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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.

The increasing quadratic variation has a limit in $[0,\infty ]$. Since $X_t$ has a finite limit, the exponentials have limits too, with value zero when the bracket is infinite. Expanding exponents provesbecause the coefficient of $X_t$ on the right isand the bracket coefficient is $-1/2$. Taking limits preserves the identity, including the zero case.

Let $A=\mathbb{E}[\mathcal{E}(X{)}_{\infty }]$, $B=\mathbb{E}[\mathcal{E}(rX{)}_{\infty }]$, and $D=\mathbb{E}[e^{r{X}_{\infty }/2}]$. The Holder inequality first yieldsWhen $D$ is finite, apply the Jensen inequality to the concave function $z\mapsto z^{2/(r+1)}$:Since $D>0$, rearrangement gives the terminal scaling inequality for stochastic exponentialsHere $2(r^2-1)/(r+1)=2(r-1)$. If the last exponential moment is infinite, the bound has no positive content; the finite-moment form is the one used below.

Use deterministic times tending to infinity and the Fatou lemma:Apply the terminal scaling inequality for stochastic exponentials to $X=aM$ and $r=1/a$. Part (c) gives $\mathbb{E}[\mathcal{E}(aM{)}_{\infty }]=1$, soLetting $a\uparrow 1$ shows that the terminal expectation is at least one. The nonnegative local martingale $\mathcal{E}(M)$ starts at one and is a supermartingale, so the Fatou lemma gives the opposite inequality. Consequently

The terminal expectation criterion for a nonnegative local martingale now closes the argument. Conditional Fatou lemma applied to the supermartingale at times tending to infinity givesThe left side has expectation one and the right side at most one, so equality holds almost surely. Thusand uniform integrability of conditional expectations proves that $\mathcal{E}(M)$ is a uniformly integrable martingale. This establishes the required stopped-moment form of the Kazamaki criterion.