 Solution

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 $\mathbb{E}{M}_{\infty }^2=\mathbb{E}{M}_0^2+\mathbb{E}[M{]}_{\infty }$.