We have the following theorem:
Let $M$ be a square integrable martingale, that is $\sup_{t\geq 0}E[M_t^2]<\infty$. Furthermore, let $H$ predictable s.t. $$E[\int_0^\infty H_u^2 d[M,M]_u]<\infty$$. Then $H\bullet M$ is a square integrable martingale.
I am wondering if the theorem is still true if we replace the condition $E[\int_0^\infty H^2 d[M,M]]<\infty$ with $E[\int_0^\infty H^2 d\langle M,M\rangle]<\infty$?
Yes, you are right. Since for square integrable martingale $M$, $\langle M,M\rangle=[M,M]^p$ is the predictable dual projection of $[M,M]$, hence for predictable $H$, $\mathsf{E}[\int_0^\infty H^2\mathrm{d}\langle M,M\rangle]=\mathsf{E}[\int_0^\infty H^2\mathrm{d}[M,M]]$.