Theorem 1.2.5 (Equivalence of Martingales). Let $(\Omega, \bracs{\cf_t|t \ge 0}, \bp)$ be a filtered probability space,
be a second-order $\bracs{\mathcal{F}_t}$-progressively measurable random differential operator, and $\bracs{X_t|t \ge 0}$ be a $\bracs{\mathcal{F}_t}$-progressively measurable process with continuous sample paths, then the following are equivalent:
For every $f \in C_{c}^{\infty}(\real^{d})$, the process
\[E_{t} = f(X_{t}) - \int_{0}^{t} (L_{r}f)(X_{r})dr\]is a $\bracs{\mathcal{F}_t}$-martingale.
For every $f \in C_{b}^{1, 2}([0, \infty) \times \real^{d})$,
\[H_{t} = f(t, X_{t}) - \int_{0}^{t} [(\partial_{r} + L_{r})f](r, X_{r})dr\]is a $\bracs{\mathcal{F}_t}$-martingale.
For each uniformly positive $f \in C_{b}^{1, 2}([0, \infty) \times \real^{d})$,
\[X_{t} = f(t, X_{t})\exp\braks{-\int_0^t \frac{(\partial_{r} + L_{r})f}{f}(r, X_r)dr}\]is a $\bracs{\mathcal{F}_t}$-martingale.
For each $x \in \real^{d}$ and $g \in C_{b}^{1, 2}([0, \infty) \times \real^{d})$, the process
\begin{align*}Y_{t}^{x, g}&= \exp\braks{\dpn{x, X_t - X_0 - \int_0^t b(r)dr}{\real^d} + g(t, X_t)}\\&\cdot \exp\braks{-\frac{1}{2}\int_0^t \dpn{x + Dg, A(r)(x + Dg)}{\real^d}(r, X_r)dr}\\&\cdot \exp\braks{\int_0^t [(\partial_r + L_r)g](r, X_r)}dr\end{align*}is a $\bracs{\mathcal{F}_t}$-martingale.
For each $\theta \in \real^{d}$,
Proof [Theorem 4.2.1, SV97]. (1) $\Rightarrow$ (2): Let $0 \le s \le t$ and $f \in C_{c}^{\infty}([0, \infty) \times \real^{d})$, then by the Fundamental Theorem of Calculus,
where by (1),
By the Fundamental Theorem of Calculus,
By (1) applied to $(\partial_{r} f)$,
Similarly, by the Fundamental Theorem of Calculus,
Since the two iterated integrals are over the same region,
Therefore $\bracs{E_t}$ is a $\bracs{\mathcal{F}_t}$-martingale.
(2) $\Rightarrow$ (3): Let $f \in C_{b}^{1, 2}([0, \infty) \times \real^{d})$ be uniformly positive. Define
and
then using the by parts formula for Riemann-Stieltjes integrals,
Since
Using the by parts formula again,
Therefore Theorem 1.2.4 implies that the above process is a $\bracs{\mathcal{F}_t}$-martingale.
(3) $\Rightarrow$ (4): Let $x \in \real^{d}$ and $g \in C_{b}^{1, 2}([0, \infty) \times \real^{d})$. Define
then (3) applied to $f$ yields that $\bracs{Y_t|t \ge 0}$ is a $\bracs{\mathcal{F}_t}$-martingale. However, since $f$ is unbounded and not uniformly positive, a direct application is ineffective.
Let $\seq{f_n}\subset C_{b}^{1, 2}([0, \infty) \times \real^{d})$ be uniformly positive such that $f_{n}|_{\bracs{|y| \le n}}= f$. Define
then $\bracsn{Y_{t \wedge \tau_n}^{x, g}}$ is a $\bracs{\mathcal{F}_t}$-martingale with $Y_{t \wedge \tau_n}^{x, g}\to Y_{t}^{x, g}$ almost surely as $n \to \infty$. In addition,
where $\ev\braks{Y_{t \wedge \tau_n}^{2x, 2g}}= f(s, X_{s})^{2} \le \exp(2\norm{g}_{u})$. Therefore $\bracsn{Y_{t \wedge \tau_n}^{x, g}}$ is bounded in $L^{2}$, uniformly integrable in $L^{1}$, and converges to $Y_{t}^{x, g}$ in $L^{1}$.$\square$