Given a Feller semigroup $(T_t)$ on locally compact, separable metric space $S$, I wish to show that, for $\lambda > 0$, the resolvent $R_{\lambda}f(x) := \int_0^{\infty}e^{-\lambda t}T_tf(x)dt$ satisfies $R_{\lambda}-R_{\mu}=(\mu-\lambda)R_{\mu}R_{\lambda}$. This is a result from Kallenberg's Foundations of Modern Probability, Theorem 17.4.
Here's what I have so far:
The LHS is $(R_{\lambda} - R_{\mu})(f)(x)=\int_0^{\infty}(e^{-\lambda t}-e^{-\mu t})T_tf(x)dt$
The RHS is $(\mu-\lambda)\int \int e^{-\lambda t - \mu s}T_sf(x)T_tf(x)dsdt = (\mu-\lambda)\int \int e^{-\lambda t - \mu s}T_{s+t}f(x)dsdt $, and I am not sure how to simplify either side to get closer to the other.