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The terminal bracket value printed as zero is incompatible with a strictly increasing bracket starting at zero. If the intended value is infinity, solution applies directly. In fact the substantive conclusion holds for every deterministic continuous bracket $a(t)=[M{]}_t$, whether its terminal value is finite or infinite, and without strict increase.

For $u\in \mathbb{R}$, the Itô formula makesa complex local martingale. Its modulus is the deterministic value $e^{u^{2}a(t)/2}$, bounded on every finite horizon, so its real and imaginary parts are true martingales there. Conditioning between $s$ and $t$ givesThis deterministic conditional characteristic function shows that the increment is independent of ${\mathcal{F}}_s$ and is normal with mean zero and variance $a(t)-a(s)$. Induction over ordered times gives independent Gaussian increments and hence jointly Gaussian coordinate vectors. Since $a(t)$ is finite at every finite time, the integrable terminal bracket criterion applied on each fixed horizon also upgrades $M$ to a true square-integrable martingale. This proves the deterministic quadratic variation characterizes a Gaussian continuous local martingale result in a form covering the corrected question.

True. The Dambis-Dubins-Schwarz theorem, with an independent continuation of the Brownian motion if $f$ is bounded, representsBecause $f$ is deterministic, every finite vector $(M_{t_1},\dots ,M_{t_n})$ is a finite vector of a Brownian motion at deterministic times and therefore has a multivariate normal distribution. Hence $M$ is a Gaussian process. This is the deterministic quadratic variation characterizes a Gaussian continuous local martingale result.