 Solution

The Novikov condition holds on every horizon, sinceTherefore $Z=\mathcal{E}(M)$ is a true martingale and the Girsanov theorem applies through each finite $T$.

The deterministic bound also gives a genuine infinite-horizon density. For any $p>1$, the stochastic exponential identity isA nonnegative local martingale is a supermartingale, soThis $L^p$ bound gives uniform integrability. Also $M$ is a L2-bounded continuous martingale, since localization and the Itô isometry give $\mathbb{E}{M}_t^2=\mathbb{E}[M{]}_t\le C$. The L2 martingale convergence theorem gives a finite limit $M_{\infty }$; hence $Z_{\infty }=\exp (M_{\infty }-[M{]}_{\infty }/2)>0$ and $\mathbb{E}{Z}_{\infty }=1$. Weighting by $Z_{\infty }$ defines an equivalent probability measure on ${\mathcal{F}}_{\infty }$. This is the bounded-bracket criterion for a stochastic exponential.