I attended a lecture today, in which the professor went through an example with a lot of tedious calculations to show something which I'd think would follow directly from the Dambis-Dubins-Schwarz theorem. I'd appreciate it if someone would tell me where my argument fails.
We first 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, with $Z_{\tau}^M:=\lim_{t\uparrow \tau}Z_t^M$ we have that
$\{\lim_{n\uparrow\infty}\langle M\rangle_{\tau_n}<\infty\}=\{ Z_{\tau}^M>0\}$,
where $\lim_{n\to\infty}\tau_n=\tau$.
Now, in lieu of numerous results on the limits and quadratic variation of local martingales (e.g. V.1.8 and V.1.16 of Revuz & Yor) this result is hardly surprising. What does surprise me is the idea that the DDS theorem, which allows us to represent a continuous local martingale as a time-changed Brownian motion, somehow would be less suited in proving the result than an entire blackboard of calculations.
The argument would go something like this:
First, note that $Z_{\tau}^M=0$ if and only if $\log(Z_{\tau}^M)=-\infty$ and that
$\log(Z_t^M)=\langle M\rangle_t\left(\frac{M_t}{\langle M\rangle_t}-\frac12\right)$
for all $t\in(0,\tau)$. So it really boils down to showing that $\lim_{t\to\tau}\frac{M_t}{\langle M\rangle_t}$ exists and is real. Now, I would argue that we could just use the DDS theorem and write
$\lim_{t\to\tau}\frac{M_t}{\langle M\rangle_t}=\lim_{t\to\tau}\frac{B_{\langle M\rangle_t}}{\langle M\rangle_t}$,
which would essentially finish off the proof. I would very much appreciate it if someone could inform me as to where I've made a mistake.