= Solution
For the <simple predictable process> from part (a), define its \b[<stochastic integral>] by
$$
\boxed{(H\cdot M)_t=\sum_i h_i\bigl(M_{t\wedge t_{i+1}}-M_{t\wedge t_i}\bigr).}
$$
The time-zero value contributes nothing. Write $D_i=M_{t\wedge t_{i+1}}-M_{t\wedge t_i}$. If $i<j$ and $t>t_j$, then $h_iD_i h_j$ is $\mathcal F_{t_j}$-<measurable> and $\mathbb E[D_j\mid\mathcal F_{t_j}]=0$. If $t\leq t_j$, then $D_j=0$. Boundedness ensures <integrability>, so all off-diagonal terms vanish by <conditional expectation>. Consequently
$$
\boxed{\mathbb E[(H\cdot M)_t^2]=\sum_i\mathbb E[h_i^2D_i^2].}
$$
The <quadratic variation> theorem says $M^2-[M]$ is a <local martingale>. A bounded continuous <martingale> is <square-integrable>, and localization gives the conditional second-moment identity
$$
\mathbb E[D_i^2\mid\mathcal F_{t_i}]=\mathbb E\bigl([M]_{t\wedge t_{i+1}}-[M]_{t_i}\mid\mathcal F_{t_i}\bigr)\qquad(t\geq t_i).
$$
Multiplication by $h_i^2$ and summation therefore give the \b[<Itô isometry>]
$$
\boxed{\mathbb E\left|\int_0^t H_s\,dM_s\right|^2=\mathbb E\int_0^t H_s^2\,d[M]_s.}
$$
For <Brownian motion>, $[M]_s=s$. More generally the <Itô isometry> holds for every <predictable process> with finite right-hand side, against a continuous <square-integrable> <martingale>. Indeed, on a fixed horizon the <quadratic-variation measure> $\mu_M(E)=\mathbb E\int_0^t\mathbf1_E\,d[M]$ is finite. Part (c) approximates the integrand by <simple predictable processes>; the displayed equality makes the elementary integrals Cauchy in $L^2$ and defines their unique limit. The finite sum is exactly the elementary case of the <Itô isometry>.
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