 Solution

Use the following form of the Girsanov theorem. If $W$ is Brownian motion, $\vartheta$ is predictable, andis a uniformly integrable martingale with $Z_0=1$, then the probability measure with terminal density $Z_{\infty }$ makes $W_t-{\int }_0^t{\vartheta }_{s}ds$ a Brownian motion. The analogous finite-horizon statement uses the terminal density at that horizon.

Here take $W_t=X_t-x$ and ${\vartheta }_s=1_{\{s<\tau \}}/X_s$. Before $\tau$, $X_s\in [\epsilon ,M]$, so the stochastic integral is well-defined. The assumed uniform integrability of the stopped stochastic exponential givesThus the proposed weighting really defines a probability measure, and the Girsanov theorem gives the Brownian motionunder that measure. In particularAll calculations are first made on this compact stopped interval. The integral involving $1/X$ is naturally defined on $[0,T_0)$; at $T_0$ its quadratic variation may diverge. The density will have a zero limiting extension there, as established below, rather than a finite stochastic logarithm at the hitting time.