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

Question

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.

Answer 1

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.