Let $(M_t^n)_{t \geq 0}$ be a sequence of continuous martingales of the form $M^n_t = \int_0^t X^n_s \, dB_s$ where $B_s$ is a Brownian motion. Let $\tau^n(t)$ be the time change associated to $M_t^n$ via the Dambis-Dubins-Schwarz theorem, i.e. $\tau^n(t) = \inf \, \{s \mid \langle M^n,M^n \rangle_s > t \}$ and we have that $M_t^n = \widetilde{B}_{\tau^n(t)}$, where $\widetilde{B}$ is another Brownian motion and $\langle M^n,M^n \rangle_t$ denotes the quadratic variation of $M^n_t$.
I'm reading a paper and can't figure out the following statement:
"In order for $\langle M^n,B \rangle_{\tau^n(t)}$ to converge to zero in probability (for $n \to \infty$), it is sufficient to show that $\langle M^n,B \rangle_{t}$ converges to zero in probability, uniformly in $t$ on compact sets."
How does one prove this? Is this true for any stopping time?
What the authors in fact show is uniform convergence in $L^p$, i.e. that $\sup_{0 \leq t \leq T} \operatorname{E} \left[ \left| \langle M^n,B \rangle_{t} \right|^p \right]$ converges to zero for $n \to \infty$, any $p \geq 2$ and any $T>0$, but this also doesn't give me an idea for a proof.
Well, I think the statement written in Italic is false. A counterexample would be this: $M^n(s) = B(s-n)I_{s \geq n}$, where $B$ is a BM, and $N^n = M^n$. In this case, $\tau^n(t) = t+n$,
$$ \langle M^n,N^n \rangle_{\tau^n(t)} = t \nrightarrow 0 , n \to \infty $$
but $\langle M^n,N^n \rangle_{t} \to 0$ uniformly on compacts even a.s., since $\langle M^n,N^n \rangle_{t} = 0\ $ whenever $n>t$.