An elliptic local martingale has a Brownian driver

If $T=0$, set $B_0=0$; the integral, bracket and finite-horizon Brownian claims are immediate. Henceforth suppose $T>0$. Form left-step approximations to $\sigma$ on deterministic dyadic partitions, retaining ${\sigma }_0$ at time zero. They are predictable by [F1]. Take their pointwise limit where it exists as a finite invertible matrix with inverse norm at most $C$, and use $C^{-1}{I}_d$ otherwise; necessarily $C>0$. The resulting predictable $\widehat{\sigma }$ agrees with $\sigma$ at all times on its full continuity event. Continuity on the compact horizon gives ${\int }_0^T\mathrm{\parallel }{\widehat{\sigma }}_s{\mathrm{\parallel }}_F^{2}ds<\mathrm{\infty }$ almost surely, and its bracket identity is the given one. Thus [F9] supplies a Brownian motion $B=\int {\widehat{\sigma }}^{-1}dM$ adapted to the original filtration, with bracket $t{I}_d$. This invocation makes no raw stopped evaluation of an unnormalized almost-surely continuous process.

This integral also has the left-step construction required in the Statement. On the augmented, normalized stopped intervals of [F9], put $H={\sigma }^{-1}$ and use its bounded deterministic left-step approximations. They converge uniformly on each continuous path and are bounded by $C$. The bracket-energy isometry and maximal estimate proved in [F9], together with dominated convergence [F7] and the integrable stopped bracket trace, give localized $L^2$ convergence of their integrals to the same $B$. Equality with the predictable reconstruction follows from uniqueness in that energy completion. Its original-adapted continuous version and Brownian increment identities are supplied by [F9]; the bracket is $[B^i,B^j{]}_t={\delta }_{ij}t$. Hence every asserted conclusion holds on the given filtration.