 Solution

At exit, continuity gives $X_T^2=1$. The preceding density identity becomesSince $|X_s|\le 1$ before exit, ${\int }_0^T{X}_s^{2}ds\le T$. On $\{T\le t\}$ it follows that $Z_T\ge e^{1/2-T}\ge e^{1/2-t}$. The definition of the changed measure now yieldsThe positive sign of the stochastic integral in the density is what introduces the positive drift $X_{t}dt$; reversing that sign would give a different measure and would not prove this inequality.