Lemma 5.2.2.label Let $\bp$ be the distribution of a strong Markov process on $D([0, T]; \real^{d})$ with transition functions $P_{s, t}(x, \cdot)$, $\bracs{X_t|t \in [0, T]}$ be the canonical process, $\bracs{\cf_t|t \in [0, T]}$ be its natural filtration, $h: (0, \infty) \times \real^{d} \to [0, \infty)$ be a function satisfying the mean-value property with respect to $\bp$, and $A_{0} = \bracs{h(0, \cdot) > 0}$, then the process
\[\one_{\bracs{X_0 \in A_0}}\frac{h(t, X_{t})}{h(0, X_{0})}\]
is a $\bracs{\mathcal{F}_t}$-martingale with respect to $\bp$.
Proof. Let $t \in [0, T]$ and $\mu$ be the initial distribution of $\bp$, then
\[\ev_{\bp}\braks{\one_{\bracs{X_0 \in A_0}} \frac{h(t, X_{t})}{h(0, X_{0})}}= \int_{A_0}\frac{1}{h(0, x)}\int_{\real^d}h(t, y)P_{0, t}(x, dy) \mu(dx) = \mu(A_{0})\]
so the given process is integrable. Now, since $h$ satisfies the mean-value property,
\begin{align*}\ev_{\bp}\braks{\one_{\bracs{X_0 \in A_0}} \frac{h(T, X_{T})}{h(0, X_{0})} \bigg | \cf_t}&= \frac{\one_{\bracs{X_0 \in A_0}}}{h(0, X_{0})}\cdot \ev_{\bp}\braks{h(t, X_t)| \cf_t}\\&= \one_{\bracs{X_0 \in A_0}}\frac{h(t, X_{t})}{h(0, X_{0})}\end{align*}
$\square$