $$X^{T_n} 1_{\{T_n >0\}}$$ is a square integrable martingale? I have seen several questions about this on StackExchange, but no real answer to this.

It seems like we would need this condition but then taking $$X^n := X^{T_n} 1_{\{T_n >0\}}$$, square integrability just means that $$E[(X^n_t)^2]<\infty$$ for each $$t$$. We don't get $$L^2$$ boundedness over all $$t$$ from this. So how do we use the corollary to Theorem 6 here?

My attempt: $$X$$ being a semimartingale is by definition in the text, $$X^t$$ being a total semimartingale for each $$t \ge 0$$.

Assuming that being locally square integrable local martingale is the same as being locally a square integrable martingale, we can find a fundamental sequence $$T_n$$ such that $$X^{T_n} 1_{\{T_n > 0\}}$$ is a square integrable martingale. For convenience denote $$M := X^{T_n} 1_{\{T_n > 0\}}$$. Then we have $$E[(M_t^n)^2]<\infty$$ for all $$t \ge 0$$.

Now in order to apply the Corollary to Theorem 6, we would need to show that $$M$$ is a semimartingale. And to do this we need to use Theorem 8.

Thus, we are done if we show that for each $$s \ge 0$$, $$M^s$$ is a $$L^2$$ - martingale.

Now, for each $$s \ge 0$$, $$M^s_t = X_{T_n \wedge s \wedge t} 1_{\{T_n > 0\}}$$. Since a stopped martingale is a martingale, $$M^s$$ is still a martingale. Moreover, $$s$$ and $$t$$ are not random, so square integrability gives us that $$M^s$$ is a $$L^2$$-bounded uniformly integrable martingale. Hence, by Theorem 8, $$M^s$$ is a total semimartingale, hence $$M$$ is a semimartingale. Finally, the Corollary to Theorem 6 applies. QED.

This proof has been bothering me for a long time now. I think my final argument assuming a single fundamental sequence that makes $$X$$, a locally square integrable local martingale, into a locally square integrable martingale, is correct but I don't know how to show this part. I would greatly appreciate any help.

• @TheBridge Right I meant the square to be inside the expectation. Sorry for the confusion in notation. Commented Nov 11, 2020 at 23:04

Let $$X$$ be cadlag and a locally square-integrable local martingale, with $$X_0=0$$ for simplcity. Because $$X$$ is locally square integrable, there is a localizing sequence $$(T'_n)$$ such that $$E[(X^{T_n'}_t)^2]<\infty$$ for each $$t>0$$ and each $$n$$. Because $$X$$ is a local martingale there is a localizing sequence $$(T''_n)$$ such that $$(X^{T_n''})$$ is a uniformly integrable martingale, for each $$n$$. Define $$T_n:=T_n'\wedge T_n''\wedge n$$. Then $$(T_n)$$ is an increasing sequence of stopping times with a.s. limit $$\infty$$. Moreover, $$X^{T_n}$$ is for each $$n$$ both square integrable and a U.I. martingale. In fact, because $$T_n\le n$$, $$X^{T_n}$$ is an $$L^2$$ martingale. Thus by Theorem 8, each $$X^{T_n}$$ is a semimartingale. Finally, by the Corollary to Theorem 6, $$X$$ is a semimartingale.
• I am not clear why $X^{T_n}$ remains square integrable and a U.I. martingale. Could you elaborate on this part? I know since $X^{T_n''}$ is a martingale, so is $X^{T_n}$, but the square integrability part relies on $X^{T_n'}$, and how does $E[(X_{t \wedge T_n'})^2] < \infty$ imply $E[(X_{t \wedge T_n' \wedge T_n'' \wedge n})^2]<\infty$? Also I don't follow why $T_n \le n$ implies that $X^{T_n}$ is $L^2$. Commented Nov 17, 2020 at 9:27
• 1. $X^{T''_n}$ is a UI martingale and $X^{T_n} = (X^{T''_n})^{T_n}$, so $X^{T_n}$ is a UI martingale — a UI martingale stopped early is still a UI martingale. 2. $E[(X^{T_n}_t)^2] = E[(X^{T'_n\wedge T''_n}_{t\wedge T_n})^2]\le E[(X^{T'_n\wedge T''_n}_{t\wedge n})^2]<\infty$, because $(X^{T'_n\wedge T''_n})^2$ is a submartingale and $T_n\le n$. 3. For Protter "$L^2$ martingale" means a UI martingale with square-integrable terminal value. Because $T_n\le n$, $X^{T_n}_\infty = X_{T'_n\wedge T''_n\wedge n}$ and this is square integrable by the argument used above in 2. Commented Nov 17, 2020 at 17:44
• In $2$, you use the optional sampling theorem on submartingales, but to ensure $(X^{T_n' \wedge T_n''})^2$ is a submartingale, we need to know that it is integrable for each $t$. How do we get that $E[(X_t^{T_n' \wedge T_n''})^2] < \infty$ from $E[(X_t^{T_n'})^2]<\infty$? Commented Nov 17, 2020 at 20:59
• Could you help me understand why we can ensure $E[(X_t^{T_n' \wedge T_n''})^2] < \infty$ given that $E[(X_t^{T_n'})^2] < \infty$? Commented Nov 20, 2020 at 21:39