= Solution
Put $M_t=\int_0^tH_s\,dB_s$ and $A_t=\int_0^tH_s^2ds=[M]_t$. The assumed path properties make $A$ continuous, strictly increasing and unbounded. Therefore $T$ is a finite <stopping time> and $A_T=\sigma^2$, even though its definition uses a strict inequality.
For $u\in\mathbb R$, the <Itô formula> gives
$$
E_t=\exp\left(iuM_{t\wedge T}+\frac{u^2}{2}A_{t\wedge T}\right),\qquad
dE_t=iuE_t\,dM_{t\wedge T}.
$$
Its real and imaginary parts are <local martingales>, and $|E_t|\le e^{u^2\sigma^2/2}$. Thus $E$ is a bounded true <martingale>. Continuity of $M$, finiteness of $T$, and <dominated convergence> yield $\mathbb E E_T=1$. Rearranging gives
$$
\mathbb E e^{iuM_T}=e^{-u^2\sigma^2/2}.
$$
The <characteristic function> uniquely identifies the distribution, proving the <Gaussian terminal value at a deterministic bracket level>:
$$
\boxed{\int_0^T H_s\,dB_s\sim N(0,\sigma^2).}
$$
This proof does not use the time-change theorem that the later part asks us to prove.
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