If $A$ is the generator of $(P_t)$, then $A+f$ is the generator of $(P_t^f)$

Question

Let $X=(X_t)_{t\geq0}$ be a Markov process on a state space $\Gamma$ (a Hausdorff topological vector space), let $A$ be the infinitesimal generator of $X$ and let $\mathcal C(\Gamma)$ the space of continuous functions on $\Gamma$. Then for each continuous function $f \in \mathcal C(\Gamma)$ the operator $A+f$ is the infinitesimal generator of the semigroup $(P_t^f)_{t\geq0}$ defined by \begin{equation*} P_t^f g(x) = \mathbb E_x\left[\mathrm \exp\left\{\int_0^t f(X_s) ds\right\} g(X_t) \right], \qquad g \in \mathcal C(\Gamma), x \in \Gamma, \end{equation*} where $\mathbb E_x$ denotes the expectation with respect to the process $X$ starting in $x \in \Gamma$.

In other words:

Show \begin{equation*} \lim_{t\rightarrow0} \frac{1}{t} \left( \mathbb E_x\left[\mathrm \exp\left\{\int_0^t f(X_s) ds\right\} g(X_t) \right] - g(x) \right) = Ag(x) + f(x) g(x).\tag1 \end{equation*}

In a proof I'm studying I found this stated as a known fact about Markov processes, but I couldn't find any suitable reference and my knowledge about continuous Markov processes is limited.

My question: Why and under which assumptions about $X$ and $\Gamma$ is (1) true?

(In the case I encountered it, $\Gamma$ is actually finite.)

On a further note, does the semigroup $(P_t^f)$ bear any deeper meaning? The definition first seemed quite arbitrary to me, but the fact that this statement should be generally known makes it seem like there's more to it than I realise...

Answer 1

Write \begin{equation*} \frac{1}{t} \left( \mathbb E_x\left[\mathrm e^{\,\int_0^t f(X_s) ds} g(X_t) \right] - g(x) \right) = \mathbb E_x\left[{\mathrm e^{\,\int_0^t f(X_s) ds} -1\over t}\, g(X_t) \right] +\mathbb{E}_x\left[{g(X_t)-g(x)\over t}\right]. \end{equation*} Letting $t\downarrow 0$, the right hand side becomes $f(x)g(x)+Ag(x).$ You will want to assume that $f,g$ are bounded continuous functions, that $g\in{\cal D}(A)$ and that $(X_t)$ has right continuous sample paths, almost surely.

Provided that $f\leq 0,$ $A+f$ generates the Markov process which is the $A$ process with "killing" function $f$.