 Solution

Fix a deterministic $t>0$ and define ${\hat{B}}_u=B_{t+u}-B_t$, ${\hat{H}}_u=H_{t+u}$ and ${\hat{\mathcal{F}}}_u={\mathcal{F}}_{t+u}$. The increment process is a standard Brownian motion in the shifted filtration, and $\hat{H}$ remains bounded, continuous and adapted. Thus part (a) applies even though the initial value ${\hat{H}}_0=H_t$ is random. For $\epsilon >0$,Here the constant-in-time integrand $H_t$ is ${\mathcal{F}}_t$-measurable, so its stochastic integral equals $H_t(B_{t+\epsilon }-B_t)$. The expectation of the absolute value of the right-hand side to the power $1/4$ tends to zero by part (a). For every $\eta >0$, Markov inequality consequently givesThereforeThe stated convergence in probability holds for each fixed deterministic time.