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Subtract $M_0$, which does not affect either the integral or its quadratic variation. Let $N=M-M_0$. Choose a localizing sequence ${\tau }_n$ for $N$. By local boundedness of the previsible process $H$, there are increasing stopping times ${\rho }_n\uparrow \infty$ and finite deterministic constants $K_n$ with $|H_s|\le K_n$ on $0<s\le {\rho }_n$. PutContinuity and finiteness on compact intervals imply ${\sigma }_n\uparrow \infty$ almost surely. The stopped process $N^{{\sigma }_n}$ is a bounded square-integrable martingale, its bracket is bounded by $n$, and its stopped integrand is bounded by $K_n$. The indicator $1_{\{s\le {\sigma }_n\}}$ is predictable, so the stopped stochastic integral identity is valid.

Fix $n$ and abbreviateThe Itô isometry gives $\mathbb{E}{J}_t^2=\mathbb{E}{Q}_t\le K_n^{2}n$, so $J$ is a square-integrable martingale. For $F\in {\mathcal{F}}_s$, apply the same isometry to the predictable integrand $1_F{1}_{(s,t]}H$, stopped at ${\sigma }_n$. This yieldsAlso $\mathbb{E}[J_t-J_s\mid {\mathcal{F}}_s]=0$. Expanding $J_t^2-J_s^2$ therefore proves that $J^2-Q$ is a true martingale. The process $Q$ is continuous adapted increasing and starts at zero. By the characterization of continuous martingale quadratic variation as its increasing square compensator, and uniqueness of an increasing square compensator, we have $[J]=Q$.

Stochastic integration and quadratic variation commute with stopping; for the latter this follows either from squared-increment sums or from the same square-compensator characterization. Hence the desired identity holds up to every ${\sigma }_n$. Since these times increase to infinity, they exhaust every finite time interval almost surely, and the continuous identities patch to giveThis is the localized isometry proof of stochastic-integral quadratic variation. No unstopped second-moment or bracket-integrability assumption has been made.