The Stochastic Integral

Square integrability is immediate, since each $(H\cdot M)_t$ is the finite sum $\sum_i h_i(M_{t_{i+1}\wedge t}-M_{t_i\wedge t})$ of $L^2$ terms, since $h_i$ is bounded and the $M$ increments lie in $L^2$, so $\norm{(H\cdot M)_t}_2\le\sum_i\norm{h_i}_\infty\,\norm{M_{t_{i+1}\wedge t}-M_{t_i\wedge t}}_2<\infty$.

It remains to check the martingale property, which is the martingale-transform identity for each summand. Fix $s<t$. For the $i$-th summand and $a=t_i\wedge t$, $b=t_{i+1}\wedge t$, we claim $\E[h_i(M_b-M_a)\mid\F_s]=h_i(M_{b\wedge s}-M_{a\wedge s})$. If $s\le t_i$ then $h_i$ is unknown at time $s$, but conditioning first on $\F_{t_i}\supseteq\F_s$ makes $h_i$ measurable and leaves $h_i\,\E[M_b-M_a\mid\F_{t_i}]=0$, so by the tower property the increment collapses to $0$, matching $M_{b\wedge s}-M_{a\wedge s}=0$. If $t_i<s\le t_{i+1}$ then $h_i$ and $M_{a\wedge s}=M_{t_i}$ are $\F_s$-measurable, so the elapsed part $h_i(M_s-M_{t_i})$ is frozen while $\E[h_i(M_b-M_s)\mid\F_s]=h_i\,\E[M_b-M_s\mid\F_s]=0$ by the martingale property of $M$. If $s>t_{i+1}$ the whole summand is $\F_s$-measurable. Summing the three cases gives $\E[(H\cdot M)_t\mid\F_s]=(H\cdot M)_s$.

Expand the square of Equation (2) at $t=T$ into diagonal and off-diagonal terms. For $i<j$ the cross term $h_i h_j(M_{t_{i+1}}-M_{t_i})(M_{t_{j+1}}-M_{t_j})$ has zero expectation, since conditioning on $\F_{t_j}$ leaves the $\F_{t_j}$-measurable factors times $\E[M_{t_{j+1}}-M_{t_j}\mid\F_{t_j}]=0$. The diagonal terms give

where the last equality uses that $M_t^2-\qv_t$ is a martingale, so $\E[(M_{t_{i+1}}-M_{t_i})^2\mid\F_{t_i}]=\E[\qv_{t_{i+1}}-\qv_{t_i}\mid\F_{t_i}]$. The final sum is $\E[\int_0^T H_s^2\,d\qv_s]$, which is Equation (3).

Let $\mu_M(d\omega,ds)=d\P\,d\qv_s$ on $(\Omega\times[0,T],\mathcal P)$, a finite measure since $\E[\qv_T]<\infty$, so $\mathcal H=L^2(\mu_M)$. The indicators $\ind_{A\times(s,t]}$ with $A\in\F_s$ are simple predictable, and their linear span is an algebra generating $\mathcal P$. By the functional monotone class theorem this span is dense in $L^2(\mu_M)$, so bounded predictable $H$ are $\norm{\cdot}_M$-approximable by simple predictable processes, and a general $H\in\mathcal H$ follows by truncating to $H\ind_{|H|\le n}$. Given $H\in\mathcal H$ choose simple $H^n\to H$ in $\norm{\cdot}_M$. The isometry Theorem 2 makes $(H^n\cdot M)$ Cauchy in $L^2$, and by Doob's inequality $\E[\sup_{t\le T}((H^n-H^m)\cdot M)_t^2]\le 4\,\norm{H^n-H^m}_M^2\to 0$, giving a.s. uniform Cauchy convergence, so the limit $H\cdot M$ exists, is independent of the approximating sequence, and is a square-integrable martingale by closure of the square-integrable martingales under $L^2$ limits. For the quadratic variation, apply the computation of Theorem 2 to the simple integrand $H\ind_{(s,t]}$ under $\E[\cdot\mid\F_s]$ rather than full expectation. The cross terms vanish by conditioning on the intermediate breakpoints $t_j\in(s,t]$, and each diagonal term reduces via $\E[(M_{t_{i+1}}-M_{t_i})^2\mid\F_{t_i}]=\E[\qv_{t_{i+1}}-\qv_{t_i}\mid\F_{t_i}]$ as before, the truncated endpoint cells handled by the same identity on the partial subintervals. For simple $H$ and $s<t$ this gives

Since $H\cdot M$ is a martingale, $\E[(H\cdot M)_t^2\mid\F_s]-(H\cdot M)_s^2=\E[((H\cdot M)_t-(H\cdot M)_s)^2\mid\F_s]$, so $(H\cdot M)_t^2-\int_0^t H_s^2\,d\qv_s$ is a martingale. The process $t\mapsto\int_0^t H_s^2\,d\qv_s$ is increasing and predictable, the integral of the predictable integrand $H^2$ against the predictable finite-variation process $\qv$, so by the uniqueness of the Doob-Meyer compensator it is $\langle H\cdot M\rangle_t$. The identity extends to general $H\in\mathcal H$ by the $L^2$ convergence above.