I have encountered a problem, which is hopefully rather easily solvable. I just can't get my head around it at the moment.
First we extend a well known notion. We call a stochastic process $M$ a continuous local martingale on $[0,\tau)$ for some predictable positive stopping time $\tau>0$ if the stopped process $M^{\tilde{\tau}}:=M_{\cdot\wedge\tilde{\tau}}$ is a continuous local martingale for any stopping time $\tilde{\tau}<\tau$. With $\tau=\infty$ we have the usual class of continuous local martingales.
Now the result I want to prove is the following:
Lemma: Fix a predictable positive stopping time $\tau>0$ and a continuous local martingale $M$ on $[0,\tau)$ and consider the exponential local martingale $Z^M=\mathcal{E}(M)=\exp(M-\langle M\rangle /2)$ on $[0,\tau)$. Then the random variable $Z_{\tau}^M:=\lim_{t\uparrow \tau}Z_t^M$ exists, is nonnegative and satisfies $\{\lim_{n\uparrow\infty}\langle M\rangle_{\tau_n}<\infty\}=\{ Z_{\tau}^M>0\}$
Now, in proving this, I essentially just try to adapt the proof of a standard convergence result for martingales (e.g. prop. 1.3.35 of Karatzas & Shreve). I'll try to go easy on formalities concerning standard upcrossing arguments.
Proof?: Let $\tau_n$ be a nondecreasing sequence of stopping times with $\lim_{n\to\infty}\tau_n=\tau$. Doob's upcrossing inequality (e.g. thm 1.3.8(iii) of Karatzas & Shreve) should yield that $Z_{\tau(\omega)}^M(\omega)$ exists for almost all $\omega\in\Omega$ . If $U$ denotes the number of upcrossings, we have
$ EU_{[0,\tau_n]}(\alpha,\beta;Z^M(\omega))\leq\frac{E[(Z_{\tau_n}^M)^+]+|\alpha|}{\beta-\alpha},$
for any $\tau_n\geq 1$ and real numbers $\alpha<\beta$. Then, letting $\tau_n\to\tau$, by monotone convergence we find
$ EU_{[0,\tau)}(\alpha,\beta;Z^M(\omega))\leq \frac{\sup_{t\geq0}E[(Z_t^M)^+]+|\alpha|}{\beta-\alpha}.$
Now, IF the supremum on the right hand side is finite, the events $A_{\alpha,\beta}:=\{ \omega | U_{[0,\tau)}(\alpha,\beta;Z^M))=\infty\}$, $-\infty<\alpha<\beta<\infty$, are $P$-null sets, and as is the set $A=\displaystyle\bigcup_{\stackrel{\alpha<\beta}{\alpha,\beta\in\mathbb{Q}}}A_{\alpha,\beta},$ which contains the set $\{ \omega | \limsup_{t\to\tau}Z_t^M(\omega)>\liminf_{t\to\tau}Z_{t}^M(\omega)\}$. Therefore $Z_{\tau}^M$ exists for each $\omega\in\Omega\setminus A$.
So, my only question is then. Am I entirely wrong in trying to adapt a proof, in which the supremum is in fact assumed to be finite? Or is there something I've missed which actually makes the assumption redundant?
Applying the lemma below shows that $Z^M$ is a supermartingale, and therefore we have in particular $\mathbb{E}(Z_t^M) \leq 1$, i.e. $$\sup_{t \geq 0} \mathbb{E}(Z_t^M) \leq 1<\infty.$$
Proof: Denote by $(\tau_n)_n$ the localizing sequence. Then, by Fatou's lemma,
$$\mathbb{E}(X_t) = \mathbb{E}\left( \liminf_{n \to \infty} X_{t \wedge \tau_n}\right) \leq \liminf_{n \to \infty} \mathbb{E}(X_{t \wedge \tau_n}) = \mathbb{E}X_0<\infty$$
where we used in the last step that $(X_{t \wedge \tau_n})_{t \geq 0}$ is a martingale. Hence, $X_t \in L^1$. Similarly, we obtain
$$\mathbb{E}(X_t \mid \mathcal{F}_s) \leq \liminf_{n \to \infty} \mathbb{E}(X_{t \wedge \tau_n} \mid \mathcal{F}_s) = \liminf_{n \to \infty} X_{s \wedge \tau_n} = X_s.$$
This finishes the proof.