= Solution
{c}
For $s<t<1$, <independence> and the Gaussian <Brownian transition density> yield the <Chapman-Kolmogorov equation>
$$
\mathbb E[p(1-t,B_t,y)\mid\mathcal F_s]
=\int_{\mathbb R}p(t-s,B_s,z)p(1-t,z,y)\,dz
=p(1-s,B_s,y).
$$
The numerator is positive and its expectation is $p(1,0,y)$, so
$$
\boxed{M_0=1,\quad\mathbb EM_t=1,\quad
\mathbb E(M_t\mid\mathcal F_s)=M_s.}
$$
These statements hold for every $t<1$, in particular on the specified closed interval $[0,T]$.
The function $h(t,z)=p(1-t,z,y)$ solves the <backward heat equation> $h_t+\tfrac12h_{zz}=0$. <Itô's formula> therefore gives
$$
dM_t=M_t\theta_t\,dB_t,\qquad
\theta_t=\frac{\partial}{\partial x}\log p(1-t,B_t,y)
=\frac{y-B_t}{1-t}.
$$
On each interval ending before one this adapted continuous integrand is locally square integrable. Thus
$$
\boxed{N_t=\int_0^t\frac{y-B_s}{1-s}\,dB_s,\qquad
\langle N\rangle_t=\int_0^t\frac{(y-B_s)^2}{(1-s)^2}\,ds}
$$
gives $M_t=\exp(N_t-\langle N\rangle_t/2)$. This also follows by applying <Itô's formula> to $\log M_t$. The source's notation $t\geq0$ for this density process is restricted by its defining formula to $t<1$.
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