= Solution
{c}
Integrating the preceding differential gives $M(h_j)_T=1+\int_0^Th_j(s)M(h_j)_s\,dB_s$. Consequently $\mathbb E F=\sum_jc_j$, and the representation is
$$
\boxed{f_s=\sum_{j=1}^Nc_jh_j(s)M(h_j)_s,\qquad
F=\sum_jc_j+\int_0^Tf_s\,dB_s.}
$$
The process is continuous and <adapted>, hence <progressively measurable>, and the finite-sum inequality together with the previous second-moment estimate gives $\mathbb E\int_0^T|f_s|^2ds<\infty$.
If $g$ is another square-integrable predictable <integrand> with the same representation, <Itô isometry> gives
$$
0=\mathbb E\left|\int_0^T(f_s-g_s)dB_s\right|^2
=\mathbb E\int_0^T|f_s-g_s|^2ds.
$$
Thus \b[uniqueness is up to $ds\otimes d\mathbb P$ <null sets> in the square-integrable <integrand> class]. Pointwise uniqueness is impossible: changing a value at one deterministic time changes no <Itô integral>. Square integrability is the natural class for this $L^2$ representation and is needed for this uniqueness argument; the mere word <adapted> does not by itself specify an integrable <stochastic integral>.
Back to article page
