[EDIT: The claim may not be true.
I misunderstood the authors' argument in proving that the expected fourth variation goes to zero as partitions become fine (Lemma 5.10 in Karatzas and Shreve). They "consider first the bounded case: $|X_s|\le K<\infty$ and $\langle X\rangle_s\le K$ for all $s\in[0,t]"$. Then they show it when $X$ is not necessarily bounded. I interpreted this to mean "the bounded case" is precisely the case in which $X$ is not bounded, and hence that uniform boundedness of quadratic variation follows from $X$ being bounded.
However, all we care about is showing the result for continuous square integrable martingales. To do so, it suffices to show it when $X$ is bounded and its quadratic variation is bounded.]
Let $X$ be a bounded square integrable martingale, so $|X_s|\le K$ for all $s$ for some $K>0$. I'd like to show that for any $t>0$, there exists $J>0$ so that $\langle X\rangle_s\le J$ for all $s\in[0,t]$.
We know $X^2_s-\langle X\rangle_s=M_s$, where $M_s$ is a martingale. So $\langle X\rangle_s=X_s^2-M_s\le K^2-M_s$, and it would suffice to show that $M_s$ is bounded. How do we rule out the possibility that $M$ and $A$ move in opposite directions such that they are unbounded but $X$ is still bounded?
Thanks in advance.