= Solution
Let $\mathcal M_c^2$ be the continuous $L^2$-bounded martingales starting at zero, modulo indistinguishability, with norm $\|N\|_{\mathcal M^2}=\|N_\infty\|_2$. For a fixed $M\in\mathcal M_c^2$, define a finite measure on $\mathcal P$ by
$$
\nu_M(C)=\mathbb E\int_0^\infty\mathbf1_C(\omega,s)\,d[M]_s,
$$
and let $L^2(M)=L^2(\mathcal P,\nu_M)$. The <Itô isometry> is the isometric extension
$$
I_M:L^2(M)\longrightarrow\mathcal M_c^2,
\qquad
H\longmapsto H\mathbin\cdot M,
$$
satisfying
$$
\mathbb E|(H\mathbin\cdot M)_\infty|^2
=\mathbb E\int_0^\infty H_s^2\,d[M]_s.
$$
For the simple process in part b, orthogonality gives the sum there. Conditional on $\mathcal F_{t_i}$, the martingale identity for $M^2-[M]$ gives
$$
\mathbb E\!\left[
H_i^2(M_{t_{i+1}}-M_{t_i})^2\right]
=\mathbb E\!\left[H_i^2([M]_{t_{i+1}}-[M]_{t_i})\right].
$$
Summing proves the isometry. Part c then supplies the unique extension to all of $L^2(M)$.
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