I have a question about the generator of a stochastic process.
$T>0$: fix Let $b: \mathbb{R} \to \mathbb{R}$ be a bounded measurable function. Let $\left( (X_{t})_{t \in [0,T]}, \left(P_{x} \right)_{x \in \mathbb{R}} \right)$ be a $\mathbb{R}$-valued Brownian motion.
By Girsanov theorem, $\hat{X}_{t}:=X_{t}-\int_{0}^{t}b(X_{s})ds$ is a Brownian motion under $Q_{x}$. where \begin{align*} \left. \frac{dQ_{x}}{dP_{x}} \right|_{\mathcal{F}_{t}}:=\exp \left(\int_{0}^{t}b(X_{s})dX_{s}-\frac{1}{2} \int_{0}^{t}b(X_{s})^{2}ds \right) \end{align*}
Question
I want to determine the generator of $(X_{t},Q_{x})$. If $b$ is continuous, I can.
By Ito's formula, for $f \in C^{2}, x \in \mathbb{R}$
\begin{align*} \frac{E_{x}^{Q}[f(X_{t})]-E_{x}^{Q}[f(X_{0})]}{t} =\frac{1}{t}\int_{0}^{t}E^{Q}_{x}\left[b(X_{s})f'(X_{s})+\frac{1}{2}f''(X_{s}) \right]ds \end{align*} By the continuity of $\displaystyle s \mapsto E^{Q}_{x}\left[b(X_{s})f'(X_{s})+\frac{1}{2}f''(X_{s}) \right] $, \begin{align*} \frac{E_{x}^{Q}[f(X_{t})]-E_{x}^{Q}[f(X_{0})]}{t} \to b(x)f'(x)+\frac{1}{2}f''(x) \end{align*} I don't know when $b$ is not continuous. Perhaps we can't determine the generator? What should I do?
Thank you in advance.