= Solution
Apply the exact discrete product identity
$$
t_{j+1}B_{j+1}-t_jB_j=t_j\Delta B_j+B_{j+1}\Delta t_j
$$
and sum over the deterministic <partition of an interval>. Since $B_0=0$,
$$
\sum_jt_j\Delta B_j=tB_t-\sum_jB_{j+1}\Delta t_j.
$$
The left side converges in <L2 space> to $\int_0^ts\,dB_s$ by the definition and <Itô isometry>, since
$$
\int_0^t|s-\eta(s)|^2ds
=\frac13\sum_j(\Delta t_j)^3\leq\frac{t|\pi|^2}{3}.
$$
The right-endpoint <Riemann sum> converges in <L2 space> to $\int_0^tB_sds$: its squared error is at most
$$
t\sum_j\int_{t_j}^{t_{j+1}}\mathbb E|B_{j+1}-B_s|^2ds
=\frac t2\sum_j(\Delta t_j)^2\longrightarrow0.
$$
Consequently,
$$
\boxed{\int_0^ts\,dB_s=tB_t-\int_0^tB_s\,ds.}
$$
As in (a)(iii), continuity extends fixed-time equality to equality of the continuous processes.
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