Proposition 1.2.4.label Let $(\Omega, \bracs{\cf_t|t \ge 0}, \bp)$ be a filtered probability space,
\[L_{t}(\omega)u = \frac{1}{2}\dpn{A_t(\omega), u}{\real^{d \times d}}+ \dpn{B_t(\omega), u}{\real^d}\]
be a second-order $\bracs{\mathcal{F}_t}$-progressively measurable random differential operator, $\bracs{X_t|t \ge 0}$ be a $\bracs{\mathcal{F}_t}$-progressively measurable process with continuous sample paths, and $f \in C_{b}^{1, 2}([0, \infty) \times \real^{d})$. If the processes
\[Y_{t} = f(t, X_{t}) - \int_{0}^{t} [(\partial_{r} + L_{r})f](r, X_{r})dr\]
and
\[W_{t} = f(t, X_{t})^{2} - \int_{0}^{t} [(\partial_{r} + L_{r})(f^{2})](r, X_{r})dr\]
are martingales, then
\[[X]_{t} = \frac{1}{2}\int_{0}^{t} \dpn{Df, A(r)(Df)}{\real^d}(r, X_{r})dr\]
Proof. Firstly, since $f$ admits bounded derivatives, the integral term is of locally bounded variation. Therefore if $Z_{t} = f(t, X_{t})$, then $[X]_{t} = \angles{Z}_{t}$.
Let $\bracs{t_j}_{0}^{n} \subset [0, t]$ be a partition, then
\[Z_{t}^{2} - Z_{0}^{2} = \sum_{j = 1}^{n}Z_{t_j}^{2} - Z_{t_{j-1}}^{2} = \sum_{j = 1}^{n} 2Z_{t_{j-1}}(Z_{t_j}- Z_{t_{j-1}}) + \sum_{j = 1}^{n} (Z_{t_j}- Z_{t_{j-1}})^{2}\]
As the above holds for every partition,
\[Z_{t}^{2} - Z_{0}^{2} = 2\int_{0}^{t} Z_{r}Z(dr) + \angles{Z}_{t}\]
Since $Z_{t} = Y_{t} + \int_{0}^{t}(\partial_{r} + L_{r})fdr$,
\begin{align*}\int_{0}^{t} Z_{r}Z(dr)&= \int_{0}^{t} Z_{r} Y(dr) + \int_{0}^{t} f(r, X_{r})(\partial_{r} + L_{r})f(r, X_{r})dr \\ Z_{t}^{2} - Z_{0}^{2}&= 2\int_{0}^{t} Z_{r} Y(dr) + 2\int_{0}^{t} f(r, X_{r})(\partial_{r} + L_{r})f(r, X_{r})dr + \angles{Z}_{t}\end{align*}
On the other hand,
\[Z_{t}^{2} - Z_{0}^{2} = W_{t} - W_{0} + \int_{0}^{t} [(\partial_{r} + L_{r})(f^{2})](r, X_{r})dr\]
Therefore
\begin{align*}\angles{Z}_{t}&= W_{t} - W_{0} + \int_{0}^{t} [(\partial_{r} + L_{r})(f^{2})](r, X_{r})dr \\&- 2\int_{0}^{t} Z_{r} Y(dr) - 2\int_{0}^{t} f(r, X_{r})(\partial_{r} + L_{r})f(r, X_{r})dr \\&= W_{t} - W_{0} - 2 \int_{0}^{t} f(r, X_{r}) Y(dr) \\&+ \int_{0}^{t} [L_{r}(f^{2})(r, X_{r}) - 2f(r, X_{r})(L_{r}f)(r, X_{r})]dr \\&= W_{t} - W_{0} - 2\int_{0}^{t} f(r, X_{r})Y(dr) + \frac{1}{2}\int_{0}^{t} \dpn{Df, A(r)(Df)}{\real^d}(r, X_{r})dr\end{align*}
where the process $t \mapsto W_{t} - W_{0} - 2\int_{0}^{t} f(r, X_{r})Y(dr)$ is of finite variation, so it must be $0$ by [Lemma 5.13, Bau14].$\square$