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For the integrable terminal bracket criterion, assume $M_0=0$, or more generally $M_0\in L^2$. This initial-value assumption is necessary if it is not already part of the notation: an integrable but non-square-integrable random constant process has zero quadratic variation and still fails square-integrability.

For the zero-starting case, choose a localizing sequence ${\tau }_n$ bounding both $|M|$ and its quadratic variation. The stopped processes are true square-integrable martingales, and the Itô isometry givesFor each fixed time these variables are uniformly integrable by their bounded second moments. Almost sure convergence as $n\to \infty$ is therefore convergence in $L^1$. Passing to the limit in the stopped conditional-expectation identity gives $\mathbb{E}[M_t\mid {\mathcal{F}}_s]=M_s$: the local martingale is a true martingale. The Fatou lemma also givesso $M$ belongs to the global ${\mathcal{M}}_c^2$ space. For general square-integrable $M_0$, apply this argument to $M-M_0$ and use $|M_t{|}^2\le 2|M_0{|}^2+2|M_t-M_0{|}^2$.

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.

Work first with $M_0=0$. More generally, the assertion needs $M_0\in L^2$ and then applies to $M-M_0$. Without this assumption, an integrable but non-square-integrable initial random variable kept constant in time has zero quadratic variation but is not bounded in L2 space.

Choose increasing stopping timesThey tend to infinity almost surely, because the continuous path and its quadratic variation are finite on every compact time interval. The stopped process is a bounded martingale, and $M_{t\wedge {\tau }_n}^2-[M{]}_{t\wedge {\tau }_n}$ is bounded and a local martingale. Taking expectations givesSuppose $C=\mathbb{E}[M{]}_{\infty }<\infty$. These squared expectations are at most $C$. For each fixed $t$, the stopped values converge almost surely to $M_t$ and are uniformly integrable, since they are uniformly bounded in L2 space. They therefore converge in $L^1$. Passing in the stopped conditional-expectation identity, using contraction of conditional expectation in $L^1$, gives $\mathbb{E}(M_t\mid {\mathcal{F}}_s)=M_s$ for $s\le t$. Thus $M$ is a true martingale. The Fatou lemma gives $\mathbb{E}{M}_t^2\le C$, uniformly in $t$, so it is an L2-bounded continuous martingale.

Conversely, suppose $M$ is a true martingale with ${\sup }_t\mathbb{E}{M}_t^2\le C$. For fixed $t$, bounded-time optional sampling theorem gives $M_{t\wedge {\tau }_n}=\mathbb{E}(M_t\mid {\mathcal{F}}_{t\wedge {\tau }_n})$. Conditional Jensen inequality then yieldsThe brackets increase to $[M{]}_t$. Applying the monotone convergence theorem first in $n$ and then in $t\uparrow \infty$ proves $\mathbb{E}[M{]}_{\infty }\le C<\infty$.

The exact identities and terminal limits can also be justified. On each fixed horizon $t$, the Doob L2 maximal inequality givesThis is an integrable dominating variable for $M_{t\wedge {\tau }_n}^2$, so the dominated convergence theorem and increasing convergence of the brackets remove the stopping:The L2 martingale convergence theorem supplies $M_{\infty }$ with $M_t\to M_{\infty }$ almost surely and in L2 space. Consequently the squares converge in $L^1$, while $[M{]}_t\uparrow [M{]}_{\infty }$. We obtain the integrable terminal bracket criterion with the precise terminal identityFor $M_0\in L^2$ rather than zero, the final identity becomes $\mathbb{E}{M}_{\infty }^2=\mathbb{E}{M}_0^2+\mathbb{E}[M{]}_{\infty }$.