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This is an exponential-moment criterion for the stochastic exponential to be a true martingale. In the infinite-horizon stopped-moment form, a zero-starting convergent continuous local martingale satisfying ${\sup }_T\mathbb{E}{e}^{M_T/2}<\infty$ has a uniformly integrable martingale $\mathcal{E}(M)$. The half-threshold for the exponential-martingale Hölder bound first controls strict scalings. The terminal scaling inequality for stochastic exponentials and terminal expectation criterion for a nonnegative local martingale then include the endpoint scaling.

Let $K={\sup }_T\mathbb{E}[e^{M_T/2}]<\infty$, with $T$ ranging over finite stopping times. To locate the half-threshold for the exponential-martingale Hölder bound, put $u=\sqrt{pq}>\sqrt{q}$. Since $u(u-1)$ is increasing for $u>1$,On the other hand, taking $q=1+\epsilon$ and $p=1+{\epsilon }^2$ gives $C(p,q)\to 1/2$ as $\epsilon \downarrow 0$. Hence

Fix $0<a<1$. Choose $p,q>1$ with $aC(p,q)<1/2$, and put $\lambda =2aC(p,q)\in (0,1)$. Part (a) and the Jensen inequality give, for every finite stopping time,Thus the entire stopped family has a uniform $L^p$ bound for some $p>1$. By uniform integrability from an Lp bound, it is uniformly integrable.

To check that this local martingale is a true martingale, stop it by a localizing sequence. At each fixed time the stopped variables are uniformly integrable by the same bound; taking limits in their conditional martingale identities proves the unstopped identity. The uniform bound over all finite stopping times then makes it a uniformly integrable martingale, with terminal expectation one. Therefore