Resolvent operators and inverses proof 
I am trying to prove for myself that $A(R_{\alpha}g)=\alpha R_{\alpha}g-g$ which is proving problematic.
The definition of $A$, the generator, is 
$\displaystyle Af(x)= \lim_{t \rightarrow 0} \frac{E^{x}[f(X_t)]-f(x)}{t}$.
However this definition does not seem to tend itself to calculations and I cannot see how it could be used on the last identity in the passage.
I believe that the $\int^{t+s}_t$ may be what is causing me difficulty.
My definition of the resolvent operator is
 
 A: Here is an alternative proof: By the Markov property, we have
$$\mathbb{E}^{x}\big( \mathbb{E}^{X_t} g(X_s) \big) = \mathbb{E}^x g(X_{s+t})$$
for any $s,t \geq 0$. Applying Fubini's theorem therefore yields
$$\mathbb{E}^{x}(R_{\alpha}g(X_t)) = \int_0^{\infty} e^{-\alpha s} \mathbb{E}^x g(X_{s+t}) \, ds.$$
Consequently, we find
$$\begin{align*} \frac{\mathbb{E}^x(R_{\alpha}g(X_t))-R_{\alpha} g(x)}{t} &=\frac{1}{t} \int_0^{\infty} e^{-\alpha s} \big( \mathbb{E}^x g(X_{s+t})-\mathbb{E}^x g(X_s) \big) \, ds \\ &= \frac{1}{t} \int_{t}^{\infty} e^{-\alpha (s-t)} \mathbb{E}^x g(X_s) \, ds - \frac{1}{t} \int_0^{\infty} e^{-\alpha s} \mathbb{E}^x g(X_s) \, ds \\ &= \frac{e^{\alpha t}-1}{t} \int_t^{\infty} e^{-\alpha s} \mathbb{E}^xg(X_s) \, ds - \frac{1}{t} \int_0^t e^{-\alpha s} \mathbb{E}^xg(X_s) \, ds \\ &\stackrel{t \to 0}{\to} \alpha \int_0^{\infty} e^{-\alpha s} \mathbb{E}^xg(X_s) \, ds - g(x). \end{align*}$$
A: Start with the formula $$E^x\left[R_\alpha g(X_t)\right]=\alpha\int^\infty_0e^{-\alpha s}\int_t^{t+s}E^x[g(X_v)]dv\,ds.$$
Plugging in $t=0$ we get 
$$R_\alpha g(x)=\alpha\int^\infty_0e^{-\alpha s}\int_0^{s}E^x[g(X_v)]dv\,ds.$$
Subtracting and dividing by $t$ gives,
$${E^x\left[R_\alpha g(X_t)\right]-R_\alpha g(x)\over t} =
\alpha\int^\infty_0e^{-\alpha s}\left({1\over t}\int^{t+s}_s E^x[g(X_v)]\,dv\right)ds-{1\over t}\int^t_0 E^x[g(X_v)]\,dv.$$
Now letting $t\to0$ we get the limit 
$$\alpha\int^\infty_0e^{-\alpha s} E^x[g(X_s)]\, ds- g(x)=\alpha R_\alpha g(x)-g(x).$$
