Use the <continuous semimartingale decomposition> $X=X_0+M+A$, where $M$ is a continuous local martingale and $A$ is a continuous adapted <finite-variation process>. Pointwise limits preserve predictability, so $H$ is predictable; it is bounded by the common bound for the $H_n$.

Localize so that $[M]_\infty$ and the total variation $V(A)_\infty$ are bounded. The <Doob L2 maximal inequality> and the <Itô isometry> give
$$
\mathbb E\sup_t\left|\int_0^t(H_n-H)\,dM\right|^2
\leq4\mathbb E\int_0^\infty(H_n-H)^2\,d[M]
\longrightarrow0
$$
by the <dominated convergence theorem>. For the finite-variation part,
$$
\sup_t\left|\int_0^t(H_n-H)\,dA\right|
\leq\int_0^\infty|H_n-H|\,dV(A)
\longrightarrow0
$$
almost surely, again by dominated convergence, now for each sample path. Hence the two integrals converge uniformly in probability after every localization. Part (b) removes the localization and proves
$$
\int H_n\,dX\longrightarrow\int H\,dX
$$
u.c.p.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 202 / 2 / c / Solution

{c}
{title2=$\mathbb E\sup_{s\leq t}|M_s|^2\leq4\mathbb E|M_t|^2$}
{wiki=Doob's_martingale_inequality}

For a square-integrable martingale starting at zero,
$$
\mathbb E\sup_{s\leq t}|M_s|^2\leq4\mathbb E|M_t|^2.
$$
For a continuous local martingale stopped so that its quadratic variation is integrable, the <Itô isometry> makes the right-hand side $4\mathbb E[M]_t$.

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 Incoming links: Doob L2 maximal inequality