If $M_t$ is continuous martingale, we know that there exists quadratic variation process which is continuous and increasing. I am interested to know if the converse is also true. To make it precise
Let $ \{A_t, \mathcal{F_t}; 0\leq t < \infty\}$ be continuous, increasing process starting at 0 a.s. Let $\{B_t, \mathcal{F_t}; 0\leq t < \infty\}$ be standard Brownian motion. Is it true that $M_t := B_{A_t}$ is a local martingale with quadratic variation $A_t$?
I was trying to use time change for Brownian motion. However I am not successful in getting the right formulation with stopping times.