# Expressing a continuous local martingale as an integral against a Brownian motion

I'm interested in the following problem.

Suppose $$X$$, $$X_0=0$$ is a continuous local martingale with quadratic variation

$$[X]_t = \int_0^t A_s\mathrm{d}s$$

for a non-negative previsible process $$(A_t)_{t\ge 0}$$. Show that there exists a Brownian motion $$B$$ such that $$X_t = \int_0^t A_s^{1/2} \mathrm{d}B_s.$$

The solution I've seen involves defining $$B_t = \int_0^t A_s^{-1/2} 1_{A_s>0} \mathrm{d}X_s+\int_0^t 1_{A_s=0} \mathrm{d}W_s$$ where $$W$$ is a Brownian motion independent of $$X$$. It is not hard to check that $$B$$ is a Brownian motion by the Levy characterisation, but I'm confused as to how we can conclude $$A_t^{1/2} \mathrm{d}B_t = \mathrm{d}X_t$$? It is immediate that $$A_t^{1/2}\mathrm{d}B_t = 1_{A_t>0} \mathrm{d}X_t$$ but why is this enough to conclude?

Intuitively it may have something to do with continuous local martingales being constant on intervals where their quadratic variation is but I can't see how to justify this rigorously.

Since $$X_t$$ is constant when $$A_t = 0$$, we have $$\int_0^t 1_{A_s = 0}dX_t = 0$$. Since you showed $$A^{1/2}_t dB_t = 1_{A_t>0}dX_t$$, we have \begin{align*} \int_0^t A^{1/2}_s dB_s &= \int_0^t 1_{A_s>0}dX_s \\ &= \int_0^t 1_{A_s>0}dX_s + \int_0^t 1_{A_s=0}dX_s \\ &= \int_0^t 1_{A_s \ge 0} dX_s \\ &= \int_0^t dX_s \\ &= X_t \end{align*}
• Thanks, but what does it mean for $X_t$ to be constant when $A_t=0$? I've since realised that $\int_0^t 1_{A_s=0} \mathrm{d}X_s$ has quadratic variation $\int_0^t 1_{A_s=0} A_s\mathrm{d}s = 0$ so we must have $\int_0^t 1_{A_s=0} \mathrm{d}X_s=0$ as you claimed. However is there an easier way to see this?
• I guess to be more precise I should say that $X_t$ is constant on any interval $(a,b)$ such that $A_t = 0$ on $(a,b)$, i.e. if $A_t = 0$ for all $t \in (a,b)$ then $X_t = X_a$ for all $t \in (a,b)$. Commented May 22 at 15:39