= Solution
Put $A_t=[M]_t$ and use the <Dambis-Dubins-Schwarz theorem> to write $M_t=B_{A_t}$, enlarging the space if necessary after the terminal clock value. In the time-changed filtration, $A_t$ is a stopping time. For $b<0$, let $S_b=\inf\{s:B_s-s=b\}$. Applying part c with $R=A_t$ gives
$$
1=\mathbb E\!\left[
\mathbf1_{\{S_b\leq A_t\}}e^{b+S_b/2}
+\mathbf1_{\{A_t<S_b\}}e^{M_t-A_t/2}
\right].
$$
On $\{S_b\leq A_t\}$, the first integrand is at most $e^be^{A_t/2}$. The assumed <Novikov condition> therefore implies
$$
\mathbb E\left[
\mathbf1_{\{S_b\leq A_t\}}e^{b+S_b/2}
\right]
\leq e^b\mathbb E e^{[M]_T/2}\longrightarrow0
$$
as $b\to-\infty$, uniformly for $t\leq T$. Meanwhile $mathbf1_{\{A_t<S_b\}}\uparrow1$, so the <monotone convergence theorem> gives $\mathbb E\mathcal E(M)_t=1$. A nonnegative local martingale with constant expectation is a martingale. Thus $\mathcal E(M^T)$ is a martingale, proving the <Novikov condition>.
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